# How can you determine the molecular formula of a substance from its percent composition?

May 19, 2014

The percent composition gives you only the empirical formula.

#### Explanation:

To get the molecular formula, you must either know the molecular mass or do an experiment to find it.

Example

An unknown compound contains $\text{85.63 % C}$ and $\text{14.37 % H}$. Its experimental molar mass is $\text{56 g/mol}$. What is its molecular formula?

Solution

Assume we have $\text{100 g}$ of the compound. Then we have $\text{85.63 g of C}$ and $\text{14.37 g of H}$.

$\text{Moles of C" = 85.63 color(red)(cancel(color(black)("g C"))) × "1 mol C"/(12.01color(red)(cancel(color(black)("g C")))) = "7.130 mol C}$

$\text{Moles of H" = 14.37 color(red)(cancel(color(black)("g H"))) × "1 mol H"/(1.008 color(red)(cancel(color(black)("g H")))) = "14.26 mol H}$

"Moles of C"/"Moles of H" = (7.130 color(red)(cancel(color(black)("mol"))))/(14.26 color(red)(cancel(color(black)("mol")))) = 1/2.000 ≈ 1/2

The empirical formula is ${\text{CH}}_{2}$.

The empirical formula is the simplest formula of a compound.

The actual formula is an integral multiple of the empirical formula.

If the empirical formula is ${\text{CH}}_{2}$, the actual formula is ${\left({\text{CH}}_{2}\right)}_{n}$ or ${C}_{n} {H}_{2 n}$, where n = 1, 2, 3, … .

Our job is to determine the value of $n$.

The empirical formula mass of ${\text{CH}}_{2}$ is $\text{14.03 u}$. The molecular mass of $\text{56 u}$ must be some multiple of this number.

n = (56 color(red)(cancel(color(black)("u"))))/(14.03 color(red)(cancel(color(black)("u")))) = 4.0 ≈ 4

∴ The molecular formula is ${C}_{n} {H}_{2 n} = {\text{C"_4"H}}_{8}$.