Formulas from combustion data

By the end of this lesson

By the end you can use the masses of $\text{CO}_2$ and $\text{H}_2\text{O}$ from complete combustion to find the empirical formula of a compound (including any oxygen it contains), then its molecular formula.

Chemical forensics

Burn an unknown organic compound in plenty of oxygen, trap the gases, and weigh them. Every carbon atom from the sample ends up in carbon dioxide; every hydrogen atom ends up in water. The products are a full confession.

Follow the atoms

element	where it goes	mass in the sample
C	all into $\text{CO}_2$	$m(\text{C}) = m(\text{CO}_2) \times \dfrac{12.0}{44.0}$
H	all into $\text{H}_2\text{O}$	$m(\text{H}) = m(\text{H}_2\text{O}) \times \dfrac{2.0}{18.0}$
O	can't be traced	$m(\text{O}) = m(\text{sample}) - m(\text{C}) - m(\text{H})$

Why the fractions? 44.0 g of $\text{CO}_2$ contains 12.0 g of carbon; 18.0 g of water contains 2.0 g of hydrogen (two H atoms per molecule).

Why is oxygen different? The O atoms in $\text{CO}_2$ and $\text{H}_2\text{O}$ come from the sample and from the oxygen gas it burned in. So find the sample's oxygen by difference.

Quick check

CalculateQuestion 1 · Recall

Multiple choiceQuestion 2 · Apply

Answer the 2 questions above to continue.