Question

combustion analysis of a hydrocarbon produced 33.01g CO2 and 13.51g H2O. Calculate the empirical formula for the hydrocarbon

Answer 1

`\rm CH_2`.

Step-by-step explanation:

Carbon and hydrogen are the only two elements in a hydrocarbon. When a hydrocarbon combusts completely in excess oxygen, the products would be
`\rm CO_2` and
`\rm H_2O`. The
`\rm C` and
`\rm H` would come from the hydrocarbon, while the
`\rm O` atoms would come from oxygen.

Look up the relative atomic mass of these three elements on a modern periodic table:

• 
  `\rm C`:
  `12.011`.
• 
  `\rm H`:
  `1.008`.
• 
  `\rm O`:
  `\rm 15.999`.

Calculate the molar mass of
`\rm CO_2` and
`\rm H_2O`:

`M(\mathrm{CO_2}) = 12.011 + 2 * 15.999 = 44.009\; \rm g \cdot mol^(-1)`.

`M(\mathrm{H_2O}) = 2 * 1.008 + 15.999 = 18.015\; \rm g \cdot mol^(-1)`

Calculate the number of moles of
`\rm CO_2` molecules in
`33.01\; \rm g` of
`\rm CO_2\!`:

`\displaystyle n(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO2})} = (33.01\; \rm g)/(44.009\; \rm g\cdot mol^(-1)) \approx 0.7501\; \rm mol`.

Similarly, calculate the number of moles of
`\rm H_2O` molecules in
`13.51\; \rm g` of
`\rm H_2O\!`:

`\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = (13.51\; \rm g)/(18.015\; \rm g\cdot mol^(-1)) \approx 0.7499\; \rm mol`.

Note that there is one carbon atom in every
`\rm CO_2` molecule. Approximately
`0.7501\; \rm mol` of
`\rm CO_2\!` molecules would correspond to the same number of
`\rm C` atoms. That is:
`n(\mathrm{C}) \approx 0.7501\; \rm mol`.

On the other hand, there are two hydrogen atoms in every
`\rm H_2O` molecule. approximately
`0.7499\; \rm mol` of
`\rm H_2O` molecules would correspond to twice as many
`\rm H\!` atoms. That is:
`n(\mathrm{H}) \approx 2 * 0.7499 \; \rm mol\approx 1.500\; \rm mol`.

The ratio between the two is:
`n(\mathrm{C}): n(\mathrm{H}) \approx 1:2`.

The empirical formula of a compound gives the smallest whole-number ratio between the elements. For this hydrocarbon, the empirical formula would be
`\rm CH_2`.