# Combustion data of an organic compound yielded 74.0%, C; 7.4%, H; 8.6%, N; and the balance oxygen. What is the empirical formula?

Sep 29, 2016

${C}_{10} {H}_{12} N O$

#### Explanation:

As with all problems of this type, we assume a $100 \cdot g$ mass of compound, and we determine the molar composition with respect to each element:

And thus:

$\text{Moles of carbon}$ $=$ $\frac{74.0 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1}$ $=$ $6.16 \cdot m o l$

$\text{Moles of hydrogen}$ $=$ $\frac{7.4 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1}$ $=$ $7.34 \cdot m o l$

$\text{Moles of nitrogen}$ $=$ $\frac{8.6 \cdot g}{14.01 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.614 \cdot m o l$

$\text{Moles of oxygen}$ $=$ $\frac{10.0 \cdot g}{16.00 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.625 \cdot m o l$

So we have the molar composition. We simply divide thru by the smallest molar quantity to gets our empirical formula:

Dividing thru by $0.614 \cdot m o l$ we get ${C}_{10} {H}_{12} N O$ (clearly, I have done some rounding off!), as the empirical formula.