The ideal gas law is pV = nRT
where:
p = pressure (Pa)
V = volume (m3)
n = amount (mol)
R = 8.31 J mol-1 K-1
T = Temperature (K)
| Question: 0.168 g of a hydrocarbon is held under conditions of 227 oC and 100 kPa and occupies a volume of 50 cm3. A seperate 2.31 g sample of the hydrocarbon is then burned completely in oxygen and found to produce 3 g of water. Determine the molecular formula of the hydrocarbon.
Answer This question involves two different types of analysis! The first sentence allows us to work out the number of moles, and therefore the Mr of the hydrocarbon. Firstly, convert p, V, T into the correct units: p = 100 kPa = 100 x 103 Pa V = 50 cm3 = 50 x 10-6 m3 T = 227 oC = 500 K Then, re-arrange the equation and substitute to find the number of moles of the hydrocarbon: pV = nRT n = pV / RT = (100 x 103)(50 x 10-6)/(8.31)(500) = 1.2 x 10-3 mol Finally, we know the mass and moles of the hydrocarbon, so we can calculate its Mr: Mr = 0.168 / 1.2 x 10-3 = 140 The second sentence allows us to work out the mass of H in the hydrocarbon. From this, we can then calculate the mass of C and then the empirical formula of the hydrocarbon. m H2O = 3 g m H = 3 x (2/18) = 0.33 g (NB: The factor of 2/18 comes from the fact that 2/18-ths of the mass of H2O is hydrogen) m C = 2.31 – 0.33 = 1.98 g
The empirical formula is therefore CH2, which has an Mr of 14. The Mr of the molecule is 140, so there must be 140 / 14 = 10 lots of the empirical formula in the molecular formula. The molecular formula is therefore C10H20. |