## Chapter 3

## Q. 3.14

The compound C_{2}H_{2}F_{4} is a propellant in the inhalers used by asthma sufferers (Figure 3.23). What is the percent composition of C_{2}H_{2}F_{4}?

## Step-by-Step

## Verified Solution

**Collect, Organize, and Analyze** We are asked to calculate the percent composition of C_{2}H_{2}F_{4}. We need the molar masses of each of the elements (C, H, and F) to calculate the molar mass of C_{2}H_{2}F_{4} and the relative contribution of each element to the molar mass. To calculate the percent composition of C_{2}H_{2}F_{4}, we must determine the percentage by mass of each element in one mole of C_{2}H_{2}F_{4}. These percentages can be calculated by dividing the mass of each element in 1 mole of C_{2}H_{2}F_{4} by the molar mass of C_{2}H_{2}F_{4}.

**Solve** The molar mass of C_{2}H_{2}F_{4} is

2(12.01 g/mol) + 2(1.008 g/mol) + 4(19.00 g/mol) = 102.04 g/mol

Thus the percent composition of this compound is

\% C=\frac{24.02 g C}{102.04 g} \times 100\%=23.54 \% C

\% H=\frac{2.016 g H}{102.04 g} \times 100\%=1.98\% H

\% F=\frac{76.00 g F}{102.04 g} \times 100\%=74.48\% F

**Think About It** Although C and F have similar molar masses, the observation that the percentage of F is nearly three times the percentage of C makes sense because there are two moles of F for every 1 mole of C in C_{2}H_{2}F_{4}. It also makes sense that the percentage of the lightest element, hydrogen, is very small .