Quantitative analysis

How much of each element?

Qualitative analysis says which elements are present. Quantitative analysis says how much. The answer is a percentage by mass for each element.

The idea is the same every time. You change a weighed sample into a product that you can weigh or measure. The product holds all of one element from the sample. Then a ratio of masses gives the element's share.

Carbon and hydrogen: Liebig's method

A weighed sample is heated strongly with copper(II) oxide in a stream of oxygen. All its carbon becomes carbon dioxide, and all its hydrogen becomes water.

• The water is absorbed first, by anhydrous calcium chloride.
• The carbon dioxide is absorbed next, by potassium hydroxide solution.

Each absorber is weighed before and after. The gain in mass is the mass of water, and the mass of carbon dioxide.

In $44\ \text{g}$ of carbon dioxide there are $12\ \text{g}$ of carbon. In $18\ \text{g}$ of water there are $2\ \text{g}$ of hydrogen. So:

• $\%\text{C} = \dfrac{12}{44} \times \dfrac{\text{mass of CO}_2}{\text{mass of compound}} \times 100$
• $\%\text{H} = \dfrac{2}{18} \times \dfrac{\text{mass of H}_2\text{O}}{\text{mass of compound}} \times 100$

An example. $0.30\ \text{g}$ of a compound gives $0.44\ \text{g}$ of carbon dioxide and $0.18\ \text{g}$ of water.

• $\%\text{C} = \dfrac{12}{44} \times \dfrac{0.44}{0.30} \times 100 = 40.0$
• $\%\text{H} = \dfrac{2}{18} \times \dfrac{0.18}{0.30} \times 100 = 6.67$

Nitrogen: the Dumas method

The sample is heated with copper(II) oxide in a stream of carbon dioxide. Its nitrogen is set free as $\text{N}_2$, mixed with some oxides of nitrogen. A hot copper gauze turns these oxides into $\text{N}_2$. The gas is collected over potassium hydroxide solution, which absorbs the carbon dioxide, and its volume is measured.

The volume is brought to standard temperature and pressure, STP. At STP, $22400\ \text{mL}$ of nitrogen weigh $28\ \text{g}$. So:

• $\%\text{N} = \dfrac{28 \times V}{22400 \times m} \times 100$

Here $V$ is the volume of nitrogen at STP in mL, and $m$ is the mass of the compound in g.

The gas was collected over water, so part of its pressure is water vapour. The pressure of the nitrogen alone is the atmospheric pressure minus the aqueous tension. Then $\dfrac{p_1 V_1}{T_1} = \dfrac{p_2 V_2}{T_2}$ gives the volume at STP.

An example. $0.20\ \text{g}$ of a compound gives $56\ \text{mL}$ of nitrogen at STP.

• $\%\text{N} = \dfrac{28 \times 56}{22400 \times 0.20} \times 100 = 35.0$

Nitrogen: the Kjeldahl method

The compound is heated with concentrated sulphuric acid, with a little potassium sulphate to raise the boiling point and copper sulphate as a catalyst. The nitrogen turns into ammonium sulphate. Then excess sodium hydroxide is added, and the ammonia released is distilled into a measured volume of standard sulphuric acid. The acid left over is titrated against standard alkali. The difference tells how much acid the ammonia used.

Each mole of ammonia takes one mole of $\text{H}^+$, and holds one mole of nitrogen, $14\ \text{g}$. With $M$ the molarity of the acid, its basicity $b$, and $V$ the volume of acid used up in mL:

• $\%\text{N} = \dfrac{1.4 \times M \times b \times V}{m}$

Sulphuric acid has a basicity of 2.

An example. The ammonia from $0.50\ \text{g}$ of a compound neutralises $20\ \text{mL}$ of $0.10\ \text{mol/L}$ sulphuric acid.

• $\%\text{N} = \dfrac{1.4 \times 0.10 \times 2 \times 20}{0.50} = 11.2$

Kjeldahl's method is simpler and cheaper than Dumas, so it is used for foods and fertilisers. But it fails for nitro and azo compounds, and for nitrogen in a ring, as in pyridine. Their nitrogen does not turn fully into ammonium sulphate.

Halogens, sulphur and phosphorus: the Carius method

A weighed sample is heated with fuming nitric acid in a sealed, thick glass Carius tube. For a halogen, silver nitrate is added too.

• A halogen becomes a silver halide, AgX. It is filtered, washed, dried and weighed. $\%\text{X} = \dfrac{\text{atomic mass of X}}{\text{molar mass of AgX}} \times \dfrac{\text{mass of AgX}}{\text{mass of compound}} \times 100$.
• Sulphur becomes sulphuric acid. Barium chloride turns it into barium sulphate. $\%\text{S} = \dfrac{32}{233} \times \dfrac{\text{mass of BaSO}_4}{\text{mass of compound}} \times 100$.
• Phosphorus becomes phosphoric acid. It is precipitated as ammonium phosphomolybdate, or as magnesium ammonium phosphate that is heated to $\text{Mg}_2\text{P}_2\text{O}_7$. $\%\text{P} = \dfrac{62}{222} \times \dfrac{\text{mass of Mg}_2\text{P}_2\text{O}_7}{\text{mass of compound}} \times 100$.

An example. $0.25\ \text{g}$ of a chlorine compound gives $0.40\ \text{g}$ of silver chloride, with a molar mass of $143.5\ \text{g/mol}$.

• $\%\text{Cl} = \dfrac{35.5}{143.5} \times \dfrac{0.40}{0.25} \times 100 = 39.6$

Oxygen

Oxygen is hard to measure, so it is usually found by difference:

• $\%\text{O} = 100 - (\%\text{C} + \%\text{H} + \%\text{N} + \%\text{S} + \text{others})$

It can also be found directly. The compound is heated in nitrogen, and its oxygen is turned into carbon monoxide over red hot coke. Iodine pentoxide then turns the carbon monoxide into carbon dioxide, which is weighed. Each $88\ \text{g}$ of this carbon dioxide stands for $32\ \text{g}$ of oxygen from the compound.

From percentages to a formula

The empirical formula gives the simplest whole number ratio of the atoms.

1. Divide each percentage by the atomic mass. This gives the relative number of moles.
2. Divide all by the smallest answer.
3. If a ratio is not a whole number, multiply all by 2 or 3 until it is.

For the example above, with $40.0\ \%$ C, $6.7\ \%$ H and $53.3\ \%$ O:

• $\text{C}: \dfrac{40.0}{12} = 3.33$, $\text{H}: \dfrac{6.7}{1} = 6.7$, $\text{O}: \dfrac{53.3}{16} = 3.33$
• Dividing by $3.33$ gives $1 : 2 : 1$, so the empirical formula is $\text{CH}_2\text{O}$.

The molecular formula is a whole number multiple of the empirical formula. If the molar mass is known, $n = \dfrac{\text{molar mass}}{\text{empirical formula mass}}$. A molar mass of $180\ \text{g/mol}$ with $\text{CH}_2\text{O}$, which weighs $30\ \text{g/mol}$, gives $n = 6$. The molecular formula is $\text{C}_6\text{H}_{12}\text{O}_6$.

Practise with new numbers

Try it in Python

The program turns the masses of carbon dioxide and water into percentages, and then into the empirical formula.

Common mistakes

• Treating all the carbon dioxide as carbon. Only 12 parts in 44 are carbon.
• Forgetting the basicity in Kjeldahl's formula. Sulphuric acid supplies two hydrogen ions for each molecule.
• Using the volume at the lab pressure in Dumas. Correct it for the aqueous tension and to STP first.
• Rounding a ratio too early. A ratio of $1.5$ means multiply by 2. Do not round it to 1 or 2.

Remember these

• Liebig: $\%\text{C} = \tfrac{12}{44} \times \tfrac{m_{CO_2}}{m} \times 100$ and $\%\text{H} = \tfrac{2}{18} \times \tfrac{m_{H_2O}}{m} \times 100$.
• Dumas: $\%\text{N} = \tfrac{28 V}{22400\, m} \times 100$, with $V$ at STP in mL.
• Kjeldahl: $\%\text{N} = \tfrac{1.4 M b V}{m}$.
• Carius gives halogens, sulphur and phosphorus from the mass of a precipitate.
• Oxygen by difference. Empirical formula: percentage over atomic mass, divided by the smallest.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. In Liebig's method, why is the water absorbed first, and not the carbon dioxide?

2. Why does the Kjeldahl method fail for nitrobenzene?

3. A $0.40\ \text{g}$ sample gives $28\ \text{mL}$ of nitrogen at STP. Find the per cent of nitrogen.

4. A hydrocarbon has $85.7\ \%$ carbon and $14.3\ \%$ hydrogen. What is its empirical formula?