Combustion Analysis Problem Answers for Examples 11, 12 and 13

Example #11: 0.487 grams of quinine (molar mass = 324 g/mol) is combusted and found to produce 1.321 g CO2, 0.325 g H2O and 0.0421 g nitrogen. Determine the empirical and molecular formulas.

Solution:

Carbon: 1.321 g x (12.011 ÷ 44.0098) = 0.3605 g
Hydrogen: 0.325 g x (2.016 ÷ 18.0152) = 0.0364 g
Nitrogen: 0.0421 g given in problem
Oxygen: 0.487 - (0.3605 + 0.0364 + 0.0421) = 0.0480 g

Carbon: 0.3605 g ÷ 12.011 g/mol = 0.030 mol
Hydrogen: 0.0364 g ÷ 1.008 g/mol = 0.036 mol
Nitrogen: 0.0421 g ÷ 14.007 g/mol = 0.003 mol
Oxygen: 0.0480 g ÷ 15.999 g/mol = 0.003 mol

Carbon: 0.030 ÷ 0.003 = 10
Hydrogen: 0.036 ÷ 0.003 = 12
Nitrogen: 0.003 ÷ 0.003 = 1
Oxygen: 0.003 ÷ 0.003 = 1

The empirical formula is C10H12NO, with an "EFW" of 162. This gives a scaling factor of 2, so the molecular formula is C20H24N2O2.


Example #12: 95.6 mg of menthol (molar mass = 156 g/mol) are burned in oxygen gas to give 269 mg CO2 and 110 mg H2O. What is menthol's empirical formula?

Solution:

Carbon: 269 mg x (12.011 ÷ 44.0098) = 73.4 mg
Hydrogen: 110 mg x (2.016 ÷ 18.0152) = 12.31 mg
Oxygen: 95.6 - (73.4 + 12.31) = 9.88 mg

Carbon: 73.4 mg ÷ 12.011 mg/mmol = 6.11 mmol
Hydrogen: 12.31 mg ÷ 1.008 mg/mmol = 12.21 mmol
Oxygen: 9.88 mg ÷ 15.999 mg/mmol = 0.617 mmol

Carbon: 6.11 ÷ 0.617 = 9.9
Hydrogen: 12.21 ÷ 0.617 = 19.8
Oxygen: 0.617 ÷ 0.617 = 1

If we round off these values, we get an empirical formula of C10H20O. However, were we correct to round the values off? For that, let us turn to the "EFW." If it equals the molar mass (or requires a whole number scaling factor), then we are justified in rounding off.

The "EFW" equals 156, which is the molar mass, so the rounding off is shown to be valid.


Example #13: 0.1005 g of menthol is combusted, producing 0.2829 g of CO2 and 0.1159 g of H2O. What is menthol's empirical formula? (Yes, the answer will be the same as #2. Just do the calculations. OK?)

Solution:

Carbon: 0.2829 g x (12.011 ÷ 44.0098) = 0.07721 g
Hydrogen: 0.1159 g x (2.016 ÷ 18.0152) = 0.01297 g
Oxygen: 0.1005 - (0.07721 + 0.01297) = 0.01032 g

Carbon: 0.07721 g ÷ 12.011 g/mol = 0.006428 mol
Hydrogen: 0.01297 g ÷ 1.008 g/mol = 0.01287 mol
Oxygen: 0.01032 g ÷ 15.999 g/mol = 0.0006452 mol

Carbon: 0.006428 mol ÷ 0.0006452 mol = 9.96
Hydrogen: 0.01287 mol ÷ 0.0006452 mol = 19.94
Oxygen: 0.0006452 mol ÷ 0.0006452 mol = 1

Rounding off gives an empirical formula of C10H20O. Note that the same answer is arrived at in two different experiments. This is the way of science: to be able to carry out an experiment many times, under as similar conditions as possible. No value is accepted in science until it has been verified by multiple experiments and published in peer-reviewed journals with full details for examination by the entire scientific community.