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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. The number listed as the atomic mass of an element equals that element’s Molar mass If the element is monatomic.

Of course, chemistry involves the making and breaking of bonds (as you find out in Chapter 5), 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.

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.

4.

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 the 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, 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.

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 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 1 atm. 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.

Q.

A.

What if you want to convert between the volume of a gaseous substance and its mass, or the mass of a substance and the number of particles it contains? You already have all the information you need! To make these kinds of conversions, simply build a chain of conversion factors, converting units step by step from the ones you have (say, liters) to the ones you want (say, grams). You’ll find that your chain of conversion factors always includes central links featuring units of moles. You can think of the mole as a family member who passes on what you’ve said, loudly barking into the ear of your nearly deaf grandmother, because you have laryngitis and can’t speak any louder. Without such a central translator, your message would no doubt be misinterpreted. "Grandma, how was your day?" would be received as "Grandma, you want to eat clay?" So, unless you’re bent on force-feeding clay to your grandmother, do not attempt to convert directly from volume to mass, from mass to particles, or any other such shortcut. Use your translator, the mole.

On her last day of work before retirement, Dr. Dentura daydreams about her newly purchased condo in Florida. Distracted, Dr. Dentura forgets to turn off the dinitrogen monoxide (laughing gas) after administering it to a root canal patient. The gas flows, filling 10 percent of the room, until her dental hygienist notices the hissing and nonchalantly turns off the gas. Being denser than air, the laughing gas settles near the floor so the dentist doesn’t notice. How many moles of dinitrogen monoxide escaped into the room? Assume that Dr. Dentura suffers from frequent hot flashes and therefore keeps her office at a chilly 0°C. Further assume that her exam room is a spacious 10-x-10-x-10-foot cube.

126 mol. The problem tells you that the exam room is at standard temperature, and because the good doctor is presumably operating somewhere on the surface of the earth, it’s safe to assume that the local pressure is somewhere around 1 atm. Having assured yourself that the room is at STP, you can safely use the 22.4L mol-1 conversion factor. To use it effectively, you must convert the volume of the room into liters. You know that the room has a volume of 10 ft X 10 ft X 10 ft = 1,000 ft3. You also know that the dinitrogen oxide gas fills 10 percent of the room, and so accounts for 100 ft3. Convert from cubic feet to liters, referring if you like to Chapter 2 for the conversion factor:

100 ft3 1

;12in) (1ft )3

(2.54 cm)

X 1mL x

1L

1in)

1cm3 1000 mL

2.83 X103L

Then convert this volume in liters to moles by using the STP conversion factor of 22.4L mol-1:

2.83 x103L 1 mol

1

22.4L

126 mol

X

Q. Supervillain Lex Luthor accidentally purchases a vessel containing 0.05 kg of krypton, mistaking it for kryptonite, from an online retailer. Kryptonite, of course, is a glowing, green, and entirely fictional solid capable of utterly destroying Luthor’s arch nemesis, Superman. Krypton, by contrast, is a relatively innocuous noble gas. What’s the volume of Lex Luthor’s unhelpful vessel, assuming that it was meticulously filled to only atmospheric pressure and is shipped on ice?

A. 13.4L. You’re given a mass and are asked to convert it to a volume. Ice keeps the vessel at a temperature near 0°C, and you’re assured that the internal pressure of the vessel is 1 atm. So, you can assume STP. First, convert from grams to moles by using the gram atomic mass (in this case, 83.8g). Then convert from moles to volume, as shown in the following equation. The answer is 13.4 liters. That’s nearly 4 gallons of useless noble gas, an embarrassing error for a supervillain.

0.05 kg x 1000g x 1mol x 22^L = 13 4L 1 1kg 83.8g 1mol

4. How many moles make up 350g of table salt, NaCl?

Solve It

5. The average volume of a human breath is roughly 500 mL. How many moles do you inhale with each breath on a day when the temperature hovers near freezing? Assuming that human exhalations eventually intermix evenly throughout the approximately 3 x 1022L of Earth’s atmosphere, calculate how many molecules from Genghis Khan’s last breath you take into your lungs each time you inhale.

Solve It

6. In pounds, what is the mass of 3 moles of platinum?

Solve It

7. If you fill a 2L soda bottle (at STP) with

Carbon dioxide, how many particles does the bottle contain?

Solve It

Giving Credit Where It’s Due: Percent Composition

Chemists often are 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. Calculating percent composition is trickier than you may think. Consider the following problem, for example.

The human body is composed of 60 to 70 percent water, and water contains twice as many hydrogen atoms as oxygen atoms. If two-thirds of every water molecule is hydrogen and water makes up 60 percent of the body, it seems logical to conclude that hydrogen makes up 40 percent of the body. Yet hydrogen is only the third most abundant element in the body By mass. What gives?

Oxygen is 16 times more massive than hydrogen. So, equating Atoms Of hydrogen and Atoms Of oxygen is a bit like equating a toddler to a sumo wrestler. When the doors of the elevator won’t close, the sumo wrestler is the first one you should kick out, weep though he may.

Within a compound, it’s important to sort out the atomic toddlers from the atomic sumo wrestlers. To do so, follow three simple steps.

1. Calculate the gram molecular mass or gram formula mass of the compound, as we explain in the previous section.

Percent compositions are completely irrelevant to gram atomic masses, because these apply only to pure monatomic substances; by definition, these substance 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.

Calculate the percent composition for each element present in sodium sulfate, Na2SO4. Na: 32.4%, S: 22.5%, O: 45.1%. The gram formula mass of sodium sulfate is

(2 X 23.0g mol-1) + (1 X 32.0g mol-1) + (4 X 16.0g mol-1) = 142g mol-1.

Of each mole of compound,

2 X 23.0g = 46.0g are sodium

1 X 32.0g = 32.0g are sulfur

4 X 16.0g = 64.0g are oxygen

Dividing each of these quantities by the molar mass of sodium sulfate (142g) yields the percent composition:

46g 32g 64g

, ° x 100 = 32.4% sodium, ° X 100 = 22.5% sulfur, and ° X 100 = 45.1% oxygen 142g 142g 142g

As a check, add the three percentages to ensure they equal 100%: 32.4% + 22.5% + 45.1% = 100%.

Moving from Percent Composition to 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% magnesium and 39.7% 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 the elements.

For example, if there is 1 nitrogen atom for every 0.5 oxygen atom 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 6).

Q. What is the empirical formula of a substance that is 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen, by mass?

A. CH2O. For the sake of simplicity, assume that you have a total of 100g of this mystery compound. Therefore, you have 40.0g of carbon, 6.7g of hydrogen, and 53.3g of oxygen. Convert each of these masses to moles by using the gram atomic masses of C, H, and O.

40.0g C 1molC

-r2— x —77-—— = 3.33 mol C

1 12g C

6.7g H x 1mol H 1 1g H

6.7 mol H

53.3g O x 1mol O 1 16g O

3.33 mol O

Notice that the carbon and oxygen mole numbers are the same, so you know the ratio of these two elements is 1:1 within the compound. Next, divide all the mole numbers by the smallest among them, which is 3.33. This division yields

3.33mol C

3.33

1mol C,

6.70mol H 3.33

2molH, and

3.33mol O

- :

3.33

1molO

The compound has the empirical formula CH2O. The actual number of atoms within each particle of the compound is some multiple of the numbers expressed in this formula.

10. Calculate the empirical formula of a compound with a percent composition of 88.9% oxygen and 11.1% hydrogen.

11. Calculate the empirical formula of a compound with a percent composition of 40.0% sulfur and 60.0% oxygen.

Solve It

Solve It

Moving from Empirical Formulas to Molecular Formulas

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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 gram molecular 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 gram molecular 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 gram formula 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 gram molecular mass by the empirical formula mass.

3. Multiply each of the subscripts within the empirical formula by the number calculated in Step 2.

Q. What is the molecular formula of a compound which has a gram molecular mass of 34g mol-1 and the empirical formula HO.

A. H2O2. The empirical formula mass is

1g mol-1 + 16g mol-1 = 17g mol-1.

Dividing the gram molecular mass by this value yields

34g mol-1/17g mol-1 = 2.

Multiplying the subscripts within the empirical formula by this number gives the molecular formula H2O2. This formula corresponds to the compound hydrogen peroxide.

12. What is the molecular formula of a compound which has a gram formula mass of 78g mol-1 and the empirical formula NaO.

Solve It

13. A compound has a percent composition of 49.5% carbon, 5.2% hydrogen, 16.5% oxygen, and 28.8% nitrogen. The compound’s gram molecular mass is 194.2g mol-1. What are the empirical and molecular formulas?

Solve It

Answers to Questions on Moles

The following are the answers to the practice problems presented in this chapter.

D 0.03 mol. Here’s how you calculate the answer (remember Chapter 1′s rules for multiplying and dividing in scientific notation):

1.5 X 1011 stars X 1.25 X 1011galaxies X 1mol stars

-1-i-X-—-X-23-

1 galaxy universe 6.022 X 10 stars

0.03 mol

Reflect on this answer. The entire universe contains only 3 percent of a mole of stars! Efl 2 X 10-6 mol. Here’s the calculation:

7 X 109 People X 2 X 108 insects X

1 mol

1

1person

6.022 X 1023 insects

0.000002 mol

Again, a number that nonchemists find unimaginably large comes out to the tiniest fraction of 1 mole.

CM 1.72 x 1015 atoms. First, find the total mass of carbon in the body by taking 23% of 150 kg, which equals 34.5 kg. Then convert those 34.5 kg to atoms by using the conversion factors given in the problem and the conversion between moles and atoms (which you already know):

34.5 kg X 83 mol X 6.022 X 1023 atoms C X 1 atom 14C

-1-x —Z—j-x-"- – x-12-■

1 1kg 1 mol C 1 X 1012 atoms C

= 1.72 X 1015 atoms

MM 6.0 mol. First, find the gram formula mass of sodium chloride using the same steps you used to calculate gram molecular mass. It’s 23g mol-1 + 35.5g mol-1 = 58.5g mol-1. Given the gram formula mass, the conversion is simple:

350g NaCl 1mol NaCl — 6 0 l I X r n r — 6.0 mol 1 58.5g NaCl

CM 0.02 mol; 0.2 particles. First, calculate how many moles of air you inhale with each breath:

0.5L X 1 mol — 0 02 mol 1 22.4L

Next, calculate the number of particles in each mole of air that were once part of Genghis Khan’s final breath. Do this by finding the fraction of the atmosphere comprised by a single breath. Then, multiply that number by the number of particles in 1 mole:

0 5L 6.022 X 1023particles 1A.. . . -x–——-— 10 particles per mole

3 1022L 1mol

That’s right — 10 particles in every mole of air on the planet were inhaled by the mighty Mongolian in his last scowling moment. Now, figure out how many of these particles you take in

When you inhale. Multiply the total number of inhaled moles by the number of Genghis-kissed particles per mole:

0.02 mol X 10 particles — 02 particles 1 1mol

This means that, on average, one out of every five times you inhale, you breathe in a piece of East Asian history!

Ff 1.3 lb. Use your conversion factor techniques from Chapter 1 to convert moles to grams to pounds as shown in the following equation:

L^LEt X 1^ X If — 1.3 lb 1 1mol Pt 453g

Gg 5.37 X 1022 particles. Use conversion factors to convert from liters to moles, and then moles to particles as shown in the following equation:

2L 1 mol 6.022 1023particles 22

X Ar. V, x-:—-— 5.37 X 10particles

1 22.4L 1mol

Hh K: 40.3%; Cr: 26.8%; O: 33.0%. First, calculate the gram molecular mass of potassium chro-mate, which comes to 194.2g mol-1:

(2 X 39.1g mol-1) + (1 X 52.0g mol-1) + (4 X 16.0g mol-1) = 194.2g mol-1

In a 100g sample, 78.2g are potassium, 52.0g are chromium, and 64.0g are oxygen. Divide each of these masses by the gram molecular mass, and multiply by 100 to get the percent composition. Note: If you’ve rounded your percentages properly, they don’t quite add to 100%, but rather 100.1%. If you had done away with rounding, you would have gotten exactly 100%. Rounding is common practice though, so don’t be too worried if your answer is off by a tenth or two.

78.2g x — 40 3% potassium 52.0g

194.2g

194.2g

X 100 — 26.8% chromium, and

64.0g 194.2g

100 —33.0% oxygen

EH C: 81.8%; H: 18.2%. The molecular mass of propane is (3 12.0g mol-1) + (8 1.0g mol-1) = 44g mol-1 The percent composition of propane is therefore:

36.0g 44.0g

X 100:

: 81.8% carbon and -80g – X 100 = 44.0g

18.2% hydrogen

Jj H2O. First, assume that you have 88.9g of oxygen and 11.1g of hydrogen in a 100g sample. Then convert each of these masses into moles by using the gram atomic masses of oxygen and hydrogen:

88.9g O 1mol O — 5 55 l I X, ~ ^ 7~" — 5.55 mol 1 16.0g O

11.1g H x 1mol H 1 1.0g H

11.1 mol

Next, divide each of these mole quantities by the smallest among them, 5.55 mol:

5.55molO, , „ ,11.1 molH „ …

1 mol O and–— 2mol H

5.55

5.55

Attach these quotients as subscripts and list the atoms properly. This yields H2O. The compound is water.

|f| SO3. Following the same procedure as in Question 10, you calculate 1.25 mol sulfur and

3.75 mol oxygen. Dividing each of these quantities by 1.25 mol (the smallest quantity) yields 1.25 / 1.25 = 1 mol of sulfur and 3.75 / 1.25 = 3 mol oxygen, or a mole ratio of 1:3. The compound is SO3, sulfur trioxide.

U Na2O2. First, find the empirical formula mass of NaO, which is

(1 x 23.0g mol-1) + (1 x 16.0g mol-1) = 39.0g mol-1

Then divide the gram formula mass of the mystery compound, 78g mol-1, by this empirical formula mass to obtain the quotient, 2. Multiply each of the subscripts within the empirical formula by this number to obtain Na2O2. You’ve just found the molecular formula for sodium peroxide.

C4H5N2O Is the empirical formula; C8H10N4O2 Is the molecular formula. You’re not directly given the empirical formula of this compound. But you Are Given the percent composition. Using the percent composition, you can calculate the empirical formula. To do this, assume that you have 100g of the substance, giving you 49.5g carbon, 5.2g hydrogen, 16.5g oxygen, and 28.8g nitrogen. Then divide these masses by the atomic mass of each element, giving you

49.5 / 12.0 = 4.125 mol carbon

5.2 / 1.0g = 5.2 mol hydrogen

16.5 / 16.0 = 1.031 mol oxygen

28.8 / 14.0 = 2.057 mol nitrogen

Finally, divide each of these mole values by the lowest among them, 1.031, giving you 4.0 mol carbon, 5.0 mol hydrogen, 1.0 mol oxygen, and 2.0 mol nitrogen, giving you the empirical formula C4H5N2O.

Here’s how you get the molecular formula: The empirical formula mass is 97.1g mol-1 (calculated by multiplying the number of atoms of each element in the compound by its atomic mass and adding them all up). Dividing the gram molecular mass you were given (194.2g mol-1) by this empirical formula mass yields the quotient, 2. Multiplying each of the subscripts in the empirical formula by 2 produces the molecular formula, C8H10N4O2. The common name for this culturally important compound is caffeine.