Topic 20 · page 1 of 15

Formulae from combustion and composition

Turn the masses of carbon dioxide and water from a combustion, or the percentages by mass of each element, into an empirical formula, then use the molecular ion to reach the molecular formula.

14 min read20.1

The first question about an unknown 20.1

Every identification problem in this topic starts the same way. You are given data about an unknown compound and asked what it is. The first thing to settle is how many atoms of each element. Pages 2 to 5 then use tests and spectra to decide how those atoms are joined.

Ethanoic acid has the molecular formula C₂H₄O₂ and the empirical formula CH₂O. The molecular formula is the empirical formula multiplied by a whole number, n.

MethodFrom data to a formula

  1. Find the mass of each element in a sample (or take the percentages as masses in 100 g).
  2. Divide each mass by its ArA_\text{r} to get moles of atoms.
  3. Divide every answer by the smallest one.
  4. Turn the ratio into whole numbers. Multiply all of it by 2, 3 or 4 if a value is near .5, .33 or .25.
  5. That is the empirical formula. Find n = MrM_\text{r} ÷ empirical mass, using the molecular ion from the mass spectrum, to get the molecular formula.

Combustion analysis 20.1

In combustion analysis a weighed sample is burned completely in excess oxygen. All of the carbon becomes carbon dioxide and all of the hydrogen becomes water. The two products are trapped one after the other and the mass gained by each trap is measured.

CX2HX5OH(l)+3 OX2(g)→2 COX2(g)+3 HX2O(l) \ce{C2H5OH(l) + 3O2(g) -> 2CO2(g) + 3H2O(l)}
A flow diagram of combustion analysis. A weighed sample burns in a stream of excess oxygen in a furnace. The gases pass first through a tube of anhydrous calcium chloride, which absorbs water, then through a tube of soda lime, which absorbs carbon dioxide. The increase in mass of each tube gives the mass of water and of carbon dioxide.
Water is absorbed first, then carbon dioxide. The mass gained by each tube is measured.

The trap that absorbs water comes first because soda lime would also absorb water, and the water mass would be lost. Neither trap collects oxygen. The oxygen in the products comes partly from the air supply, so you cannot find the oxygen in the sample from the products. Oxygen is found by difference.

All the carbon in the sample ends up in CO₂ and all the hydrogen in H₂O:

Worked exampleA compound of carbon, hydrogen and oxygen

Illustrative data. 0.230 g of compound X, containing C, H and O only, burns completely to give 0.440 g of CO₂ and 0.270 g of H₂O. Find the empirical formula.

  1. n(CO₂) = 0.440 ÷ 44.0 = 0.0100 mol, so n(C) = 0.0100 mol. Mass of C = 0.0100 × 12.0 = 0.120 g.
  2. n(H₂O) = 0.270 ÷ 18.0 = 0.0150 mol, so n(H) = 2 × 0.0150 = 0.0300 mol. Mass of H = 0.0300 × 1.0 = 0.0300 g.
  3. Mass of O by difference = 0.230 − 0.120 − 0.0300 = 0.0800 g, so n(O) = 0.0800 ÷ 16.0 = 0.00500 mol.
  4. Divide by the smallest (0.00500): C : H : O = 2 : 6 : 1.
  5. The empirical formula is C₂H₆O. Its mass is 46.0, so if MrM_\text{r} = 46 the molecular formula is also C₂H₆O.

C₂H₆O fits two compounds: ethanol, CH₃CH₂OH, and methoxymethane, CH₃OCH₃. Page 2 uses tests and page 4 uses NMR to tell them apart.

When there is no oxygen

If the masses of carbon and hydrogen add up to the mass of the sample, there is no oxygen in the compound.

Worked exampleA hydrocarbon

Illustrative data. 0.420 g of a hydrocarbon gives 1.32 g of CO₂ and 0.540 g of H₂O. The molecular ion is at m/z = 42. Find the molecular formula.

  1. n(CO₂) = 1.32 ÷ 44.0 = 0.0300 mol, so n(C) = 0.0300 mol (0.360 g).
  2. n(H₂O) = 0.540 ÷ 18.0 = 0.0300 mol, so n(H) = 0.0600 mol (0.0600 g).
  3. Check: 0.360 + 0.0600 = 0.420 g, the mass of the sample. There is no oxygen.
  4. C : H = 0.0300 : 0.0600 = 1 : 2, so the empirical formula is CH₂, with mass 14.0.
  5. n = 42 ÷ 14.0 = 3, so the molecular formula is C₃H₆.

Percentage composition 20.1

When you are given the percentage of each element, treat the percentages as masses in a 100 g sample and follow steps 2 to 5.

Worked examplePercentages and the molecular ion

Illustrative data. Compound Y contains 54.5% C, 9.1% H and 36.4% O by mass. Its mass spectrum has its molecular ion peak at m/z = 88. Find the empirical and molecular formulae.

  1. In 100 g: n(C) = 54.5 ÷ 12.0 = 4.54 mol, n(H) = 9.1 ÷ 1.0 = 9.1 mol, n(O) = 36.4 ÷ 16.0 = 2.275 mol.
  2. Divide by 2.275: C : H : O = 2.0 : 4.0 : 1.0.
  3. The empirical formula is C₂H₄O, with mass 44.0.
  4. n = 88 ÷ 44.0 = 2, so the molecular formula is C₄H₈O₂.

Compound Y comes back on page 5, where its spectra show that it is ethyl ethanoate.

A ratio that is not close to whole numbers is a signal, not a rounding error.

Worked exampleA ratio of 1 : 1.5

Illustrative data. A compound contains 48.6% C, 8.1% H and 43.2% O, with MrM_\text{r} = 74. Find the molecular formula.

  1. n(C) = 48.6 ÷ 12.0 = 4.05, n(H) = 8.1 ÷ 1.0 = 8.1, n(O) = 43.2 ÷ 16.0 = 2.70.
  2. Divide by 2.70: C : H : O = 1.5 : 3.0 : 1.
  3. Multiply by 2 to clear the 1.5: C : H : O = 3 : 6 : 2.
  4. The empirical formula is C₃H₆O₂, with mass 74.0. n = 74 ÷ 74.0 = 1, so the molecular formula is C₃H₆O₂.

Check yourself

  1. 0.360 g of a compound containing C, H and O burns to give 0.528 g of CO₂ and 0.216 g of H₂O. Find the empirical formula.
  2. The molecular ion of the compound in question 1 is at m/z = 60. Give its molecular formula.
  3. 1.40 g of a hydrocarbon burns to give 4.40 g of CO₂ and 1.80 g of H₂O, and its molecular ion is at m/z = 56. Show that it contains no oxygen and find its molecular formula.
  4. A compound contains 38.7% C, 9.7% H and 51.6% O by mass, and its molecular ion is at m/z = 62. Find its molecular formula.
  5. A compound contains 22.0% C, 4.6% H and 73.4% Br by mass. Find its empirical formula.
  6. Explain why the mass of oxygen in a sample cannot be found from the masses of CO₂ and H₂O.

Answers cover these until you have tried the questions

  1. n(C) = 0.528 ÷ 44.0 = 0.0120 mol (0.144 g). n(H) = 2 × 0.216 ÷ 18.0 = 0.0240 mol (0.0240 g). Mass of O = 0.360 − 0.144 − 0.0240 = 0.192 g, so n(O) = 0.192 ÷ 16.0 = 0.0120 mol. C : H : O = 0.0120 : 0.0240 : 0.0120 = 1 : 2 : 1. The empirical formula is CH₂O.
  2. The empirical mass of CH₂O is 30.0. n = 60 ÷ 30.0 = 2, so the molecular formula is C₂H₄O₂.
  3. n(C) = 4.40 ÷ 44.0 = 0.100 mol (1.20 g). n(H) = 2 × 1.80 ÷ 18.0 = 0.200 mol (0.200 g). 1.20 + 0.200 = 1.40 g, the mass of the sample, so no oxygen is present. C : H = 0.100 : 0.200 = 1 : 2, so the empirical formula is CH₂ (14.0). n = 56 ÷ 14.0 = 4, so the molecular formula is C₄H₈.
  4. Divide by ArA_\text{r}: C 38.7 ÷ 12.0 = 3.225; H 9.7 ÷ 1.0 = 9.7; O 51.6 ÷ 16.0 = 3.225. Divide by 3.225: C : H : O = 1 : 3 : 1. The empirical formula is CH₃O (31.0). n = 62 ÷ 31.0 = 2, so the molecular formula is C₂H₆O₂.
  5. C 22.0 ÷ 12.0 = 1.83; H 4.6 ÷ 1.0 = 4.6; Br 73.4 ÷ 79.9 = 0.919. Divide by 0.919: C : H : Br = 2.0 : 5.0 : 1.0. The empirical formula is C₂H₅Br, bromoethane.
  6. Some of the oxygen in CO₂ and H₂O comes from the excess oxygen supplied for the combustion, not from the sample. The oxygen in the sample is found by subtracting the masses of carbon and hydrogen from the mass of the sample.