## Chapter 3

## Q. 3.15

Nitrous oxide, the laughing gas used in dentists’ offices, is 63.65% N. What is the empirical formula of nitrous oxide?

## Step-by-Step

## Verified Solution

**Collect and Organize** We are given the percent composition by mass of N (63.65%) and asked to determine the simplest whole-number ratio of nitrogen and oxygen in the chemical formula.

**Analyze** We can obtain the percent composition by mass of oxygen in nitrous oxide by subtracting the percent N from 100%. We can use the four steps described earlier to determine the empirical formula.

**Solve** The percent O in the sample is:

\%O = 100.00\%-\%N = 100.00\%- 63.65\% = 36.35\%

Now we have the data we need to determine the empirical formula of nitrous oxide.

1. Convert percent composition values into masses by assuming a sample size of exactly 100 g:

63.65\% N = 63.65 g N in 100 g

36.35\% O = 36.35 g O in 100 g

2. Convert masses into moles:

63.65 \sout{g N} \times \frac{1 mol N}{14.01 \sout{g N}}=4.543 mol N

36.35 \sout{g O} \times \frac{1 mol O}{16.00 \sout{g O}}=2.272 mol O

3. Simplify the mole ratio 4.543:2.272 by dividing by the smaller value:

\frac{4.543 mol N}{2.272}=2.000 mol N \frac{2.272 mol O}{2.272}=1.000 mol O

The mole ratio N:O is 2:1.

4. Since the mole ratio of N to O is already a ratio of whole numbers, we have the empirical formula of nitrous oxide: N_{2}O.

**Think About It** The molar masses of N and O are similar, so a ≈2:1 ratio of the mass of N to the mass of O is consistent with an empirical formula that is 2:1 nitrogen to oxygen. Note that “nitrous oxide”—whose molecular formula is also N_{2}O—is formally named “dinitrogen monoxide” under the rules we learned in Chapter 2.