Originally Posted by

**o_O** It's just a matter of logic really. Also, sticking that $\displaystyle 10 \tfrac{1}{2}$ wouldn't balance the equation at all since we haven't compensated for the carbons and hydrogens!

Notice that there's 7 carbons and 14 hydrogens on the left. We can stick a 7 in front of carbon dioxide to balance the carbons (it's our only source!) and stick a 7 in front of our dihydrogen monoxide (since this is our only source as well).

Now, let's compensate for the oxygens. On the right hand side, we would have 14 oxygens from $\displaystyle \text{CO}_2$ and 7 oxygens from $\displaystyle \text{H}_2\text{O}$ giving us a total of 21 oxygens.

This suggests sticking in $\displaystyle \tfrac{21}{2}$ in front of $\displaystyle O_2$ on the left hand side and we'll be done.

$\displaystyle \text{C}_7\text{H}_{14} + {\color{red}\tfrac{21}{2}}\text{O}_2 \rightarrow {\color{red}7}\text{CO}_2 + {\color{red}7}\text{H}_2\text{O}$

If you don't like fractions, multiply both sides by 2 and you'll get integer values.