Mole Conversions Practice

Two relationships do all the work on this page, and there is no third one hiding anywhere:

$n=\frac{m}{M}N=n\times 6.022\times 10^{23}$

with $n$ the amount in moles, $m$ the mass in grams, $M$ the molar mass in grams per mole, and $N$ the number of particles.

Molar masses to two decimal places from the periodic table. Carry full precision through the working and round once, at the end, to the number of significant figures the data allow. Every line of working carries units.

1. How many moles are there in 25.0 g of sodium chloride?

Answer 1

Molar mass first, because every one of these starts there.

$M(\mathrm{NaCl})=22.99+35.45=58.44\text{ g/mol}$

$n=\frac{m}{M}=\frac{25.0\text{ g}}{58.44\text{ g/mol}}=0.4278\text{ mol}$

0.428 mol, to three significant figures, because the mass was given to three and the molar mass to four. The weakest measurement sets the ceiling.

Check the units did what they should: grams divided by grams per mole leaves moles. If your units had come out as grams squared per mole, you divided the wrong way round, and the units told you before the answer did.

2. What is the mass of 2.50 mol of carbon dioxide?

Answer 2

$M(\mathrm{CO}{}_2)=12.01+2(16.00)=44.01\text{ g/mol}$

Rearranging $n=\frac{m}{M}$ gives $m=nM$:

$m=(2.50\text{ mol})(44.01\text{ g/mol})=110.025\text{ g}$

$1.10\times 10^2$ g, to three significant figures.

That form is worth using rather than “110 g”, which is ambiguous about whether the trailing zero counts. Scientific notation makes the claim explicit: three figures, no more.

3. How many molecules are there in 4.00 g of water? How many atoms?

Answer 3

$M(\mathrm{H}{}_2\mathrm{O})=2(1.01)+16.00=18.02\text{ g/mol}$

$\begin{array}{rl}n& =\frac{4.00\text{ g}}{18.02\text{ g/mol}}=0.22198\text{ mol}\\ N& =(0.22198)(6.022\times 10^{23})=1.3367\times 10^{23}\end{array}$

$1.34\times 10^{23}$ molecules, to three significant figures.

For the atoms, notice what the formula says: each molecule contains three atoms, two hydrogens and one oxygen. So

$3\times 1.3367\times 10^{23}=4.01\times 10^{23}$ atoms.

Keep those two numbers apart in your head. “How many particles” is never a complete question until somebody says particles of what, and a great many marks are lost in the gap between molecules and atoms.

4. A sample of calcium nitrate, $\mathrm{Ca}(\mathrm{NO}{}_3{)}_2$, has a mass of 50.0 g. (a) How many moles is that? (b) How many nitrate ions does it contain?

Answer 4

(a) Build the molar mass carefully — the subscript outside the bracket multiplies everything inside it.

$M=40.08+2[14.01+3(16.00)]=40.08+2(62.01)=164.10\text{ g/mol}$

$n=\frac{50.0\text{ g}}{164.10\text{ g/mol}}=0.30469\text{ mol}$

0.305 mol, to three significant figures.

(b) One formula unit contains two nitrate ions, so

$\begin{array}{rl}n(\mathrm{NO}{}_3{}^{-})& =2\times 0.30469=0.60938\text{ mol}\\ N& =(0.60938)(6.022\times 10^{23})=3.6697\times 10^{23}\end{array}$

$3.67\times 10^{23}$ nitrate ions.

Sanity check: the answer should be a bit more than half of Avogadro’s number times one, and it is. If you had got $1.8\times 10^{23}$ you would have forgotten the 2 — and multiplying by 2 at the end is easier to remember if you write $n(\mathrm{NO}{}_3{}^{-})=2n(\text{compound})$ as its own line rather than doing it in your head.

5. A copper sample contains $1.505\times 10^{22}$ atoms. What is its mass?

Answer 5

This one runs the chain backwards: particles to moles to mass.

$n=\frac{N}{6.022\times 10^{23}}=\frac{1.505\times 10^{22}}{6.022\times 10^{23}}=0.024992\text{ mol}$

$m=nM=(0.024992\text{ mol})(63.55\text{ g/mol})=1.5882\text{ g}$

1.588 g, to four significant figures, because the particle count was given to four.

Worth a moment: fifteen thousand million million million atoms weigh about as much as a paperclip. The mole exists precisely so that you never have to hold that sentence in your head while doing arithmetic.

6. Which contains more atoms — 10.0 g of carbon or 10.0 g of lead? By what factor?

Answer 6

Same mass, different atoms, so the comparison is entirely about molar mass.

Carbon: $n=\frac{10.0}{12.01}=0.83264\text{ mol}$, so $N=(0.83264)(6.022\times 10^{23})=5.01\times 10^{23}$ atoms.

Lead: $n=\frac{10.0}{207.2}=0.048263\text{ mol}$, so $N=(0.048263)(6.022\times 10^{23})=2.91\times 10^{22}$ atoms.

The carbon contains more, by a factor of about 17.

And here is the shortcut, which is worth more than the arithmetic. You never needed Avogadro’s number at all. For equal masses, the ratio of the counts is just the inverse ratio of the molar masses:

$\frac{N_{\mathrm{C}}}{N_{\mathrm{Pb}}}=\frac{M_{\mathrm{Pb}}}{M_{\mathrm{C}}}=\frac{207.2}{12.01}=17.3$

Divide the two rounded counts above instead and you get 17.2, not 17.3. Neither is a mistake — it is what rounding twice does to you, and it is the reason the rule is to round once, at the end.

A lead atom is about seventeen times heavier than a carbon atom, so ten grams of lead buys you about a seventeenth as many of them. If you can see that before reaching for the calculator, you understand what molar mass is.

7. A tablet contains 325 mg of acetylsalicylic acid, $\mathrm{C}{}_9\mathrm{H}{}_8\mathrm{O}{}_4$. (a) How many moles is that? (b) How many molecules?

Answer 7

Convert the mass to grams before anything else — the molar mass is in grams per mole and mixing milligrams into it is the commonest slip on a question that looks easy.

$325\text{ mg}=0.325\text{ g}$

$M=9(12.01)+8(1.01)+4(16.00)=108.09+8.08+64.00=180.17\text{ g/mol}$

(a) $n=\frac{0.325\text{ g}}{180.17\text{ g/mol}}=1.8039\times 10^{-3}\text{ mol}$

$1.80\times 10^{-3}$ mol, to three significant figures.

(b) $N=(1.8039\times 10^{-3})(6.022\times 10^{23})=1.0863\times 10^{21}$

$1.09\times 10^{21}$ molecules.

A dose small enough to swallow without noticing contains around a thousand million million million molecules. This is the arithmetic behind why dosage matters and why “a tiny amount” is not a chemical argument — see Chemicals We Live With.

8. Four statements from a study group. Each is wrong or incomplete. Fix each one. (a) “There are $6.022\times 10^{23}$ atoms in one mole of water.” (b) “0.5 mol of oxygen has a mass of 8 g.” (c) “To find moles you multiply the mass by the molar mass.” (d) “A mole of lead weighs more than a mole of carbon, so a mole of lead contains more atoms.”

Answer 8

(a) Molecules, not atoms. One mole of water contains $6.022\times 10^{23}$ molecules, and each molecule holds three atoms, so the atom count is $3\times 6.022\times 10^{23}=1.81\times 10^{24}$ atoms. The mole counts whatever entity you name, and the sentence is only complete once you have named it.

(b) It depends on what “oxygen” means, and the usual meaning makes this wrong. If the student meant oxygen atoms, then $M=16.00$ g/mol and $0.5\times 16.00=8.0$ g, which is what they wrote. But “oxygen” in a chemical context ordinarily means the substance $\mathrm{O}{}_2$, for which $M=32.00$ g/mol, giving $0.5\times 32.00=16.0$ g. The safe habit is to write the formula rather than the element’s name whenever there is any doubt — and there is doubt for every diatomic element.

(c) Divide, do not multiply. $n=\frac{m}{M}$. The units settle it without any memory being involved: grams divided by grams per mole gives moles, whereas grams multiplied by grams per mole gives grams squared per mole, which is not a quantity that exists. Check the magnitude too — 25.0 g of sodium chloride is a spoonful, so a fraction of a mole is plausible and 1461 mol is not.

(d) The first half is right and the conclusion does not follow. A mole of lead does have a greater mass — 207.2 g against 12.01 g. But a mole is a count, and both samples contain exactly $6.022\times 10^{23}$ atoms, by definition. That is the whole point of the unit: it fixes the number and lets the mass vary. If “mole” could be replaced by “gram” in a sentence you have written and the sentence still seemed to make sense, something has gone wrong.

Reference: The Mole and Molar Mass and Composition. How many figures you are entitled to: Significant Figures and Units.

Curriculum connection

D2.1

use appropriate terminology related to quantities in chemical reactions, including, but not limited to: stoichiometry, percentage yield, limiting reagent, mole, and atomic mass [C]

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D2.3

solve problems related to quantities in chemical reactions by performing calculations involving quantities in moles, number of particles, and atomic mass [AI]

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