# Phenylbutazone can be synthesised, in a multi-step process, starting from compound A. Elemental microanalysis showed that compound A has a composition, by mass, of 50∙0% C; 5∙60% H; 44∙4% O Calculate the empirical formula of compound A?

May 8, 2018

${\text{C"_3"H"_4"O}}_{2}$

#### Explanation:

An $100 \textcolor{w h i t e}{l} \text{g}$ sample of compound A would, by the mass ratio, contain

• $50.0 \textcolor{w h i t e}{l} \text{g}$ of carbon $\text{C}$
• $5.60 \textcolor{w h i t e}{l} \text{g}$ of carbon $\text{H}$
• $44.4 \textcolor{w h i t e}{l} \text{g}$ of carbon $\text{O}$

which corresponds to approximately

• $\frac{50.0}{12} \approx \frac{25}{6} \textcolor{w h i t e}{l} \text{mol}$ of $\text{C}$
• $\frac{5.60}{1} \approx \frac{28}{5} \textcolor{w h i t e}{l} \text{mol}$ of $\text{H}$
• $\frac{44.4}{16} \approx \frac{25}{9} \textcolor{w h i t e}{l} \text{mol}$ of $\text{O}$

where $12$, $1$, and $16$ the respective relative atomic mass of the three elements (rounded to the nearest whole number).

Coefficients in an empirical formula are expected to be whole numbers. Thus it would be necessary eliminate any fraction in these expressions. Multiplying the mass of the sample by $90$- the least common multiple of the three denominators- can serve this purpose. That is, there are

• $375 \textcolor{w h i t e}{l} \text{mol}$ of $\text{C}$,
• $504 \textcolor{w h i t e}{l} \text{mol}$ of $\text{H}$, and
• $250 \textcolor{w h i t e}{l} \text{mol}$ of $\text{O}$

in a $9 \cdot {10}^{3} \textcolor{w h i t e}{l} \text{g}$ sample of compound A.

Dropping the last digit of $50 \textcolor{\mathrm{da} r k b l u e}{4}$ gives the approximation $504 \approx 500 = 4 \cdot 125$, such that it shares the common divisor $125$ with the rest two figures.

Factoring $125$ out of the ratio gives

$n \left(\text{C"):n("H"):n("O}\right) = 375 : \textcolor{\mathrm{da} r k b l u e}{504} : 250$
$\textcolor{w h i t e}{n \left(\text{C"):n("H"):n("O}\right)} \approx \left(3 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{125}}}\right) : \left(\textcolor{\mathrm{da} r k b l u e}{4 \cdot \textcolor{red}{\cancel{\textcolor{\mathrm{da} r k b l u e}{125}}}}\right) : \left(2 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{125}}}\right)$
$\textcolor{w h i t e}{n \left(\text{C"):n("H"):n("O}\right)} = 3 : 4 : 2$

Hence the empirical formula would be

${\text{C"_3"H"_4"O}}_{2}$