# Question e67fa

Jun 8, 2017

${C}_{4} {H}_{10} O$

#### Explanation:

As with all with these problems, we assume a mass of $100 \cdot g$, and we address the elemental makeup by dividing thru by the $\text{atomic mass}$ of each component.

$\text{Moles of carbon} = \frac{64.65 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1} = 5.38 \cdot m o l$.

$\text{Moles of hydrogen} = \frac{13.52 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1} = 13.41 \cdot m o l$.

$\text{Moles of oxygen} = \frac{21.62 \cdot g}{15.999 \cdot g \cdot m o {l}^{-} 1} = 1.351 \cdot m o l$.

They are spoon-feeding you a bit here, in that combustion analysis gives %C,H,N. %O would normally NOT be measured, and determined by the difference (100-%C-%H-%N)%#. And now we divide thru by the SMALLEST MOLAR QUANTITY, that of oxygen to give $C , H , O$

$C : \frac{5.38 \cdot m o l}{1.351 \cdot m o l} = 3.98 \cong 4$

$H : \frac{13.41 \cdot m o l}{1.351 \cdot m o l} = 9.93 \cong 10$

$O : \frac{1.351 \cdot m o l}{1.351 \cdot m o l} = 1$

And thus an $\text{empirical formula}$, the simplest whole number defining constituent atoms in a species of ${C}_{4} {H}_{10} O$.