# What would be considered the molecular formula for the compound that has an empirical formula of CH_4 but a molar mass of 80 g/mol?

Jan 22, 2017

Empirical formula means the ratio $C \div H = 1 \div 4$

#### Explanation:

If we translate this to atomic mass units, the ratio goes to:
$1 \times 12 \div 4 \times 1 = 12 \div 4$

The sum of the ratios is $16$ and this goes into $80$ five times.

So the molecular formula would be ${C}_{1 \times 5} {H}_{4 \times 5} = {C}_{5} {H}_{20}$

Check : $5 \times 12 + 20 \times 1 = 80$

Note : This is a non-existing compound, as $5 C$ atoms can at most hold $12 H$ atoms (matter of valencies).