# Microanalysis of an organic compound containing carbon, hydrogen, and oxygen, gives 62.6%*C, and 4.21%*H. What is the empirical formula of the compound?

Sep 7, 2017

${C}_{5} {H}_{4} {O}_{2}$ is the $\text{empirical formula.....}$

#### Explanation:

We assume a $100 \cdot g$ mass of compound, and thus....

$\text{Moles of carbon} = \frac{62.6 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1} = 5.21 \cdot m o l \cdot C$

$\text{Moles of hydrogen} = \frac{4.21 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1} = 4.16 \cdot m o l \cdot H$

$\text{Moles of oxygen} = \frac{33.3 \cdot g}{15.999 \cdot g \cdot m o {l}^{-} 1} = 2.08 \cdot m o l \cdot H$

In each case, we have divided the mass of element by the ATOMIC mass of each element to give an elemental ratio.... How did I know that the compound was 33.3% by mass with respect to oxygen?

And now we divide thru by the LOWEST molar quantity, that of oxygen to get a trial empirical formula of ${C}_{2.50} {H}_{2} O$. But by definition, the empirical formula is the lowest WHOLE number ratio defining constituent atoms in a species...so we gots ${C}_{5} {H}_{4} {O}_{2}$.

We cannot access the molecular formula inasmuch as we need a measurement of molecular mass. See here.

$\text{Molecular formula"="empirical formula} \times n$....