Question

0.2 mol of hydrocarbons undergo complete combustion to give 35.2 of carbon dioxide and 14.4g of water as the only product. What is the molecular formula of hydrocarbon

Answer 1

`\rm C_4 H_8`.

Step-by-step explanation:

Look up the relative atomic mass of
`\rm C`,
`\rm H`, and
`\rm O` on a modern periodic table:

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

Calculate the molecular 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)`.

Given the mass
`m` of
`\rm CO_2` and
`\rm H_2O` produced, calculate the number of moles of molecules that were produced:

`\displaystyle n(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO_2})} \approx 0.80\; \rm mol`;

`\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} \approx 0.80\; \rm mol`.

Calculate the number of moles of
`\rm C` atoms and
`\rm H` atoms in these
`\rm CO_2` and
`\rm H_2O` molecules.

• Each
  `\rm CO_2` molecule contains one
  `\rm C` atom. Therefore, that
  `0.80\; \rm mol` of
  `\rm CO_2\!` contains
  `0.80\; \rm mol\!` of
  `\rm C\!` atoms.
• Each
  `\rm H_2O` molecule contains two
  `\rm H` atoms. Therefore, that
  `0.80\; \rm mol` of
  `\rm H_2O\!` contains
  `2 * 0.80\; \rm mol = 1.60\; \rm mol` of
  `\rm H\!` atoms.

The combustion reaction here include two reactants: the hydrocarbon and
`\rm O_2`.

As the name suggests, hydrocarbons contain only
`\rm C` atoms and
`\rm H` atoms. On the other hand,
`\rm O_2` contains only
`\rm O` atoms.

Therefore, all the
`\rm C` and
`\rm H` atoms in those
`\rm CO_2` and
`\rm H_2O` molecules are from the unknown hydrocarbon. (With a similar logic, all the
`\rm O` atoms in those combustion products are from
`\rm O_2`.)

In other words, that
`0.20\; \rm mol` of this unknown hydrocarbon molecules contains:

• 
  `0.80\; \rm mol` of
  `\rm C` atoms, and
• 
  `1.60\; \rm mol` of
  `\rm H` atoms.

Hence, each of these hydrocarbon molecules would contain
`(0.80\; \rm mol) / (0.20\; \rm mol) = 4` carbon atoms and
`(1.60\; \rm mol) / (0.20\; \rm mol) = 8` hydrogen atoms.

The molecular formula of this hydrocarbon would be
`\rm C_(4) H_(8)`.