# Chemical analysis shows that citric acid contains 37.51% C, 4.20% H, and 58.29% O. What is the empirical formula for citric acid?

Jan 12, 2017

${C}_{6} {H}_{8} {O}_{7}$

#### Explanation:

As with all empirical formula problems of this kind, it is useful to assume that there $100 \cdot g$ of compound, and we work out the atomic ratios on this basis.

$\text{Moles of carbon}$ $=$ $\frac{37.51 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1} = 3.12 \cdot m o l$

$\text{Moles of hydrogen}$ $=$ $\frac{4.20 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1} = 4.17 \cdot m o l$

$\text{Moles of oxygen}$ $=$ $\frac{58.29 \cdot g}{15.999 \cdot g \cdot m o {l}^{-} 1} = 3.64 \cdot m o l$.

We divide thru by the SMALLEST molar quantity, and then normalize the result (i.e. convert it to whole numbers!):

$C {H}_{1.34} {O}_{1.17}$

To normalize, we mulitply this formula by a factor of $6$ to give
${C}_{6} {H}_{8} {O}_{7}$. And there we have it: the empirical formula, which is the simplest WHOLE number ratio defining constituent atoms in a species.

We could find out from mass spectroscopy that citric acid gives a molecular ion at $192 \cdot \text{amu}$. Can you tell me the MOLECULAR formula of $\text{citric acid}$? Remember the molecular formula is always a multiple of the empirical formula, and the mulitple MAY be $1$.