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?

(A)	65%
(B)	70%
(C)	75%
(D)	80%
(E)	85%

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

(A)	1.
(B)	2.
(C)	3.
(D)	4.
(E)	5.

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

(A)	CH4
(B)	C3H8
(C)	C5H12
(D)	C6H14
(E)	C7H16

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

Ular formula?
(A)	C2H4O2
(B)	C3H6O3
(C)	C4H8O4
(D)	C5H10O5
(E)	C6H12O6

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.