IGCSE Chemistry 0620 Paper 4 Practice Paper: Identification of Ions and Gases, Chromatography and Quantitative Analysis

Level: Year 10-11 (IGCSE) | Subject: Chemistry 0620 | Time: 60 minutes | Total: 100 marks | Parts: 32

Paper at a glance

Paper	Time	Marks	Calculator
Paper 4 (Extended, structured)	1 h 15 min	75	Not allowed

Questions

Question 1 (16 marks) – Flame tests and identification of cations

(a) Describe how a flame test is used to identify a metal ion in solution, and state the treatment the nichrome wire needs before each test. (2 marks)

(b) State the flame colour of lithium, sodium, potassium and copper ions in a Bunsen flame. (3 marks)

(c) Explain why the flame colour of potassium ions is often masked by that of sodium ions, and state how the problem is overcome. (3 marks)

(d) 25.0 cm3 of 0.0100 mol/dm3 potassium chloride solution is diluted with water to a total volume of 100 cm3. Calculate the concentration of potassium ions in mol/dm3 and the mass concentration in mg/dm3 (Ar: K = 39.1). (4 marks)

(e) A 2.50 g sample of a mixture is dissolved in water and gives a yellow flame; silver nitrate solution is added and 0.7170 g of white precipitate forms. Calculate the mass of chlorine in the sample and its percentage by mass (Mr: AgCl = 143.4, Ar: Cl = 35.5). (4 marks)

Question 2 (16 marks) – Tests for gases and identification of anions

(a) Describe the test for carbon dioxide and the observation. (2 marks)

(b) Describe the tests and observations used to identify chloride ions and sulfate ions in solution. (3 marks)

(c) 25.0 cm3 of 0.0500 mol/dm3 silver nitrate solution reacts completely with 0.100 mol/dm3 sodium chloride solution. Calculate the volume of sodium chloride solution needed and the mass of silver chloride formed (Mr: AgCl = 143.4). (4 marks)

(d) Explain why dilute nitric acid must be added before silver nitrate solution. (3 marks)

(e) State the test and observation for ammonia, hydrogen, oxygen and chlorine. (4 marks)

Question 3 (18 marks) – Paper chromatography and Rf values

(a) Explain why the start line on the chromatography paper must be drawn in pencil and not with a ballpoint pen. (2 marks)

(b) In a chromatogram the solvent front is 8.40 cm above the start line. Spot A is 6.30 cm and spot B is 2.94 cm above the start line. Calculate the Rf values of A and B to three significant figures. (4 marks)

(c) Under the same conditions sample X gives two spots in exactly the same positions as A and B. Pure substance A is known to be glycine. State the composition of X and explain your conclusion. (4 marks)

(d) If in the same experiment the solvent front only rises to 6.00 cm above the start line, state where the spot of A would be and explain whether the Rf value of A changes. (3 marks)

(e) State two factors that would change the Rf value of the same substance. (3 marks)

(f) Explain why the chromatography tank must be sealed and why the start line must not dip into the solvent. (2 marks)

Question 4 (16 marks) – Separation, purification and extraction

(a) Describe how dry sodium chloride and pure water can both be obtained from brine, and name the separation technique used. (3 marks)

(b) 200 g of brine contains 12.0% sodium chloride by mass. Calculate the mass of salt obtained on evaporation and the mass of water in the solution. (4 marks)

(c) State one difference between fractional distillation and simple distillation, and give one industrial use where fractional distillation is necessary. (3 marks)

(d) A solid sample melts between 76 and 80 degrees C, while the pure substance melts at 80 degrees C. Deduce whether the sample is pure and explain. (3 marks)

(e) 1.00 g of a substance is dissolved in 100 cm3 of water and extracted with ether; the partition coefficient (ether/water) is 4.0. Calculate the fraction extracted using 50 cm3 of ether in one portion and using 25 cm3 in two portions, and state which method is more effective. (3 marks)

Question 5 (18 marks) – Gravimetric and volumetric analysis

(a) Explain why the precipitate is washed with distilled water after filtration. (2 marks)

(b) 1.250 g of a sample containing barium chloride is dissolved and treated with excess sodium sulfate solution, giving 1.165 g of barium sulfate. Calculate the mass of barium chloride in the sample and its percentage by mass (Mr: BaSO4 = 233.4, BaCl2 = 208.2). (3 marks)

(c) Describe how to check that the washing water is free of the ions being removed. (3 marks)

(d) Explain what is meant by heating to constant mass and state the criterion for reaching it. (3 marks)

(e) If 0.020 g of the barium sulfate was lost during washing, recalculate the percentage in part (b) and state whether the result is too high or too low. (4 marks)

(f) 25.0 cm3 of 0.100 mol/dm3 sodium hydroxide solution is neutralised by 0.125 mol/dm3 hydrochloric acid. Calculate the volume of acid required in cm3. (3 marks)

Question 6 (16 marks) – Instrumental analysis and data interpretation

(a) In the mass spectrum of an organic compound the peak at the highest m/z value is 46. State the relative molecular mass and explain why this value alone does not determine the molecular formula. (3 marks)

(b) The infrared spectrum of a compound shows a strong absorption at 1700 cm^-1 and a broad strong absorption at 3300 cm^-1. Deduce which bonds or functional groups are present. (3 marks)

(c) The mass spectrum of copper shows two peaks: m/z 63 with a relative abundance of 69.2% and m/z 65 with 30.8%. Calculate the average relative atomic mass of copper to two decimal places. (4 marks)

(d) State two advantages of flame emission spectroscopy over the traditional flame test. (3 marks)

(e) In gas chromatography two reference substances have retention times of 3.41 min and 5.80 min; a sample gives only one peak at 3.42 min. State the conclusion and its limitation. (3 marks)

Answers and marking notes

Question 1 – Flame tests and identification of cations

(a) Dip the nichrome wire in concentrated hydrochloric acid and heat it in the Bunsen flame until no colour appears (one mark); then dip it in the sample and hold it at the edge of the flame to observe the colour (one mark). [2 marks]

(b) Lithium ions give a crimson flame (one mark); sodium ions give a yellow flame and potassium ions a lilac flame (one mark); copper ions give a blue-green flame (one mark). [3 marks]

(c) The yellow flame from sodium is intense and long-lasting (one mark); the flame is viewed through cobalt blue glass, which absorbs the yellow light (one mark); the lilac colour of potassium is then seen (one mark). [3 marks]

(d) n(K+) = 0.0250 x 0.0100 = 2.50 x 10^-4 mol (one mark); concentration = 2.50 x 10^-4 / 0.100 = 2.50 x 10^-3 mol/dm3 (one mark); mass concentration = 2.50 x 10^-3 x 39.1 = 0.0978 g/dm3 (one mark); that is 97.8 mg/dm3 (one mark). [4 marks]

(e) n(AgCl) = 0.7170 / 143.4 = 5.00 x 10^-3 mol (one mark); n(Cl) = 5.00 x 10^-3 mol (one mark); m(Cl) = 5.00 x 10^-3 x 35.5 = 0.1775 g (one mark); percentage = 0.1775 / 2.50 x 100 = 7.10% (one mark). [4 marks]

Question 2 – Tests for gases and identification of anions

(a) Bubble the gas through limewater, calcium hydroxide solution (one mark); the limewater turns milky or cloudy (one mark). [2 marks]

(b) Chloride: add dilute nitric acid then silver nitrate solution and a white precipitate forms (one mark); sulfate: add dilute hydrochloric acid then barium chloride solution and a white precipitate forms (one mark); both solutions are acidified first to remove carbonate interference (one mark). [3 marks]

(c) n(Ag+) = 0.0250 x 0.0500 = 1.25 x 10^-3 mol (one mark); the ratio is 1 to 1 so n(NaCl) = 1.25 x 10^-3 mol (one mark); V = 1.25 x 10^-3 / 0.100 = 0.0125 dm3 = 12.5 cm3 (one mark); m(AgCl) = 1.25 x 10^-3 x 143.4 = 0.179 g (one mark). [4 marks]

(d) Carbonate ions in the solution would also give a white precipitate of silver carbonate (one mark); dilute nitric acid decomposes the carbonate and removes the carbon dioxide (one mark); the white precipitate can then only come from a halide ion, so the conclusion is reliable (one mark). [3 marks]

(e) Ammonia turns damp red litmus paper blue (one mark); hydrogen gives a squeaky pop with a lighted splint (one mark); oxygen relights a glowing splint (one mark); chlorine turns damp litmus paper red then bleaches it (one mark). [4 marks]

Question 3 – Paper chromatography and Rf values

(a) The ink of a ballpoint pen is soluble in the running solvent (one mark); it would travel up the paper, spoil the chromatogram and give extra spots (one mark). [2 marks]

(b) Rf(A) = 6.30 / 8.40 (one mark) = 0.750 (one mark); Rf(B) = 2.94 / 8.40 (one mark) = 0.350 (one mark). [4 marks]

(c) Sample X contains glycine (A) together with a second substance (B) (two marks); under the same solvent, temperature and paper the same substance always gives the same Rf value, so identical spot positions indicate the same substance (two marks). [4 marks]

(d) The spot would be 0.750 x 6.00 = 4.50 cm above the start line (two marks); the Rf value is unchanged because Rf is a ratio of two distances (one mark). [3 marks]

(e) A different solvent or solvent ratio (one mark); a different temperature (one mark); also acceptable: a different type of paper, an unsaturated tank, or an overloaded spot (one mark). [3 marks]

(f) Sealing keeps the air saturated with solvent vapour so the front stays level instead of evaporating from the edges (one mark); if the start line dipped into the solvent the sample would simply dissolve into it and no separation would occur (one mark). [2 marks]

Question 4 – Separation, purification and extraction

(a) Use simple distillation: heat so that water distils over and is condensed and collected (one mark); evaporate the remaining brine to crystallise the solid salt (one mark); the technique is simple distillation (one mark). [3 marks]

(b) m(NaCl) = 200 x 0.120 = 24.0 g (two marks); m(water) = 200 – 24.0 = 176 g (two marks). [4 marks]

(c) A fractionating column allows repeated evaporation and condensation so liquids with close boiling points can be separated (one mark); simple distillation only separates one volatile liquid from a non-volatile solute (one mark); example: fractional distillation of crude oil (one mark). [3 marks]

(d) The sample is impure (one mark); impurities lower the melting point (one mark); the melting range is also wide, about 4 degrees, whereas a pure substance melts over a very narrow range (one mark). [3 marks]

(e) One portion: 4.0 x 50 / (4.0 x 50 + 100) = 0.667, so 66.7% is extracted (one mark); two portions: each gives 4.0 x 25 / (4.0 x 25 + 100) = 0.500 and 0.500 x 0.500 = 0.250 remains, so 75.0% is extracted (one mark); two smaller portions are more effective (one mark). [3 marks]

Question 5 – Gravimetric and volumetric analysis

(a) Washing removes soluble impurities adsorbed on the precipitate (one mark); otherwise the mass after drying would be too high and the result inaccurate (one mark). [2 marks]

(b) n(BaSO4) = 1.165 / 233.4 = 4.99 x 10^-3 mol (one mark); n(BaCl2) is the same, m = 4.99 x 10^-3 x 208.2 = 1.04 g (one mark); percentage = 1.04 / 1.250 x 100 = 83.1% (one mark). [3 marks]

(c) Collect the last portion of washings (one mark); add barium chloride solution and the absence of a white precipitate shows sulfate has been removed (one mark); silver nitrate solution is used instead when checking for chloride ions (one mark). [3 marks]

(d) Heat, cool and weigh repeatedly (one mark); until two successive masses agree (one mark); in practice the difference should not exceed 0.002 g (one mark). [3 marks]

(e) n(BaSO4) = (1.165 – 0.020) / 233.4 = 4.91 x 10^-3 mol (one mark); m(BaCl2) = 4.91 x 10^-3 x 208.2 = 1.02 g (one mark); percentage = 1.02 / 1.250 x 100 = 81.7% (one mark); the result is too low by about 1.4 percentage points (one mark). [4 marks]

(f) n(NaOH) = 0.0250 x 0.100 = 2.50 x 10^-3 mol (one mark); the ratio is 1 to 1 so n(HCl) = 2.50 x 10^-3 mol (one mark); V = 2.50 x 10^-3 / 0.125 = 0.0200 dm3 = 20.0 cm3 (one mark). [3 marks]

Question 6 – Instrumental analysis and data interpretation

(a) Relative molecular mass = 46 (one mark); different combinations give the same Mr, for example C2H6O and NO2 are both 46 (one mark); high resolution mass spectrometry or chemical tests are needed to fix the structure (one mark). [3 marks]

(b) The strong band at 1700 cm^-1 indicates a C=O group (one mark); the broad band at 3300 cm^-1 indicates an O-H bond (one mark); together they indicate a carboxylic acid, or a molecule with both groups (one mark). [3 marks]

(c) 63 x 0.692 = 43.60 (one mark); 65 x 0.308 = 20.02 (one mark); the sum is 63.62 (one mark); the answer is 63.62 (one mark). [4 marks]

(d) It detects very small amounts of ions, so it is more sensitive (one mark); the instrument splits the light into a spectrum so ions with similar colours can be distinguished (one mark); the result is recorded by the instrument, which is more objective and repeatable (one mark). [3 marks]

(e) The sample contains the first reference substance (one mark); it does not contain the second (one mark); limitation: a matching retention time is only tentative identification, since another substance could have the same value, so mass spectrometry is needed to confirm (one mark). [3 marks]

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