How would you find the molecular formula for a compound with molar mass 180 amu, that is composed of 40% carbon 6,67% hydrogen 53.3 oxygen?

Jan 3, 2016

${C}_{6} {H}_{12} {O}_{6}$

Explanation:

We assume 100 g of compound.

Elemental composition is divided thru by the atomic mass of each element:

$C : \frac{40 \cdot \cancel{g}}{12.011 \cdot \cancel{g} \cdot m o {l}^{-} 1}$ $=$ $3.33 \cdot m o l$.

$H : \frac{6.67 \cdot \cancel{g}}{1.00794 \cdot \cancel{g} \cdot m o {l}^{-} 1}$ $=$ $6 , 62 \cdot m o l$.

$O : \frac{53.3 \cdot \cancel{g}}{15.99 \cdot \cancel{g} \cdot m o {l}^{-} 1}$ $=$ $3.33 \cdot m o l$.

So the empirical formula is $C {H}_{2} O$, after we divide thru by the lowest quotient.

Now we know that the molecular formula is always a whole number mulitple of the empirical formula:

i.e. ${\left(12.011 + 2 \times 1.0074 + 15.99\right)}_{n} \cdot g \cdot m o {l}^{-} 1$ $=$ $180 \cdot g \cdot m o {l}^{-} 1.$

Thus $n$ $=$ $6$, and the molecular formula is ${C}_{6} {H}_{12} {O}_{6}$.