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## Q. 3.15

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

## 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.