# What is the molecular formula of a material with percentage composition C:40.5%, H:6.67%, O:53.33%, and molecular mass of 180.0*g*mol^-1?

May 7, 2017

This is a standard question, and as a routine we ASSUME a mass of $100 \cdot g$ of unknown compound.......and get a molecular formula of ${C}_{6} {H}_{12} {O}_{6}$.

#### Explanation:

So if there are $100 \cdot g$ of unknown compound, we can access the empirical formula given the percentage composition:

$\text{Moles of carbon} = \frac{40.0 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1} = 3.33 \cdot m o l$

$\text{Moles of hydrogen} = \frac{6.67 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1} = 6.62 \cdot m o l$

$\text{Moles of oxygen} = \frac{53.33 \cdot g}{16.00 \cdot g \cdot m o {l}^{-} 1} = 3.33 \cdot m o l$

We divide thru by the LOWEST molar quantity, that of oxygen/carbon to give an empirical formula of:

$C {H}_{2} O$, which represents the simplest whole number ratio defining constituent atoms in a species, i.e. the $\text{empirical formula}$.

But we also have the molecular mass, $180 \cdot g \cdot m o {l}^{-} 1$.

Now it is a fact that the molecular formula is always a whole number multiple of the empirical formula, and thus..............

$180 \cdot g \cdot m o {l}^{-} 1 = n \times \left(12.011 + 2 \times 1.00794 + 15.999\right) \cdot g \cdot m o {l}^{-} 1$,....................... and so we solve for $n$. Clearly, $n = 6$, and the molecular formula is ${C}_{6} {H}_{12} {O}_{6}$.

Just on a practical issue, an analyst would normally NEVER give you the percentage oxygen by mass. Most of the time you get %C,H,N, and if their sum is not 100%, then the BALANCE is assumed to be due to $\text{oxygen}$.