# Question #ef14a

Jun 8, 2015

${C}_{4} {H}_{8} {O}_{2}$

#### Explanation:

A compound's empirical formula tells you what the smallest mole ratio that exists between the atoms that form that compound is.

Your compound's empirical formula is ${C}_{2} {H}_{4} O$, which means that, regardless of how many carbon atoms the actual compound has, it will have 2 times more hydrogen atoms and 4 times fewer oxygen atoms.

To determine exactly how many atoms your compound contains, you can use its molar mass, which tells you what the mass of 1 mole of a substance is.

In your case, the molar mass is equal to 88 g/mol. This means that, if you add the molar masses of each atom that makes up your compound, you'll end up with 88 g/mol.

You know that your compound contains at least 2 carbons, 4 hydrogens, and 1 oxygen. Use their molar masses to get

${\left({C}_{2} {H}_{4} O\right)}_{\textcolor{b l u e}{n}} = \text{88 g/mol}$

$\left(2 \cdot 12.0 + 4 \cdot 1.01 + 1 \cdot 16.0\right) \cancel{\text{g/mol") * color(blue)(n) = 88cancel("g/mol}}$

$44.04 \cdot \textcolor{b l u e}{n} = 88 \implies \textcolor{b l u e}{n} = \frac{88}{44.04} = 1.998 \cong 2$

Thus, your compound's molecular formula, which contains all the atoms present in the molecule, will be

${\left({C}_{2} {H}_{4} O\right)}_{2} \implies {C}_{4} {H}_{8} {O}_{2}$