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Questions on empirical and molecular formulae
The empirical formula as the simplest whole-number ratio of atoms, finding it from masses or from percentage composition by dividing by Ar and then by the smallest, ratios such as 1 : 1.5 that must be multiplied up rather than rounded, the molecular formula from Mr, including an Mr from pV = nRT, combustion analysis to find the masses of carbon and hydrogen from carbon dioxide and water and why combustion must be complete, and how empirical, molecular, structural and displayed formulae differ, with structural isomers.
17 original questions · 55 marks · the empirical and molecular formulae notes · Amount of substance · Flashcards for this unit
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Describe how combustion analysis is used to find the empirical formula of a compound containing carbon, hydrogen and oxygen. Include the order of the absorbers and the conditions that make the result reliable.
Mark scheme
Indicative content
- a weighed, dry sample is burned in a stream of excess oxygen, so that all its carbon forms carbon dioxide and all its hydrogen forms water
- hot copper(II) oxide downstream of the sample oxidises any carbon monoxide to carbon dioxide, so that no carbon escapes unmeasured
- the gases pass first through a weighed tube of anhydrous calcium chloride, which absorbs only water, then through a weighed tube of soda lime
- soda lime absorbs carbon dioxide and water, so it must come second or its gain in mass would include water and the carbon would be too high
- mass of carbon = gain of the soda lime × 12.0 ÷ 44.0, mass of hydrogen = gain of the calcium chloride × 2.0 ÷ 18.0, and oxygen is the sample mass minus both
- each mass is divided by its Ar and by the smallest result to give the simplest whole-number ratio, which is the empirical formula
Levels
- Level 3 (5–6 marks) combustion in excess oxygen, the copper(II) oxide to convert carbon monoxide, the absorbers in the correct order with the reason the order matters, oxygen found by difference and the conversion of the three masses to the simplest ratio are all present; each stage is linked to why the result can be trusted
- Level 2 (3–4 marks) the two absorbers are named and their gains give carbon and hydrogen, with oxygen by difference; the order of the absorbers is not justified, the copper(II) oxide is absent, or the ratio is not completed
- Level 1 (1–2 marks) the compound is burned and the products absorbed, typically with the alkali first so that it takes up the water too, and the calculation is not set out
Calculate the molecular formula of a hydrocarbon that is 82.8% carbon and 17.2% hydrogen by mass and has a relative molecular mass of 58.0.
Mark scheme
- M1 in 100 g: C 82.8 ÷ 12.0 = 6.90 and H 17.2 ÷ 1.0 = 17.2
- M1 dividing by 6.90 gives 1 : 2.49, which is doubled to 2 : 5, so the empirical formula is C₂H₅; reject CH₂ or CH₃ from rounding 2.49
- M1 the empirical formula mass is 29.0, and 58.0 ÷ 29.0 = 2
- A1 the molecular formula is C₄H₁₀; allow error carried forward from a wrong empirical formula
Calculate the empirical formula of a compound of carbon, hydrogen and oxygen, given that 0.180 g of it burns completely to give 0.264 g of carbon dioxide and 0.108 g of water.
Mark scheme
- M1 mass of C = 0.264 × 12.0 ÷ 44.0 = 0.0720 g and mass of H = 0.108 × 2.0 ÷ 18.0 = 0.0120 g
- M1 mass of O = 0.180 − 0.0720 − 0.0120 = 0.0960 g
- M1 moles: C 0.00600, H 0.0120, O 0.00600, a ratio of 1 : 2 : 1
- A1 the empirical formula is CH₂O
Calculate the empirical and molecular formulae of a compound found in rum flavouring, which contains 48.6% carbon, 8.1% hydrogen and 43.2% oxygen by mass and whose mass spectrum has a molecular ion peak at m/z = 74.
Mark scheme
- M1 in 100 g: C 48.6 ÷ 12.0 = 4.05; H 8.1 ÷ 1.0 = 8.1; O 43.2 ÷ 16.0 = 2.70
- M1 dividing by 2.70 gives C 1.5, H 3.0, O 1.0, and doubling clears the half: 3 : 6 : 2
- A1 the empirical formula is C₃H₆O₂; reject C₂H₃O or any formula from rounding 1.5
- A1 empirical formula mass = 74.0, equal to Mr, so the molecular formula is also C₃H₆O₂
Calculate the molecular formula of a hydrocarbon with the empirical formula C₂H₅ from the gas syringe results given in the table. R = 8.31 J K⁻¹ mol⁻¹.
Measurement Value Mass of hydrocarbon 0.145 g Volume of vapour 62.0 cm³ Temperature 298 K Pressure 100 kPa Mark scheme
- M1 n = pV ÷ RT = (1.00 × 10⁵ × 6.20 × 10⁻⁵) ÷ (8.31 × 298) = 2.504 × 10⁻³ mol
- M1 M = 0.145 ÷ (2.504 × 10⁻³) = 57.9 g mol⁻¹
- M1 empirical formula mass of C₂H₅ = 29.0, and 57.9 ÷ 29.0 = 2.00, so the multiple is 2
- A1 the molecular formula is C₄H₁₀
Give the empirical formula of glucose, C₆H₁₂O₆, of ethane, C₂H₆, and of butane, C₄H₁₀.
Mark scheme
- B1 the empirical formula of glucose is CH₂O
- B1 the empirical formula of ethane is CH₃
- B1 the empirical formula of butane is C₂H₅
Calculate the empirical formula of the oxide formed when 1.20 g of magnesium combines with 0.80 g of oxygen. Use Mg 24.3, O 16.0.
Mark scheme
- M1 n(Mg) = 1.20 ÷ 24.3 = 0.0494 mol and n(O) = 0.80 ÷ 16.0 = 0.0500 mol
- M1 dividing by the smaller amount gives Mg : O = 1 : 1.01
- A1 the empirical formula is MgO
Calculate the empirical formula of a compound containing 52.2% carbon, 13.0% hydrogen and 34.8% oxygen by mass.
Mark scheme
- M1 in 100 g: C 52.2 ÷ 12.0 = 4.35; H 13.0 ÷ 1.0 = 13.0; O 34.8 ÷ 16.0 = 2.175
- M1 dividing by 2.175 gives C 2.00, H 5.98, O 1.00
- A1 the empirical formula is C₂H₆O
Calculate the empirical formula of an iron oxide that is 70.0% iron and 30.0% oxygen by mass. Use Fe 55.8, O 16.0.
Mark scheme
- M1 Fe 70.0 ÷ 55.8 = 1.254 and O 30.0 ÷ 16.0 = 1.875
- M1 dividing by 1.254 gives 1 : 1.50, which is multiplied by 2 to give 2 : 3
- A1 the empirical formula is Fe₂O₃
Explain why, in combustion analysis, the gases pass through the water absorber before they pass through the soda lime.
Mark scheme
- B1 the drying agent, such as anhydrous calcium chloride, absorbs only water
- B1 soda lime absorbs carbon dioxide but also absorbs water
- B1 with soda lime first, its mass gain would include the water, so the mass of carbon would be too high and the mass of hydrogen too low
Suggest two compounds that could have the empirical formula CH₂, and suggest how an analyst could tell which one a sample is.
Mark scheme
- B1 any compound CₙH₂ₙ, for example ethene, C₂H₄
- B1 a second example, such as propene, C₃H₆, or cyclohexane, C₆H₁₂
- B1 measure the relative molecular mass, for example from the molecular ion peak in a mass spectrum, and divide it by 14.0 to find the molecular formula
Compare the information given by the empirical, molecular and structural formulae of ethanoic acid.
Mark scheme
- B1 the empirical formula, CH₂O, gives only the simplest ratio of atoms
- B1 the molecular formula, C₂H₄O₂, gives the actual number of atoms of each element in one molecule
- B1 the structural formula, CH₃COOH, also shows how the atoms are arranged, which separates it from isomers such as methyl methanoate
Evaluate a student's decision to round an analysis result of C 1.00 : H 2.40 to the empirical formula CH₂.
Mark scheme
- B1 2.40 is 0.40 from a whole number, far more than the variation expected from experimental error
- B1 2.40 is two and two fifths, so multiplying by 5 gives 5 : 12
- B1 the empirical formula is C₅H₁₂, pentane, so the student's rounding is not justified
Explain how incomplete combustion of a sample in combustion analysis affects the masses of carbon and of oxygen calculated for it.
Mark scheme
- B1 some carbon leaves as carbon monoxide or stays as soot, and is not absorbed by the soda lime
- B1 the mass of carbon dioxide, and so the mass of carbon calculated, is too low
- B1 oxygen is found by difference, so the mass of oxygen comes out too high by the same amount
Explain why the mass of oxygen in an organic compound is found by difference in combustion analysis, not from the products.
Mark scheme
- B1 the oxygen in the carbon dioxide and water comes partly from the oxygen gas used for burning, not only from the sample
- B1 so the mass of oxygen in the sample is the mass of the sample minus the masses of carbon and hydrogen
Suggest the empirical formula of a hydrocarbon whose analysis gives a mole ratio of C 1.00 : H 1.33, and justify your answer.
Mark scheme
- B1 1.33 is one and a third, so both numbers are multiplied by 3, giving 3.00 : 3.99
- B1 the empirical formula is C₃H₄; rounding 1.33 to 1 would discard a third of the hydrogen
Suggest which of ethyl ethanoate (boiling point 77 °C) and butanoic acid (boiling point 164 °C), both C₄H₈O₂, could have its Mr measured in a gas syringe held at 100 °C, and why.
Mark scheme
- B1 ethyl ethanoate, because 100 °C is above its boiling point so all of the injected liquid becomes vapour
- B1 butanoic acid boils above 100 °C, so it would not all vaporise and the volume measured would not correspond to the mass injected
Practise empirical and molecular formulae one question at a time