Блог Анкара

In This Chapter

Going through what you can’t forget ^ Trying out some sample questions ^ Checking out your answers

Chapter 21 gave a broad overview of the relationships between differences in energy and the rates and equilibria observed for chemical reactions. These kinds of relationships are fundamental to chemistry and are ripe pickings for the AP chemistry exam. Do yourself a favor: Be sure you know the principles at work behind kinetics and thermodynamics and can apply those principles to problems. The best way to ensure that you know what you need to know about kinetics and thermodynamics is to review the tips and suggestions in this chapter and then practice the questions included.

Picking through the Important Points

For many students, the most difficult thing about kinetics and thermodynamics is telling the two apart. Both subjects describe things about chemical reactions. Both use a wealth of mathematical equations. Both have something to do with energy. Before launching into this summary of the highlights of Chapter 21, make sure you understand that kinetics and thermodynamics, though interrelated, deal with entirely separate issues:

Kinetics Deals with the rate at which reactants convert to products and how that rate depends on concentrations during a reaction.

Thermodynamics Deals with differences In different types of energy (enthalpy and entropy) Between reactants and products and how those differences relate to concentrations at equilibrium.

Important points about rates

The rules for measuring and describing the speed of a reaction are expressed in equations. Luckily, the equations are pretty simple and easy to interpret. Seeing all of them together helps make clear how all the equations relate to each other. Instead of spending time trying to memorize all these equations, spend time trying to understand how they all relate to one another.

In summary, here are the important points about rate made in Chapter 21:

The rate of a reaction does Not Determine the final extent of a reaction, Nor Does the final extent of a reaction determine the rate.

For the reaction: A + B — C

• Average rate = – A[A]/At = – A[B]/At = A[C]/At.

• Instantaneous rate = -d[A]/dt = -d[B]/dt = D[C]/dt. I For the reaction: A + 2B — C

• Instantaneous rate = – d[A]/dt = -1/2d[B]/dt = d[C]/dt.

You can extract rates from the shapes of reaction progress curves, which plot the concentration of a reactant or product versus time.

I Concentrations of reactants or products can be measured spectrophotometrically, using Beer’s law: A = Abc

I For a reaction with two reactants in its rate-determining step: Rate = K [reactant A]m[reactant B]n.

• K Is the Rate constant And depends on reaction conditions.

• Exponents M And N Are Reaction orders And depend on reaction conditions.

• The sum, M + N, Is the Overall reaction order. Zero-order reactions: Rates don’t depend on any concentration.

• Rate = K.

First-order reactions: Rates depend on the concentration of a single species.

• Rate = K[A].

Second-order reactions: Rates may depend on the concentration of one or two species:

• Rate = K[A]2 or Rate = K[A][B].

Integrated rate equations Describe the rate of a reaction with respect to the initial concentration of a reactant.

IU Half-life Describes the time it takes for the concentration of a reactant to drop to one half of its initital value.

• For a first-order reaction: t1/2 = (ln 2) / k.

I Reaction rates tend to increase with temperature.

I Increasing concentration tends to increase reaction rate.

I Catalysts increase reaction rates, but do not alter concentrations or final amounts produced at equilibrium.

• Catalyts decrease Activation energy, The energetic hill reactants must climb to reach a Transition state.

The Arrhenius equation relates the rate constant, k, to activation energy, Ea, And temperature, T, and a constant, A (the "frequency factor"), related to reactant orientation and frequency of collisions:

For a first-order reaction: [A] = [A]0-e"kt. or equivalently ln[A] = – kt + lr

Ln[A]0.

K = Ae

Ea /RT

I The slowest elementary step in a reaction mechanism is the Rate-determining step.

Important points about thermodynamics

As described in Chapter 21, thermodynamics is about big ideas and (usually) small equations. Thermodynamic concepts can seem a bit abstract, so as you peruse the highlights in this summary, anchor your brain to the idea that thermodynamics has nothing whatever to do with how fast a reaction proceeds — only how far it goes. Thermodynamics cares only about beginning and end states.

Re-energize your grasp of thermodynamics by perusing these points:

Laws of Thermodynamics:

• First Law: In any process, the total energy of the universe remains constant. Universe = system + surroundings.

• Second Law: In any spontaneous process, the overall entropy of the universe increases.

• Third Law: As temperature approaches absolute zero, the entropy of a system approaches a constant minimum.

The equilibrium constant, Keq, is related to the difference in free energy, AG, between products and reactants:

AG = – RT ln Keq

This equation can be rearranged to the following form:

K = e -AG /RT

Favorable (spontaneous) reactions possess negative values for AG and unfavorable (nonspontaneous) reactions possess positive values for AG.

The Gibbs equation Relates the difference in free energy (AG) between products and reactants to the differences in enthalpy (AH) And entropy (AS) And the temperature of the system in Kelvin:

• AG = AH – TAS

Under standard conditions (usually latm, pure substances at 1Mconcentration), the Gibbs equation reflects "standard state" quantities:

• AG°= AH° – TAS°

IU Nonstandard AG relates to standard AG° via corrections concentration (as reflected in Q, the reaction quotient):

• AG = AG°+ RT Ln Q

Heat capacity Is the amount of heat required to raise the temperature of a system by 1K.

• Molar heat capacity Is the heat capacity of 1 mole of a substance.

• Specific heat capacity Is the heat capacity of 1 gram of a substance.

Constant-pressure calorimetry Directly measures an enthalpy change (AH) for a reaction because it monitors heat flow at constant pressure: AH = qP.

• q = mCpAT

Constant-volume calorimetry Directly measures a change in internal energy, AE (not AH) For a reaction because it monitors heat flow at constant volume.

Q = – Ccal xAT

Exothermic reactions Release heat as a product, AH<0. Endothermic reactions Consume heat as a reactant, AH>0.

I Hess’s Law (enthalpies are additive):

• For the set of coupled reactions,

A — B — C — D the following is true:

AHAD = AHAB + AHBC + AHCD

• Before adding reaction equations, you may need to manipulate those equations, and doing so has consequences on the associated AH values:

Reaction: A — B Enthalpy change: AHAB compare that to:

Reaction: 2A — 2B Enthalpy change: 2AHAB compare that to:

Reaction: B — A Enthalpy change: – AHAB

Testing Your Knowledge

Concentration, rate, and equilibrium. Free energy, enthalpy, and entropy. Heat, mass, and temperature. Time spent on these kinds of problems is never wasted.

Multiple choice

Questions 1 through 4 refer to the following reactions: Reaction 1: A + B — C Rate = K[A]2

Reaction 2: D + E — F + 2G Rate = K[D][E]

1. What are the overall reaction orders for Reaction 1 and Reaction 2, respectively?

(A) First-order, Second-order

(B) Second-order, First-order

(C) First-order, First-order

(D) Second-order, Second-order

(E) Not enough information

2. For Reaction 1, how will the initial rate change if the concentration of A is doubled and the concentration of B is halved?

(A) Twofold increase

(B) Threefold increase

(C) Fourfold increase

(D) Twofold decrease

(E) Depends on the concentration of C

3. For Reaction 2, how will the initial rate change if the concentration of D is doubled and the concentration of E is tripled?

(A) Twofold increase

(B) Threefold increase

(C) Fivefold increase

(D) Sixfold increase

(E) Depends on the concentrations of F and G

4. For Reaction 2, what is the relationship between the rates of change in [D], [E], [F], and [G]?

(A) Rate = d[D]/dt = d[E]/dt = – d[F]/dt = -2d[G]/dt

(B) Rate = d[D]/dt = d[E]/dt = – d[F]/dt = 2d[G]/dt

(C) Rate = d[D]/dt = d[E]/dt = – d[F]/dt = -0.5d[G]/dt

(D) Rate = d[D]/dt = d[E]/dt = – d[F]/dt = 0.5d[G]/dt

(E) Rate = 2d[D]/dt = 2d[E]/dt = – d[F]/dt = -0.5d[G]/dt

5. Methane combusts with oxygen to yield carbon dioxide and water vapor: CH4 + 2O2 — CO2 + 2H2O

If methane is consumed at 2.79mol s-1, what is the rate of change in the concentrations of carbon dioxide and oxygen?

(A) +2.79mol s-1 CO2 and +5.58mol s-1 O2

(B) -2.79mol s-1 CO2 and +5.58mol s-1 O2

(C) +5.58mol s-1 CO2 and -5.58mol s-1 O2

(D) +2.79mol s-1 CO2 and -2.79mol s-1 O2

(E) +2.79mol s-1 CO2 and -5.58mol s-1 O2

6. You study the following reaction: D + E — F + 2G

You vary the concentration of reactants D and E, and observe the resulting rates:

[D], M [E], M Rate, M s-1 Trial 1: 2.7 x 10-2 2.7 x 10-2 4.8 x 106 Trial 2: 2.7 x 10-2 5.4 x 10-2 9.6 x 106 Trial 3: 5.4 x 10-2 2.7 x 10-2 9.6 x 106

At what rate will the reaction occur in the presence of 1.3 x 10-2 M Reactant D and 9.2 x 10-3 M Reactant E?

(A) 7.9 x 105 MS-1

(B) 1.2 x 10-4 MS-1

(C) 6.6 x 109 MS-1

(D) 8.6 x 107 MS-1

(E) 6.1 x 107 MS-1

7. For the reaction: A + 2B 2C

If the KEq = 1.37 x 103, what is the free energy change for the reaction at 25 °C? (Note: R = 8.314 J K-1mol-1)

(A) -1.50 x 103 Jmol-1

(B) +1.50 x 103 Jmol-1

(C) -1.79 x 104 Jmol-1

(D) +1.79 x 104 Jmol-1

(E) Not enough information

8. Which of the following statements is correct?

(I) The entropy of the system can decrease during a spontaneous reaction.

(II) The free energy of the system can increase during a spontaneous reaction.

(III) The enthalpy of the system can increase during a spontaneous reaction.

(A) I only

(B) II only

(C) III only

(D) I and II only

(E) I and III only

9. Which of the following statements is incorrect?

(I) Rates typically vary with time during a reaction.

(II) Rate constants typically vary with time during a reaction.

(III) Activation energies typically vary with time during a reaction.

(A) I only

(B) II only

(C) III only

(D) I and II only

(E) II and III only

Questions 10 through 14 refer to the following choices:

10. Chemical bonds store this due to the relative positions of electrons and nuclei

11. Sum of work and heat flow

12. Equal and opposite to pressure multiplied by a change in volume

13. Translation, vibration, and rotation give evidence of this

14. Remains constant in the universe

15. A 375g plug of lead is heated and placed into an insulated container filled with 0.500L water. Prior to the immersion of the lead, the water is at 293K. After a time, the lead and the water reach the same maximum temperature, 297K. The specific heat capacity of lead is 0.127 J g-1 K-1 and the specific heat capacity of water is 4.18 J g-1 K-1. How hot was the lead before it entered the water? (Note: Density of water = 1.00 kg / L)

16. What is the molar reaction enthalpy for the following reaction? C(s) + H2O(g) — CO(g) + H2(g) Use the following data:

Reaction 1: C(s) + O2(g) — CO2(g); AH = -605 kJ Reaction 2: 2CO(g) + O2(g) — 2CO2(g); AH = -966 kJ Reaction 3: 2H2(g) + O2(g) — 2H2O(g); AH = -638 kJ

(A) 999 kJ

(B) 197 kJ

(C) -1407 kJ

(D) 394 kJ

(E) 500 kJ

Answers with Explanations

Having worked through all those problems on rates, energy, work, and heat, you may be feeling a bit depleted of energy yourself. Hang in there long enough to check your answers.

1. (D). Rate laws include a concentration factor for each reactant whose concentration affects the rate. Each concentration factor has an exponent, even if that exponent is simply 1. The exponent on each factor is the individual reaction order for that component. The overall reaction order is the sum of the individual reaction orders.

2. (C). The rate observed for Reaction 1 depends on the rate law for the reaction: Rate = K[A]2. Component B doesn’t even show up in the rate law, so cutting in half the concentration of B has no effect on the rate. However, doubling the concentration of A has a major effect on the rate because the reaction is second order in A (meaning that the concentration factor for A has an exponent of two). You can calculate the rate change associated with doubling A by substituting in simple numbers for A within the rate law. If the original concentration of A is 1 and the doubled concentration is 2, then the original and final rates are K And 4K, Respectively. So, doubling the concentration of A increases the rate fourfold.

3. (D). As described in the answer to question 2, the rate observed for Reaction 2 depends on the rate law for the reaction: Rate = K[D][E]. The reaction is first order in D and first order in E (and second order overall). You can calculate the effect of changing the concentrations of D and E by substituting simple numbers. If the original concentrations of D and E are 1 and 1, respectively, then the new concentrations are 2 and 3, respectively. So, the original rate is K(1)(1) = K, And the new rate is K(2)(3) = 6K. The changes in concentration result in a sixfold increase in rate.

4. (C). You can figure out the relationship between the rates of change of all the reactants and products by inspecting the balanced reaction equation. D + E — F + 2G. First, because D and E are reactants and F and G are products, it’s clear that the rates of change in D and E have opposite sign to those of F and G — as D and E decrease in concentration, F and G increase in concentration. Second, you must account for the stoichiometric coefficients in the reaction equation. Two moles of product G are made for every one mole made of product F and for every one mole consumed of reactants D and E. So, the rate of change in D, E, and F is one half the rate of change in G, as indicated by the coefficient of 0.5. Note that in answer C, whether each term is positive or negative has nothing whatever to do with whether a reac-tant or product is gained or lost during the reaction; positive versus negative values simply indicate that the rates occur in opposite directions.

5. (E). The reaction equation makes clear that for each mole of methane consumed, 2 moles of oxygen gas are consumed and 1 mole of carbon dioxide is produced. So

D[CH4]/dt = 0.5d[O2]/dt = – d[CO2]/dt

This means that the disappearance of 2.79mol s-1 of methane corresponds to the appearance (positive change) of 2.79mol s-1 of CO2 and the disappearance (negative change) of 2 -2.79 = 5.58mol s-1 of O2. The concentration of methane changes at half the rate of oxygen, so the concentration oxygen changes at twice the rate of methane.

6. (A). Doubling the concentration of either reactant (D or E) doubles the rate. So, the data are consistent with the rate law, Rate = K[D][E]. Solve for K By substituting known values of rate [D] and [E] from any of the trial reactions:

K = Rate / ([D][E]) = 4.8 x 106 M S-1 / (2.7 x 10-2 M X 2.7 x 10-2 M) = 6.6 x 109 M -1 s-1.

Use this calculated value of K To determine the rate in the presence of 1.3 x 10-2 MReactant D and 9.2 x 10-3 M Reactant E:

Rate = (6.6 x 109 M -1 s-1)(1.3 x 10-2 M)(9.2 x 10-3 M) = 7.9 x 105 M S-1.

7. (C). You can calculate the free energy change directly from the equilibrium constant and the temperature by using the equation: AG = -RTLnKEq. But to use the equation properly, you must make sure that all your units match. Specifically, notice that the temperature given to you in the problem has units of Celsius degrees, while the gas constant given has temperature units of Kelvin. Convert the temperature to Kelvin first: 25 °C = (25 + 273)K = 298K. Then, substitute into AG = -RTLnKEq. You are told that KEq = 1.37 x 103, so:

AG = (-1)(8.314 J K-1mol-1)(298K)ln(1.37 x 103) = -1.79 x 104 J mol1

8. (E). Although the free energy of a system must decrease during a spontaneous process (otherwise the process would not be spontaneous), both etropy and enthalpy can either decrease or increase during that process. The reason for this flexibility in entropy and enthalpy is made clear by the Gibbs equation:

AG = AH – TAS

Decreases in enthalpy are favorable and increases in entropy are favorable, it is true. However, an unfavorable change in enthalpy can be compensated by a sufficiently favorable change in entropy and vice versa, such that the overall free energy change for the process is favorable, and the process is spontaneous. Note that, although entropy may decrease within a system, entropy always increases for the universe as a whole (system + surroundings = universe).

9. (E). Rates can certainly change during the course of a reaction because rates are frequently dependent on concentrations, as expressed by rate laws. Because concentrations change during the reaction, so do the rates. Rate constants and activation energies are a different matter, however. For a given reaction at a given temperature, the rate constant and activation energy are related by the Arrhenius equation:

K = Ae – Ea /RT

The Arrhenius equation contains no terms for concentration of reactants or products, so the rate constant and activation energy typically remain constant for a given reaction.

10. (C). Potential energy is the portion of total energy due to the position of particles. Chemical energy is essentially potential energy because energy is stored within chemical bonds as a result of the favorable position of electrons between adjacent nuclei.

11. (B). The change in total energy of a system is the sum of heat flow and work:

AE = Efinal – Einitial = Q + W

Positive heat flow (+q) is heat flow into the system, which increases the total energy of the system. Positive work (+w) is work done on the system (not By The system), and also increases the total energy of the system.

12. (E). Pressure-volume work is one kind of work that can be done on a system or by a system, especially by expanding or contracting gases: W = -PAV. If gas within a system expands against the surroundings (increasing the system’s volume), then the system has done work on the surroundings and the sign of work is negative. If the surroundings expand against the system (decreasing the system’s volume), then the surroundings have done work on the system and the sign of work is positive.

13. (D). Kinetic energy is the portion of total energy due to the motion of particles. In an ideal gas, kinetic energy consists entirely of translating particles. In real substances, other forms of motion contribute to the kinetic energy, like vibration and rotation. The average kinetic energy of the particles of a system is proportional to the temperature of that system.

14. (A). Transform as it may between different forms and slip as it may between system and surroundings, the total energy of the universe remains constant.

15. (A). The key to setting up this problem is to realize that whatever heat flows out of the lead flows into the water, so: QLead = -qWater. Calculate each quantity of heat by using Q = mCPAT. Recall that AT = Tfinal – Tinitial. The unknown in the problem is the initial temperature of lead.

Qlead = (375g)(0.127J g-1 K-1)(297K – Tinitial)

To calculate QWater, you must first calculate the mass of 0.500L water by using the density of water: 0.500L x (1.00kg L-1) = 0.500kg = 500g:

QWater = (500g)(4.18J g-1 K-1)(297K – 293K) = 8.36 x 103J

Setting QLead equal to -qWater and solving for TInitial yields 473K: (375g)(0.127J g-1 K-1)(297K – TInitial) = -8.36 x 103J 297K – TInitial = -176K

Tinitial = 473K

16. (B). Reverse Reaction 2 and divide it by 2 to get Reaction 2′ (AH = +483kJ). Reverse Reaction 3 and divide it by 2 to get Reaction 3′ (AH = +319kJ). Add Reactions 1, 2′, and 3′, yielding AH = + 197kJ. Reactions 2 and 3 need reversing to put the right compounds on the reactant and product sides of the equation. Reaction 1 already has the right orientation. Reactions 2 and 3 must be divided by 2 so that the stoichiometry of the final equation matches the stoi-chiometry of the target equation (namely, one mole each of H2O, CO and H2). Even though this division temporarily creates noninteger coefficients for O2 in these equations, the non-integer coefficients cancel out when you add reactions 1, 2′, and 3′.

In This Chapter

► Keeping the facts in mind

► Digging into some practice questions

► Seeing how you did

Chapter 25 is chock-full of juicy little organic chemistry tidbits. Make sure that you have everything filed away in your brain properly by reviewing the concepts in this chapter and attempting the practice problems we include near the end of the chapter. Remember, although there are relatively few questions that directly test organic chemistry on the AP exam, a familiarity with the names and structures of organic compounds can help you to answer many other types of questions.

Making the Main Points Stick

To master the organic chemistry portion of the AP curriculum you need to memorize, memorize, memorize! Specifically, you must remember how to identify the 13 types of organic compounds, and 3 types of isomer listed in the sections below.

Important organic compounds

Listed below are the 13 types of organic compound most likely to appear on the AP exam, together with their naming rules. Make sure that you have a thorough understanding of what makes each unique before you move on.

Alkanes Are hydrocarbons containing only single bonds, Alkenes Contain double bonds and Alkynes Contain triple bonds.

Alkanes, alkenes, and alkynes are named by first numbering the longest carbon chain. Numbering starts at the end closest to the nearest multiple bond for alkenes and alkynes, while alkanes are numbered starting at the end that gives its branches the lowest numbers.

Substituents On all organic compounds are named with the suffix Yl And are preceded by the number of the main chain carbon atom from which they branch.

Cyclic aliphatic hydrocarbons Contain rings of carbon atoms and can contain multiple bonds. Aromatic hydrocarbons Also contain rings of carbon atoms, but the bonds between carbons alternate between double or triple bonds and single bonds.

Alcohols Are hydrocarbon chains that include a hydroxide (OH) group. They are named with the ending – ol.

Ethers Contain an oxygen atom in the midst of their carbon chain. They are named by naming the substituents on either side of the oxygen and then adding the ending Ether, As in ethyl methyl ether.

The final carbon atom in a Carboxylic acid Is double bonded to an oxygen atom as well as being single bonded to a hydroxide group. In condensed structural formulas, this configuration is written as R-COOH. Carboxylic acids are named by attaching the suffix -oic acid.

I Esters Are related to carboxylic acids in that they too have a carbon atom that is

Double bonded to one oxygen. This time, however, rather than also being bonded to a hydroxide group, the final carbon is bonded to a lone oxygen atom. This oxygen is itself bonded to another hydrocarbon chain, giving esters the condensed formula R-COOR. You name an ester by first naming the lower-priority hydrocarbon chain (the one not containing a carbon double bonded to oxygen) as a substituent and then attaching the suffix -oate To the high-priority chain.

I Aldehydes Are also similar to carboxylic acids in that their final carbon atom is double bonded to an oxygen atom, however in this case the fourth available carbon bond is taken up by a simple hydrogen atom rather than a hydroxide group, giving it the formula R-CHO. Aldehydes are named by attaching the suffix -al.

I Ketones Have the same structure as aldehydes, except that the double-bonded oxygen has infiltrated its way up to a midchain carbon. These compounds have the condensed formula ROR and are named with the suffix – one. Unlike esters, ketones contain uninterrupted carbon chains, so you need not worry about prefixes.

I Halocarbons Are hydrocarbon chains with a halogen substituent. The hydrocarbon portion is named normally, with the prefix Fluoro, chloro, bromo, Or Iodo Attached, depending on which halogen it contains.

I Amines Are hydrocarbon chains that contain an NH2 group and have the basic form R-NH2. They are named by naming the carbon chain as a substituent then adding the ending Amine.

Isomers

You will also be expected to be able to identify the following three types of isomer on the AP exam. Keep in mind that all isomers contain the same number and type of atoms, but their connectivities may be different.

I Structural isomers Contain the same number of atoms of carbon, hydrogen, oxygen, etc., but in different configurations.

I Stereoisomers Must have the same connectivities (in other words, all of their bonds need to be in the same order), and are divided into two categories — compounds that are nonsuperimposable mirror images of one another (enantiomers) and compounds that are superimposable mirror images of one another (diasteriomers).

I Geometric, Or Cis-trans, Isomers are stereoisomers that differ in the location of sub-stituent groups about a double bond. When the two groups are on the same side of the double bond, you are dealing with the cis isomer, while groups on opposite side of the double bond indicate a trans isomer.

Testing Your Knowledge

Apply your knowledge of the concepts outlined in Chapter 25 with these practice questions. Make sure that you are confident in your ability to distinguish between all of the different types of organic compounds and isomers before you attempt them, though! Remember that organic chemistry in particular involves a lot of memorization.

Questions 1 through 4 refer to the following types of organic compounds

1. These organic compounds do not contain any oxygen atoms.

2. You may need to use the prefix Di In naming one of these organic compounds.

3. These organic compounds are known for having pleasant aromas.

4. These organic compounds are most closely related to ketones.

5. Of the following organic compounds, which is MOST soluble in water at 298K?

(A) Hexane

(B) Hexene

(C) Ethyl Methyl Ether

(D) Octanol

(E) Methanol

6. (a) Write out the condensed structural formula of 2-pentanone.

(b) Draw the complete structural formula of 2-pentanone, showing all atoms and bonds explicitly.

(c) Draw a complete structural formula for an isomer of pentanone.

Answers with Explanations

1. (E). The halocarbon is the only one of these organic compounds that doesn’t contain an oxygen atom.

2. (C). Ethers are named by naming the substituents on either side of the oxygen atom. If the groups on either side are the same, then the prefix Di – Is used as in CH3OCH3 (dimethyl ether).

3. (B). Esters are known for their pleasant aromas.

4. (D). Aldehydes and ketones both share a double C-O bond. In an aldehyde, the carbon double bonded to the oxygen is also bonded to a hydrogen atom, ending the carbon chain. In a ketone, the carbon chain continues beyond the double bond with oxygen.

5. (E). Following the "like dissolves like" rule (review solubility in Chapter 10 for more details), alcohols should be the most soluble in water among the compounds on this list because they have the same intermolecular forces as water. Octanol’s long nonpolar hydrocarbon chain, however, balances out the polar end of the molecule to some extent, making methanol the more soluble alcohol.

6. (a) CH3CH2CH2COCH3, 2-Pentanone is a five-carbon ketone, with the double-bonded oxygen on the second carbon. This means it has a condensed structural formula of CH3CH2CH2COCH3.

(b) Translate the condensed structural formula into a complete structural drawing, giving you

H H H O H I I I II I

H — C — C — C — C — C — H III I H H H H

(c) Any five-carbon ketone or aldehyde would be an acceptable isomer, as long as it still has the formula C5H10O. 3-pentanone and pentanal are two possibilities, both shown below.

Pentanae 3-pentanone

HHHHO HHOHH I I I I II I I II I I

H — C — C — C — C — C — H H — C — C — C — C — C — H

I I I I II II

H H H H H H H H

In This Chapter

^ Pointing out what you need to know

^ Getting in some practice

^ Going through the answers and explanations

I\ Cids and bases are extremely common participants of chemical reactions, and even have their own special class of reaction (neutralization). Because acidity/basicity is such an important property of aqueous solutions, this topic travels with a whole entourage of concepts and equations. There are different ways to define acids and bases. There are different ways to express acid and base concentrations. There are special quantities for expressing just how acidic or basic a particular compound is. This chapter summarizes key points and skills from Chapter 17, and provides practice questions that test your ability to put acid-base concepts to use.

Soaking Up the Main Points

The acid-base nitty-gritty comes down to two major ideas. First, what are acids and bases, and how can you recognize them? Second, how do acids and bases react with each other? This section organizes highlights of Chapter 17 around these two categories.

Knowing and measuring acids and bases

There are several common ways to define acids and bases, all overlapping to some extent. Having defined them, you can measure their concentrations and express those measurements in different ways, as summarized in the following points.

Three important methods for defining acids and bases are as follows:

• Arrhenius acids are those that dissociate to form hydrogen ions, while bases dissociate to form hydroxide ions in solution.

• Bronsted-Lowry acids donate protons (hydrogen ions), which are accepted by the Bronsted-Lowry base or proton acceptor.

• Lewis bases donate a pair of electrons to form a covalent bond with a Lewis acid, which accepts the pair of electrons to form a coordinate covalent bond.

The acid dissociation constant Ka gives an indication of the acidity of a substance through the expression

K a

[ H3O +]x[ A -]

I HA I

Similarly, the base dissociation constant Kb indicates the basicity of a substance through the expression

[H + ][CH3COO ] [CH 3COOH ]

The pH scale ranges from 0 to 14 and measures acidity, with low numbers indicating acidic solutions, high numbers indicating basic solutions, and 7 indicating a perfectly neutral solution. It is calculated using the equation pH = – log[H+].

The pOH scale also ranges from 0 to 14, but measures basicity instead of acidity. Low numbers mean basic solutions and high numbers mean acidic solutions. It is calculated using the equation pOH = – log[OH-].

I PH and pOH are related through the simple expression pH + pOH = 14. In a related way, [H+] x [OH-] = 10-14.

_

Reacting acids and bases

Split H2O and you can get H+ and OH-, acid and base respectively. Add acid to base, and you regain water, neutralizing the acid and base reactants. The idea is simple, but neutralization reactions can be tricky. The following points lay out the core principles.

Neutralization reactions occur when acids and bases are mixed together. The amount of acid needed to neutralize a base or vice versa depends not only on the molarity of the acidic solution, but on the number of hydrogen atoms that the acid can donate to neutralize the hydroxide ions of the base.

Equivalents of acid and base are considered rather than molarities, because different compounds can contribute different numbers of H+ or OH-. Equivalents are calculated by multiplying the number of moles of compound by the number of hydrogen or hydroxide ions being contributed by each molecule of compound.

I Titrations utilize neutralization reactions in order to determine the concentration of a mystery acid or base. They’re accomplished through the following six steps:

1. Measure out a small volume of the mystery acid or base.

2. Add a pH indicator such as phenolphthalein (see Chapter 28 for details about how to choose the best pH indicator for a reaction). Phenolphthalein, for example, is colorless in acidic solutions but pink or purple in basic solutions.

3. Neutralize by dropping the acid or base of known concentration into the solution until the indicator shows that it’s neutral (by changing color), keeping careful track of the volume added.

4. Calculate the number of moles added by multiplying the number of liters of acid or base added by the molarity of that acid or base to get the number of moles added.

5. Account for equivalents.

6. Solve for molarity. Divide the number of moles of the mystery acid or base by the number of liters measured out in Step 1, giving you the molarity.

IU Buffered solutions are made of a weak acid and its conjugate base, or a weak base and its conjugate acid, and resist changes in pH as long as only a small amount of acid or base is added. The best buffers are those that have a pKa value equal or nearly equal to the desired pH of the solution. Note that pKa = – logKa and pKb = – logKb.

PH and pKa are related through the Henderson-Hasselbach equation as follows:

Nnom 0.315mol CH3COOH 1molNaOH 1L N mi M R>u

0.080Lx-;—3-x–, _„ x -,,, „„ = 0.10LNaOH

L 1mol CH3COOH 0.250mol NaOH

3

Practice Questions

With so many concepts and equations swirling about, its clear that acids and bases just aren’t the kind of chemistry you simply read about and leave be. You are guaranteed to encounter them on the AP exam, and you are equally guaranteed to be confused by them — unless you practice, that is. Here are some targeted questions to help give you that practice.

Multiple choice

1. A solution prepared by mixing 15mL of 1.0M H2SO4 and 15mL of 2.0M KOH has a pH of

(A) 6.

(B) 7.

(C) 8.

(D) 9.

(E) 10.

2. H2O < NH3 < Ca(OH)2 < LiOH < KOH < OH-

Six bases are listed above in order of increasing base strength. Which of the following reactions must have an equilibrium constant with a value less than 1?

(A) Ca(OH)2 + K2CO3 — CaCO3 + 2KOH

(B) NH3 + H2O — NH4 + OH-

(C) 2LiOH + CO2 — Li2CO3 + 3H2O

(D) LiOH + H2O — LiH + KOH

(E) KOH + H2O — KH + OH-

3. Which of the following equations represents the reaction between solid magnesium hydroxide and aqueous hydrochloric acid?

(A) Mg(OH)2(s) + 2HCl(/) — MgCl2(a<7) + H2O(I)

(B) MgOH(s) + HCl(a<7) — MgCl(ag) + 2H2O(I)

(C) Mg(OH)2(s) + 2HCl(a<7) — MgCl2(ag) + 2H2O(I)

(D) Mg(OH)2(s) + HCl(a<7) — MgCl2(ag) + H2O(I)

(E) MgOH(s) + 2HCl(ag) — MgCl(ag) + 2H2O (I)

4. What is the ideal pKa for an indicator in a titration when the pOH at the equivalence point is 9.8?

(A) 2.1

(B) 4.2

(C) 4.9

(D) 9.8

(E) 10

Free response

5. CH3COOH(a< ) — H+(a< ) + CH3COO-(a< )

Acetic acid, CH3COOH, is a weak acid that dissociates in water according to the equation above.

(a) Calculate the pH of the solution if the H+ concentration is 3.98×10-4M.

(b) A solution of NaOH is titrated into a solution of CH3COOH.

(i) Calculate the volume of 0.250M NaOH needed to reach the equivalence point when titrated into an 80mL sample of 0.315M CH3COOH.

(c) Calculate the number of moles of CH3COO-Na+ that would have to be added to 150mL of 0.025M CH3COOH to produce a buffered solution with [H+] = 4.25 x 10-7M if the Ka Of acetic acid is 1.8×10-5. Assume that the volume change is negligible.

6. C5H5N(a<7) + H2O(I) — C5H5NH+(a<7) + OH"(aq)

Pyridine (C5H5N), a weak base, reacts with water according to the reaction above.

(a) Write the equilibrium constant expression, KB, for the reaction above.

(b) A sample of pyridine is dissolved in water to produce 50mL of a 0.20M solution. The pH of the solution is 8.50. Calculate the equilibrium constant, KB, for this reaction.

Answers with Explanations

Have any doubts about your answers to the practice questions? Neutralize those doubts by inspecting the answers given here. Even if you find that you’ve answered all the questions correctly, it’s a good idea to read the explanations to solidify your acid-base genius and because, well, you might have been simply lucky on one or two.

1. (B). Problems like this are really just limiting reagent problems. The acid and base will react with one another until one or the other is used up and the amount of remaining reac-tant will determine the pH. In this case, however, the 1.0M diprotic sulfuric acid and the 2.0M potassium hydroxide exactly balance one another:

0.015L H2SO4 1.0mo/ H2SO4 2H + nnon, ,,+

-t^2—1 x—, T,, *—1 x, 2 " = 0.030mo/ H+

1 1L H2SO4 1H2SO4

0.015L KOH x 2.0mo/ K°H x |_OH_ = 0.030mo/ OH-1 1L KOH 1 KOH

The acid has two protons to donate per molecule and the base is twice as concentrated. They’re mixed in equal amounts, so there is no leftover acid or base. The resulting pH is perfectly neutral.

2. (C). If the equilibrium constant is less than 1, then at equilibrium the concentration of reactants must be greater than the concentration of products. Because the information given is regarding base strength, the problem must be dealing with the tendency of dissociation (and therefore concentration) to increase with the strength of the base. You’re looking for a reaction where the stronger base appears on the lefthand side of the reaction arrow. The only reaction where this is the case is C.

3. (C). You’re looking for a balanced reaction, which shows the proper phases for all of the products and reactants. C is the only reaction that fulfills both of those criteria and contains molecules with atoms combining in the proper ratios for charge balance.

4. (B). The ideal PKA Is equal to the pH at the equivalence point. Use the given pOH to determine this value using the equation pH + pOH = 14. This equation gives you 14 – 9.8 = 4.2.

5. (a) Plug the given H+ concentration into the equation pH = – log[H+], which gives you pH = 3.40.

^uaaaoai 0.315molCH3COOH 1molNaOH 1L Aiaim^h

(b) 0.080Lx-R—3-x, x „ »,.,,.,,,,„„ = 0.10LNaOH

L 1mol CH3COOH 0.250mol NaOH

3

(c) 159mL. You’re asked to solve for the number of moles of CH3COO-Na+ to be added. The easiest way to do this is to go through the CH3COO- concentration, which is equivalent to the CH3COO-Na+ concentration because it dissociates in water. Begin by writing out the expression for the acid dissociation constant.

Then, solve for the concentration of CH3COO-:

0080, x 0.315 MolCH3COOH x 1 mol NaOH x 1L = 0i0iNaOH

0.080Lx-l-x 1 mol CH3COOH x 0.250 molNaOH = 0.10LNaOH

Finally, multiply this molarity by the volume to get the number of moles added.

6. (a) The equilibrium constant is the concentration of the products over the concentration of the reactants, excluding water. This results in the expression

[C5H5NH+]x[0H]

KB =-CH N-

(b) 2.3 x 10-9. Use the given pH to find the pOH and from there the OH- concentration. Because pH = 9.33, the pOH = 14 – 9.33 = 4.67. Use this to solve for [OH-].

[OH-]= 10 467 = 2.14x 10 5

This concentration should be equivalent to the [C5H5NH+] concentration. Plug these and the given C5H5N concentration into the equation derived in part (a) to get your KB, giving you

(2.14 x 10-5 )2 Kb =±-0~20— = x 10 9

Favorable AH cannot overcome unfavorable TAS

Favorable TAS cannot overcome unfavorable AH

AH

TAS

But wait! The second law of thermodynamics states that all spontaneous processes occur with a positive change in entropy — how can a chemical reaction be driven forward by enthalpy alone? Good question. Answer: The second law refers to the entropy of the Universe. The Gibbs equation refers to the enthalpy and entropy of the System. A decrease in the enthalpy of a system means an increase in the entropy of the surroundings, which means an increase in the entropy of the universe. The second law holds true. What a relief.

Under standard conditions (typically, 1atm, pure substances at 1M concentration), the state functions are said to be in standard state and the state functions of the Gibbs equation are annotated with the ° symbol:

AG°= AH° – TAS°

However, while most free energy changes are quoted at 298K, technically this is not defined by the "standard conditions." Texts will provide tables where the temperature is clearly indicated, as in AGo298. Some texts may omit this, in which case 298K temperature must be assumed.

Changes in standard free energy, standard enthalpy, and standard entropy for a reaction can be calculated directly from the stoichiometry of the reaction along with tabulated values. The tabulated values are those for the Standard free energies, enthalpies, or entropies of formation Of the reactants and products, AG/1, AHf°and AS/1, respectively:

AG°= X aAGf°(products) – X bAGf°(reactants) AH° = X aAHf°(products) – X bAHf°(reactants) AS° = X aAS/°(products) – X bASf°(reactants)

Where X refers to the sum of products or reactants and where A And B Refer to the stoichio-metric coffiecients of each product or reactant. Standard enthalpies and entropies of formation are the enthalpies and entropies associated with forming a compound under standard conditions (1atm, 298K, 1mol L-1) from its component elements, in their elemental forms under standard conditions.

The relationship between the standard free energy and the free energy under nonstandard conditions is:

AG = AG°+ RT Ln Q

Where Q Is the reaction quotient (see Chapter 15). Compare this equation to Keq = E ~hGIBJ And be sure you understand the differences.

Measuring Heat: Thermochemistry and Calorimetry

Energy shifts between many forms. It may be tricky to detect, but energy is always conserved. Sometimes energy reveals itself as heat. Thermodynamics explores how energy moves from one form to another. Thermochemistry Investigates changes in thermal energy that accompany chemical reactions. To understand how thermochemistry is done, you need to first understand how the particular form of energy called thermal energy and the elusive term "heat" fit into the overall dance of energy and matter. In this section, we discuss how thermal energy fits into the overall energy picture, and we describe how you can measure heat transfer during chemical reactions by using a concept called heat capacity and a technique called calorimetry.

Seeing heat within the bigger energy picture

You’re probably familiar with the idea that objects can have greater or lesser amounts of energy associated with them. One of the kinds of energy an object can possess is thermal energy. In this section, we show how thermal energy relates to the overall energy of a system (like a chemical system), and how thermal energy relates to heat.

Energy itself can be divided into

Potential energy (PE) Is energy due to position. Chemical energy Is a kind of potential energy, arising from the positions of particles within systems.

Kinetic energy (KE) Is the energy of motion. Thermal energy Is a kind of kinetic energy, arising from the movement of particles within systems.

The total Internal energy Of a system (E) Is the sum of its potential and kinetic energies. When a system moves between two states (as it does in a chemical reaction), the internal energy may change as the system exchanges energy with the surroundings. The difference in energy (AE) Between the initial and final states derives from heat (q) Added to or lost from the system, and from work (w) Done by the system or on the system. "Heat" in physics and chemistry is strictly defined as a quantity of thermal energy being transferred between one system and another. As a result nothing can "contain" heat!

We can summarize these energy explanations with the help of a couple of handy formulas:

Etotal = KE + PE

AE = Efinal – Einitial = Q + W

What kind of "work" can atoms and molecules do in a chemical reaction? Remember (maybe) from your physics class that work is defined in terms of motion against a force. No motion, no work! One kind of work that is easy to understand is Pressure-volume work. Consider the following reaction:

CaCO3(s) — CaO(s) + CO2(g)

Solid calcium carbonate decomposes into solid calcium oxide and carbon dioxide gas. At constant pressure (P), This reaction proceeds with a change in volume (V). The added volume comes from the production of carbon dioxide gas. As gas is made, it expands, pushing against the surroundings. The carbon dioxide gas molecules do work as they push into a greater volume:

W = -PAV

The negative sign in this equation means that the system loses internal energy due to the work it does on the surroundings. If the surroundings did work on the system, thereby decreasing the system’s volume, then the system would gain internal energy. Note that if there were no pressure (resistive force), then P = 0, so W = 0 . . . so there is no work! So, expanding into a vacuum a gas does no work!

So, pressure-volume work can partly account for changes in internal energy during a reaction. When pressure-volume work is the only kind of work involved, any remaining changes come from heat. As mentioned in the previous section, Enthalpy (H) Corresponds to the thermal energy content of a system at constant pressure. An Enthalpy change (AH) In such a system corresponds to heat. The enthalpy change equals the change in internal energy minus the energy used to perform pressure-volume work:

AH = AE – (-PA V) = AE + PAV

Although E, P, V, And H Are state functions, heat (q) Is Not A state function, but is simply a quantity of thermal energy that transfers from a warmer object to a cooler object.

Now breathe. The practical consequences of all this theory are the following: W Chemical reactions involve an amount of heat, Q.

W Chemists monitor changes in thermal energy by measuring changes in temperature.

W At constant pressure, the change in thermal energy content equals the change in enthalpy, AH.

W Knowing AH values helps to explain and predict chemical behavior.

Using heat capacity in calorimetry

Heat is a quantity of thermal energy that transfers from warmer objects to cooler objects. But how much thermal energy can an object hold? If objects have the same thermal energy content, does that mean they are the same temperature? You can measure temperature changes, but how do these temperatures relate to heat? These kinds of questions revolve around the concept of Heat capacity. Heat capacity is the amount of heat required to raise the temperature of a system by 1K (or 1°C). In this section, we describe how the concept of heat capacity is used in an experimental technique called calorimetry.

It takes longer to boil a large pot of water than a small pot of water. With the burner set on high, the same amount of heat transfers into each pot, but the larger pot of water has a higher heat capacity. So, it takes more heat transfer to increase the temperature of the larger pot.

You’ll encounter heat capacity in different forms, each of which is useful in different scenarios. Any system has a heat capacity. To best compare heat capacities between chemical systems you use one of the following:

W Molar heat capacity, Which is the heat capacity of 1 mole of a substance

W Specific heat capacity, Or just Specific heat, Which is the heat capacity of 1 gram of a substance

How do you know whether you’re dealing with heat capacity, molar heat capacity, or specific heat capacity? Look at the units:

W Heat capacity: Energy / K

W Molar heat capacity: Energy / (molK)

W Specific heat capacity: Energy / (g°K)

Fine, but what are the units of energy? Well, that depends. The SI unit of energy is the Joule (J) (see Chapter 3 for more about the International System of units), but the units of Calorie (cal) And Liter-atmosphere (Latm) May also be used. Here’s how the joule, the calorie, and the liter-atmosphere are related (AP will stick to joules!):

1 J = 0.2390 cal 101.3 J = 1 Latm

Note that in everyday language about nutrition, a "calorie" actually refers to a "Calorie" or a kilocalorie — a calorie of cheesecake is 1,000 times larger than you think it is. Why do you think the American food industry wants to keep it that way? In Europe and Canada, candy bars are rated in kilojoules, so a 100 calorie bar is 418 kJ of energy.

Calorimetry Is a family of techniques that puts all this thermochemical theory to use. When chemists do calorimetry, they initiate a reaction within a defined system, and then measure any temperature change that occurs as the reaction progresses. There are a few variations on the theme of calorimetry, each measuring heat transfer under different conditions:

W Constant-pressure calorimetry Directly measures an enthalpy change (AH) For a reaction because it monitors heat transferred at constant pressure: AH= qP.

Typically, heat is observed through changes in the temperature of a reaction solution. If a reaction warms a solution, then that reaction must have released thermal energy into the solution. In other words, the change in thermal energy content of the reaction (qReaction) has the same magnitude as the change in thermal energy for the solution (qSolution) so the heat into one is the same as the heat out of the other, but has opposite

Sign: qsolution = – qreaction.

So, measuring QSolution allows you to calculate QReaction, but how can you measure QSolution? You do so by measuring the difference in temperature (AT) Before and after the reaction:

Qsolution = (mass of solution) x (specific heat of solution) xAT

In other words,

Q = MCPAT

Here, "m" is the mass of the solution and CP is the specific heat capacity of the solution at constant pressure. AT Is equal to TFinal – TInitial.

When you use this equation, be sure that all your units match. For example, if your CP has units of J g-1 K-1, don’t expect to calculate heat flow in kilocalories.

W Constant-volume calorimetry Directly measures a change in internal energy, AE (not AH) For a reaction because it monitors heat flow at constant volume. Often, AE and AH are very similar values, especially if no gases are involved to do pressure-volume work.

A common variety of constant-volume calorimetry is Bomb calorimetry, A technique in which a reaction (often, a combustion reaction) is triggered within a sealed vessel called a bomb. The vessel is immersed in a water bath of known volume. The vessel and water together are considered the "calorimeter" and are enclosed by an insulated container. The temperature of the water is measured before and after the reaction. Because the heat capacity of the calorimeter (CCal) is known, you can calculate heat from the change in temperature:

Q = -CCal xAT

Dealing with Heat by Using Hess’s Law

So, as you find out in the previous section, you can monitor heat by measuring changes in temperature. But what does any of this have to do with chemistry? Chemical reactions transform both matter and energy. Though reaction equations usually list only the matter components of a reaction, you can consider thermal energy as a reactant or product as well. When chemists are interested in heat appearing during a reaction (and when the reaction is run at constant pressure), they may list an enthalpy change (AH) To the right of the reaction equation. As we explain in the previous section, at constant pressure, heat equals AH:

QP = AH= Hfinal – Hinitial

If the AHListed for a reaction is negative, then that reaction releases heat as it proceeds — the reaction is Exothermic. If the AHListed for the reaction is positive, then that reaction absorbs heat as it proceeds — the reaction is Endothermic. In other words, exothermic reactions release heat as a product, and endothermic reactions consume heat as a reactant.

The sign of the AH tells us the direction of heat, but what about the magnitude? The coefficients of a chemical reaction represent molar equivalents. So, the value listed for the AH refers to the enthalpy change for one molar equivalent of the reaction. Here’s an example:

CH4(g) + 2O2(g) — CO2(g) + 2H2O(g) AH = -802 kJ

The reaction equation shown describes the combustion of methane, a reaction you might expect to release heat. The enthalpy change listed for the reaction confirms this expectation: For each mole of methane that combusts, 802 kJ of heat are released. The reaction is highly exothermic. Based on the stoichiometry of the equation, you can also say that 802 kJ of heat are released for every 2 moles of water produced.

So, reaction enthalpy changes (or reaction "heats") are a useful way to measure or predict chemical change. But they’re just as useful in dealing with physical changes, like freezing and melting, evaporating and condensing, and others. For example, water (like most substances) absorbs heat as it melts (or Fuses) And as it evaporates:

Molar enthalpy of fusion: AHfus = 6.01 kJ

Molar enthalpy of vaporization: AHvap = 40.68 kJ

The same sorts of rules apply to enthalpy changes listed for chemical changes and physical changes. Here’s a summary of the rules that apply to both:

W The heat absorbed or released by a process is proportional to the moles of substance that undergo that process. Two moles of combusting methane release twice as much heat as does one mole of combusting methane.

W Running a process in reverse produces heat flow of the same magnitude but of opposite sign as running the forward process. Freezing one mole of water releases the same amount of heat that is absorbed when one mole of water melts.

Yikes. Earlier in the chapter we described that chemical reactions can be complicated, multi-step sorts of things. Now we’re telling you that both chemical and physical processes can be associated with heat. How can you possibly keep track of all these heats? The answer is simple: Hess’s Law.

Imagine that the product of one reaction serves as the reactant for another reaction. Now imagine that the product of the second reaction serves as the reactant for a third reaction. What you have is a set of coupled reactions, connected in series like the cars of a train:

A — B and B — C and C — D

So,

A —B —C —D

You can think of these three reactions adding up to one big reaction, A — D. What is the overall enthalpy change associated with this reaction, AHAD? Here’s the good news:

AHAD = AHAB + AHBC + AHCD

Enthalpy changes are additive. Hess’s Law declares that the total enthalpy change for the overall reaction is the sum of the enthalpy changes for each of the steps, as summarized in Figure 21-8.

But the good news gets even better. Imagine that you’re trying to figure out the total enthalpy change for the following multistep reaction:

Multi-step reaction: A — B — C — D

Equivalent to Figure 21-8:

Schematic Single-step reaction: illustration a — D of Hess’s Law.

AH.,

A

AH„,

AH„,

Here’s a wrinkle: For technical reasons, you can’t measure this enthalpy change, AHXZ, directly, but must calculate it from tabulated values for AHXY and AHYZ. No problem, right? You simply look up the tabulated values and add them. But here’s another wrinkle: When you look up the tabulated values, you find the following:

AHXY = -37.5 kJmol"1 AHZY = -10.2 kJmol-1

Gasp! You need AHYZ, but you’re provided only AHZY! Relax: Reactions are reversible. The enthalpy change for a reaction has the same magnitude and opposite sign as the enthalpy change for the reverse reaction. So, if AHZY = -10.2 kJmol-1, then AHYZ = 10.2 kJmol-1. It really is that simple. So,

AHXZ = AHXY + (-AHZY) = -37.5 kJmol-1 + 10.2 kJmol-1 = -27.3 kJmol-1 Thanks be to Hess.

In This Chapter

^ Going over important points to remember

^ Putting in some practice with example questions

^ Seeing how your answers stack up

/f we had to name one single thing that is key to acing the AP chemistry exam, a solid understanding of stoichiometry would be it. Stoichiometry is chemistry’s bread and butter. Make sure that you understand all of the concepts outlined in Chapter 13 in detail and you will score some major points on exam day. To ensure you nail the concepts of stoichiometry, we not only include the major points that are essential to your success, but we also include practice questions near the end of the chapter so you can sharpen your test-taking skills on stoichiometry.

Making Sure You React to Questions With the Right Answers

Stoichiometry and the skills that surround it might seem mundane, but they are central skills that let you grapple with the very heart of chemistry: reactions. It’s well worth your time to make sure you know these basics back and forth.

Knowing reactions when you see them

These tips and reminders are all about recognition — knowing how to decode chemical formulas and reaction equations to understand what reaction is taking place, and even how to predict the products of a reaction when all you’re given is the reactants.

Reaction equations use symbols to indicate details of the reaction. Review the symbols in Chapter 13 in Table 13-1 and make sure that you’re familiar with all of them.

Make sure that you’re familiar with the seven common reaction types and are able to identify them and predict products if given only reactants. It is not so important to memeorize the names of the types as it is to be able to predict products and balance

Equations. Many reactions on tests expect "familiar" compounds as products, not truly exotic ones, so don’t get hung up on memorizing lots of exotic compounds!

• Synthesis/Combination reactions: A + B — C. Two or more reactants combine to form a single product.

• Decomposition reactions: A — B + C. Four types of decomposition reaction are seen quite commonly on the AP chemistry exam. These are

Metal carbonate — metal oxide + CO2(g)

Metal chlorate — metal chloride + O2(g)

Metal hydroxide — metal oxide + H2O

Acid — nonmetal oxide + H2O

• Single replacement reactions: A + BC — B + AC. Single replacement reactions are all redox reactions! The likelihood of single replacement reactions occurring relies on the reactivity series of metals outlined in Table 13-2. A will only displace B if it is a more-reactive species, which means it appears higher in the reactivity series.

• Double replacement reactions: AB + CD — AD + CB. A and D will not necessarily combine in the same ratios as A and B did. Take careful account of the charges of each species and make sure to balance the equation in the end.

• Combustion reactions: These contain oxygen among the reactants. The most common types of combustion reactions are those where hydrocarbons are burned to form water and carbon dioxide as products, but a familiar trap is to forget that substances that already contain oxygen, such as ethanol (C2H6O), also "burn" to form carbon dioxide and water. All combustion reactions are also redox reactions.

• Neutralization, or acid/base reactions: Acid + Base — Water + Salt. Chapter 17 explains these in greater detail.

• There are also a number of acid-base reaction types in addition to neutralization reactions. These are

Nonmetal oxide + H2O — acid

Metal oxide + H2O — base

Metal + H2O(l) — base + H2(g)

Active metal + acid — salt + H2(g)

Carbonate + acid — CO2(g) + H2O(l) + salt

Base + salt — metal hydroxide precipitate + salt

Strong base + salt containing NH4 — NH3 + H2O + salt

• Redox reactions are outlined in detail in Chapter 19. They typically involve the transfer of electrons between one reactant and the other. The species that loses the electrons is oxidized and that which gains the electrons is reduced. The mnemonics "LEO and GER" or "OIL RIG" will help you remember this information.

Dealing with the numbers that flow from stoichiometry

Coming up with accurate formulas and reaction equation isn’t the end of the story — it’s usually just the beginning. Once you’ve got an equation, you’ve got to make sure that it’s balanced, and then use the correct stoichiometry to solve problems. These tips deal with that process.

*e Avogadro’s number, 6.022 x 1023, is equivalent to the number of particles in one mole of a substance. By "particles" we mean atoms, ions, or molecules — whichever is relevant for the circumstance.

*e The gram atomic, molecular, or formula mass of a substance is calculated by adding up the atomic masses of all of its components (multiplying by the number of atoms of each substance present where applicable). It’s equivalent to the mass in grams of one mole of particles of that substance.

*e At standard temperature and pressure, one mole of an ideal gas would occupy 22.4 liters volume. No gases are truly ideal (see Chapter 9 on gas laws for further details). However, this approximation is a very good one in many cases and is very useful in converting volume to moles of a gas and vice versa.

E The percent composition of a substance takes into account the mass ratios of the atoms that make it up as well as the ratios in which they combine. To calculate a percent composition of a molecular or ionic compound, calculate the gram molecular or gram formula mass of the compound. Then, calculate the mass of each type of atom in the substance (by multiplying the mass by the number of atoms present in each molecule) and divide by the total mass of the compound.

E You can also use percent compositions to find the formula of a compound, though it is a somewhat complicated process. If you follow the five following steps, you’ll get it every time.

1. Assume that you have 100g of the unknown compound.

2. Convert the assumed masses from Step 1 into moles by using gram atomic masses.

3. Divide each of the element-by-element mole quantities from Step 2 by the lowest among them.

4. If any of your mole ratios aren’t whole numbers, multiply all numbers by the smallest possible factor that produces whole-number mole ratios for all of the elements.

5. Write the empirical formula by attaching these whole-number mole ratios as subscripts to the chemical symbol of each element. Order the elements according to the general rules for naming ionic and molecular compounds (described in Chapter 5).

E Given an empirical formula and either a formula or molar mass for a compound, you can also determine its molecular formula by dividing the gram formula mass by the empirical formula mass and then by taking this whole number value and multiplying the subscripts for each atom in the empirical formula by that number.

E Before you take the AP exam, make sure you’ve mastered the method for balancing equations. Unless the AP exam tells you that an equation is balanced, always check. To check that an equation is properly balanced, simply count the total number of each type of atom on each side of the equation and make sure that they match. Also be sure to check that any total charges on each side are balanced.

E Always consider the possibility of limiting reagents in reactions. Limiting reagents determine theoretical yields. Percent yield is 100 percent times the actual yield divided by the theoretical yield.

Testing Your Knowledge

Now that you’ve grounded yourself in a solid knowledge of reactions and stoichiometry, see if you can keep your balance through these practice questions.

Multiple choice

1. Which of the following represents a process in which a species is reduced?

(A) Mg — Mg2+

(B) Cl2 — Cl-

(C) Ni3+— Ni4+

(D) CO — CO2

(E) NO2- — NO3-5%

2. A sample of Li2SO4 (molar mass 110g) is reported to be 10.2% lithium. Assuming that none of the impurities contain any lithium, what percentage of the sample is pure lithium sulfate?

3. When the following equation for the acid base reaction above is balanced and all of the coefficients are reduced to lowest whole-number terms, the coefficient on the H2O is

_H2SO4 +_Ca(OH)2 —_CaSO4 +_H2O

4. What is the simplest formula for a compound containing only carbon and hydrogen and containing 16.67% H?

5. A substance has an empirical formula of CH2O and a molar mass of 180.0. What is its molec -

6. In the reaction

2Fe2O3 + 3C — 4Fe + 3CO2 if 500.g of Fe2O3 reacts with 75.0g of C, how many grams of Fe will be produced?

(A) 56.4g

(B) 75.0g

(C) 316.g

(D) 349.g

(E) 465.g

7. The mass of 5 atoms of lead is

(A) 3.44 x 10-20g.

(B) 1.72 x 10-21g.

(C) 8.60 x 10-21g.

(D) 1.20 x 10-23g.

(E) 6.02 x 10-23g.

8. AgNO3(aq) + KCl(aq) —_+_

What are the missing products?

(A) AgNO3(aq) + KCl(aq) (no reaction)

(B) AgCl2(s) + K2NO3(aq)

(C) AgCl(s) + KNO3(aq)

(D) Ag2+(aq) + Cl-(aq) +KNO3 (aq)

(E) AgCl2(s) + K+(aq) + Cl – (aq)

9. HCl + NaOH — NaCl + H2O

If 30.0g of HCl is mixed with excess NaOH according to the reaction above and 41.0g of NaCl is produced, what is the difference between the actual and the theoretical yields for this reaction?

(A) 7.1g

(B) 8.3g

(C) 15.5g

(D) 41.0g

(E) 48.1g

Free response

10. Do each of the following conversions based on the reaction below. SiO2 (s) + 6HF(g) — H2SiF6(aq) + 2H2O(l)

(a) 5.25g HF to moles

(b) 6.00 moles SiO2 to grams of H2SiF6

(c) 10.0 moles HF to liters HF (at STP)

(d) 3.20 mol SiO2 to molecules H2SiF6

11. Write the formulas to show the reactants and the products for the reactions described below. Assume that the reaction occurs and that solutions are aqueous unless otherwise indicated. Represent substances in solution as ions if the substances are extensively ionized. Omit spectator ions or molecules. You do not need to balance the equations.

(a) Iron (II) sulfide is added to a solution of hydrochloric acid.

(b) The combustion of methane in large excess of air.

(c) Calcium carbonate is strongly heated (hint: carbon dioxide is one of the products).

(d) A solution of potassium hydroxide is added to solid ammonium chloride.

(e) A strip of iron is added to a solution of copper (II) sulfate.

(f) Chlorine gas is bubbled into a solution of sodium fluoride.

(g) Dilute sulfuric acid is added to a solution of strontium nitrate.

(h) Copper ribbon is burned in oxygen.

12. For each of the following three reactions, in part one for each reaction write a balanced equation for the reaction and in part two for each reaction answer the question about the reaction. In the balanced equation, coefficients should be in terms of lowest whole numbers. Assume that solutions are aqueous unless otherwise indicated. Represent substances in solution as ions if the substances are extensively ionized. Omit formulas for any ions or molecules that are unchanged by the reaction.

(a) Excess nitric acid is added to solid sodium bicarbonate.

(i) Balanced Equation:

(ii) What is the minimum mass of sodium bicarbonate that must be added in order to react completely with 25.0g of nitric acid?

(b) A solution containing the magnesium ion (an oxidizing agent) is mixed with a solution containing lead(II) (a reducing agent).

(i) Balanced Equation:

(ii) If the contents of the reaction mixture described above are filtered, what sub-stance(s), if any, would remain on the filter paper?

(c) A solution of sodium hydroxide is added to a solution of copper (II) sulfate.

(i) Balanced Equation:

(ii) What is the percent composition of oxygen in sodium hydroxide?

13. For each of the following, use appropriate chemical principles to explain the observation.

(a) Lithium metal will react with water, but copper will not.

(b) The human body is made of 60% to 70% water, but hydrogen is only the third most abundant element in the body by mass.

(c) Water can act as both an acid and a base.

14. A compound is 36.5% sodium, 25.4% sulfur, and 38.1% oxygen.

(a) What is its empirical formula?

(b) If the compound in (a) has a molar mass of 366.3g, what is its molecular formula?

15. Calculate the percent composition of each element in caffeine C8H10N4.

Checking Your Work

Still on your feet? Good. Now check with the judges and receive your score. If you faltered, find out why. If you triumphed, enjoy it.

1. (B). This reaction is the only one where electrons are gained by the reactant to form the product, a sure sign of a reduction reaction. The negative sign on the chlorine is an indicator that extra electrons are present, and the fact that the reactant is of neutral charge cements the reaction as reduction.

2. (D). Begin by finding the percent composition of lithium in pure lithium sulfate, 14.0g * 110g = 12.7%. Impurities, however, have driven the lithium concentration down to 10.2% lithium. Dividing this percentage by the first will give you the percent of the pure compound in the sample. 10.2 * 12.7 = 80.3%, which is closest to answer choice D.

3. (B). There are two tricks to balancing an equation like this. The first is to begin by balancing the polyatomic ions, in this case SO4 and OH-. But where is the OH- among the products? This point is the second important one. In balancing equations, particularly neutralization reactions like this one, it is often extremely useful to write H2O as HOH. The fully balanced equation here is H2SO4 + Ca(OH)2 — CaSO4 + 2H2O. That’s right. Water is the only component that takes a subscript greater than 1, and that subscript is 2.

4. (C). Here you’re given a percent composition and are asked to derive an empirical formula. Follow the steps outlined in Chapter 13 to accomplish this task. Or, since the answers are right in front of you and percent composition is really a fairly easy calculation to make, you can simply calculate the percent composition of hydrogen in each of the compounds among the answer choices and see which one matches the question. You will find that this answer is choice C, pentane.

5. (E). Calculate the empirical formula mass of CH2O (30g per mole) and divide the molar mass by it, giving you 180 * 30 = 6. Multiply each of the subscripts in the empirical formula to give you the molecular formula C6H12O6, answer choice E.

6. (D). First, identify the limiting reagent in the reaction. Converting grams of iron (III) oxide to grams of carbon reveals that 56.4g of carbon are needed to react completely with 500.g of iron (III) oxide.

500g Fe2O3 1mol Fe2O3

3molC x 12.0gC

1

159.7gFe2O3 2mol Fe2O3 1mol C

56.4gC

This discovery means that there is excess carbon among the reactants and iron oxide should be used as the limiting reagent. With this established, all that remains is to convert grams of iron (III) oxide to grams of the iron product according to the calculation

500gFe2O3 x 1moFe2O3 x 4molFe x 55.8gFe = 3494gFe 1 159.7gFe2O3 2mol Fe2O3 1mol Fe

This matches answer choice D.

7. (B). Remember that you must convert to moles first and cannot convert directly from particles to grams. Set up your three conversion factors as follows and you get an answer matching answer choice B.

5atoms Pb x_1mol Pb_x 207.2g Pb = 1 72 x 10-21 „

1 6.022 x 1023 atoms Pb 1mol Pb

8. (C). This choice represents a double replacement reaction. The first thing that you will need to do is deduce the charge on the silver. As a diligent AP chemistry student, you should already have memorized the charges on the common polyatomic ions and know that NO3 carries a charge of -1. Your excellent powers of deduction should then lead you to the conclusion that you are dealing with Ag+, whose charge balances that of NO3- in a 1:1 ratio. When silver and potassium switch places, the silver and chlorine and potassium and nitrate will all combine in 1:1 ratios, giving the products AgCl and KNO3, which is consistent with answer choice C.

9. (A). By telling you that NaOH is present in excess, the problem saves you from having to do a limiting reagent calculation. All you need to do now is to convert grams of HCl, your limiting reagent, to grams of the NaCl product as in the calculation

30:0gHCl x x 1m±NaCi x 58.5gNaCl — 48.1gNaCl

1 36.5gHCl 1mol HCl 1mol NaCl

Be careful, though, that you’ve read the problem carefully. It does not ask for the theoretical yield or the percent yield, but the Difference Between the theoretical and actual yields. Subtract the two to get 48.1g – 41.0g = 7.1g, answer choice A.

10. For all of the conversions in this problem, be careful to set up your conversion factors so that the units cancel correctly.

Here is the answer for (a):

5.25gHF x j^HF = 0.263molHF 1 20.0gHF

Here is the answer for (b):

6.00mol SiO2 x 1mol H2SiF6 x 144.1gH2SiF6 = h SiF 1 1molSiO2 1molH2SiF6 g 2 6

Here is the answer for (c):

10.0mol HF x 22.4LHF — 224i HF 1 1mol HF

Here is the answer for (d):

3.20molSiO2 1molH2SiF6 6.022 x 1023molecules H2SiF6 = 3 1024 l l H

-1-x —"- — x-"- — -— 1.93 x 10 molecules H2SiF6

1 1mol SiO 2 1mol H 2SiF6

11. (a) The complete, balanced reaction here is FeS(s) + 2HCl(aq) — FeCl2(aq) + H2S(g). However, the problem has asked you to eliminate spectators and also allows you to ignore balancing. Chlorine is the only spectator here, so eliminating it from both sides leaves you with FeS(s) + H+(aq) — Fe2+(aq) + H2S(g).

(b) This reaction does not take place in solution, so there will not be any spectators. Recall that all combustion reactions require the reactant O2 and that the products are always water and carbon dioxide. With this knowledge, you are left with CH4 + O2 — CO2 + H2O, which you do not need to balance. In minimal air, some CO will form.

(c) Since this substance is "strongly heated" you should be able to identify it as a decomposition reaction immediately. If one of the products is CO2, the remaining product must be CaO, giving you the reaction CaCO3 — CaO + CO2.

(d) This is a double replacement reaction of the pattern. KOH (aq) + NH4Cl(s) — NH4OH(aq) + KCl(aq). You know that both of the products are soluble because one contains ammonium and the other contains an alkali metal, salts of both of which are always soluble. Eliminating spectators, this leaves you with the equation NH4Cl(s) — NH4+ (aq) + Cl- (aq).

(e) This is both a single replacement reaction and a redox reaction. Iron is more reactive than copper and so displaces it in solution. (Recall that you are to assume that a reaction always occurs in this type of problem, so even if you didn’t remember that iron was higher than copper on the reactivity series, you could still treat it as such.) Eliminating spectators, your net equation is Fe(s) + Cu2+(aq) — Cu(s) + Fe2+(aq).

(f) Nothing happens here — fluorine is a stronger oxidizing agent than chlorine.

(g) This double replacement reaction follows the equation H2SO4(aq) + Sr(NO3)2(s) — HNO3(aq ) + SrSO4(s), where you know that strontium sulfate is a precipitate because it is insoluble. Eliminating spectators, you are left with the equation SO42-(aq) + Sr(NO3)2(s) — NO3-(aq) + SrSO4(s).

(h) This is a combustion/combination reaction in which a metal is burned to create a metal oxide according to the equation. Cu(s) + O2(g) — CuO(s).

12. (a) (i) This is one of those seven acid/base reactions outlined earlier in the chapter in which an acid and carbonate mix to form three products: water, carbon dioxide, and a salt. In this case, the reaction is therefore NaHCO3(s) + HNO3(aq) — CO2(g) + H2O(l) + NaNO3(aq). Both the nitric acid and the sodium nitrate product will be significantly dissociated in water, leaving you with the equation NaHCO3(s) + H+(aq) + NO3- (aq) — CO2(g) + H2O(l) + Na+(aq) + NO3-(aq). NO3- is the only spectator ion here, so the final equation is NaHCO3 + H+ — CO2 + H2O + Na+. Note that you do not need to include phases in your final answer.

(ii) Convert mass of nitric acid to mass of sodium bicarbonate as follows. 25.0gHNO3 1molHNO3 1molNaHCO3 84.0gNaHCO3 „„ „ … ,,„„

_2_Iv_3 V_ly_2_L— XX Xrf NaT-IPO

(b) (i) Magnesium is the oxidizing agent, so it must be reduced (gain electrons) in the reaction. Lead (II) is the reducing agent, so it must be oxidized (lose electrons) in the reaction. This leaves you with the net reaction. Mg2+ + Pb2+ — Mg + Pb4+.

(ii) As a solid, magnesium will be on the paper. If the lead salt formed is insoluble then it may also be on the paper. Either way, Mg(s) will precipitate.

1

63.0g HNO

1mol HNO

1mol NaHCO

(c) (i) This reaction also follows one of the seven acid/base reaction types listed earlier, this time the base + salt — metal hydroxide precipitate + salt. In other words, this is a relatively simple double replacement reaction. The complete, balanced reaction is 2NaOH(aq) + CuSO4(aq) — Cu(OH)2(s) + Na2SO4(aq). Sodium and sulfate both appear in aqueous solution on both sides of the equation, so eliminating them as spectators, you are left with 2OH – + Cu2+ — Cu(OH)2.

(ii) 40%. The percent composition of oxygen can be calculated by dividing the total number of grams per mole of oxygen by the molecular mass of sodium hydroxide. This results in 16g440gx100 = 40%.

13. (a) Lithium is a very reactive metal and is significantly higher than hydrogen on the activity series and is therefore bound to displace it in a reaction. Copper, however, is lower than hydrogen on the activity series and therefore will not replace it.

(b) Both carbon and oxygen are more abundant in the human body than hydrogen by mass because even though there are many more hydrogen atoms than either carbon or oxygen atoms, hydrogen is the lightest of the elements. Carbon is 12 times more massive and oxygen is 16 times more massive than hydrogen.

(c) Simply speaking, water can act as both an acid and a base because it contains both an acid’s characteristic hydrogen and the hydroxide ion common to most bases. Another way to look at it is that water can accept or donate a proton.

14. (a) Na2SO3.To find the empirical formula from the percent composition, follow the steps outlined in Chapter 13. Begin by assuming that you have a total of 100g of the substance, which means that you have 36.5g sodium, 25.4g sulfur and 38.1g oxygen. Divide each of these by their gram atomic masses as follows:

36.5gNa x 1mol Na = 1 59mol Na 1 23.0gNa

25.4gS 1mol S —7s— X "7 „ = 0.79mol S 1 32.1gS

38lgOximoLO = 2.38molO 1 16.0gO

Next, divide each of these by the lowest among them (1.12), giving 1.59mol Na 4 0.79 = 2.0mol Na, 0.79mol S 4 0.79 = 1.0mol S, and 2.38mol O 4 0.79 = 3.0mol O. Attach each of these as subscripts on your empirical formula Na2SO3. This is your empirical formula.

(b) Na6S3O9. You are given the molar mass and will need to divide it by the empirical formula mass in order to find your conversion factor from the empirical to molecular formulae. The empirical formula mass of Na2SO3 is 122.1g per mole. 366.3 divided by 122.1 is 3. Multiply each of the subscripts in the empirical formula by this number to arrive at the molecular formula of Na6S3O9.

15. 59.3% C, 6.2%H and 34.6%N. Begin all percent composition calculations by finding the mass per mole of each type of atom in the substance by multiplying the number of each atom present by its gram atomic mass. In this case, you get 8 12.0g = 96g C, 10 1.0g = 10.0g H, and 4 14.0g = 56.0g N. Next calculate the molar mass of the compound by adding these three values together (96.0 + 10.0 + 56.0 = 162.0g). Finally, divide each of the individual masses by this molar mass and multiply by 100 to give you the percent composition. You should get 59.3% C, 6.2% H, and 34.6% N.

In This Chapter

^ Tackling 75 multiple-choice questions

^ Responding to six free-response questions

I IN the AP exam, you will be given a list of formulas and constants. You may use your

Cheat Sheet (which mimics the AP formula list) and a periodic table for this exam, but do not use your calculator until the section specifically says that you may. Make sure to time yourself to get a sense of your pacing.

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Multiple-Choice Questions (90 Minutes)

CALCULATORS MAY NOT BE USED

Questions 1 through 3 refer to the following compounds:

(A) PI3

(B) NO2

(C) Fe2(SO4)3

(D) MnO2

(E) Li2SO4

1. Contains metal with oxidation number +1

2. Contains metal with oxidation number +3

3. Contains metal with oxidation number +4

Questions 4 through 8 refer to the following compounds, which may be used only once:

(A) C2H6

(B) C2H2

(C) C2H5NH2

(D) C2H4

(E) C2H4O

4. Alkene

5. Aldehyde

6. Alkane

7. Amine

8. Alkyne

Questions 9 through 11 refer to the following types of bonding:

(A) Ionic bonding

(B) Covalent bonding

(C) Metallic bonding

(D) Sigma bonding

(E) Pi bonding

9. Characterized by orbital overlap that is symmetrical to a plane

10. Is the most polar form of bonding

11. Characterized by highly mobile electrons

Questions 12 through 14 refer to the following gases:

(A) HCl

(B) O2

(C) NO

(D) NO2

(E) CO

12 Diffuses at the slowest rate

13. Has the most weakly interacting molecules

14. At any given temperature and pressure, has particles with the highest average velocity

15. What volume is closest to that occupied by 56.0g of carbon monoxide gas at standard temperature and pressure?

(A) 2.00L

(B) 11.2L

(C) 34.0L

(D) 44.8L

(E) 68.0L

16. What is the conjugate base of ascorbic acid in the following reaction?

C5H7O4COOH + OH – <rҐ H2O + C5H7O4COO-

(A) C5H7O4COOH

(B) OH-

(C) H2O

(D) C5H7O4COO-

(E) C5H7O4COOH-

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17. The following data list the results for three trials in which compounds W and X reacted to produce compound Y. Given these data, what is the overall reaction order?

Trial [W]/M [X]/M

0.25 0.25 0.50

0.50 1.00 1.00

Initial rate, d[Y]/dt / M s-1

0.20

0.40

1.60

(A) 2

(B) 3

(C) 4

(D) 7

(E) 12

18. An aqueous solution of which of the following compounds is the best conductor of electricity?

(A) Glucose, C6H12O6

(B) Ethanol, C2H6O

(C) Sodium hydroxide, NaOH

(D) Carbon monoxide, CO

(E) Carbon dioxide, CO2

19. Which species does NOT Have the following electron configuration?

1s22s22p63s23p63J104s24p6

(A) Rb+1

(B) Sr+2

(C) Kr

(D) Br-1

(E) Se

20. A buffer solution of acetic acid and sodium acetate is prepared at pH 5.76. The pATaof acetic acid is 4.76. What is the ratio of acetate to acetic acid (acetate:acetic acid) in the buffer solution?

(A) 1:1

(B) 1:2

(C) 2:1

(D) 1:10

(E) 10:1

21. Consider the following balanced reaction equation:

Cu + 2HCl — CuCl2 + H2

How does the oxidation number of copper change?

(A) 0 to +1

(B) 0 to +2

(C) +1 to -1

(D) +2 to -2

(E) -2 to 0

22. Which of the following is the formula for iron (III) sulfate?

(A) FeS

(B) Fe2S3

(C) FeSO4

(D) Fe3(SO4)2

(E) Fe2(SO4)3

23. What is the approximate mass of nitrogen gas, N2, that occupies a 45L container at 0.50atm and 273K?

(A) 7g

(B) 14g

(C) 22g

(D) 28g

(E) 56g

24. Which of the following is the shape of a molecule whose central atom is surrounded by two atoms and two lone pairs?

(A) Linear

(B) Bent

(C) Trigonal planar

(D) Trigonal bipyramidal

(E) Tetrahedral

3

25. In the following figure of an electrochemical cell, what best represents events occurring at electrode #1 (in Figure 30-1)?

Figure 30-1:

Electrochemical cell.

C(s)

C(s)

ZnSO(aq)

4 n

26. What volume of 0.25 H3PO4 neutralizes 500mL 1.5M KOH?

(A) 600mL

(B) 900mL

(C) 1L

(D) 3L

(E) 6L

27. Prepared at 0.5 molar concentration in water, which of the following solutions has the highest boiling point?

(A) Propanol

(B) Propanoic acid

(C) Iron (II) nitrate

(D) Iron (III) nitrate

(E) Nitric acid

1

2

(A) Electrons travel out of electrode toward voltage source.

(B) Electrons travel out of electrode toward aqueous sulfate ions.

(C) Zinc metal electroplates onto electrode.

(D) Zinc metal donates electrons to electrode.

(E) Oxygen gas bubbles from electrode.

28. Which of the following statements about the energy diagram shown in Figure 30-2 is incorrect?

Figure 30-2: Energy diagram.

Reaction Progress

(A) The overall reaction is endothermic.

(B) The reaction has two intermediates.

(C) Point X represents a transition state.

(D) Raising the energy of point Y decreases the equilibrium concentration of product.

(E) Raising the energy of point X decreases the rate of product formation.

29. In the following reaction, which species are the Lewis acid and Lewis base, respectively?

BF3 + N(CH3)3 — BF3N(CH3)3

(A) BF3, BF3N(CH3)3

(B) BF3, N(CH3)3

(C) BF3N(CH3)3, BF3

(D) BF3N(CH3)3, N(CH3)3

(E) N(CH3)3, BF3

30. The free energy change for a reaction at 273K and 1atm is -35.7 kJ. If the equilibrium constant for the reverse of that reaction is ex, then what is x?

(A) -35700 / (8.314 x 273)

(B) 35700 / (8.314 x 273)

(C) (8.314 x 273) / -35700

(D) (8.314 x 273) / 35700

(E) (ln 35700) / (8.314 x 273)

31. If the ratio of the rate of effusion of an unknown gas to the rate of effusion for helium gas, He(g), is 0.35, what is the approximate molar mass of the unknown

Gas?

(A) 2.9g/mol

(B) 11g/mol

(C) 33g/mol

(D) 1.3 x 102g/mol

(E) 1.4 x 102g/mol

X

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32. Which of the following compounds is least soluble in water?

(A) Sodium propanoate

(B) Propanoic acid

(C) Propanediol

(D) Propanone

(E) Propene

33. What is the EMF of the voltaic cell driven by the following reaction?

Zn(s) + Cu(NO3)2(agJ — Zn(NO3)2(aq) + Cu(s)

The following data may help you arrive at your answer:

Cu2+(aq) + 2e – — Cu(s)E0red = 0.34V

NO3-(aq) + 2H+(aq) + e – — NO2(g) + H2O(I)E0red = 0.78V

Zn2+(aq) + 2e – — Zn(s)E0red = -0.76V

(A) -0.42V

(B) 1.10V

(C) 0.36V

(D) -1.20V

(E) 0.44V

34. Which of the following indicators would be the best choice to monitor a change that occurs at pH = 5.0?

(A) Bromophenol blue, PKa = 4.0

(B) Phenolphthalein, pKa = 7.9

(C) Thymol blue, pKa = 9.3

(D) Methyl red, pKa = 5.1

(E) Methyl orange, pKa = 3.7

35. Which element can have a mass number of 40 and 21 neutrons?

(A) Calcium

(B) Potassium

(C) Argon

(D) Neon

(E) Promethium

36. The net ionic equation for the reaction of lead (II) nitrate with potassium iodide is

(A) KI(aq) + Pb(NO3)2(aq) — KNO3(aq) + PbI2(aq).

(B) 2KI(aq) + Pb(NO3)2(aq) — 2KNO3(aq) + PbI2(a< ).

(C) 2KI(a< ) + Pb(NO3)2(a< ) — 2KNO3(a< ) + PbI2(s).

(D) 2I-(aq) + Pb2+(aq) — PbI2(s).

(E) 2K+(a< ) + 2I-(a< ) + Pb2+(a< ) + 2NO3-(a< ) — 2K+(aq) + 2NO3-(aq) + PbI2(s).

37. A sulfide of copper is found to contain 20% sulfur. What is the formula of the compound?

(A) CuS

(B) CuS2

(C) Cu2S

(D) Cu2S2

(E) Cu4S

38. At standard temperature and pressure, how many liters of H2S are produced by the complete reaction of 45.4g of nickel (II) sulfide?

NiS(s) + 2HCl(aq) — NiCl2(aq) + H2S(g)

(A) 0.500L

(B) 11.2L

(C) 41.0L

(D) 49.0L

(E) 91.0L

39. Which metal has the lowest ionization energy?

(A) Lithium

(B) Sodium

(C) Calcium

(D) Strontium

(E) Cesium

40. Which set of coefficients properly balances the equation for the combustion of ethane?

_ethane(g) +_oxygen(g) —_carbon

Dioxide(g) +_water(g)

(A) 1, 7, 2, 3

(B) 1, 2, 1, 2

(C) 2, 7, 4, 6

(D) 1, 4, 1, 2

(E) 1, 7, 4, 6

41. Which linear compound may exhibit cis-trans isomerism?

(A) Butane

(B) 1-butyne

(C) 1-butene

(D) 2-butene

(E) 2-butyne

42. You dilute 100.mL of a stock NaCl solution by adding 400.mL water, creating a 0.200M NaCl working solution. What is the concentration of the stock solution?

(A) 10.0M

(B) 4.00M

(C) 1.00M

(D) 0.800M

(E) 0.400M

43. At 298K, the Ksp for CuCO3 is 2.5 x 10-10 and the Ksp for BaCrO4 is 2.0 x 10-10. Which of the following statements is true?

(A) In saturated CuCO3, the molar concentration of Cu2+ is 2.5 x 10-5.

(B) In saturated BaCrO4, the molar concentration of Ba2+ is 2.0 x 10-10.

(C) In saturated CuCO3, addition of Cu2+ increases CuCO3 solubility.

(D) BaCrO4 is more soluble in water than CuCO3.

(E) In saturated BaCrO4, the molar concentration of Ba2+ is 1.4 x 10-5.

44. Which of the following compounds is possible?

(I) Fe2(SO4)3

(II) FeSO4

(III) Fe(OH)3

(A) I only

(B) II only

(C) III only

(D) I and III only

(E) I, II, and III

45. 48g methane reacts with fluorine gas to produce 48g 1-fluoroethane and hydrogen gas. What is the percent yield? Percent yield = 100% x (actual yield / theoretical yield)

(A) 33%

(B) 66%

(C) 50%

(D) 100%

(E) 300%

46. In the compound 2,3-dichloropropene-1-ol what are the hybridizations of carbons 1, 2, and 3, respectively?

(A) sp, sp2, Sp3

(B) sp3, sp2, Sp

(C) Sp2, sp2, sp3

(D) Sp2, sp3, sp3

(E) Sp, sp, sp2

47. Based on the standard reduction potentials listed below, which is the strongest

(A) Ni2+

(B) Fe2+

(C) Zn2+

(D) Sn2+

(E) H+

48. The following reaction achieves equilibrium 51. at 400K:

4H2(g) + 2NO(g) 2H2O(g) + N2(g)

At constant temperature, increasing the volume of the reaction vessel has what effect?

(A) Reaction rates increase.

(B) Water condenses into liquid.

(C) More mass shifts into the form of 52. hydrogen gas.

(D) Less mass shifts into the form of nitrogen monoxide.

(E) There is no effect.

49. 15g of an unreactive crystalline compound are massed and set aside for later use. When the same sample is massed once more before dissolving into a solution, their

New mass is 15.9g. What is the most likely 53. explanation?

(A) The compound degrades.

(B) The compound reacts with itself.

(C) The compound is amphoteric.

(D) The compound is hygroscopic.

(E) The compound is hydrophobic.

50. At 200.K, a certain reaction is spontaneous, having a free energy change of -500. J mol-1 and an enthalpy change of -1.50 x 103 J mol-1. At what temperature does the reaction become nonspontaneous?

(A) 197K

(B) 203K

(C) 300K

(D) 500K

(E) The reaction is spontaneous at all temperatures.

The pKa of a certain acid is 4.00. What is the Kb of that substance?

(A) 1 x 10-10

(B) 1 x 10-4

(C) 1 x 10-3

(D) 1 x 103

(E) 1 x 1010

An electromagnetic wave travels through air with a wavelength of 300. nm. What is the frequency of the wave?

(A) 6.00 s-1

(B) 9.00 s-1

(C) 9.00 x 1010 s-1

(D) 1.00 x 10-15s-1

(E) 1.00 x 1015 s-1

What amount of water is in a 2.5 molal NaCl solution made with 292.3g NaCl?

(A) 8.6kg

(B) 8.6kg

(C) 5.0L

(D) 2.0kg

(E) 2.0L

54. Moving from left to right, which of the following describes events in segments 2 and 4, respectively, of the heating curve shown in Figure 30-3?

Figure 30-3:

Heating

Heat Added

3

4

2

Curve.

(A) Kinetic energy remains constant, gas expands.

(B) Kinetic energy increases, gas expands.

(C) Kinetic energy increases, kinetic energy remains constant.

(D) Solid melts, gas expands.

(E) Kinetic energy remains constant in both segments.

55. Which of the following statements is false?

(A) In a closed system, energy can flow from system to surroundings.

(B) In a closed system, the amount of internal energy can change.

(C) The total difference in energy between system and surroundings is constant.

(D) In an adiabatic system, energy does not flow between system and surroundings.

(E) The energy of the universe minus the energy of the system can change.

56. Which of the following processes cannot cause solid water to become liquid?

(A) Adding energy

(B) Decreasing molecular motion

(C) Increasing pressure

(D) Decreasing pressure

(E) Decreasing potential energy

57. Which of the following is most likely to cause precipitation in a solution of Ca(OH)2? The molar solubility of MgCl2 is 5.7mol L-1. The molar solubility of Ca(OH)2 is 0.025mol L-1.

(A) Add heat

(B) Add MgCl2

(C) Add 2% CH3OH(a< )

(D) Add base

(E) Add acid

58. The conjugate base of a weak acid is which of the following?

(A) A weaker base

(B) A stronger base

(C) A stronger acid

(D) Neutral

(E) A salt

59. At 300.K, a 100.L sample of He(g) at 2.00atm pressure consists of hown many moles?

(A) 6.00 / R

(B) 1.50 x R

(C) 0.667 x R

(D) 0.667 / R

(E) 1.50 / R

60. Which of the following properties of water will be affected by dissolved NaCl?

(I) Boiling point

(II) Vapor pressure

(III) Osmotic potential

(A) I only

(B) II only

(C) I and II only

(D) II and III only

(E) I, II and III

61. For how long must a current of 2.00 amperes flow in order to transport 2.00 moles of charge?

(A) 9.65 s

(B) 9.65 x 104 s

(C) 38.6 s

(D) 38.6 x 104 s

(E) Not enough information is given.

62. When mixed with 100.mL of 0.100M H2SO4, which of the following produces a solution with pH ~ 7?

(A) 50.0mL 0.2M NaOH

(B) 100.mL 0.2M Mg(OH)2

(C) 50.0mL 0.4M CH3OH

(D) 50.0mL 0.4M KOH

(E) 50.0mL 0.4M Na2SO4

63. The Keq for the following reaction is 0.01: A + B 2C

If the concentrations of A and C are 5.0M and 1.0M, respectively, what must be the approximate concentration of B in an equilibrium mixture?

(A) 5.0M

(B) 4.0M

(C) 1.0M

(D) 20.M

(E) 0.050M

64. Which of the following manipulations represents good laboratory practice?

(I) Adding 10mL 2M HCl to 1L water

(II) Adding 1L water to 10mL 2MHCl

(III) Adding 10mL 10M HCl to 10mL 10M NaOH

(A) I only

(B) II only

(C) III only

(D) I and II only

(E) 1, II, and III

65. The thermodynamic critical point of a CO2 is 304K and 73atm. At 310K and 80atm, what is the phase of CO2?

(A) Solid

(B) Liquid

(C) Gas

(D) Mixture of solid, liquid, and gas

(E) Supercritical fluid

66. A 1.0L container holding He(g) at 2.0atm pressure is joined to a 2.0L container holding CO2(g) at 3.0atm pressure. Assuming constant temperature, what will the mole fraction of helium be when the gases equilibrate in the new volume?

67. Solid zinc and silver nitrate undergo a

Single displacement reaction. If 4.00 moles of solid silver is formed, what mass of solid zinc was consumed?

(A) 32.7g

(B) 65.4g

(C) 108g

(D) 131g

(E) 216g

Go on to next page

68. After resolving the components of a sample by using thin layer chromatography (TLC), a compound is visible on the TLC plate at retention factor (Rf) value 0.36. The solvent front had mobility of 8.4cm. What was the mobility of the compound?

(A) 4.3 x 10-2 cm

(B) 4.3 cm

(C) 3.0 cm

(D) 5.4 cm

(E) 8.0 cm

69. Which molecule is most polar?

(A) Carbon dioxide

(B) Carbon tetrachloride

(C) Ammonia

(D) Ammonium cation

(E) Sulfur hexafluoride

70. 400.mL of 0.025MNaOH are added to 100.mL of a strong monoprotic acid solution. The pH of the final solution is 7.0. What was

The pOH of the 100.mL solution?

71. What are the oxidation numbers of

Chromium in chromate and dichromate anions, respectively?

(A) +8, +14

(B) +8, +7

(C) +7, +7

(D) +6, +6

(E) +4, +7

72. The pressure of a gaseous recation container is increased. The container holds hydrogen gas, nitrogen gas, and ammonia gas, which all participate in the reaction shown below. What are the possible effects?

3H2(g) + N2(g) <rҐ 2NH3(g)

(I) Reaction rates increase.

(II) Reaction shifts toward products.

(III) Reaction shifts toward reactants.

(A) I only

(B) II only

(C) III only

(D) I and II only

(E) I, II, and III

73. Under equivalent conditions of pressure, temperature, and volume, what is the most likely ranking of the number of moles present in samples of methane(g), helium(g), and water(g)?

(A) Helium > methane > water

(B) Water > methane > helium

(C) Helium > water > methane

(D) Methane > water > helium

(E) Water > helium > methane

74. Which of the following is an alkaline earth metal?

(A) Mass number 88, atomic number 38

(B) Mass number 86, atomic number 37

(C) Mass number 88, atomic number 39

(D) Mass number 84, atomic number 36

(E) Mass number 137, atomic number 57

75. If uranium-238 emits an alpha particle, what will be the product?

(A) Neptunium-234

(B) Neptunium-238

(C) Uranium-236

(D) Thorium-234

(E) Radium-236

STOP

DO NOT TURN THE PAGE UNTIL TOLD TO DO SO. DO NOT RETURN TO A PREVIOUS TEST.

Chapter Practice Test 335

Free-Response Questions

Part A (55 minutes)

YOU MAY USE YOUR CALCULATOR

CLEARLY SHOW ALL STEPS YOU TAKE TO ARRIVE AT YOUR ANSWER. It is to your advantage to do this, since you will receive partial credit for partially correct responses. Make sure to pay attention to significant figures.

Answer questions 1, 2, and 3. The

Score weighting for each question is 20 percent.

1. Formic acid is a significant component of bee venom. Also known as methanoic acid, formic acid has an acid dissociation constant, Ka, of 1.80 x 10-4.

(a) Calculate the pOH of a 0.25MSolution of formic acid.

(b) Calculate the percent dissociation of the solution in part (a).

(c) Calculate the pH of a solution prepared by mixing equal 1.00L volumes of 0.25M Formic acid and 0.20MSodium methanoate.

(d) Using only compounds already mentioned, what should be added to the solution in part (c) to produce a solution with maximum capacity to resist change in pH? Mention

(i) The compound to be added.

(ii) The mass of the compound to be added

2. A chemist dissolves 51.2g of an unknown, nonelectrolyte compound into 750.g of water. The freezing point depression constant (Kf) For water is 1.86.

(a) Attempting to characterize the unknown, the chemist observes a 1.41 °C difference between the freezing point of the solution and that of pure water. What is the melting point of the solution?

(b) Calculate the molal concentration of the unknown compound in the solution.

(c) Calculate the molar mass of the unknown compound.

(d) A sample of the solid compound is subjected to analysis for percent composition, yielding the following results: 40.0% carbon, 6.70% hydrogen, and 53.3% oxygen. What is the molecular formula of the unknown compound?

(e) What is one possible Lewis structure for the unknown compound? Write the structure in the space below.

3. Electrical current is passed through a 1.0M Solution of HCl(a<7) by means of two nonre-active electrodes immersed into the solution, with the electrodes connected to opposing terminals of a voltage source.

(a) Sketch and label the diagram of the electrolytic cell, including the direction of electron flow, the half-reaction occurring at each electrode, labels for anode and cathode, and the voltage source.

(b) At 4.5 amperes, how many Coulombs pass through the cell in 30. minutes?

(c) How many moles of electrons pass through the cell in the period described in part (b)?

(d) What number of moles of gas bubble pass from each electrode during the time described in part (b)?

(e) Calculate the volume of gas (at standard temperature and pressure) that would bubble from the cell under a current of 3.0 amperes for 60 minutes.

(f) What happens to the pH of the solution as current passes through it within the cell?

STOP

DO NOT TURN THE PAGE UNTIL TOLD TO DO SO. DO NOT RETURN TO A PREVIOUS TEST.

Chapter Practice Test 337

Part B (40 minutes)

CALCULATORS MAY NOT BE USED

Answer question 4 below. The score weighting for this question is 10 percent.

4. For each of the following three reactions, in part (i) write a balanced equation for the reaction and in part (ii) answer the question about the reaction. In part (i), coefficients should be in terms of lowest whole numbers. Assume that solutions are aqueous unless otherwise indicated. Represent substances in solution as ions if the substances are extensively ionized. Omit formulas for any ions or molecules that are unchanged by the reaction.

(a) Potassium hydroxide is added to a solution of iron (III) sulfate.

(i) Write the balanced equation.

(ii) What would you observe about the reactants and/or products during this reaction?

(b) Current is passed through molten nickel chloride.

(i) Write the balanced equation.

(ii) What occurs at the anode and cathode, respectively?

(c) Magnesium oxide is added to water.

(i) Write the balanced equation.

(ii) What happens to the pH of the resulting solution as the reaction proceeds?

Answer questions 5 and 6 below. The score weighting for each question is 15 percent.

5. Give plausible scientific explanations for each of the following observations.

(a) Oxygen concentrations in deep waters are sometimes higher than those in shallower waters.

(b) Dissolving potassium hydroxide in water heats the solution while dissolving ammonium nitrate in water cools the solution.

(c) Cu(s) conducts electricity well but CuCl2(s) conducts electricity poorly.

6. A set of three vials contains three different organic compounds. Each compound contains only one kind of functional group, and each functional group is different from the others. None of the compounds has an ester or amide linkage, and none is an alkene or alkyne.

(a) All of the compounds possess a car-bonyl group. What kinds of compounds are these three?

(b) Assuming that each of the three compounds contains four carbon atoms, is and is linear (not branched), draw Lewis structures for the three compounds.

(c) Ethanol is added to each of the three vials. With which of the three compounds is ethanol most likely to react to produce an ester?

(d) Draw the Lewis structure of the ester that would be produced in the reaction described in part (c).

Answers to Practice

Test Two

Multiple choice

1. (E). As a group IA metal (that is, an alkali metal), lithium exclusively forms a +1 cation, the better to achieve a full valence shell.

2. (C). Although phosphorous has an oxidation number of +3 in PI3, phosphorous isn’t a metal. In iron (III) sulfate, the iron metal ions each have +3 charge, offsetting the (3)(-2) = -6 charge brought by the sulfate anions.

3. (D). Although nitrogen has a +4 oxidation number in NO2, nitrogen isn’t a metal. In manganese (IV) oxide, the manganese metal ion has +4 charge, offsetting the (2)(-2) = -4 charge brought by the oxygens.

4. (D). As 2-carbon hydrocarbon (too short for branching) with the general formula CnH2n, C2H4 can only be an alkene, H2C = CH2.

5. (E). As the only listed hydrocarbon with an oxygen-containing functional group, C2H4O must be an aldehyde. In this case, C2H4O has the structure H3C-CHO.

6. (A). As 2-carbon hydrocarbon with the general formula CnH2n+2, C2H6 can only be an alkane, H3C-CH3.

7. (C). As the only listed hydrocarbon with an amine-containing functional group, C2H5NH2 must be the amine. In this case, the compound has the structure H3C-CHNH2.

8. (B). As a 2-carbon hydrocarbon with the general formula CnH2n-2, C2H2 can only be an alkyne, HC-CH.

9. (E). Although sigma bonding is completely symmetrical around the axis of a sovalent bond, pi bonding is symmetrical about only a single plane through that axis, with overlapping P Orbitals above and below the plane.

10. (A). In pure ionic bonding, electrons are entirely transferred from one atom or group to another. Because the separation of charge is so complete, the bond is extremely polar.

11. (C). Metallic bonding is characterized by a lattice of positively charged metal atom nuclei, around and through which migrates a "sea" of mobile electrons.

12. (D). Because the molar mass of NO2 is the greatest among the compounds listed, its rate of diffusion is the smallest.

13. (B). Because the two atoms of molecular oxygen are identical, the bonds that connect them are entirely nonpolar. This apolarity results in very weak interactions between O2 molecules.

14. (E). As the compound with the lowest molar mass among those listed, the average kinetic energy associated with a given temperature and pressure derives more from velocity in the case of CO than is the case with the other gases. Because kinetic energy — equal between gases at the same temperature — equals (1/2)(mass)(velocity)2, the kinetic energy of the other gases derives more from mass.

15. (D). Because the molar mass of carbon monoxide is 28.0g/mol, the 56.0g described represents 2 moles CO. At STP, 1 mole of (ideal) gas occupies 22.4L, so 2 moles occupy 44.8L.

16. (D). Typically, conjugate acid/conjugate base pairs include a common molecular "chunk," differing only in the presence or absence of an H+. In the given reaction, for example, H2O is the conjugate acid of OH-, a conjugate base. However, the conjugate base of ascorbic acid, C5H7O4COOH, is ascorbate, C5H7O4COO-.

17. (B). Doubling the concentration of X doubles the reaction rate, suggesting the reaction is first-order in X. Doubling the concentration of W further quadruples the reaction rate, suggesting the reaction is second-order in W. These observations suggest the rate law,

Rate = /e[W]2[X]. The overall reaction order is the sum of the individual exponents, 2 + 1 = 3.

18. (C). Dissociated ions are necessary to carry current through a solution. Although many of the listed compounds are soluble in water, and a few dissociate into ions, only NaOH strongly dissociates into ions when dissolved in water.

19. (E). The given electron configuration is most obviously that of the noble gas, krypton. All the listed ions either shed or gain electrons to achieve the same stable, filled valence shell as krypton. Selenium, though it might gain two electrons to achieve the given configuration, is shown in its neutral, elemental form.

20. (E). Use the Henderson Hasselbach equation, pH = PKa + log([A-]/[HA]). The pKa of 4.76 is

1 pH unit lower than the desired pH of 5.76. So, the term log([A-]/[HA]) needs to equal +1. This, in turn, means that the ratio of A – to HA must equal 10:1.

21. (B). On the reactant side, copper is shown in elemental form, so its oxidation number is 0. On the product side, copper participates in the compound CuCl2. Because the oxidation number of chlorine within compounds is -1, and because the compound has two chloride anions, you know that the oxidation number of copper within the compound must be +2 to offset the negative charge.

22. (E). The Roman numeral within parentheses alerts you to the fact that iron in this compound has +3 charge. Next, recognize the sulfate anion, SO42-. Finally, find the right numerical combination of iron cations and sulfate anions to result in a neutral ionic compound: Fe2(SO4)3.

23. (D). Before finding the mass of nitrogen gas, you must find the number of moles. To do so, first recognize that at 273K, one mole of an ideal gas (like N2) occupies 22.4L at 1atm. But the given pressure is half that — so one mole of gas occupies Twice That volume, which is

2 x 22.4L = 44.8L ~ 45L. So, you’ve got about one mole of N2 on your hands. The molar mass of N2 is 28.0g.

24. (B). The four electron pairs represented by the two bound atoms and two lone pairs suggest a tetrahedral geometry. Because only two of the four orbital axes include actual bonds to actual atoms, the overall molecular shape resulting from the tetrahedral orbital geometry is "bent," just like water.

25. (C). The signs on the terminals of the voltage source clue you into the direction of electron flow within this electrolytic cell. Electrons flow away from the negative terminal, down into electrode 1, through the electrolyte solution and up through electrode 2 toward the positive terminal. So, at electrode 1, Zn2+ cations are reduced by the flow of electrons, electroplating solid zinc onto the electrode surface.

26. (C). The key idea in a neutralization reaction is that the total moles of acid must equal the total moles of base. So, you must provide enough acid to counteract (0.500L)(1.5MKOH) = 0.75mol KOH, where each KOH contributes a single OH-. Because each mole of H3PO4 contributes three equivalents of acid, 0.25MH3PO4 effectively acts as 0.75MAcid. Adding 1000mL (that is, 1.00L) of H3PO4 therefore neutralizes the base.

27. (D). This is a colligative properties question, requiring you to know that the more moles of solute you dissolve into a given mass of water, the more you’ll elevate the boiling point. Because iron (III) nitrate, Fe(NO3)3, dissociates into four ionic particles, the given concentration of that solute contributes the most particles and leads to the highest boiling point.

28. (D). All the listed statements are correct except for the statement about point Y, which alters the energy level of an intermediate state. The equlibirium concentration of product depends on the energy difference between initial (reactant) and final (product) states.

29. (B). Lewis acids are electron acceptors, and Lewis bases are electron donors. The boron of BF3 has two spots still empty in its valence shell. The nitrogen of N(CH3)3 has two available electrons in a lone pair. So, the nitrogen donates its lone pair to the boron, creating a coordinate covalent bond in a Lewis acid/base reaction.

30. (A). First, recognize that the free energy change for a reverse reaction has the same magnitude and opposite sign as the free energy change for the forward reaction. So, the free energy change for the reverse reaction is +35.7kJ = +35700 J. Second, recall the following relationship between the equilibrium constant and the free energy change:

K = E – AG /RT

This means that X = -(-AG/RT) = -35700 / (8.314 x 273).

31. (C). Recall Graham’s law of diffusion and effusion: The ratio of the square roots of the molar masses equals the inverse of the rates of effusion or diffusion. Because the ration of the rates of effusion is 0.35, the ratio of the molar masses is (1 / 0.35)2 = 8.3. Helium has a molar mass of 4.0g/mol. Solve for the molar mass of the unknown by multiplying 8.3 x 4.0g/mol = 33 g/mol.

32. (E). Choices A and B are both strongly polar because of the carboxylate/carboxylic acid group. Choices C and D are both moderately polar because of the alcohol and ketone groups, respectively. Choice E, an alkene, is simply nonpolar, and is therefore the least soluble in water.

33. (B). To understand the redox reaction involved here, view the reaction as a net ionic equation:

Zn(s) + Cu2+(aqJ — Zn2+(aq) + Cu(s)

Now you can see that the first and third listed standard reduction potentials are the relevant ones. Because copper is more easily reduced than zinc (as revealed by its more positive standard reduction potential), you know that copper reduction occurs at the cathode and zinc oxidation occurs at the anode. Finally, apply the equation for calculating a standard cell potential:

Ј°cell = Ј°red(cathode) – Ј°red(anode) = 0.34V – (-0.76V) = +1.10V

34. (D). Methyl red has a pKA of 5.1, the closest to the target pH, and is therefore the most suitable.

35. (B). With a mass number of 40 and 21 neutrons, the element in question must have 19 protons, and therefore an atomic number of 19. Unambiguously, then, the element is potassium.

36. (D). Choice A is the balanced reaction equation. Choice E is the total ionic equation. But only choice D is the net ionic equation.

37. (C). Percent composition is all about mass. The only formula listed in which the mass of sulfur is 20% of the total mass of the compound is Cu2S. Choice E is a red herring, because in that compound sulfur is 20% of the compound by atom count, not by mass.

38. (B). First, concentrate on the moles of H2S produced by the given reaction. The molar mass of nickel (II) sulfide is 90.8g/mol, so 45.4g represents 0.500 mole of the compound. The stoi-chiometry of the reaction tells you that you’ll therefore produce 0.500 mole of hydrogen sul-fide. At standard temperature and pressure, 1 mole gas occupies 22.4L volume. One half of that quantity is 11.2L.

39. (E). Ionization energy increases overall upward and to the right within the periodic table. So, the element with the lowest ionization energy should hold the lowest, most leftward position. Of those elements listed, cesium best fits the bill.

40. (C). To answer this question, you must know the formulas for the common compounds listed, and you must be able to balance a reaction equation:

2C2H6 + 7O2 — 4CO2 + 6H2O

41. (D). Each compound listed shares a 4-carbon skeleton. In linear (not cyclic) compounds, geometric (cis-trans) isomerism doesn’t occur about single bonds, so choice A is out. Both butyne compounds create sp-hybridized carbons with purely linear geometry. Double bonds are classic sites for cis-trans isomerism, but 1-butene is disqualified because one of the carbons participating in the double bond is a terminal carbon attached to two identical hydrogen atoms. So, only 2-butene, with its central double bond, offers the possibility of cis-trans isomerism.

42. (C). This problem requires you to use the dilution equation:

C1 x V1 = C2 X V2

Because you add 400mL to 100mL, notice that your final volume is 500mL, not 400mL.

43. (E). To answer this question, you must remember the definition of the solubility product constant. For the dissolution reaction XY(s) — X(aq) + Y(aq), Ksp = [X]x[Y]. Each of the listed compounds undergoes simple dissociation to 2 counterions (Cu2+ and CO32- on the one hand, Ba2+ and CrO42- on the other hand), [X] = [Y] in each case. So, the concentration of either ion in a saturated solution is simply the square root of the KSp. This fact rules out choices A and B. The common ion effect rules out choice C. The given KSp values reveal that choice D is wrong. Thus, by process of elimination and because the math works out properly, choice E is correct.

44. (E). Ionic compounds have neutral charge. Iron can take on variable charge, as Fe (II) cation or Fe (III) cation. Sulfate anion has -2 charge. Hydroxide anion has -1 charge. So, all three compounds are possible.

45. (B). To answer this question you must first generate a balanced reaction equation:

4CH4 + F2 — 2C2H5Fl + 3H2

If 48g of CH4 react, then 48g / 16g / mol-1 = 3 moles methane react. Because only 2 moles of fluoroethane product are made for every 4 moles of ethane reactant, the theoretical yield of fluoroethane is 1.5 moles. The molar mass of fluoroethane is 48g/mol. So, the theoretical yield of fluoroethane is 1.5 mol x 48g/mol = 72g. The actual yield is 48g. Percent yield = 100% x (actual/theoretical) = 66%.

46. (C). Remember to take bonds to hydrogen into account. Carbons 1 and 2 each have three bond axes, and so are Sp2 Hybridized. Carbon 3 has four bond axes, and so is sp3 hybridized.

47. (D). Standard reduction potentials reflect ease of reduction. Tin (II) cation is most easily reduced, and therefore is mostly likely to oxidize other atoms. So, tin (II) cation is the strongest oxidizing agent listed.

48. (C). Equilibrium shifts to oppose any perturbation. At constant temperature, increasing volume decreases temperature. So, the equilibrium shifts mass in the way that increases pressure — in other words, it shifts to increase the total moles of gas. Because the reactant side has 6 moles gas to the product side’s 3 moles of gas, equilibrium shifts to the left. Choice A is wrong because lower pressures decrease gas-phase reaction rates. Choice B is wrong because lower pressure favors the water vapor, not liquid water. Choice D is wrong because the equilibrium shifts to the left for the reasons already described. Choice E is just plain wrong.

49. (D). Because the compound is decribed as unreactive, choices A, B, and E are unlikely. Choice C may or may not be true, but has little relevance to the observed increase in mass. Hygroscopic compounds are those which interact so strongly with water that they absorb water vapor from the air, thereby increasing the overall mass of the sample.

50. (C). Begin this problem with the Gibbs equation: AG = AH - TAS. Plugging in the given values for AG, AH and T, you can solve for AS, getting -5 J mol-1 K-1. Next, recognize that the temperature at which the reaction moves from spontaneous to nonspontaneous is the one at which the free energy change is zero, so plug in "0" for AG, plug in the values for AH and AS, and solve for T. Doing so gives you 300K.

51. (A). Because pKA = – logKA, it follows that KA = 10-pKa. Using this equation, solve for the acid dissociation constant, KA = 1 x 10-4. Next, remember that KA x KB = KW = 1 x 10-14. Using that fact, solve for the base dissociation constant, KB = 1 x 10-10.

52. (E). For any wave, velocity = frequency x wavelength. For electromagnetic waves in air, velocity = 3.00 x 108 m/s. So, to solve for the frequency, simply divide 3.00 x 108 m/s by the wavelength, 300. x 10-9 m. Doing so gives you 1.00 x 1015.

53. (D). Molality = (moles solute particles) / (kilograms solvent). Start by determining the moles of NaCl present in the 292.3g sample. Using the molar mass of NaCl (58.45g/mol), this figure comes to 5.00 moles. To achieve 2.5 molality, you require 2kg water.

54. (E). Segments 2 and 4 are flat with respect to temperature — though heat is added, that heat goes into accomplishing the phase change of melting or vaporizing, not into increasing the temperature of the sample. So, within neither of these segments does kinetic energy (the underlying cause of temperature) change. Choice D is tempting because solid does melt in segement 2, but segment 4 does not include gas expansion, but rather the conversion of liquid at the boiling point into gas.

55. (C). All the statements are true except choice C. What remains constant is the Sum Of energy between system and surroundings.

56. (B). Water has some pretty anomalous properties, including regions of its phase diagram in which either increasing pressure or decreasing pressure could result in a change from solid to liquid state. However, nowhere in the phase diagram does decreasing temperature (that is, reducing the kinetic energy of molecular motions) cause a shift from solid to liquid.

57. (D). For most solid solutes, adding heat increases solubility. Adding acid increases calcium hydroxide solubility by converting dissociated hydroxide ions into water. Choices B and C could conceivably reduce solubility, but by far the best choice is D, because adding base (that is, effectively adding OH-) reduces calcium hydroxide solubility by the common ion effect.

58. (B). Descriptive chemistry at its finest. The conjugate base of a weak acid is a stronger base. That is why the weak acid is weak.

59. (D). Use the ideal gas law for this one. Rearranged to solve for moles, n, the equation becomes N = (PV) / (RT). Substituting in known values (and leaving R Unsubstituted) gives you (2.00atm x 100.L) / (300.K x R) = 0.667 / R.

60. (E). Boiling point, vapor pressure, and osmotic potential are all colligative properties, properties that change as solute is added to solution..

61. (B). Current equals charge divided by time (I = Q / T), So to solve for time, you must divide the total charge by the current. There are 2.00 moles of charge, which means there are 2 x F Coulombs, or 2(9.65 x 104 coulombs). Divide this quantity by 2.00 amperes (2.00 coulombs per second) to get 9.65 x 104 seconds.

62. (D). A neutral solution (pH = 7) emerges when total moles of base and total moles of acid are equal. Each mole of sulfuric acid contributes two moles acid, so the original solution contains (0.100L)(0.100mol L-1)(2) = 0.0200 moles acid. Using similar calculations on the choices, you’ll find that the 50.0mL of 0.400M KOH offered by choice D does the trick.

63. (D). First, you must be able to construct the expression for the equilibrium constant of the reaction: Keq = [C]2 / ([A]x[B]). Next plug in the known values for Keq, [A] and [C], and solve for [B]. Doing so gives you 20.M.

64. (A). When diluting concentrated acid or concentrated base, always add the concentrated compound to a large volume of water, and not the other way around. The dilution may generate a lot of heat, and by adding to the large volume of water, that heat can be adequately dispersed.

65. (E). Temperatures and pressures above the critical point produce a supercritical fluid.

66. (B). The question is asking about a mole fraction — that of helium within the final mixture, to be precise. From the ideal gas law, you can calculate for the separate helium container that the moles of helium, NHe = (2.0atm)(1.0L) / (RT). Second, you can calculate the moles of carbon dioxide, NCO2 = (3.0atm)(2.0L) / (RT). Because each question shares RT, You can set up the proportion (3.0 / nCO2) = (1.0 / nHe). In other words, there are three times the moles of CO2 as there are moles He. So, the moles fraction of He = 1.0 / (1.0 + 3.0) = 0.25.

67. (D). Start by writing a balanced reaction equation

Zn(s) + 2AgNO3(a< ) — 2Ag(s) + Zn(NO3)2(a< )

From the equation you can see that two moles of solid silver form for every mole of solid zinc consumed. So, to produce four moles of solid silver required two moles of solid zinc. Because each mole of zince has 65.4g mass, the two mole total has mass 131g.

68. (C). To attack this problem, you need to recall that

Rf = (sample mobility) / (solvent mobility)

Substitute in the known values of Rf = 0.36 and solvent mobility = 8.4cm, and you can solve for the sample mobility, which is 3.0cm.

69. (C). Although many of the other compounds have polar covalent bonds, only ammonia has a geometric distribution of bonds and a lone pair whose polarities don’t entirely cancel out.

70. (E). If the pH of the final solution is 7.0 (that is, neutral), then the moles of base added equal the moles of acid originally present in the strong acid solution. The moles of base added were (0.400L)(0.025mol/L) = 0.010mol OH-. Because acid must have been equal to base for the final solution to be neutral, there must have been 0.010mol H+ in the original 0.100L acid solution. In other words, [H+] = 0.010mol / 0.100L = 0.10M. This means that the pH of that acid solution was pH = – log[0.10] = 1.00. Because pH + pOH = 14.00, you know that the pOH of the acid solution was 13.00.

71. (D). You must know the formulas of these polyatomic anions (CrO42- and Cr2O72-) to figure out the oxidation numbers of the metals. In each case, start with the oxygen atoms, each of which carries an oxidation number of -2. In order to offset much of this negative charge while still leaving a -2 formal charge, each chromium atom must carry on oxidation number of +6.

72. (D). Increased pressure can increase reaction rates as particles collide more frequently. Increasing pressure perturbs the equilibrium, which shifts to create lower pressure. In the case of the reaction shown, shifting toward products results in lower pressure because the products contain fewer moles of gas than the reactants.

73. (B). Water molecules have the greatest degree of intermolecular attraction, particularly from hydrogen bonding. Methane molecules lag far behind water in this respect, but can still experience a small intermolecular attraction due to London forces. Of the three substances listed, the atomic noble gas helium has the fewest intermolecular attractions. All other things being equal, the gas with the greatest attractive forces will have the greatest tendency to contract, so achieving any given pressure (at a given temperature) will require more molecules for that gas than for the others.

74. (A). Focus on the atomic number here — the mass numbers can only mislead you because they may represent different isotopes of different elements. Alkaline earth metals are in group IIA, and only atomic number 38, strontium, fits that bill.

75. (D). Alpha particles are helium nuclei, containing two protons and two neutrons. So,

A nuclide that decays by alpha emission loses four mass number units and two atomic number units. By losing two atomic number units, the element is transformed into a different element entirely. The new mass number is 238 – 4 = 234. The new atomic number is 92 – 2 = 90. The new element is thorium, Th.

Free response

1. (a) pOH = 11.83. Make sure you understand the meaning of the KA for this system by writing down the acid dissociation reaction. If the common name, formic acid, doesn’t help you with the structure, the systematic name, methanoic acid, should do so.

CHOOH CHOO- + H+

Ka = ([CHOO-]x[H+]) / [CHOOH] = 1.80 x 10-4

Making a 0.25mol/L formic acid solution begins the process of acid dissociation, in which some concentration, x, shifts from CHOOH form to equivalent concentrations, x, of each of the dissociated forms, CHOO – and H+:

Initial concentration, mol/L Change, mol/L Final concentration, mol/L

CHCOOH, 0.25 -x 0.25-x

CHCOO-, 0 +x x

H+, 0 +x x

Substituting into the expression for KA KA = (x2) / (0.25-x) = 1.80 x 10-4

This equation is a somewhat messy quadratic to solve under time pressure, and it is reasonable to assume that, because formic acid is a weak acid, 0.25-x ~ 0.025. So, the equation simplifies to

1.80 x 10-4 = (x2) / (0.25)

Solving for x gives you 6.7 x 10-3 = x = [H+]. pH = – log[H+] = 2.17 pH + pOH = 14.00, so pOH = 11.83

(b) Percent dissociation = 2.6%. If the total concentration of formic acid and formate together is 0.25mol/L, and the concentration of formate alone is 6.7 x 10-3 mol/L, then the percent dissociation is

100% x (6.7 x 10-3) / (6.7 x 10-3 + 0.25) = 2.6%

(c) pH = 3.64. This is a buffer problem, for which you should use the Henderson-Hasselbach equation:

PH = pKa + log([formate]/[formic acid])

To calculate the pKA: pKA = – logKA = 3.74 pH = 3.74 + log(0.20 / 0.25) = 3.64

(d) (i) Sodium formate. Maximum buffering capacity occurs when the conjugate base and conjugate acid are present in equal concentrations. The initial concentrations are 0.20mol formate / 2.00L = 0.10mol/L formate; 0.25mol formic acid / 2.00L = 0.13mol/L formic acid. You must add enough solid sodium formate, CHCOONa, to result in 0.13mol/L formate, CHCOO-.

(ii) 11g sodium formate

(0.10mol + x mol) / 2.00L = 0.13mol/L formate

Solving for x gives you 0.16mol formate (added as sodium formate, 68.0g/mol) 0.16mol x 68.0g/mol = 11g sodium formate

2. (a) -1.41 °C. Freezing point and melting point are the same temperature. The normal melting point of water is 0 °C, and adding solute lowers the freezing point.

(b) 0.758mol/kg

A7f = Molality X Kf

Substituting in the known values gives you 1.41 °C = Molality X 1.86. Solving for molality gives you 0.758mol/kg.

(c) 90.1g/mol

Molality = (moles solute particles) / (kilograms solvent)

Because the compound is a nonelectrolyte, you needn’t worry that a mole of compound dissolves into multiple moles of ionic particles. So, you can substitute in the known values:

0.758mol/kg = (moles solute) / (0.750kg water)

Moles solute = 0.569 moles

Because the chemist added 51.2g of the compound, you can calculate the apparent molar mass as (mass solute added) / (moles solute) = 51.2g / 0.569 moles = 90.0g/mol.

(d) C3H6O3. Begin by determining the empirical formula. Assume a 100.g sample of the compound. At the given percent composition, the sample will contain 40.0g carbon, 6.70g hydrogen, and 53.3g oxygen. Based on the gram atomic mass of each element, these values correspond to 3.33mol carbon, 6.67mol hydrogen, and 3.33mol oxygen. Dividing all mole values through by the smallest among them (3.33), you come to 1 mole carbon, 2 moles hydrogen, and 1 mole oxygen. These values correspond to the empirical formula CH2O. Next, divide the apparent molar mass of the compound calculated in part (c) by the molar mass of the empirical formula: (90.0g/mol) / (30.0g/mol) = 3. The molecular formula of the compound is a multiple of 3 times the empirical formula:

3 x CH2O = C3H6O3

(e) Figure 30-4 is a possible Lewis structure for the compound.

Figure 30-4:

Lewis dot structure

HO

OH

Dihydroxyacetone, C3H6O.

363

3. (a) Figure 30-5 is a labeled diagram of the electrolytic cell.

Emf

- +

E-

2H+(ao) + 2e – — H,(o)

Figure 30-5:

Electrolytic cell.

N

Cathode

E-

Anode

1.0Af HCl(aq)

2Cl (ao) — 2Cl2(«) + 2e-

2

(b) 8.1 x 103 coulombs. Use the equation Q = It (charge = current x time). Charge = (4.5 coulombs/s)(1.8 x 103 s) = 8.1 x 103 coulombs

(c) 8.4 x 10-2 mol electrons. To calculate the moles of charge, you must use Faraday’s constant: 96485 C/mol e-:

Moles of electrons = (8.1 x 103 coulombs)(1mol / 96485 coulombs) = 8.4 x 10-2 mol

(d) 4.2 x 10-2 mol gas per electrode. Based on the half-reactions shown for the electrodes in part (a), 2 moles of electrons are required to liberate 1 mole of H2(g) and 1 mole of Cl2(g), by reduction and exidation, respectively.

8.4 x 10-2 mol electrons x (1mol gas / 2mol electrons) = 4.2 x 10-2 mol gas

So, 4.2 x 10-2 mol gas are liberated at each electrode, for a total of 8.4 x 10-2 mol gas.

(e) 2.5L total gas. Again, start with Q = It, then convert to moles charge by using Faraday’s constant, and then to moles gas by using the half-reaction equations (noting that 2 moles electrons frees 2 moles gas, if you count both electrodes), and finally to volume of gas by using 22.4L/mol.

(3.0C/s)(3.6 x 103 s)(1mol e-/96485C)(2mol gas/2mol e-)(22.4L/1mol gas) = 2.5L gas (total)

(f) As the electrolysis continues, dissolved hydrochloric acid leaves the solution as H+ is converted into H2 and as Cl – is converted into Cl2. Because the concentration of acid decreases over time, the pH rises.

4. (a) (i) 3OH-(aq) + Fe3+(aq) — Fe(OH)3(s). Because the potassium cation and sulfate anions are ionized and unchanged on both sides of the reaction equation, they are omitted from the net ionic equation.

(ii) A solid precipitate forms as the reaction proceeds. The precise colors of the iron (III) sulfate solution and iron (III) hydroxide precipitate depend on their concentrations. The iron (III) sulfate solution typically has a greenish color, whereas the iron (III) hydroxide product would appear as a red-brown precipitate.

(b) (i) Ni2+(l) + 2Cl-(/) — Ni(s) + Cl2(g). Molten salts are liquid, not aqueous. All reactants and products remain because each changes during the reaction.

(ii) During the course of this salt hydrolysis, solid nickel would deposit at the cathode and chlorine gas would bubble from the anode.

Cathode: Ni2+(l) + 2e – — Ni(s)

Anode: 2Cl-(l) — Cl2(g) + 2e-

(c) (i) MgO(s) + H2O(l) — Mg(OH)2(s) . All reactants and products are shown because all change during the reaction.

(ii) During the course of the reaction, the metal oxide, initially a white powder, would dissolve into a clear aqueous solution and undergo the reaction shown in part (c). Although magnesium hydroxide is only sparingly soluble, some hydroxide ion would dissociate into solution. As the reaction progressed, the pH of the solution would progressively increase due to the appearance of hydroxide ion product.

5. (a) Although atmospheric oxygen occurs at the surface of a deep body of water, two important parameters change as you move to greater depths. First, the temperature decreases. Second, the pressure increases. Decreasing temperatures increase the solubility of gaseous solutes (like oxygen); at lower temperatures, the solutes have decreased kinetic energy and are less able to disrupt the solute-solvent interactions that hold them in solution. At increased pressures, gaseous solutes are more soluble in accordance with Henry’s law.

(b) The tendency of a solute to heat or cool a solution while dissolving is quantified by the parameter, AHsoln, the molar heat of solution. If this parameter is negative, the solute releases heat into solution as it dissolves. If the parameter is positive, the solute absorbs heat from solution as it dissolves. The physical basis for differences in heats of solution lies in the collection of solute-solute, solute-solvent, and solvent-solvent interactions that govern the dissolution process. If, in dissolving, an excess of enthalpically favorable interactions must be disrupted, dissolution is accompanied by an absorption of heat — in other words, by cooling. If, in dissolving, an excess of enthalpically favorable interactions becomes available, dissolution is accompanied by a release of heat — in other words, by heating. Even when dissolution is enthalpically unfavorable, it may nevertheless occur if it is driven by entropic effects.

(c) Solid copper, Cu(s), participates in metallic bonding. Solid copper (II) chloride, CuCl2(s) participates in ionic bonding. In metallic bonding, metal atoms pack into a crystalline lattice in which the electropositive nuclei sit amidst a sea of mobile electrons. The electrons move freely throughout the lattice because the energetic penalty for hopping from one nucleus to the next is negligible. So, when motivated to flow by an applied voltage, the mobile electrons do so. In ionic bonding, the crystalline lattice is composed of alternating ectropositive and electronegative bonding partners. One partner releases electrons and the other partner binds them. In order for electrons to flow under an applied voltage, the favorable electron-electronegative partner interactions would have to be disrupted, which is unfavorable.

6. (a) The three compounds are an aldehyde, a ketone, and a carboxylic acid.

(b) Here are the 4-carbon compounds:

O O O

Butanal butanone butanoic acid

(c) Ethanol is most likely to react with the carboxylic acid compound to form an ester. Esters are the typical products of dehydration reactions between carboxylic acids and alcohols.

(d) Here is a figure of the ester that would form:

O

O

Ethyl butanoate

In This Chapter

^ Picturing the particles that make up liquids and solids

^ Predicting what happens when you change temperature and pressure

^ Seeing what happens when you mix particles to form solutions

Hen asked, young children often report that solids, liquids, and gases are made up of different kinds of matter. This (mistaken) idea, not usually shared by older people, is understandable given the striking differences in the properties of these three states. But the AP chemistry exam leaves no room for charming misconceptions. In this chapter, we discuss how the properties of the condensed states — liquids and solids — emerge from the interactions between particles, and how these properties can change as temperature and pressure change. We also explore the kinds of interactions that occur between particles within homogenous mixtures called Solutions. Like the properties of pure substances, the properties of solutions depend on the interactions of the particles that make them up.

Restricting Motion: Kinetic Molecular Theory of Liquids and Solids

Kinetic theory (as described in greater detail in Chapter 9) explains the properties of matter in terms of the energetic motions of its particles. Well, not All Of the motions. Really, kinetic theory usually limits itself to a simplified version of matter, one in which infinitely small particles do nothing but zip around quickly and bump into one another or into the sides of a container. This description isn’t ever literally true, but is sometimes a very good approximation of reality, especially for gases in large volumes.

Adding energy to a sample increases its temperature and increases the kinetic energy of the particles, which means that they move about more quickly and bump into things more vigorously. Removing energy from a sample (cooling it) has the opposite effect.

When atoms or molecules have less kinetic energy, or when that energy competes with other effects (like high pressure or strong attractive forces), the matter ceases to be in the diffuse, gaseous state and comes together into one of the Condensed states:

Liquid: The particles within a liquid are much closer together than those in a gas. As a result, applying pressure to a liquid does very little to change the volume. The particles still have an appreciable amount of kinetic energy associated with them, so they may undergo various kinds of twisting, stretching, and vibrating motions. In addition, the particles can slide past one another (translate) fairly easily, so liquids are fluid, though less fluid than gases. Fluid matter assumes the shape of anything that contains it, as shown in Figure 11-1. The particles of a liquid have Short-range order, Meaning that they tend to exhibit some degree of organization over short distances.

^ Solid: The state of matter with the least amount of obvious motion is the solid. In a solid, the particles are packed together quite tightly and undergo almost no long-range translation. Therefore, solids are not fluid. Matter in the solid state still vibrates in place or undergoes other types of motion, depending on its temperature (in other words, on its kinetic energy). The particles of a solid have Long-range order, Meaning that they tend to exhibit organization over long distances. However, matter at a certain temperature must contain a specific amount of energy, regardless of its state. Temperature has no meaning without motion energy.

R

Fluid

Condensed

Gas

Liquid

Increasing order among particles

Solid

Decreasing average kinetic energy

The temperatures and pressures at which different types of matter switch between states depend on the unique properties of the atoms or molecules within that matter. But be careful! It is easy to get fooled by trying to compare different substances at different temperatures. Typically, particles that are very attracted to one another and have easily stackable shapes tend to be in condensed states (at a fixed temperature). Particles with no mutual attraction (or that have mutual repulsion) and with not-so-easily stackable shapes tend toward the gaseous state. Think of a football game between fiercely rival schools. When fans of either school sit in their own section of the stands, the crowd is orderly, sitting nicely in rows. Put rival fans into the same section of the stands, however, and they’ll repel each other with great energy. But be sure you are comparing fans with the same amount of energy. Water as ice has less energy than as a liquid because it has a lower temperature. Technically if you could have a mole of water vapor, a mole of liquid water, and a mole of solid water (ice) all at the same temperature, the water molecules would all have the same kinetic energy!

Getting a Firm Grip on Solids

Solids all have less-apparent motion than their liquid or gaseous counterparts, but that doesn’t mean all solids are alike. The forces between particles within solids as well as the degree of order in the packing of particles within solids vary greatly, giving each solid different properties. The sections that follow shed some light on both the forces that affect solids as well as the packing order that helps to determine a solid’s properties.

Different types of solids and their properties

The properties of a solid depend heavily on the forces between the particles within it. The easiest property to compare is the melting point — that temperature at which the kinetic energy overcomes the strong forces of attraction holding the particles vibrating tightly in a solid.

IU Several different forces determine different melting points of a solid:

• Ionic solids Are held together by an array of very strong ionic bonds (see Chapter 5 for more about these bonds), and, therefore, tend to have high melting points — it takes a great deal of energy to pull apart the particles.

• Molecular solids Consist of packed molecules that are less strongly attractive to each other, so molecular solids tend to have lower melting points.

• Some solids consist of many particles that are covalently bonded to one another in an extensive array. These Covalent solids Tend to be exceptionally strong due to the strength of their extensive covalent network. One example of a covalent solid is diamond. Covalent solids have Very High melting points. Ever try to melt a diamond? (Chapter 5 has details about covalent bonds.)

I Metallic solids Are made up of closely packed metal atoms. These atoms bond to one another more strongly than the particles of most molecular solids, but less strongly than the particles of covalent solids. Because metal atoms so easily give up valence electrons, the atoms within the lattice of a metallic solid seem to exist in a shared "sea" of mobile electrons. The positively charged metal nuclei are held together by their attraction to this negatively charged sea. Metallic solids can be be soft or relatively hard, are ductile and malleable, and are good conductors of heat and electricity. The orderly array of atoms in many metals can allow "sheets" of atoms to slide over one another easily, hence the ease with which metals can be made into wires (ductility) Or beaten into thin foils (malleability).

Packing order in solids

The degree of order in the packing of particles within a solid can vary tremendously. How ordered the particles are within a solid determines how well defined its melting point is. If the particles are well ordered, then the whole sample tends to melt at the same temperature, but if different regions of the sample have different degrees of order, then those regions melt at different temperatures.

I Most solids are highly ordered, packing into neat, repeating patterns called Crystals. The smallest packing unit, the one that repeats over and over to form the Crystalline solid, Is called the Unit cell. Crystalline solids tend to have well-defined melting points.

The particles in crystalline solids tend to organize themselves into arrangements that make the most of the attractive forces between them. Usually, this means packing the particles as closely together as possible.

IU Amorphous solids Are those solids that lack an ordered packing structure. Glass and plastic are examples of amorphous solids. Amorphous solids tend to melt over a broad range of temperatures because some parts of the structures are more easily pulled apart than others.

When cooling a liquid through a phase transition into a solid, the rate of cooling can have a significant impact on the properties of the solid. The particles may need time to move into the extreme order with which they are packed together in crystalline solids. So, substances that are capable of forming crystalline solids may nevertheless freeze into amporphous solids if they are cooled rapidly. The particles may become trapped in disordered packing arrangements. Sometimes this adds considerable strength to a substance, so steels may be "hardened" by heating and sudden cooling.

A collection of different types of forces is very important in determining liquid-solid phase behavior. These forces are more important for the liquid-solid phase than in gases because liquids and solids are condensed states; the molecules within these states are in very close proximity.

In molecular solids, dipole-dipole forces, London dispersion forces, and hydrogen bonds play prominent roles (see Figure 11-2). At the same time, these forces are relatively weak compared to those that dominate in other kinds of solids. Because of the weakness of these forces, molecular solids are relatively soft and tend to have much lower melting points than other solids. In the list below, we describe how these forces work.

Forces at work in condensed states

The forces at work between the particles in a solid (or liquid) largely determine the properties of the substance. For the AP exam, you should definitely know each of the kinds of forces at work in solids and liquids, and be able to predict which forces are most important within a sample of a given compound. These forces include relatively weaker forces (dipole-dipole, London dispersion, and hydrogen bonding) and relatively stronger forces (ionic and covalent bonds).

Here are the intermolecular forces you should know:

IU Dipole-dipole forces (see Figure 11-2) take place between molecules with permanent dipoles (separated regions of opposite charge). Oppositely charged parts of different molecules attract and regions with same type of charge repel. These forces tend to order the molecules.

IU London dispersion forces (shown in Figure 11-2) take place when the positively charged nucleus of one atom attracts the electron cloud of another atom while the electron clouds of both atoms mutually repel one another. In other words, the two atoms induce dipoles in each other, and these Induced dipoles (temporary dipoles created by the nearness of electron clouds) attract one another. It is more easy to redistribute the electrons of some molecules into an induced dipole than it is with others. In other words, some molecules are more Polarizable (capable of having their electrons redistributed) than others. Polarizable molecules tend to take part more strongly in London dispersion forces.

IU Hydrogen bonds Are specific kinds of dipole-dipole attractions that take place between a hydrogen atom in a polar bond and a lone pair of electrons on an electronegative atom (see Chapter 5 for a refresher on electronegativity). Because it participates in a polar bond, the hydrogen has a partial positive charge, 8+. Because it is electronegative, the atom that contributes the lone pair has a partial negative charge, 8-. These partial charges attract. Hydrogen atoms that bond with fluorine, oxygen, and nitrogen are particularly prone to engage in hydrogen bonds. When these interactions take place Between molecules, They significantly increase melting and freezing points. Water hydrogen bonds avidly to itself and to other molecules, as shown in Figure 11-2.

In addition to the relatively weak forces described above, ionic and covalent bonds (discussed in detail in Chapter 5) are strong forces that greatly affect the melting point of a compound:

In ionic solids, Ionic bonds (electrostatic interactions) provide a major source of attraction between particles. These types of solids tend to be hard but brittle (the ionic lattice can crack) and have very high melting points. Lattice energy Is a measure of the strength of the interactions between ions in the lattice of an ionic solid. The larger the lattice energy, the stronger the ion-ion interactions.

^ In covalent solids, particles are bound to each other within strong networks of Covalent bonds. These solids are often exceptionally hard and have very high melting points.

Figure 11-2:

Inter-molecular intera ctions include

(a) dipole-dipole interactions,

(b) London dispersion

Forces, and

(c) hydrogen Attraction- 8- S+ 8-8+

Bonds. Repulsion…..

Moving Through States with Phase Diagrams

The previous sections described how microscopic interactions between particles can lead to large-scale differences in the properties of a sample, especially by causing the sample to be in different states (solid, liquid, or gas). This section describes some tools you can use to track the state of a sample as it moves through different regions of temperature, pressure, or added heat energy.

Phases and phase diagrams

M&l Each state (solid, liquid, gas) is called a Phase. When matter moves from one phase to another due to changes in temperature and/or pressure, that matter is said to undergo a Phase transition. The way a particular substance moves through states as temperature and pressure vary is summarized by a Phase diagram. Phase diagrams usually display pressure on the vertical axis and temperature on the horizontal axis. Lines drawn within the temperature-pressure field of the diagram represent the equilibrium boundaries between phases. A representative phase diagram is shown in Figure 11-3. Refer back to this figure as you read through the section.

Moving from liquid to gas is called Boiling, And the temperature at which boiling occurs is called the Boiling point. The normal boiling point is when this transition occurs at 1 atmosphere of pressure. Moving from solid to liquid is called Melting, And the temperature at which melting occurs is called the Melting point. The melting point temperature is the same as the Freezing point Temperature, but freezing implies matter moving from liquid to solid phase. The melting point a substance has at 1atm pressure is called the Normal melting point. Just as freezing and melting points are the same, condensation points and boiling points are the same temperature. For this reason published tables are of freezing points and boiling points.

Temperature

At the surface of a liquid, molecules can enter the gas phase more easily than elsewhere within the liquid because the motions of those molecules aren’t as constrained by the molecules around them. So, these surface molecules can enter the gas phase at temperatures below the liquid’s characteristic boiling point. This low-temperature phase change is called Evaporation And is very sensitive to pressure. Low pressures allow for greater evaporation, while high pressures encourage molecules to re-enter the liquid phase in a process called Condensation. The pressure of the gas over the surface of a liquid is called the Vapor pressure. It is important not to confuse this with atmospheric pressure due to other gases in the air. For example a sample of liquid water at 20°C has a vapor pressure of 0.023atm while the atmosphere has a pressure of l. Oatm. Understandably, liquids with low boiling points tend to have high vapor pressures; particles in liquids with low boiling points are weakly attracted to each other. At the surface of a liquid, weakly interacting particles have more of a chance to escape into vapor phase, thereby increasing the vapor pressure. See how kinetic theory helps make sense of things?

In addition to having high vapor pressure and low boiling points, substances with weakly interacting molecules tend to have low surface tension and low cohesion. Surface tension Is a measure of the amount of energy it takes to spread out a substance over a larger surface; the weaker the interactions between molecules, the easier this is to do. Cohesion Is the tendency of the molecules of a substance to attract one another. Adhesion Is the tendency of those molecules to bond to the molecules of another substance. Substances that are both adhesive and cohesive display Capillary action, The ability to pull themselves through narrow tubes.

At the right combination of pressure and temperature, matter can move directly from solid to a gas or vapor. This type of phase change is called Sublimation, And is the kind of phase change responsible for the white mist that emanates from dry ice, the common name for solid carbon dioxide. Movement in the opposite direction, from gas directly into solid phase, is called Deposition.

For any given type of matter there is a unique combination of pressure and temperature at the nexus of all three states. This pressure-temperature combination is called the Triple point. At the triple point, all three phases coexist. In the case of good old H2O, going to the triple point produces ice-water vapor. Take a moment to bask in the weirdness.

Other weird phases include the following:

Plasma Is a gaslike state in which electrons pop off gaseous atoms to produce a mixture of free electrons and cations (which are atoms or molecules with positive charge). For most types of matter, achieving the plasma state requires very high temperatures, very low pressures, or both. Matter at the surface of the sun, for example, exists as plasma.

^ Supercritical fluids (SCF) exist under high temperature-high pressure conditions. For a given type of matter, there is a unique combination of temperature and pressure called the Critical point. At temperatures and pressures higher than those at this point, the phase boundary between liquid and gas disappears, and the matter exists as a kind of liquidy gas or gassy liquid. Supercritical fluids can diffuse through solids like gases do, but can also dissolve things like liquids do. SCFs are being used in some areas as extraction agents in dry cleaning.

Changing phase and temperature along heating curves

Starting from the solid phase, you can add heat energy to a sample, causing it to progress through liquid and vapor phases. If you measure the temperature of the sample as you do this, you’ll find that it has a staircase pattern. This pattern is called a Heating curve, And results from the fact that it takes energy simply to move particles from solid to liquid and from liquid to vapor (moving in the opposite direction releases energy). When a substance is at its melting point or freezing point, added heat energy goes toward disrupting attractive forces between the molecules instead of increasing the temperature (the average kinetic energy) of the molecules. Like phase diagrams, the exact shape of a heating curve varies from one substance to another. The heating curve for water is shown in Figure 11-4. Heating curves usually assume a constant rate of energy input.

Figure 11-4:

A heating curve for water.

125 100 75 50 25 0

-25

Heat Added

Dissolving With Distinction: Solubility and Types of Solutions

Compounds can form mixtures. When compounds mix completely, right down to the level of individual molecules, we call the mixture a Solution. Each type of compound in a solution is called a Component. The component of which there is the most is usually called the Solvent. The other components are called Solutes. Although most people think "liquid" when they think of solutions, a solution can be a solid, liquid, or gas. The only criterion is that the components are completely intermixed. We explain what you need to know in this section. By far the most important solutions (99% on the AP exam) are those where water is the solvent. Master those before worrying about other solutions.

Forces in solvation

For gases, forming a solution is a straightforward process. Gases simply diffuse into a common volume (see Chapter 9 for more about diffusion). Things are a bit more complicated for condensed states like liquids and solids. In liquids and solids, molecules or ions are crammed so closely together that Intermolecular forces (forces between molecules) are very important. Examples of these kinds of forces include dipole-dipole, hydrogen bonding, and London dispersion forces as discussed previously. In addition, ion-dipole forces can be important in solutions, as shown for water in Figure 11-5.

Figure 11-5:

Water molecules participate in ion-dipole intera ctions with a cation.

V

O 8-

O

8-

+

©

8-

O

8-

O

Introducing a solute into a solvent initiates a tournament of forces. Attractive forces between solute and solvent compete with attractive solute-solute and solvent-solvent forces, as depicted in Figure 11-6. A solution forms only to the extent that solute-solvent forces dominate over the others. The process in which solvent molecules compete and win in the tournament of forces is called Solvation Or, in the specific case where water is the solvent, Hydration. Solvated solutes are surrounded by solvent molecules. When solute ions or molecules become separated from one another and surrounded in this way, we say they are Dissolved.

Imagine that the members of a ridiculously popular band exit their hotel to be greeted by an assembled throng of fans and the media. The band members attempt to cling to each other, but are soon overwhelmed by the crowd’s ceaseless, repeated attempts to get closer. Soon, each member of the band is surrounded by his own attending shell of reporters and hyperventilating fans. So it is with dissolution.

Figure 11-6:

An ionic compound dissolves in water.

The tournament of forces plays out differently among different combinations of components. In mixtures where solute and solvent are strongly attracted to one another, more solute can be dissolved. One factor that always tends to favor solvation is Entropy, A kind of disorder or "randomness" within a system. Dissolved solutes are less ordered than undissolved solutes. Beyond a certain point, however, adding more solute to a solution doesn’t result in a greater amount of solvation. At this point, the solution is in dynamic equilibrium; the rate at which solute becomes solvated equals the rate at which dissolved solute Crystallizes, Or falls out of solution. A solution in this state is Saturated. By contrast, an Unsaturated Solution is one that can accommodate more solute. A Supersaturated Solution is a temporary one in which more solute is dissolved than is necessary to make a saturated solution. A supersaturated solution is unstable; solute molecules may crash out of solution given the slightest perturbation. The situation is like that of Wile E. Coyote, who runs off a cliff and remains suspended in the air until he looks down — at which point he inevitably falls.

To dissolve or not to dissolve: Solubility

The concentration of solute (the amount of solute relative to the amount of solvent or the total amount of solution) required to make a saturated solution is the Solubility Of that solute. Solubility varies with the conditions of the solution. The same solute may have different solubility in different solvents, and at different temperatures, and so on.

When one liquid is added to another, the extent to which they intermix is called Miscibility. Typically, liquids that have similar properties mix well — they are Miscible. Liquids with dissimilar properties often don’t mix well — they are Immiscible. This pattern is summarized by the phrase, "like dissolves like." Alternately, you may understand lack of miscibility in terms of the Italian Salad Dressing Principle. Inspect a bottle of Italian salad dressing that has been sitting in your refrigerator. Observe the following: The dressing consists of two distinct layers, an oily layer and a watery layer. Before using, you must shake the bottle to temporarily mix the layers. Eventually, they will separate again because water is polar and oil is nonpolar.

(See Chapter 5 if the distinction between polar and nonpolar is lost on you.) Polar and non-polar liquids mix poorly, though occasionally with positive gastronomic consequences.

Similarity or difference in polarity between components is often a good predictor of solubility, regardless of whether those components are liquid, solid, or gas. This rule is often described as "like dissolves like." Why is polarity such a good predictor? Because polarity is central to the tournament of forces that underlies solubility. So, solids held together by ionic bonds (the most polar type of bond) or polar covalent bonds tend to dissolve well in polar solvents, like water. For a refresher on ionic and covalent bonding, visit Chapter 5.

Heat effects on solubility

Increasing temperature magnifies the effects of entropy on a system. Because the entropy of a solute is usually increased when it dissolves, increasing temperature usually increases solubility — for solid and liquid solutes, anyway. Another way to understand the effect of temperature on solubility is to think about heat as a reactant in the dissolution reaction:

Solid solute + solvent + heat — Dissolved solute

Heat is usually absorbed when a solute dissolves. Increasing temperature corresponds to added heat. So, by increasing temperature you supply a needed reactant in the dissolution reaction. (In those rare cases where dissolution releases heat, increasing temperature can decrease solubility.) NaCl is an example where the solubility changes very little with temperature change.

Gaseous solutes behave differently than do solid or liquid solutes with respect to temperature. Increasing the temperature tends to decrease the solubility of gas in liquid. To understand this pattern, recall the concept of vapor pressure from earlier in the chapter. Increasing temperature increases vapor pressure because added heat increases the kinetic energy of the particles in solution. With added energy, these particles stand a greater chance of breaking free from the intermolecular forces that hold them in solution. A classic, real-life example of temperature’s effect on gas solubility is carbonated soda. Which goes flat (loses its dissolved carbon dioxide gas) more quickly: warm soda or cold soda?

Pressure effects on gas solubility

The comparison of gas solubility in liquids with the concept of vapor pressure highlights another important pattern: Increasing pressure increases the solubility of a gas in liquid. Just as high pressures make it more difficult for surface-dwelling liquid molecules to escape into vapor phase, high pressures inhibit the escape of gases dissolved in solvent, as highlighted by Figure 11-7. The relationship between pressure and gas solubility is summarized by Henry’s law:

SA = k x PA

Where SA is the solubility of A, PA is the partial pressure of A in the vapor over the solution, and K Is Henry’s constant. The value of Henry’s constant depends on the gas, solvent, and temperature, and is accurate for small concentrations of solute. A particularly useful form of Henry’s law relates the change in solubility (S) That accompanies a change in pressure (P) Between two different states:

S1 / P1 = S2 / P2

According to this relationship, tripling the pressure triples the gas solubility, for example.

Figure 11-7:

Pressure alters the

Equilibrium of vapor

Molecules at a liquid interface.

O o o

Lid

Measuring solute concentration

It seems that different solutes dissolve to different extents in different solvents in different conditions. How can anybody keep track of all these differences? Chemists do so by measuring Concentration. Qualitatively, a solution with a large amount of solute is said to be Concentrated. A solution with only a small amount of solute is said to be Dilute. As you may suspect, simply describing a solution as concentrated or dilute is usually about as useful as calling it "pretty" or naming it "Fifi." We need numbers. Two important ways to measure concentration are Molarity And Percent solution.

Molarity relates the amount of solute to the volume of the solution:

Molarity (M )= molessolute liters solution

In order to calculate molarity, you may have to use conversion factors to move between units. For example, if you are given the mass of a solute in grams, use the molar mass of that solute to convert the given mass into moles. If you are given the volume of solution in cm3 or some other unit, you’ll need to convert that volume into liters.

The units of molarity are always mol L-1. These units are often abbreviated as M And referred to as "molar." Thus, 0.25M KOH(aq) is described as "Point two-five molar potassium hydroxide" and contains 0.25 moles of KOH per liter of solution. Note that this does Not Mean that there are 0.25 moles KOH per liter of Solvent (water, in this case) — only the final volume of the solution (solute plus solvent) is important in molarity. Why? Because often volumes just don’t add up when two substances are mixed. (It is kinetic theory again!) Molar concentrations of a substance are often denoted by brackets, as in [KOH] = 0.25. Like other units, the unit of molarity can be modified by standard prefixes, as in millimolar (mM, 10-3 mol L-1) and micromolar (uJM, 10-6 mol L-1).

One important quantity that is measured in units of molarity is the Solubility product constant, Ksp. The solubility product is useful for measuring the dynamic equilibrium of ionic compounds in any given solvent. Once a saturated solution of the compound has been made, further addition of that compound has no effect on the concentration of dissolved solute. The concentrations of the component ions of the compound therefore remain constant and

Reflect the characteristic solubility of the compound in that solvent. The Ksp measures this solubility. For the the ionic compound XAYB,

Ksp = [X]A x [Y]B

Percent solution Is another common way to express concentration. The precise units of percent solution typically depend on the phase of each component. For solids dissolved in liquids, mass percent is usually used:

Mass % = 100% x-masssolute.

Total mass solution

This kind of measurement is sometimes called a mass-mass percent solution because one mass is divided by another. Very dilute concentrations (as in the concentration of a contaminant in drinking water) are sometimes expressed as a special mass percent called Parts per million (ppm) Or Parts per billion (ppb). In these metrics, the mass of the solute is divided by the total mass of the solution, and the resulting fraction is multiplied by 106 (ppm) or by 109 (ppb).

Clearly, it’s important to pay attention to units when working with concentration. Only by observing which units are attached to a measurement can you determine whether you are working with molarity, mass percent, or with mass-mass, mass-volume, or volume-volume percent solution.

Real-life chemists in real-life labs don’t make every solution from scratch. Instead, they make concentrated Stock solutions (starting solutions) and then make Dilutions (solutions in which solvent is added to stock solution) of those stocks as necessary for a given experiment.

To make a dilution, you simply add a small quantity of a concentrated stock solution to an amount of pure solvent. The resulting solution contains the amount of solute originally taken from the stock solution, but disperses that solute throughout a greater volume. So, the final concentration is lower; the final solution is less concentrated and more dilute.

But how do you know how much of the stock solution to use, and how much of the pure solvent to use? It depends on the concentration of the stock and on the concentration and volume of the final solution you want. You can answer these kinds of pressing questions by using the dilution equation, which relates concentration (C) And volume (V) Between initial and final states:

C1 x V1 = C2 x V2

This equation can be used with any units of concentration, provided the same units are used throughout the calculation. Because molarity is such a common way to express concentration, the dilution equation is sometimes expressed in the following way, where M1 and M2 refer to the initial and final molarity, respectively:

M1 x V1 = M2 x V2

Dissolving with Perfection: Ideal Solutions and Colligative Properties

If you’ve read the rest of this chapter, you may consider yourself a recently minted expert in solubility and molarity, ready to write off solutions as another chemistry topic mastered. Don’t. You, as a chemist worth your salt, must be aware of another piece to the puzzle: Colligative properties. Colligative properties are the properties of a solution compared to a

Pure solvent that change as a function of the number of solute particles in solution, regardless of what kind of particles. The presence of extra particles in a formerly pure solvent has a significant impact on some of that solvent’s characteristic properties, such as vapor pressure, freezing point, and boiling point.

Understanding ideal solutions

Understanding how solute particles affect the properties of a solution requires you to know first whether you’re dealing with an "ideal solution." An Ideal solution Is one in which properties change proportionally (that is, in a linear way) with the amount of added solute. Thankfully, only ideal solutions are on the AP test!

Two kinds of solutions tend to approach ideal behavior:

Very dilute solutions

Solutions in which solute-solvent interactions are about the same strength as solvent-solvent and solute-solute interactions

Ideal solutions obey Raoult’s law. Raoult’s law states that the vapor pressure over the surface of an ideal solution should be the sum of the vapor pressures of the pure components multi-pled by their mole fraction in the solution. In other words, each solution component should contribute exactly its fair share to the total vapor pressure, no more, no less. For a two-component solution

Ptotal = PA X XA + PB X XB

Where Ptotal is the total vapor pressure over the solution, PA And PB Are the vapor pressures of pure samples of components A and B, and - A and - B are the mole fractions of components A and B in the solution.

Mole fraction Is the ratio of the number of moles of one component in a solution to the QfBER Number of moles of all the components in the solution.

In general, the mole fractions of a two-component solution are expressed as

X A = ——— and XB = ———

At"I-1-T"I B „ i „

N A + nB nA + n B

Where nA is the number of moles of component A (like a solute) and nB is the number of moles of component B (like a solvent).

Raoult’s law makes a prediction: If you add a nonvolatile solute (one that contributes no vapor pressure of its own) to a solvent, the vapor pressure of the resulting solution should be lower than the vapor pressure of the pure solvent. If you’ve got a solution that seems to obey Raoult’s law, then you’ve got a solution for which you can make useful predictions about colligative properties.

Using molality to predict colligative properties

To correctly account for the effects of solute particles on some colligative properties, you need a new way to measure solution concentration: Molality. No, that’s not a typo. Molality is different from molarity.

Like the difference in their names, the difference between molarity and molality is subtle. Whereas molarity measures the moles of solute per liter of solution, molality measures the Moles of solute particles per kilogram of solvent:

,, , w / \ moles solute particles

Molality (M ) = ——–

V ‘ kilogram solvent

Notice that the numerator of the fraction for calculating molality includes "solute particles" and not just "solute." What’s the difference? When one mole of the ionic compound NaCl dissolves into one liter of aqueous solution, it produces 1 molar sodium chloride, 1M NaCl(aq). But when NaCl dissolves, it becomes one mole of Na+ cation and one mole of Cl – anion — two moles of solute particles. So, when one mole of NaCl dissolves into one kilogram of water, it produces 2 molal sodium chloride, 2m NaCl(aq).

Calculating molality is no more or less difficult than calculating molarity, so you may be asking yourself, "Why all the fuss?" Is it even worth adding another quantity and another variable to memorize? Yes! Although molarity is exceptionally convenient for calculating concentrations and working out how to make dilutions in the most efficient way, molality is useful for predicting important colligative properties, including the boiling point of a solution. When solute particles are added to a solvent, the boiling point of the resulting solution tends to increase relative to the boiling point of the pure solvent. This phenomenon is called Boiling point elevation. The more solute you add, the greater you elevate the boiling point.

Boiling point elevation is directly proportional to the molality of a solution, but chemists have found that some solvents are more susceptible to this change than others. The formula for the change in the boiling point (TB) of a solution therefore contains a proportionality constant, Kb (not to be confused with an equilibrium constant!). The Kb is determined by experiment and in practice you usually look it up in a table such as Table 11-1. The formula for the boiling point elevation is

A7b = Kb x M

Note the use of the Greek letter delta (A) in the formula to indicate that you are calculating a Change In boiling point, not the boiling point itself. You’ll need to add this number to the boiling point of the pure solvent to get the boiling point of the solution. The units of KB are given in degrees Celsius per molality (°C m"1).

Boiling point elevations are a result of the attraction between solvent and solute particles in a solution. Adding solute particles increases these intermolecular attractions because there are more particles around to attract one another. Solvent particles must therefore achieve a greater kinetic energy to overcome this extra attractive force and boil off the surface. Greater kinetic energy means a higher temperature, and therefore a higher boiling point. An alternative explanation is that there are simply fewer solvent molecules on the surface to escape as some surface locations are occupied by solute particles.

The second of the important colligative properties you can calculate by using molality is the freezing point (TF) of a solution. When solute particles are added to a solvent, the freezing point of the solution tends to decrease relative to that of the pure solvent. This phenomenon is called Freezing point depression. The more solute you add, the more you decrease the freezing point. This is the reason, for example, that you sprinkle salt on icy sidewalks. The salt mixes with the ice and lowers its freezing point. If this new freezing point is lower than the outside temperature, the ice melts, eliminating the spectacular wipeouts so common on salt-free sidewalks. The colder it is outside, the more salt is needed to melt the ice and lower the freezing point to below the ambient temperature.

Like boiling point elevation, freezing point depression is directly proportional to the molality of the solution. So, the formula for freezing point depression contains a constant of proportionality, KF, that depends on the solvent in question. The formula for freezing point depression is

Tf = Kf x M

To calculate the new freezing point of a compound, you must Subtract The change in freezing point from the freezing point of the pure solvent. Table 11-2 lists several common KF values.

Freezing point depressions are the result of solute particles interrupting the crystalline order of a frozen solid. In order to reach a solid, frozen state, the solution must achieve an even lower average kinetic energy. Lower kinetic energy means lower temperature, and therefore a lower freezing point.

In summary, adding solute particles to a solvent increases the stability of the liquid phase. Boiling point elevation and freezing point depression mean that greater changes in energy are required to shift the solution out of liquid phase into solid or vapor phase. This effect can be seen in a phase diagram that overlays the behavior of a solution with that of the pure solvent, as shown in Figure 11-8.

By carefully measuring boiling point elevations and freezing point depressions, you can determine the molar mass of a mystery compound that is added to a known quantity of solvent. To pull off this trick you must know the total mass of the pile of mystery solute that has been dissolved, the mass of solvent into which the compound was dissolved, and either the change in the freezing or boiling point or the new freezing or boiling point itself. From this information you then follow this set of simple steps to determine the molecular mass:

1. If you know the boiling point of the solution, calculate the ATJ, By subtracting the boiling point of the pure solvent from that value. If you know the freezing point of the solution, subtract the freezing point of the pure solvent to that value to get the ATF.

2. Look up the Kb Or Kf Of the solvent (in a place like Table 11-1 or 11-2).

3. Solve for the molality of the solution using the equation for A7B Or ATF.

4. Calculate the number of moles of solute in the solution by multiplying the molality calculated in Step 3 by the mass of solvent, in kilograms.

5. Divide the mass of the pile of mystery solute that has been dissolved by the number of moles calculated in Step 4. The result of this calculation is the molar mass (number of grams per mole) of the mystery compound. From this value, you can often make an educated guess about the identity of the compound.

Osmotic pressure Is another property of a solution that depends on the number of solute particles. To understand osmotic pressure, it helps to understand Osmosis, The movement of solvent molecules through a semipermeable barrier (one which lets some things pass, but not others) from areas of low solute concentration to areas of high solute concentration. Osmosis is a result of the more general process of Diffusion, The net movement of a substance from where it is more concentrated to where it is less concentrated. In solutions with higher solute concentration, solvent is less concentrated. In solutions with lower solute concentration, solvent is more concentrated. So, given the opportunity, solvent diffuses toward solutions with higher solute concentration.

If solvent and a solution made with that solvent are separated by a semipermeable membrane in a setup like the one shown in Figure 11-9, solvent molecules will move by osmosis into the solute-containing chamber. This movement of solute will alter the surface height within each chamber until the difference in height causes enough pressure to prevent a further net movement of solvent into the solute-containing chamber. This pressure difference is equal to the osmotic pressure, n, of the solution. Osmotic pressure depends on the moles of solute particles per unit volume of solution; that is, in contrast to boiling points and freezing points, osmotic pressure depends on molarity, not molality:

N = I N Soiae I Rt = MRT

{ VSoUion J

Where Nsolute Is the moles of solute particles, Vsolutoon Is the volume of solution, M Is molarity of particles, R Is the gas constant and T Is temperature.

■ Semipermeable membrane

Pressure

Figure 11-9:

Osmosis across a semipermeable membrane.

O O O

O o o

Solvent

O

O

O °Q

©o Qo

O ___.

Osmosis

Solute

Osmosis and osmotic pressure are important in biology because cell membranes are semi-permeable, allowing transport of solvent water molecules while restricting transport of many solutes. So, if a cell finds itself in a Hypertonic Environment (one with higher solute concentration than that of the cell), osmosis may cause the cell to shrivel as water diffuses outward. If the cell finds itself in a Hypotonic Environment (one with lower solute concentration than that of the cell), osmosis may cause the cell to swell (or explode!) as water diffuses inward. So, even if you yawn at osmotic pressure, the cells that compose you consider it a life-or-death kind of thing.

Dissolving with Reality: Nonideal Solutions

Many solutions come very close to ideal behavior. But some solutions deviate pretty significantly from ideal behavior. And for sensitive applications, sometimes even very close isn’t close enough. Situations like these force us to deal with Nonideal solutions, Ones that don’t obey Raoult’s law or Henry’s law, and whose properties aren’t proportional to the amount of added solute.

Nonideal solutions occur when solute concentrations are very high and/or when solute-solvent interactions are significantly more attractive or repulsive than solute-solute interactions and solvent-solvent interactions.

When solute-solvent interactions are especially attractive, the solute is effectively more dissolved (that is, more intermixed) than an ideal solute. The vapor pressure of the solvent is lower than predicted by Raoult’s law. The partial pressure of a dilute solute is higher than that predicted by Henry’s law.

When solute-solvent interactions are especially repulsive, the solute is effectively less intermixed than an ideal solute. The vapor pressure of the solution is higher than predicted by Raoult’s law. The partial pressure of a dilute solute is lower than that predicted by Henry’s law.

To account for nonideal behavior, chemists use the concept of Activity, The effective concentration of a component in solution. When the activity of a component differs significantly from its formal concentration, chemists often use an experimentally determined Activity coefficient In their calculations. The concentration of the nonideal component (its molarity, molality or mole fraction) is multiplied by an activity coefficient, y, to produce an effective concentration for the calculation:

Activity = yx Concentration

In This Chapter

^ Reviewing the seven common reaction types ^ Mastering the chemical balancing act

^ Familiarizing yourself with the mole and how to use it in conversions

^ Using net ionic equations

^ Counting what counts with limiting reagents

^ Quantifying success with percent yield

^^toichiometry. Such a complicated word for such a simple idea. The Greek roots of the

Word mean "measuring elements," which doesn’t sound nearly as intimidating. Moreover, the ancient Greeks couldn’t tell an ionic bond from an ionic column, so just how technical and scary could stoichiometry really be? Simply stated, stoichiometry (stoh-ick-eee-AH-muh-tree) is the quantitative relationship between atoms in chemical substances.

We will begin this chapter by walking you through each of the types of chemical reaction you will encounter on the AP chemistry exam. Several of them will be discussed in greater detail in later chapters. Then, we will show you how to use your understanding of chemical equations to do AP-type Chemistry problems, and Chapter 14 will allow you to practice your newfound skills.

The first half of the chapter describes the basics of stoichiometry, then delves into the seven types of reaction that you’d do well to recognize (notice how their names tell you what happens in each reaction). By recognizing the patterns of these seven types of reaction, you can often predict reaction products given only a set of reactants. There are no perfect guidelines, and predicting reaction products can take what is called "chemical intuition," a sense of what reaction is likely to occur based on knowing the outcomes of similar reactions. Still, if you are given both reactants and products, you should be able to tell what kind of reaction connects them, and if you are given reactants and the type of reaction, you should be able to predict likely products. Figuring out the formulas of products often requires you to apply knowledge about how ionic and molecular compounds are put together. To review these concepts, see Chapter 5.

The remainder of the chapter covers how to use your knowledge of these reaction types to calculate the quantities that allow you to quantify the success of a reaction.

Stoichiometry Lingo: How to Decipher Chemical Reactions

Chemists have a nice, ordered set of rules for describing chemical reactions. Reactants are always on the lefthand side of the equation, and products on the right, for example. Chemists would rather stick their hands in a vat of concentrated hydrochloric acid than reverse this order. There are also a nice, ordered set of chemical symbols that chemists use to describe reactions, and each symbol has one precise meaning. These fundamentals of stoichiometry are described in this section and, since they form the basis of all of stoichiometry, which is a huge point-gainer on the AP exam, it would behoove you to memorize them before moving on.

In compound formulas and reaction equations, you express stoichiometry by using subscripted numbers on atoms and coefficients in front of groups of atoms. Just like an accountant can’t afford to make a mistake when tracking the dollars and cents in your parents’ bank account, you can’t afford to lose track of any atoms on the AP exam. The road to poor scores in AP Chemistry is littered with students who couldn’t master stoichiometry!

In general, all chemical equations are written in the basic form:

Reactants — Products

Where the arrow in the middle means Yields Or "turns into." The basic idea is that the reactants react, and the reaction produces products. By Reacting, We simply mean that bonds within the reactants are broken, to be replaced by new and different bonds within the products.

Chemists fill chemical equations with symbols because they think it looks cool and, more important, because the symbols help pack a lot of meaning into a small space. Table 13-1 summarizes the most important symbols you’ll find in chemical equations.

After you understand how to interpret chemical symbols, compound names (see Chapter 5), and the symbols in Table 13-1, there’s not a lot you can’t understand. You’re equipped, for example, to decode a chemical equation into an English sentence describing a reaction. Conversely, you can translate an English sentence into the chemical equation it describes. When you’re fluent in this language, you regrettably won’t be able to talk to the animals; you will, however, be able to describe their metabolism in great detail.

Building and Breaking: Synthesis and Decomposition Reactions

The two simplest types of chemical reactions involve the joining of two compounds to form one compound or the breaking apart of one compound to form two. We describe both of these simple reactions in the sections that follow.

Synthesis

In synthesis (or combination) reactions, two or more reactants combine to form a single product, following the general pattern

A + B — C

For example,

2Na(s) + Cl2(g) — 2NaCl(s)

The combining of elements to form compounds (like NaCl) is a particularly common kind of combination reaction. Here is another such example:

2Ca(s) + O2(g) — 2CaO(s)

Compounds can also combine to form new compounds, such as in the combination of sodium oxide with water to form sodium hydroxide:

Na2O(s) + H2O(l) — 2NaOH(ag)

Decomposition

In a decomposition reaction, a single reactant breaks down (decomposes) into two or more products, following the general pattern

A —B + C

For example,

2H2O(l) — 2H2(g) + O2(g)

Notice that combination and decomposition reactions are the same reaction in opposite directions.

Many decomposition reactions produce gaseous products, such as in the decomposition of carbonic acid into water and carbon dioxide:

H2CO3(ag) — H2O(l) + CO2(g)

Swapping Spots: Displacement Reactions

Many common reactions involve the swapping of one or more atoms or polyatomic ions in a compound. You can check these reactions out in the following sections.

Single replacement

In a single replacement reaction, a single, more-reactive element or group replaces a less-reactive element or group, following the general pattern

A + BC — AC + B

For example,

Zn(s) + CuSO4(a<7) — ZnSO4(ag) + Cu(s)

Single replacement reactions in which metals replace other metals are especially common. You can determine which metals are likely to replace which others by using the Metal activity series, A ranked list of metals in which ones higher on the list tend to replace ones lower on the list. Table 13-2 presents the metal activity series.

Double replacement

Double replacement is a special form of Metathesis reaction (that is, a reaction in which two reacting species exchange bonds). Double replacement reactions tend to occur between ionic compounds in solution. In these reactions, cations (atoms or groups with positive

Charge) from each reactant swap places to form ionic compounds with the opposing anions (atoms or groups with negative charge), following the general pattern

AB + CD — AD + CB

For example,

KCl(a<7) + AgNO3(ag) — AgCl(s) + KNO3(ag)

Of course, ions dissolved in solution move about freely, not as part of cation-anion complexes. So, to allow double replacement reactions to progress, one of several things must occur:

One of the product compounds must be insoluble, so it precipitates (forms an insoluble solid) out of solution after it forms.

One of the products must be a gas that bubbles out of solution after it forms.

One of the products must be a solvent molecule, such as H2O, that separates from the ionic compounds after it forms.

Burning Up: Combustion Reactions

Oxygen is always a reactant in combustion reactions, which often release heat and light as they occur. Combustion reactions frequently involve hydrocarbon reactants (like propane, C3H8(g), the gas used to fire up backyard grills), and yield carbon dioxide and water as products. For example,

C3H8(g) + 5O2(g) — 3CO2(g) + 4H2O(l)

Combustion reactions also include combination reactions between elements and oxygen, such as:

S(s) + O2(g) — SO2(g)

So, if the reactants include oxygen (O2) and a hydrocarbon or an element, you’re probably dealing with a combustion reaction. If the products are carbon dioxide and water, you’re almost certainly dealing with a combustion reaction.

Getting Sour: Acid-Base Reactions

Acids and bases are complicated enough to deserve their own chapter, so flip forward to Chapter 17 for the nitty-gritty details of this reaction type. In the meantime, and so you have a comprehensive list of reaction types which the AP chemistry exam may quiz you on in one place, we’ll provide a summary here.

Neutralization reactions, which occur when an acid and a base are mixed together to form a salt and water, are the bread and butter of acid-base reactions. The best way to spot a reaction of this type is to look for water among the products of a reaction. If the water came from a double replacement reaction in which a hydrogen from one reactant (the acid) joined up with a hydroxide from the other (the base), then you are definitely dealing with an acid-base reaction.

Not all acid-base reactions are so simple, however. Bases do not necessarily contain hydroxide, and the products are not always water and a salt. See Chapter 17 for the details of the myriad of acid-base reactions.

Getting Charged: Oxidation-Reduction Reactions

Oxidation-reduction (or redox) reactions, like acid-base reactions, have an entire chapter (Chapter 19) devoted to them. Redox reactions are concerned with the transfer of electrons between reactants. Because they involve the transfer of charged particles, ions are a staple of redox reactions and are the things to look out for when trying to spot one. The species in a redox reaction that loses electrons to the other reactant is Oxidized In the reaction, while the species that gains the electrons is Reduced. There are two easy mnemonics commonly used to remember this, so take your pick or make up your own:

LEO And GER: Lose electrons, oxidation. Gain electrons, reduction. OIL RIG: Oxidation is lost, reduction is gained.

Oxidation and reduction reactions always occur together, but may be written separately as half reactions in a chemistry problem. In these half reactions, electrons being transferred are written explicitly, so if you see them, take it as a dead giveaway that you are dealing with a redox reaction.

The Stoichiometry Underground: Moles

Chemical equations are all well and good, but what do they actually mean? For translating chemical equations into useful chemical data, you can find no better or more fundamental tool than the mole. As an AP chemistry student, you’re probably already exhaustively familiar with the mole, but in the following sections, we provide you with a comprehensive review of the mole and its usefulness in case you forgot any of the basics along the way.

Counting your particles: The mole

Chemists routinely deal with hunks of material containing trillions of trillions of atoms, but ridiculously large numbers can induce migraines. For this reason, chemists count particles (like atoms and molecules) in multiples of a quantity called the Mole. Initially, counting particles in moles can be counterintuitive, similar to the weirdness of Dustin Hoffman’s character in the movie Rainman, When he informs Tom Cruise’s character that he has spilled 0.41 fraction of a box of 200 toothpicks onto the floor. Instead of referring to 82 individual toothpick particles, he refers to a fraction of a larger unit, the 200-toothpick box. A mole is a very big box of toothpicks — 6.022 x 1023 toothpicks, to be precise.

If 6.022 x 1023 strikes you as an unfathomably large number, then you’re thinking about it correctly. It’s larger, in fact, than the number of stars in the sky or the number of fish in the sea, and is many, many times more than the number of people who have ever been born throughout all of human history. When you think about the number of particles in something as simple as, say, a cup of water, all your previous conceptions of "big numbers" are blown out of the water, as it were.

The number 6.022 x 1023, known as Avogadro’s number, Is named after the 19th century Italian scientist Amedeo Avogadro. Posthumously, Avogadro really pulled one off in giving his name to this number, because he never actually thought of it. The real brain behind Avogadro’s number was that of a French scientist named Jean Baptiste Perrin. Nearly 100 years after Avogadro had his final pasta, Perrin named the number after him as an homage. Ironically, this humble act of tribute has misdirected the resentment of countless hordes of high school chemistry students to Avogadro instead of Perrin.

Avogadro’s number is the conversion factor used to move between particle counts and numbers of moles:

1 mole / 6.022 x 1023 particles

Like all conversion factors, you can invert it to move in the other direction, from moles to particles.

Assigning mass and volume to your particles

Chemists always begin a discussion about moles with Avogadro’s number. They do this for two reasons. First, it makes sense to start the discussion with the way the mole was originally defined. Second, it’s a sufficiently large number to intimidate the unworthy.

Still, for all its primacy and intimidating size, Avogadro’s number quickly grows tedious in everyday use. More interesting is the fact that one mole of a pure Monatomic Substance (in other words, a substance that always appears as a single atom) turns out to possess exactly its atomic mass’s worth of grams. In other words, one mole of monatomic hydrogen weighs about 1 gram. One mole of monatomic helium weighs about 4 grams. The same is true no matter where you wander through the corridors of the periodic table. If we define Molar mass As the mass of one mole of any substance, then the number listed as the atomic mass of an element equals that element’s molar mass If the element is monatomic. Guess what — chemists actually set up the value of the mole to make that happen, though it took years of arguments and an international commission to get the final details down to the last decimal place just right.

Of course, chemistry involves the making and breaking of bonds, so talk of pure monatomic substances gets you only so far. How lucky, then, that calculating the mass per mole of a complex molecule is essentially no different than finding the mass per mole of a monatomic element. For example, one molecule of glucose (C6H12O6) is assembled from 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms. To calculate the number of grams per mole of a complex molecule (such as glucose), simply do the following:

1. Multiply the number of atoms per mole of the first element by its atomic mass.

In this case, the first element is carbon, and you’d multiply its atomic mass, 12, by the number of atoms, 6.

2. Multiply the number of atoms per mole of the second element by its atomic mass.

Here, you multiply hydrogen’s atomic mass of 1 by the number of hydrogen atoms, 12.

3. Multiply the number of atoms per mole of the third element by its atomic mass. Keep going until you’ve covered all the elements in the molecule.

The third element in glucose is oxygen, so you multiply 16, the atomic mass by 6, the number of oxygen atoms.

4. Finally, add the masses together.

In this example, (12g mol-1 x 6) + (1g mol-1 x 12) + (16g mol-1 x 6) = 180g mol-1.

This kind of quantity, called the Gram molecular mass, Is exceptionally convenient for chemists, who are much more inclined to measure the mass of a substance than to count all of millions or billions of individual particles that make it up.

If chemists don’t try to intimidate you with large numbers, they may attempt to do so by throwing around big words. For example, older chemists may distinguish between the molar masses of pure elements, molecular compounds, and ionic compounds by referring to them as the Gram atomic mass, gram molecular mass, And Gram formula mass, Respectively. Don’t be fooled! The basic concept behind each is the same: molar mass. Fortunately these other terms are never used by those modern folks who write AP chemistry exams.

It’s all very good to find the mass of a solid or liquid and then go about calculating the number of moles in that sample. But what about gases? Let’s not engage in phase discrimination; gases are made of matter, too, and their moles have the right to stand and be counted. Fortunately, there’s a convenient way to convert between the moles of gaseous particles and their Volume. Unlike gram atomic/molecular/formula masses, this conversion factor is constant No matter what kinds of molecules make up the gas. Every gas assumed to behave ideally has a volume of 22.4 liters per mole, regardless of the size of the gaseous molecules.

Before you start your hooray-chemistry-is-finally-getting-simple dance, however, understand that certain conditions apply to this conversion factor. For example, it’s only true at Standard temperature and pressure (STP), Or 0° Celsius and 1atm. Also, the figure of 22.4L mol-1 applies only to the extent that a gas resembles an ideal gas, one whose particles have zero volume and neither attract nor repel one another. Ultimately, no gas is truly ideal, but many are so close to being so that the 22.4L mol-1 conversion is very useful.

Giving credit where it’s due: Percent composition

Chemists are often concerned with precisely what percentage of a compound’s mass consists of one particular element. Lying awake at night, uttering prayers to Avogadro, they fret over this quantity, called Percent composition. Percent composition was mentioned in Chapter 5, but now that you have reviewed the basics of stoichiometry, we will explain in detail how to calculate it.

Follow these three simple steps:

1. Calculate the molar mass of the compound, as we explain in the previous section.

Percent compositions are completely irrelevant to molar masses for pure monatomic substances, which by definition have 100 percent composition of a given element.

2. Multiply the atomic mass of each element present in the compound by the number of atoms of that element present in one molecule.

3. Divide each of the masses calculated in Step 2 by the total mass calculated in Step 1. Multiply each fractional quotient by 100%. Voila! You have the Percent composition By mass of each element in the compound.

Checking out empirical formulas

What if you don’t know the formula of a compound? Chemists sometimes find themselves in this disconcerting scenario. Rather than curse Avogadro (or perhaps After Doing so), they analyze samples of the frustrating unknown to identify the percent composition. From there, they calculate the ratios of different types of atoms in the compound. They express these ratios as an Empirical formula, The lowest whole-number ratio of elements in a compound.

To find an empirical formula given percent composition, use the following procedure:

1. Assume that you have 100g of the unknown compound.

The beauty of this little trick is that you conveniently gift yourself with the same number of grams of each elemental component as its contribution to the percent composition. For example, if you assume that you have 100g of a compound composed of 60.3 percent magnesium and 39.7 percent oxygen, you know that you have 60.3g of magnesium and 39.7g of oxygen.

2. Convert the assumed masses from Step 1 into moles by using gram atomic masses.

3. Divide each of the element-by-element mole quantities from Step 2 by the lowest among them.

This division yields the mole ratios of the elements of the compound.

4. If any of your mole ratios aren’t whole numbers, multiply all numbers by the smallest possible factor that produces whole-number mole ratios for all of the elements.

For example, if there is one nitrogen atom for every 0.5 oxygen atoms in a compound, the empirical formula is not N1O0.5. Such a formula casually suggests that an oxygen atom has been split, something that would create a small-scale nuclear explosion. Though impressive-sounding, this scenario is almost certainly false. Far more likely is that the atoms of nitrogen to oxygen are combining in a 1:0.5 Ratio, But do so in groups of 2 x (1:0.5) = 2:1. The empirical formula is thus N2O.

Because the original percent composition data is typically experimental, expect to see a bit of error in the numbers. For example, 2.03 is probably within experimental error of 2.

5. Write the empirical formula by attaching these whole-number mole ratios as subscripts to the chemical symbol of each element. Order the elements according to the general rules for naming ionic and molecular compounds (described in Chapter 5).

Differentiating between empirical formulas and molecular formulas

Many compounds in nature, particularly compounds made of carbon, hydrogen, and oxygen, are composed of atoms that occur in numbers that are multiples of their empirical formula. In other words, their empirical formulas don’t reflect the actual numbers of atoms within them, but only the ratios of those atoms. What a nuisance! Fortunately, this is an old nuisance, so chemists have devised a means to deal with it. To account for these annoying types of compounds, chemists are careful to differentiate between an empirical formula and a Molecular formula. A molecular formula uses subscripts that report the actual number of atoms of each type in a molecule of the compound (a Formula unit Accomplishes the same thing for ionic compounds).

Molecular formulas are associated with molar masses that are simple whole-number multiples of the corresponding Empirical formula mass. In other words, a molecule with the empirical formula CH2O has an empirical formula mass of 30.0g mol-1 (12 for the carbon + 2 for the two hydrogens + 16 for the oxygen). The molecule may have a molecular formula of CH2O, C2H4O2, C3H6O3, and so on. As a result, the compound may have a molar mass of 30.0g mol-1, 60.0g mol-1, 90.0g mol-1, and so on.

You can’t calculate a molecular formula based on percent composition alone. If you attempt to do so, Avogadro and Perrin will rise from their graves, find you, and slap you 6.022 x 1023 times per cheek (Ooh, that smarts!). The folly of such an approach is made clear by comparing formaldehyde with glucose. The two compounds have the same empirical formula, CH2O, but different molecular formulas, CH2O and C6H12O6, respectively. Glucose is a simple sugar, the one made by photosynthesis and the one broken down during cellular respiration. You can dissolve it into your coffee with pleasant results. Formaldehyde is a carcinogenic component of smog. Solutions of formaldehyde have historically been used to embalm dead bodies. You are not advised to dissolve formaldehyde into your coffee. In other words, molecular formulas differ from empirical formulas, and the difference is important.

To determine a molecular formula, you must know the molar mass of the compound as well as the empirical formula (or enough information to calculate it yourself from the percent composition; see the previous section for details). With these tools in hand, calculating the molecular formula involves a three-step process:

1. Calculate the empirical formula mass.

2. Divide the molar mass by the empirical formula mass.

3. Multiply each of the subscripts within the empirical formula by the number calculated in Step 2.

Keeping the See-Saw Straight: Balancing Reaction Equations

Equations that simply show reactants on one side of the reaction arrow and products on the other are Skeleton equations, And are perfectly adequate for a qualitative description of the reaction: who are the reactants and who are the products. But if you look closely, you’ll see that those equations just don’t add up quantitatively. As written, the mass of one mole of each of the reactants doesn’t equal the mass of one mole of each of the products. The skeleton equations break the Law of Conservation of Mass, Which states that all the mass present at the beginning of a reaction must be present at the end. To be quantitatively accurate, these equations must be Balanced So the masses of reactants and products are equal.

In chemistry this is easy — the requirement is met as mentioned earlier by assuring that the numbers of the same types of atoms are equal on each side. All the equations we have so far written have met this balance requirement. (The crime of losing or splitting atoms is punishable by an Avogadro number of lashes!).

How to use coefficients

To balance an equation, you use Coefficients To alter the number of moles of reactants and/or products so the mass on one side of the equation equals the mass on the other side. A Coefficient Is simply a number that precedes the symbol of an element or compound, multiplying the number of moles of that Entire Compound within the equation. Coefficients are different from Subscripts, Which multiply the number of atoms or groups within a compound. Consider the following:

4Cu(NO3)2

The number 4 that precedes the compound is a coefficient, indicating that there are four moles of copper (II) nitrate. The subscripted 3 and 2 within the compound indicate that each nitrate contains three oxygen atoms, and that there are two nitrate groups per atom of copper. Coefficients and subscripts multiply to yield the total number of moles of each atom:

4mol Cu(NO3)2 x (1mol Cu/mol Cu(NO3)2) = 4mol Cu

4mol Cu(NO3)2 x (2mol NO3/mol Cu(NO3)2) x 1mol N/mol NO3 = 8mol N

4mol Cu(NO3)2 x (2mol NO3/mol Cu(NO3)2) x 3mol O/mol NO3 = 24mol O

When you balance an equation, You change only the coefficients. Changing subscripts alters the chemical compounds themselves. If your pencil were equipped with an electrical shocking device, that device would activate the moment you attempted to change a subscript while balancing an equation.

Here’s a simple recipe for balancing equations:

1. Given a skeleton equation (one that includes formulas for reactants and products), count up the number of each kind of atom on each side of the equation.

If you recognize any polyatomic ions, you can count these as one whole group (as if they were their own form of element). See Chapter 5 for information on recognizing polyatomic ions.

2. Use coefficients to balance the elements or polyatomic ions, one at a time.

For simplicity, start with those elements or ions that appear only once on each side.

3. Check the equation to ensure that each element or ion is balanced.

Checking is important because you may have "ping-ponged" several times from reactants to products and back — there’s plenty of opportunity for error.

4. When you’re sure the reaction is balanced, check to make sure it’s in lowest terms.

For example,

4H2(g) + 2O2(g) —4H2O(l) should be reduced to

2H2(g) + O2(g) — 2H2O(l)

You can’t begin to wrap your brain around the unimaginably large number of possible chemical reactions. It’s good that so many reactions can occur, because they make things like life and the universe possible. From the perspective of a mere human brain trying to grok all these reactions, we have another bit of good news: A few categories of reactions pop up over and over again. After you see the very basic patterns in these categories, you’ll be able to make sense of the majority of reactions out there.

Conversions

Mass and energy are conserved. It’s the law. Unfortunately, this means that there’s no such thing as a free lunch, or any other type of free meal. Ever. On the other hand, the conservation of mass makes it possible to predict how chemical reactions will turn out.

Balancing equations can seem like a chore, like taking out the trash. But a balanced equation is far better than any collection of coffee grounds and orange peels because such an equation is a useful tool. After you’ve got a balanced equation, you can use the coefficients to build Mole-mole conversion factors. These kinds of conversion factors tell you how much of any given product you get by reacting any given amount of reactant. This is one of those calculations that makes chemists particularly useful, so they needn’t get by on looks and charm alone.

Consider the following balanced equation for generating ammonia from nitrogen and hydrogen gases:

N2(g) + 3H2(g) — 2NH3(g)

For every mole of nitrogen molecules reactanting, a chemist expects two moles of ammonia product. Similarly, for every three moles of hydrogen molecules reacting, the chemist expects two moles of ammonia product. These expectations are based on the coefficients of the balanced equation and are expressed as mole-mole conversion factors as shown in Figure 13-1.

2mol NH,

Figure 13-1:

Building mole-mole conversion factors from a balanced equation.

N2 + 3 H2 —*- 2 NH,

2 2 3

3mol H

The mole is the beating heart of stoichiometry, the central unit through which other quantities flow. Real-life chemists don’t have magic mole vision, however. A chemist can’t look at a pile of potassium chloride crystals, squint her eyes, and proclaim: "That’s 0.539 moles of salt." Well, she could proclaim such a thing, but she wouldn’t bet her pocket protector on it. Real reagents (reactants) tend to be measured in units of mass or volume, and occasionally even in actual numbers of particles. Real products are measured in the same way. So, you need to be able to use Mole-mass, mole-volume, And Mole-particle conversion factors To translate between these different dialects of counting. Figure 13-2 summarizes the interrelationship between all these things and serves as a flowchart for problem solving. All roads lead to and from the mole.

Figure 13-2:

A problem-solving flowchart showing the use of mole-mole, mole-mass, mole-volume, and mole-particle conversion factors.

Particles

Mole Particle

Mole

Mass

Volume

Reactants -

Mole Mole

Mole

Volume

Particle Mole

P roducts

Mass Mole

Volume Mole

Particles

Mass

Volume

Mass

*JJIBE*

Getting rid of mere spectators: Net ionic equations

Chemistry is often conducted in aqueous solutions. Soluble ionic compounds dissolve into their component ions, and these ions can react to form new products. In these kinds of reactions, sometimes only the cation or anion of a dissolved compound reacts. The other ion merely watches the whole affair, twiddling its charged thumbs in electrostatic boredom. These uninvolved ions are called Spectator ions.

Because spectator ions don’t actually participate in the chemistry of a reaction, you don’t need to include them in a chemical equation. Doing so leads to a needlessly complicated reaction equation. So, chemists prefer to write Net ionic equations, Which omit the spectator ions. A net ionic equation doesn’t include every component that might be present in a given beaker. Rather, it includes only those components that actually react.

Here is a simple recipe for making net ionic equations of your own:

1. Examine the starting equation to determine which ionic compounds are dissolved, as indicated by the (aq) symbol following the compound name.

Zn(s) + HCl(aq) — ZnCl2(aq) + H2(g)

2. Rewrite the equation, explicitly separating dissolved ionic compounds into their component ions.

This step requires you to recognize common polyatomic ions, so be sure to familiarize yourself with them.

Zn(s) + H+(aq) + GT(aq) — Zn2+(aq) + 3GT(aq) + H2(g)

3. Compare the reactant and product sides of the rewritten reaction. Any dissolved ions that appear in the same form on both sides are spectator ions. Cross out the spectator ions to produce a net reaction.

Zn(s) + H+(aq) + Cl – (aq) — Zn2+(aq) + 2Cl – (aq) + H2(g)

Net reaction:

Zn(s) + H+(aq) — Zn2+(aq) + H2(g)

As written, the preceding reaction is imbalanced with respect to the number of hydrogen atoms and the amount of positive charge.

4. Balance the net reaction for mass and charge.

Zn(s) + 2H+(aq) — Zn2+(aq) + H2(g)

If you wish, you can balance the equation for atoms and charge first (at Step 1). This way, when you cross out spectator ions at Step 3, you’ll be crossing out equivalent numbers of ions. Either method produces the same net ionic equation in the end. Some people prefer to balance the starting reaction equation, but others prefer to balance the net reaction because it is a simpler equation.

Running Out Early: Limiting Reagents

In real-life chemistry, not all of the reactants present convert into product. That would be perfect and convenient. Does that sound like real life to you? More typically, one reagent is completely used up, and others are left behind, perhaps to react another day.

The situation resembles that of a horde of Hollywood hopefuls lined up for a limited number of slots as extras in a film. Only so many eager faces react with an available slot to produce a happily (albeit pitifully) employed actor. The remaining actors are in excess, muttering quietly all the way back to their jobs as waiters. In this scenario, the slots are the limiting reagent.

Those standing in line demand to know, how many slots are there? With this key piece of data, they can deduce how many of their huddled mass will end up with a gig. Or, they can figure out how many will continue to waste their film school degrees serving penne with basil and goat cheese to chemists on vacation.

Chemists demand to know, which reactant will run out first? In other words, which reactant is the Limiting reagent? Knowing that information allows them to calculate how much product they can expect, based on how much of the limiting reagent they’ve put into the reaction. Also, identifying the limiting reagent allows them to calculate how much of the excess reagent they’ll have left over when all the smoke clears. Either way, the first step is to figure out which is the limiting reagent.

In any chemical reaction, you can simply pick one reagent as a candidate for the limiting reagent, calculate how many moles of that reagent you have, and then calculate how many grams of the other reagent you’d need to react both to completion. You’ll discover one of two things. Either you have an excess of the first reagent or you have an excess of the second reagent. The one you have in excess the the excess reagent. The one that is not in excess is the limiting reagent.

Having Something to Show for Yourself: Percent Yield

In a way, reactants have it easy. Maybe they’ll make something of themselves and actually react. Or maybe they’ll just lean against the inside of the beaker, flip through a back issue of People Magazine, and sip a caramel macchiato.

Chemists don’t have it so easy. Someone is paying them to do reactions. That someone doesn’t have time or money for excuses about loitering reactants. So you, as a fresh-faced chemist, have to be concerned with just how completely your reactants react to form products. To compare the amount of product obtained from a reaction with the amount that should have been obtained, chemists use Percent yield. You determine percent yield with the following formula:

Percent yield = 100% x (actual yield) / (theoretical yield)

Lovely, but what is an actual yield and what is a theoretical yield? An Actual yield Is. . . well. . . the amount of product actually produced by the reaction. A Theoretical yield Is the amount of product that could have been produced had everything gone perfectly, as described by theory — in other words, as predicted by your painstaking calculations.

Things never go perfectly. Reagents stick to the sides of flasks. Impurities sabotage reactions. Chemists attempt to dance. None of these ghastly things are Supposed To occur, but they do. So, actual yields fall short of theoretical yields.