Question 3.15: Nitrous oxide, the laughing gas used in dentists’ offices, i...

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.