# Lactic acid - Elementary analysis

Lactic acid is a hydroxy carboxylic acid that can form in the muscles and lead to soreness. The molecular weight of the lactic acid is $\ce{90.1 g/mol}$. Elemental analysis shows that the substance contains $\ce{40.00 \% w/w C}$, $\ce{6.71 \% w/w H}$, and $\ce{53.3\% w/w O}$. Determine the molecular formula.

My method:

$\ce{C}$ has an atomic mass of $\ce{12.0107}$. $\ce{H}$ has an atomic mass of $\ce{1.008}$. $\ce{O}$ has an atomic mass of $\ce{15.994}$. We start by determining the moles:

$\ce{C=40/{12.0107}=3.33}$

$\ce{H=6.71/{1.008}=6.66 mol}$

$\ce{O=53.3/{15.5994}=3.33}$

We determine the molar ratio:

$\ce{C=3.33/{3.33}=1}$ $\ce{H=6.66/{3.33}=2}$ $\ce{O=3.33/{3.33}=1}$

But the answer is $\ce{C_3H_6O_3}$. What am I doing wrong?

• The molar ratio is correct but you need to consider that the molecular formula can be some multiple of C1:H2:O1, not just CH2O – Waylander May 20 '18 at 21:04

You get $\pu{30g/mol}$ for $\ce{C1H2O1}$ and
$(\pu{90g/mol})/(\pu{30g/mol})=3$
$\ce{C3H6O3}$