# Suppose I am trying to determine a compound that has 15.87% carbon (C), 2.15 % hydrogen (H), 18.5% nitrogen (N), & 63.41% oxygen (O) (n.b., this does not equal exactly 100%): what is the empirical formula of the compound?

Jan 17, 2015

For my piece of mind, I'll add something to each percentage to make the total 100% - 0.02% to the first three, and 0.01% to oxygen.

So, you know that your compound has $\text{15.89% C}$, $\text{2.17% H}$, $\text{18.52% N}$, and $\text{63.42% O}$. The first step in determining it empirical formula is to divide each element's percentage by its molar mass

"For C": (15.89%)/(12.0) = 1.34

"For H": (2.17%)/(1.00) = 2.17

"For N": (18.52%)/(14.0) = 1.32

"For O": (63.42%)/(16.0) = 3.96

The next step is to divide all these numbers by the smallest one; this is done in order to get the ratios of the four elements in the molecule

$\text{For C} : \frac{1.34}{1.32} \cong 1.00$

$\text{For H} : \frac{2.17}{1.32} = 1.64$

$\text{For N} : \frac{1.32}{1.32} = 1.00$

$\text{For O} : \frac{3.96}{1.32} = 3$

The empirical formula must only contain subscripts that are integers, which means that your actual empirical formula is

${C}_{3} {H}_{5} {N}_{3} {O}_{9}$