Question

what is the empirical formula of a compound containing 49.97% carbon, 10.51% hydrogen, and 39.52% fluorine by mass?
a. c₂h₅f
b. c₄h₁₀f₂
c. c₄h₁₀f₄
d. c₂h₇f₂
e. c₃h₂f

Explanation:

Step1: Assume 100g of the compound

If we assume 100g of the compound, then the mass of carbon ($C$) is $m(C)=49.97g$, the mass of hydrogen ($H$) is $m(H) = 10.51g$, and the mass of fluorine ($F$) is $m(F)=39.52g$.

Step2: Calculate the number of moles of each element

The molar mass of $C$ is $M(C)=12.01g/mol$, the molar mass of $H$ is $M(H) = 1.008g/mol$, and the molar mass of $F$ is $M(F)=18.998g/mol$.
The number of moles of $C$: $n(C)=\frac{m(C)}{M(C)}=\frac{49.97g}{12.01g/mol}\approx4.16mol$
The number of moles of $H$: $n(H)=\frac{m(H)}{M(H)}=\frac{10.51g}{1.008g/mol}\approx10.43mol$
The number of moles of $F$: $n(F)=\frac{m(F)}{M(F)}=\frac{39.52g}{18.998g/mol}\approx2.08mol$

Step3: Find the mole - ratio

Divide each number of moles by the smallest number of moles (in this case, $n(F)\approx2.08mol$)
For $C$: $\frac{n(C)}{n(F)}=\frac{4.16mol}{2.08mol} = 2$
For $H$: $\frac{n(H)}{n(F)}=\frac{10.43mol}{2.08mol}\approx5$
For $F$: $\frac{n(F)}{n(F)} = 1$

Answer:

A. $C_{2}H_{5}F$