# What is the empirical formula of a substance with 74.0% C , 8.65% H, and 17.4% N?

Jun 25, 2016

${C}_{5} {H}_{7} N$

#### Explanation:

First, assume that there is 100 g of substance.

Next is to calculate the number of moles or each element:

$74.0 g C \cdot \frac{1 m o l C}{12.01 g} \approx 6.16 m o l C$
$8.65 g H \cdot \frac{1 m o l H}{1.008 g} \approx 8.58 m o l H$
$17.4 g N \cdot \frac{1 m o l N}{14.01 g} \approx 1.24 m o l N$

After getting the mol of each element, you can calculate now the mole ratio. Do this by dividing each number of moles by the lowest number of moles obtained.

$C = \left(\frac{6.16}{1.24}\right) \approx 5$
$H = \left(\frac{8.58}{1.24}\right) \approx 7$
$N = \left(\frac{1.24}{1.24}\right) \approx 1$

Therefore, the empirical formula of a substance with 74.0% C , 8.65% H, and 17.4% N is ${C}_{5} {H}_{7} N$