Question

The equation for the burning of octane is:

$2{\mathrm{C}}_8{\mathrm{H}}_{18}+25{\mathrm{O}}_2\to 16{\mathrm{CO}}_2+18{\mathrm{H}}_2\mathrm{O}$

1. How many moles of carbon dioxide are produced when $1$mole of octane burns?
2. What volume, at STP's is occupied by the number of moles determined in (i)?
3. If the relative molecular mass of carbon dioxide is$44$, what is the mass of carbon dioxide produced by burning two moles of octane?
4. What is the empirical formula of octane?

Answer 1

• The balanced equation:
  $\underset{\underset{2\mathrm{vol}}{\mathrm{octane}}}{2{\mathrm{C}}_8{\mathrm{H}}_{18}\left(\mathrm{g}\right)}+\underset{\underset{25\mathrm{vol}}{\mathrm{oxygen}}}{25{\mathrm{O}}_2\left(\mathrm{g}\right)}\to \underset{\underset{16\mathrm{vol}}{\mathrm{carbon}\mathrm{dioxide}}}{16{\mathrm{CO}}_2\left(\mathrm{g}\right)}+\underset{\underset{18\mathrm{vol}}{\mathrm{water}}}{18{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$

(i) We can see from the balanced equation that,

$2$moles of $C_8{H}_{18}$ produced $16$moles of ${\mathrm{CO}}_2$
also, $1$ moles of $C_8{H}_{18}$ produced $\frac{16}{2}=8$ moles of ${\mathrm{CO}}_2$

(ii) Under standard condition of temperature and pressure we know $1$ mole occupies a volume of $22.4litre$

Similarly, $8$ mole occupies $$\begin{gathered} 8\times 22.4 \\ =179.2litre \end{gathered}$$

(iii) $1$ mole of ${\mathrm{CO}}_2$ weighs $44g$

Similarly, $16$ mole of ${\mathrm{CO}}_2$ weighs $$\begin{gathered} 44\times 16 \\ =704g \end{gathered}$$

(iv) The empirical formula of any chemical compound is defined as the simplest positive integer ratio (or whole number ratio) of atoms present in a compound.

So on dividing this by $2$ we will get the simplest ratio which is $4:9$

So the empirical formula is${\mathrm{C}}_4{\mathrm{H}}_9$.