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## 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 $C_{2}H_{2}F_{4}$? ## 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 .