# What is the empirical formula of a compound that is 75% carbon and 25% hydrogen?

Oct 27, 2015

$C {H}_{4}$

#### Explanation:

For this kind of problem, you can assume that the weight of the unknown sample is 100g ( since 75% $C$ + 25% $H$ = 100%).

Thus,

weight of $C$ = 75g
weight of $H$ = 25g

Since chemical formulas deal with number of moles rather than weight in grams, you need to convert each element by multiplying it to their respective atomic masses.

atomic mass of $C = 12.01 \frac{g}{\text{mol}}$

atomic mass of $H = 1.01 \frac{g}{\text{mol}}$

Thus,

mol $C$ = 75 $\cancel{\text{grams}}$ x "1 mol"/ (12.01 cancel "grams") = 6.245 mol $C$

mol $H$ = 25 $\cancel{\text{grams}}$ x "1 mol"/ (1.01 cancel "grams") = 24.752 mol $H$

Now we can see from the computation above that every

6.245 mol $C$ = 24.752 mol $H$

Dividing both sides by the smallest number of moles, (in this case 6.245 mol $C$)

$C$ = $\left(6.245 \cancel{\text{mol")/(6.245 cancel "mol}}\right)$ = 1
$H$ = $\left(24.752 \cancel{\text{mol")/(6.245 cancel "mol}}\right)$ = 3.98 $\approx$ 4

Therefore, empirical formula is ${C}_{1} {H}_{4}$ or simply, $\textcolor{red}{C {H}_{4}}$.