# Question 33b75

Jul 24, 2017

We gets a molecular formula......of ${C}_{4} {H}_{8} {O}_{2}$.....

#### Explanation:

AS for all these sorts of problems, we assume a $100 \cdot g$ mass of unknown compound, and we access the empirical formula......

$\text{Moles of C} = \frac{54.5 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1} = 4.54 \cdot m o l .$

$\text{Moles of H} = \frac{9.3 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1} = 9.23 \cdot m o l .$

$\text{Moles of O} = \frac{36.2 \cdot g}{15.999 \cdot g \cdot m o {l}^{-} 1} = 2.26 \cdot m o l .$

Note that normally %O would never be measured OR QUOTED. For a ${C}_{n} {H}_{m} O$ compound, you would assume that the missing percentage i.e. 100%-%C-%H=%O#.

And now we divide thru by the smallest molar quantity, that of oxygen, to get an empirical formula of......

$C : \frac{4.54 \cdot m o l}{2.26 \cdot m o l} = 2.00$

$H : \frac{9.23 \cdot m o l}{2.26 \cdot m o l} = 4.08$

$O : \frac{2.26 \cdot m o l}{2.26 \cdot m o l} = 1.00$

......to get an empirical formula, the simplest whole number ratio defining constituent atoms in a species, of......

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

Now the molecular formula is a whole number mulitple of the empirical formula. And thus........

$\text{(empirical formula)"xxn="molecular formula}$....but we were quoted a molecular mass, and so.....

$n \times \left(2 \times 12.011 + 4 \times 1.00794 + 1 \times 16.00\right) \cdot g \cdot m o {l}^{-} 1 = 88 \cdot g \cdot m o {l}^{-} 1.$

Clearly, $n = 2$, and so our molecular formula is.....

$2 \times \left\{{C}_{2} {H}_{4} O\right\} = {C}_{4} {H}_{8} {O}_{2}$.....