What is the empirical formula of a compound containing C, H, and O if combustion of 3.69 g of the compound yields 5.40 g of CO_2 and 2.22 g of H_2O?

Sep 9, 2016

We get finally an empirical formula of $C {H}_{2} O$; I think the question is suspect.

Explanation:

ONLY the $C$ and the $H$ of the combustion can be presumed to derive from the unknown. (Why? Because the analysis is performed in air and typically an oxidant is added to the combustion.)

$\text{Moles (i) and mass (ii) of carbon}$ $=$ $\frac{5.40 \cdot g}{44.01 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.123 \cdot m o l$ $\equiv$ $1.47 \cdot g \cdot C$

$\text{Moles (i) and mass (ii) of hydrogen}$ $=$ $2 \times \frac{2.22 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.247 \cdot m o l$ $\equiv$ $0.249 \cdot g \cdot H$. Note that the hydrogen in the compound was combusted to water; this is why we multiply the molar quantity by 2.

And now, finally, we work out the percentage composition of $C , H , O$ with respect to the original sample:

%C $=$ (1.47*g)/(3.69*g)xx100%=40.00%

%H $=$ (0.249*g)/(3.69*g)xx100%=6.75%

%O $=$ (100-39.84-6.75)%=53.25%

Note that we cannot (usually) measure the percentage of oxygen in a microanalysis as extra oxidant is typically added. Thus O% is the percentage balance.

After all this wrok we start again. We assume that there were $100 \cdot g$ of compound. And from this we work out the empirical formula.

$\text{Moles of carbon}$ $=$ $\frac{40.0 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1}$ $=$ $3.33 \cdot m o l \cdot C$.

$\text{Moles of hydrogen}$ $=$ $\frac{6.75 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1}$ $=$ $6.70 \cdot m o l \cdot H$.

$\text{Moles of oxygen}$ $=$ $\frac{53.41 \cdot g}{15.999 \cdot g \cdot m o {l}^{-} 1}$ $=$ $3.34 \cdot m o l \cdot H$.

If we divide thru by the smallest molar quantity, we get, an empirical formula of $C {H}_{2} O$. I am not terribly satisifed with this question. A molecular mass should have been quoted. This is a lot of work for simple sugar.