# Exercise 5 | Chemical Calculations

Combustion analysis is used to determine the formula of a compound containing carbon, hydrogen and oxygen.

1.250 g of this compound is burned in a combustion analysis apparatus. The mass of CO_{2 }produced is 3.350 g and the mass of H_{2}O is 0.823 g.

Determine the empirical formula of the compound.

Chemical equation:

C_{x}H_{y}O_{z} + O_{2} → CO_{2} + H_{2}O (unbalanced)

C_{x}H_{y}O_{z} + (X + $\frac{\mathrm{Y}}{4}$ - $\frac{\mathrm{Z}}{2}$) O_{2} → X CO_{2} + $\frac{\mathrm{Y}}{2}$ H_{2}O (balanced)

__Determine the mass of carbon and hydrogen in the original sample:__

mass of carbon in original sample = mass of carbon in the carbon dioxide formed

m_{C} = m_{CO2} x $\frac{{\mathrm{M}}_{\mathrm{C}}}{{\mathrm{M}}_{\mathrm{CO}2}}$

m_{C} = 3.350 x $\frac{12.0}{44.0}$ = 9.14 x 10^{-1} g

mass of hydrogen in original sample = mass of hydrogen in the water formed

m_{H} = m_{H2O} x $\frac{{\mathrm{M}}_{\mathrm{H}}}{{\mathrm{M}}_{\mathrm{H}2\mathrm{O}}}$

m_{C} = 0.823 x $\frac{2.02}{18.0}$ = 9.24 x 10^{-2} g

__Determine mass % of carbon and hydrogen in cpd:__

mass % of C = $\frac{{\mathrm{m}}_{\mathrm{C}}}{{\mathrm{m}}_{\mathrm{compound}}}$ x 100 = $\frac{0.914}{1.250}$ x 100 = 73.1%

mass % of H = $\frac{{\mathrm{m}}_{\mathrm{H}}}{{\mathrm{m}}_{\mathrm{compound}}}$ x 100 = $\frac{0.0924}{1.250}$ x 100 = 7.39 %

__Determine mass % of oxygen in cpd:__

mass % of O = 100 - mass % of C - mass % of H = 19.5 %

__Determine the empirical formula:__

In a 100 g sample: 73.1 g of C, 7.39 g of H and 19.5 g of O

⇒ 6.09 mol of C, 7.32 mol of H and 1.219 mol of O

⇒ For 1 mole of O, we have 5 moles of C and 6 moles of H.

**The empirical formula of the compound is C _{5}H_{6}O**