# Combustion analysis gives 52.14% carbon, and 13.15% hydrogen. The molecular is 46.06*g*mol^-1. What are the empirical and molecular formulae of the compound?

Nov 19, 2017

$\text{Molecular formula} = {C}_{2} {H}_{6} O$; I have assumed that 34.73% is due to oxygen....

#### Explanation:

AS with all these problems we assume a starting mass of $100 \cdot g$ with respect to the unknown.

We then calculate the molar composition with respect to its constituent elements:

$\text{Moles of carbon} = \frac{52.14 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1} = 4.34 \cdot m o l$

$\text{Moles of hydrogen} = \frac{13.15 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1} = 13.04 \cdot m o l$

$\text{Moles of oxygen} = \frac{34.73 \cdot g}{16.00 \cdot g \cdot m o {l}^{-} 1} = 2.17 \cdot m o l$

Note that NORMALLY an analyst would NEVER give you percentage oxygen, as its analysis is non-routine. Normally you would be expected to work it out by difference. An analyst would provide %C;%H;%N, whose analysis is routine, and the missing percentage would be assumed to be due to oxygen.

And now we normalize these molar quantities by DIVIDING THRU by the smallest molar quantities, i.e. dividing thru by $\text{moles of oxygen}$....

And so we gets the $\text{empirical formula}$, the simplest ratio defining constituent atoms in a species as...

${C}_{\frac{4.34 \cdot m o l}{2.17 \cdot m o l}} {H}_{\frac{13.04 \cdot m o l}{2.17 \cdot m o l}} {O}_{\frac{2.17 \cdot m o l}{2.17 \cdot m o l}} = {C}_{2} {H}_{6} O$.

Now it is a fact that the $\text{molecular formula}$ is always a whole number multiple of $\text{empirical formula}$.

And thus "molecular formula"-={"empirical formula"}xxn.

And so we use the quoted data and the atomic masses of each constituent element....and solve for $n$.

$46.06 \cdot g \cdot m o {l}^{-} 1 = n \times \left\{2 \times 12.011 + 6 \times 1.00794 + 16.00\right\} \cdot g \cdot m o {l}^{-} 1$

Clearly $n = 1$, and the $\text{molecular formula}$ is the same as the $\text{empirical formula}$, i.e. ${C}_{2} {H}_{6} O$...

Note that normally, if you presented a liquid to an analyst (and this LOW molecular mass species would definitely be a liquid; dimethyl ether, or ethanol), he would tell you to take a running jump, and might even use ruder terms. The question is not chemically reasonable.