# Question #df0c6

Feb 11, 2015

If you're dealing with a hydrocarbon, which is a compound that only contains carbon and hydrogen, then it should actually have $\text{14.29%}$ hydrogen, rather than $\text{13.29%}$.

Because a hydrocarbon only contains carbon and hydrogen, the percentages must add to 100%. I'll show you how to solve this using $\text{14.29%}$ for hydrogen.

So, you know from the percentages given that a 100-g sample of this compound will contain 85.71 g carbon and 14.29 hydrogen. The first step in determining a compound's empirical formula is to divide each element's sample mass by its atomic mass. This wil ltell you how many moles of each element the sample contains

$\text{For C": "85,71 g"/"12.0 g/mol" = "7.14 moles carbon}$

$\text{For H": "14.29 g"/"1.01 g/mol" = "14.1 moles hydrogen}$

Now divide both these numbers by the smallest one to get the mole ratio of the two elements in the compound

$\text{For C": "7.14"/"7.14} = 1$

$\text{For H": "14.1"/"7.14} = 1.97 \cong 2$

Therefore, the empirical formula for your compound is ${C}_{1} {H}_{2}$, or $C {H}_{2}$.

SIDE NOTE If you use $\text{13.29%}$ for hydrogen, you'll get the ratio for hydrogen equal to $\frac{13.29}{7.14} = 1.87$, which could be considered equal to $2$.