Question 3.14: The compound C2H2F4 is a propellant in the inhalers used by ...
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}?
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 .