Chemistry 0620 | Paper 3 Mock 01 | Theory (Core)
Question 1 Report
Fig. 1.1 is an outline of the Periodic Table. Three blocks of elements are marked with the letters J, K and L.
(a) State the letter of the block that contains the transition elements. [1]
(b) Name one transition element. [1]
(c) State two physical properties of a typical transition element that are different from those of a Group I metal. [2]
(d) A transition element or its compound is often added to a reaction to make it faster while remaining unchanged in mass. State the term for such a substance. [1]
(e) Name the transition metal used to speed up the manufacture of ammonia. [1]
This question uses an outline of the Periodic Table to locate the transition elements and then tests their characteristic properties.
(a) The transition elements form the central block that sits between Groups I to II and Groups III to 0, so the correct block is K. [1] Block J on the far left is Groups I to II and block L on the right is Groups III to 0.
(b) Any one named transition element, for example iron (copper, nickel, zinc, manganese or vanadium are also accepted). [1]
(c) Any two physical properties in which a typical transition metal differs from a Group I metal: higher density [1]; higher melting point [1]; harder or stronger [1] (maximum [2]). Group I metals such as sodium are, by contrast, soft, low in density and low melting.
(d) A substance that speeds up a reaction while remaining unchanged in mass is a catalyst. [1]
(e) The transition metal that catalyses the manufacture of ammonia (the Haber process) is iron. [1]
Answer Details
This question uses an outline of the Periodic Table to locate the transition elements and then tests their characteristic properties.
(a) The transition elements form the central block that sits between Groups I to II and Groups III to 0, so the correct block is K. [1] Block J on the far left is Groups I to II and block L on the right is Groups III to 0.
(b) Any one named transition element, for example iron (copper, nickel, zinc, manganese or vanadium are also accepted). [1]
(c) Any two physical properties in which a typical transition metal differs from a Group I metal: higher density [1]; higher melting point [1]; harder or stronger [1] (maximum [2]). Group I metals such as sodium are, by contrast, soft, low in density and low melting.
(d) A substance that speeds up a reaction while remaining unchanged in mass is a catalyst. [1]
(e) The transition metal that catalyses the manufacture of ammonia (the Haber process) is iron. [1]
Question 2 Report
The table compares three nitrogen fertilisers. Use these relative atomic masses: H = 1, C = 12, N = 14, O = 16, S = 32.
| Fertiliser | Formula | Mr | % nitrogen by mass |
|---|---|---|---|
| ammonium nitrate | NH4NO3 | 80 | 35.0 |
| ammonium sulfate | (NH4)2SO4 | to be found | to be found |
| urea | CO(NH2)2 | 60 | to be found |
(a) Show that the relative formula mass (Mr) of ammonium sulfate, (NH4)2SO4, is 132. [2]
(b) Calculate the percentage by mass of nitrogen in ammonium sulfate. [2]
(c) Calculate the percentage by mass of nitrogen in urea, CO(NH2)2. [2]
(d) Using the completed table, state which fertiliser is the best source of nitrogen per kilogram. [1]
(e) Explain why this fertiliser is cheaper to transport than ammonium sulfate for the same mass of nitrogen. [2]
(f) Name the element present in ammonium sulfate that is not present in ammonium nitrate. [1]
(g) State why a soluble fertiliser is more useful to a plant than an insoluble one. [1]
(h) Write the balanced symbol equation for making ammonium nitrate from ammonia and nitric acid. [2]
This question tests relative formula mass and percentage composition, then uses the results to compare fertilisers. The completed table is shown first, followed by the working.
| Fertiliser | Formula | Mr | % nitrogen by mass |
|---|---|---|---|
| ammonium nitrate | NH4NO3 | 80 | 35.0 |
| ammonium sulfate | (NH4)2SO4 | 132 | 21.2 |
| urea | CO(NH2)2 | 60 | 46.7 |
(a) For (NH4)2SO4, add the masses inside the brackets first, then double, then add S and 4 O:
\[ M_r = 2\times(14 + 4\times1) + 32 + 4\times16 = 2\times18 + 32 + 64 = 36 + 32 + 64 = 132 \]
1 mark for the working, 1 mark for 132.
(b) There are 2 nitrogen atoms, mass \( 2\times14 = 28 \) [1]. Then:
\[ \%\,N = \frac{28}{132}\times100 = 21.2\% \]
1 mark for 21.2%.
(c) Urea CO(NH2)2 also has 2 nitrogen atoms, mass \( 2\times14 = 28 \), and Mr = 60:
\[ \%\,N = \frac{28}{60}\times100 = 46.7\% \]
1 mark for the 28/60 setup, 1 mark for 46.7%.
(d) Comparing the percentages (35.0, 21.2, 46.7), the best source of nitrogen per kilogram is urea [1].
(e) Urea is cheaper to transport for the same mass of nitrogen because it has a higher percentage of nitrogen [1], so a smaller mass of urea carries the same mass of nitrogen, which lowers the transport cost [1].
(f) The element in ammonium sulfate but not in ammonium nitrate is sulfur [1].
(g) A soluble fertiliser is more useful because it dissolves in soil water so its ions can be absorbed by the roots [1]; an insoluble one stays out of reach of the plant.
(h) The equation for making ammonium nitrate is:
\[ NH_3 + HNO_3 \rightarrow NH_4NO_3 \]
1 mark for correct formulae, 1 mark for balancing (it is already balanced as written).
Exam tip: percentage of an element = (total mass of that element divided by Mr) times 100; the higher the % nitrogen, the more nitrogen you carry per kilogram.
Answer Details
This question tests relative formula mass and percentage composition, then uses the results to compare fertilisers. The completed table is shown first, followed by the working.
| Fertiliser | Formula | Mr | % nitrogen by mass |
|---|---|---|---|
| ammonium nitrate | NH4NO3 | 80 | 35.0 |
| ammonium sulfate | (NH4)2SO4 | 132 | 21.2 |
| urea | CO(NH2)2 | 60 | 46.7 |
(a) For (NH4)2SO4, add the masses inside the brackets first, then double, then add S and 4 O:
\[ M_r = 2\times(14 + 4\times1) + 32 + 4\times16 = 2\times18 + 32 + 64 = 36 + 32 + 64 = 132 \]
1 mark for the working, 1 mark for 132.
(b) There are 2 nitrogen atoms, mass \( 2\times14 = 28 \) [1]. Then:
\[ \%\,N = \frac{28}{132}\times100 = 21.2\% \]
1 mark for 21.2%.
(c) Urea CO(NH2)2 also has 2 nitrogen atoms, mass \( 2\times14 = 28 \), and Mr = 60:
\[ \%\,N = \frac{28}{60}\times100 = 46.7\% \]
1 mark for the 28/60 setup, 1 mark for 46.7%.
(d) Comparing the percentages (35.0, 21.2, 46.7), the best source of nitrogen per kilogram is urea [1].
(e) Urea is cheaper to transport for the same mass of nitrogen because it has a higher percentage of nitrogen [1], so a smaller mass of urea carries the same mass of nitrogen, which lowers the transport cost [1].
(f) The element in ammonium sulfate but not in ammonium nitrate is sulfur [1].
(g) A soluble fertiliser is more useful because it dissolves in soil water so its ions can be absorbed by the roots [1]; an insoluble one stays out of reach of the plant.
(h) The equation for making ammonium nitrate is:
\[ NH_3 + HNO_3 \rightarrow NH_4NO_3 \]
1 mark for correct formulae, 1 mark for balancing (it is already balanced as written).
Exam tip: percentage of an element = (total mass of that element divided by Mr) times 100; the higher the % nitrogen, the more nitrogen you carry per kilogram.
Question 3 Report
When a hydrocarbon fuel burns, the products depend on how much oxygen is available.
(a) State the two products formed when a hydrocarbon undergoes complete combustion. [2]
(b) Incomplete combustion occurs when the supply of oxygen is limited. Name two products, other than water, that can be formed by incomplete combustion of a hydrocarbon. [2]
(c) Explain why carbon monoxide is dangerous to people. [2]
(d) Carbon monoxide is colourless and has no smell. Suggest why these properties make it especially dangerous in the home. [1]
(e) Complete the symbol equation for the incomplete combustion of methane.
2CH4 + 3O2 → 2.......... + 4.......... [2]
(f) Name one appliance in the home that could release carbon monoxide if it is faulty. [1]
(g) State one way of reducing the risk from carbon monoxide in the home. [1]
This question compares complete and incomplete combustion and focuses on carbon monoxide, the dangerous product of burning a hydrocarbon in too little oxygen. Complete combustion needs plenty of oxygen and gives only carbon dioxide and water; a limited oxygen supply gives carbon monoxide and/or soot instead.
(a) Complete combustion of a hydrocarbon gives two products: carbon dioxide [1] and water. [1]
(b) Two products of incomplete combustion, other than water: carbon monoxide [1] and carbon (soot). [1] These form because there is not enough oxygen to fully oxidise all the carbon to CO2.
(c) Carbon monoxide is dangerous because it is toxic (poisonous); [1] it combines with haemoglobin in the blood, so the blood carries less oxygen around the body, starving the tissues of oxygen. [1]
(d) Because it is colourless and has no smell, a person cannot see or smell it and is not aware of the leak, so they breathe it in without realising and take no action to escape. [1]
(e) Completing the incomplete-combustion equation of methane:
2CH4 + 3O2 → 2CO + 4H2O
[1] for CO and [1] for H2O. Count the atoms to check: 2 C, 8 H and 6 O on each side. The carbon ends up as carbon monoxide (not CO2) because oxygen is limited.
(f) One home appliance that could release carbon monoxide if faulty: a gas boiler (also acceptable: gas heater, gas cooker or gas fire). [1]
(g) One way to reduce the risk: fit a carbon monoxide detector (also acceptable: provide good ventilation, or have gas appliances serviced regularly). [1]
Exam tip: for why carbon monoxide is dangerous you need both marks, "toxic" and the haemoglobin/oxygen-carrying point. Just writing "it is poisonous" scores only one.
Answer Details
This question compares complete and incomplete combustion and focuses on carbon monoxide, the dangerous product of burning a hydrocarbon in too little oxygen. Complete combustion needs plenty of oxygen and gives only carbon dioxide and water; a limited oxygen supply gives carbon monoxide and/or soot instead.
(a) Complete combustion of a hydrocarbon gives two products: carbon dioxide [1] and water. [1]
(b) Two products of incomplete combustion, other than water: carbon monoxide [1] and carbon (soot). [1] These form because there is not enough oxygen to fully oxidise all the carbon to CO2.
(c) Carbon monoxide is dangerous because it is toxic (poisonous); [1] it combines with haemoglobin in the blood, so the blood carries less oxygen around the body, starving the tissues of oxygen. [1]
(d) Because it is colourless and has no smell, a person cannot see or smell it and is not aware of the leak, so they breathe it in without realising and take no action to escape. [1]
(e) Completing the incomplete-combustion equation of methane:
2CH4 + 3O2 → 2CO + 4H2O
[1] for CO and [1] for H2O. Count the atoms to check: 2 C, 8 H and 6 O on each side. The carbon ends up as carbon monoxide (not CO2) because oxygen is limited.
(f) One home appliance that could release carbon monoxide if faulty: a gas boiler (also acceptable: gas heater, gas cooker or gas fire). [1]
(g) One way to reduce the risk: fit a carbon monoxide detector (also acceptable: provide good ventilation, or have gas appliances serviced regularly). [1]
Exam tip: for why carbon monoxide is dangerous you need both marks, "toxic" and the haemoglobin/oxygen-carrying point. Just writing "it is poisonous" scores only one.
Question 4 Report
A student investigates the reactions of halogens with halide solutions. The tubes are shown in Fig. 11.1 and the results are recorded in Table 11.1.
Fig. 11.1
| tube | halide solution | halogen added | observation |
|---|---|---|---|
| A | potassium chloride | bromine water | no change |
| B | potassium bromide | chlorine water | |
| C | potassium iodide | chlorine water | |
| D | potassium iodide | bromine water |
(a) Complete Table 11.1 by describing the observation (colour change) in tubes B, C and D. [3]
(b) Name the type of reaction being investigated. [1]
(c) State which halogen is shown to be the most reactive by these results. [1]
(d) Explain how the results show that iodine is the least reactive of the three halogens. [2]
(e) The student shakes each tube with an organic solvent to see the colours more clearly. Suggest why the colours are easier to see in the solvent. [1]
(f) State one safety precaution the student should take during this experiment. [1]
(g) Write the word equation for the reaction in tube B. [1]
(h) Bromine water is added to potassium chloride solution (tube A). Predict the observation and explain your answer. [2]
This is a halogen displacement investigation. A halogen displaces any halogen below it in the reactivity order Cl > Br > I, and the new halogen formed gives a colour change.
(a) The completed observations are [3] (one mark each):
| tube | halide solution | halogen added | observation |
|---|---|---|---|
| A | potassium chloride | bromine water | no change |
| B | potassium bromide | chlorine water | colourless turns orange / brown (bromine formed) |
| C | potassium iodide | chlorine water | colourless turns dark brown (iodine formed) |
| D | potassium iodide | bromine water | colourless turns dark brown (iodine formed) |
(b) The reaction type is displacement [1].
(c) Chlorine displaces both bromine and iodine, so chlorine is the most reactive halogen shown [1].
(d) Iodine does not displace chlorine or bromine from their solutions [1], yet both chlorine and bromine displace iodine from potassium iodide, which shows iodine is the least reactive of the three [1].
(e) Shaking with an organic solvent concentrates the halogen into the small solvent layer, so the colour is more concentrated and clearer to see [1].
(f) A safety precaution: work in a fume cupboard (also acceptable: wear eye protection, or wear gloves) [1], because chlorine and bromine are toxic and irritating.
(g) Word equation for tube B: chlorine + potassium bromide → potassium chloride + bromine [1].
(h) Tube A shows no change / no reaction [1], because bromine is less reactive than chlorine and so cannot displace chloride from potassium chloride [1].
Exam tip: a colour change means a reaction happened (more reactive halogen added); "no change" means the added halogen is less reactive than the one already in the salt.
Answer Details
This is a halogen displacement investigation. A halogen displaces any halogen below it in the reactivity order Cl > Br > I, and the new halogen formed gives a colour change.
(a) The completed observations are [3] (one mark each):
| tube | halide solution | halogen added | observation |
|---|---|---|---|
| A | potassium chloride | bromine water | no change |
| B | potassium bromide | chlorine water | colourless turns orange / brown (bromine formed) |
| C | potassium iodide | chlorine water | colourless turns dark brown (iodine formed) |
| D | potassium iodide | bromine water | colourless turns dark brown (iodine formed) |
(b) The reaction type is displacement [1].
(c) Chlorine displaces both bromine and iodine, so chlorine is the most reactive halogen shown [1].
(d) Iodine does not displace chlorine or bromine from their solutions [1], yet both chlorine and bromine displace iodine from potassium iodide, which shows iodine is the least reactive of the three [1].
(e) Shaking with an organic solvent concentrates the halogen into the small solvent layer, so the colour is more concentrated and clearer to see [1].
(f) A safety precaution: work in a fume cupboard (also acceptable: wear eye protection, or wear gloves) [1], because chlorine and bromine are toxic and irritating.
(g) Word equation for tube B: chlorine + potassium bromide → potassium chloride + bromine [1].
(h) Tube A shows no change / no reaction [1], because bromine is less reactive than chlorine and so cannot displace chloride from potassium chloride [1].
Exam tip: a colour change means a reaction happened (more reactive halogen added); "no change" means the added halogen is less reactive than the one already in the salt.
Question 5 Report
A student investigated how the concentration of sodium thiosulfate affects the rate of its reaction with dilute hydrochloric acid, using the disappearing cross method. Water was used to change the concentration while keeping the total volume the same.
| Experiment | Volume of thiosulfate / cm3 | Volume of water / cm3 | Volume of acid / cm3 | Time for cross to disappear / s |
|---|---|---|---|---|
| 1 | 40 | 0 | 5 | 25 |
| 2 | 30 | 10 | 5 | 34 |
| 3 | 20 | 20 | 5 | 52 |
| 4 | 10 | 30 | 5 | 105 |
(a) Explain why water is added in experiments 2 to 4. [1]
(b) Describe the relationship between the volume of thiosulfate used and the time for the cross to disappear. [1]
(c) Calculate the rate (1/time) for experiment 3. [2]
(d) Explain, in terms of collisions, why a lower concentration of thiosulfate gives a longer time. [3]
(e) State two variables that must be kept constant for a fair test. [2]
(f) Name the observation used to decide that the reaction has finished. [1]
(g) The cloudiness is caused by a solid spread through the liquid. Name the solid and name this type of mixture. [2]
This is a full concentration-versus-rate investigation using the disappearing cross method, testing fair-test design, a rate calculation, and collision theory.
(a) Water is added in experiments 2 to 4 because it lowers (changes) the concentration of the thiosulfate while keeping the total volume the same, so only the concentration is being varied. [1]
(b) Reading the table, as the thiosulfate volume falls from 40 to 10 cm³ the time rises from 25 to 105 s: the smaller the volume of thiosulfate (the lower the concentration), the longer the time for the cross to disappear. [1]
(c) Rate is \(1/\text{time}\). For experiment 3 the time is 52 s:
\[ \text{rate} = \frac{1}{52\ \text{s}} = 0.019\ \text{s}^{-1} \](accept \(0.0192\ \text{s}^{-1}\)). One mark for \(1/52\), one for the value. [2]
(d) A lower concentration of thiosulfate gives a longer time because: a lower concentration has fewer thiosulfate particles in the same volume; so collisions happen less frequently; and the rate is slower, so the cross takes longer to disappear. Three marks. [3]
(e) For a fair test keep any two constant, for example the temperature, the volume (concentration) of acid, the total volume, the size of the cross, or using the same observer. [2]
(f) The reaction is judged finished when the cross can no longer be seen (the mixture has become opaque). [1]
(g) The solid causing the cloudiness is sulfur, and a solid spread through a liquid like this is a suspension (accept precipitate). One mark each. [2]
Exam takeaway: diluting with water while fixing the total volume is the standard way to change concentration alone; it is what keeps the experiment a fair test.
Answer Details
This is a full concentration-versus-rate investigation using the disappearing cross method, testing fair-test design, a rate calculation, and collision theory.
(a) Water is added in experiments 2 to 4 because it lowers (changes) the concentration of the thiosulfate while keeping the total volume the same, so only the concentration is being varied. [1]
(b) Reading the table, as the thiosulfate volume falls from 40 to 10 cm³ the time rises from 25 to 105 s: the smaller the volume of thiosulfate (the lower the concentration), the longer the time for the cross to disappear. [1]
(c) Rate is \(1/\text{time}\). For experiment 3 the time is 52 s:
\[ \text{rate} = \frac{1}{52\ \text{s}} = 0.019\ \text{s}^{-1} \](accept \(0.0192\ \text{s}^{-1}\)). One mark for \(1/52\), one for the value. [2]
(d) A lower concentration of thiosulfate gives a longer time because: a lower concentration has fewer thiosulfate particles in the same volume; so collisions happen less frequently; and the rate is slower, so the cross takes longer to disappear. Three marks. [3]
(e) For a fair test keep any two constant, for example the temperature, the volume (concentration) of acid, the total volume, the size of the cross, or using the same observer. [2]
(f) The reaction is judged finished when the cross can no longer be seen (the mixture has become opaque). [1]
(g) The solid causing the cloudiness is sulfur, and a solid spread through a liquid like this is a suspension (accept precipitate). One mark each. [2]
Exam takeaway: diluting with water while fixing the total volume is the standard way to change concentration alone; it is what keeps the experiment a fair test.
Question 6 Report
Hydrated copper(II) sulfate forms blue crystals. A student is given a warm, dilute aqueous solution of copper(II) sulfate and is asked to obtain a good sample of the crystals from it by crystallisation.
(a) Explain why the student first gently heats the solution to remove about half of the water. [2]
(b) The hot concentrated solution is then left to cool slowly. Describe what happens to the copper(II) sulfate as the solution cools. [2]
(c) Name the method used to separate the crystals that form from the liquid left in the dish. [1]
(d) Describe how the wet crystals can be dried without spoiling them. [1]
(e) Explain why the student should not heat the solution all the way to dryness when making crystals. [2]
(f) State the colour of the hydrated copper(II) sulfate crystals obtained. [1]
This question tests the crystallisation method used to obtain good crystals of a soluble salt, and the idea that solubility falls as a solution cools.
(a) Why heat off about half the water first [2]. Removing water makes the remaining solution more concentrated, closer to saturated [1], so that crystals will actually form when it is later cooled [1]. A dilute solution could be cooled a long way without depositing any solid.
(b) What happens as the solution cools [2]. The solubility of copper(II) sulfate falls as the temperature drops [1], so the excess solid can no longer stay dissolved and comes out of solution as crystals [1]. Slow cooling gives larger, well formed crystals.
(c) Separating crystals from the liquid [1]. The crystals are filtered off from the remaining solution (filtration); decanting is also acceptable [1].
(d) Drying the crystals [1]. Dab or press them gently between sheets of filter paper, or leave them in a warm place to dry [1]. Strong heating is avoided so the crystals are not damaged.
(e) Why not heat to dryness [2]. Heating all the way to dryness would drive off the water of crystallisation (or decompose the crystals) [1], and would leave only a mass of tiny powdery crystals instead of good regular ones [1]. The blue colour and shape depend on that trapped water.
(f) Colour of the crystals [1]. Hydrated copper(II) sulfate crystals are blue [1].
Exam tip: for crystallisation always state that solubility decreases on cooling, and that you stop heating at the point of saturation (a hot saturated solution), never taking it to dryness.
Answer Details
This question tests the crystallisation method used to obtain good crystals of a soluble salt, and the idea that solubility falls as a solution cools.
(a) Why heat off about half the water first [2]. Removing water makes the remaining solution more concentrated, closer to saturated [1], so that crystals will actually form when it is later cooled [1]. A dilute solution could be cooled a long way without depositing any solid.
(b) What happens as the solution cools [2]. The solubility of copper(II) sulfate falls as the temperature drops [1], so the excess solid can no longer stay dissolved and comes out of solution as crystals [1]. Slow cooling gives larger, well formed crystals.
(c) Separating crystals from the liquid [1]. The crystals are filtered off from the remaining solution (filtration); decanting is also acceptable [1].
(d) Drying the crystals [1]. Dab or press them gently between sheets of filter paper, or leave them in a warm place to dry [1]. Strong heating is avoided so the crystals are not damaged.
(e) Why not heat to dryness [2]. Heating all the way to dryness would drive off the water of crystallisation (or decompose the crystals) [1], and would leave only a mass of tiny powdery crystals instead of good regular ones [1]. The blue colour and shape depend on that trapped water.
(f) Colour of the crystals [1]. Hydrated copper(II) sulfate crystals are blue [1].
Exam tip: for crystallisation always state that solubility decreases on cooling, and that you stop heating at the point of saturation (a hot saturated solution), never taking it to dryness.
Question 7 Report
The apparatus below is used to investigate the products formed when a hydrocarbon fuel burns in air. Gases are drawn through the apparatus by a pump. A colourless liquid collects in tube A and the gas then passes through the limewater in tube B.
(a) Suggest why tube A is surrounded by ice. [1]
(b) Describe a chemical test to show that the liquid collected in tube A is water. [2]
(c) State what would be observed in tube B during the experiment. [1]
(d) Name the gas that causes the change seen in tube B. [1]
(e) State the two elements present in a hydrocarbon. [1]
(f) Write the word equation for the complete combustion of a hydrocarbon fuel. [2]
This question tests the two products of complete combustion of a hydrocarbon (water and carbon dioxide) and how you prove each one. Tube A is a cold trap that catches the water vapour; the remaining gas is bubbled through limewater to test for carbon dioxide.
(a) Tube A is surrounded by ice to cool the vapours so that the water condenses and collects as a liquid. [1] Without cooling, the water would stay as vapour and pass straight through, so none could be collected or tested.
(b) A chemical test for water, worth two marks: add the liquid to anhydrous copper(II) sulfate (or use anhydrous cobalt(II) chloride paper). [1] With water present the anhydrous copper(II) sulfate turns from white to blue (cobalt(II) chloride paper turns from blue to pink). [1] These colour changes are caused by water becoming chemically combined into the crystal structure.
(c) In tube B the limewater turns milky / cloudy. [1] This is the standard positive result for carbon dioxide (a white precipitate of calcium carbonate forms).
(d) The gas causing this change is carbon dioxide. [1]
(e) A hydrocarbon contains only two elements: carbon and hydrogen. [1] That is the definition of a hydrocarbon, and it is why complete combustion gives only CO2 (from the carbon) and H2O (from the hydrogen).
(f) The word equation for complete combustion, worth two marks:
hydrocarbon + oxygen → carbon dioxide + water [1] for the correct reactants and [1] for the correct products. Complete combustion needs a plentiful supply of oxygen; if oxygen is limited you would also get carbon monoxide and soot.
Exam tip: the water test and the carbon dioxide (limewater) test together confirm a hydrocarbon has burned completely. Learn both positive results word for word.
Answer Details
This question tests the two products of complete combustion of a hydrocarbon (water and carbon dioxide) and how you prove each one. Tube A is a cold trap that catches the water vapour; the remaining gas is bubbled through limewater to test for carbon dioxide.
(a) Tube A is surrounded by ice to cool the vapours so that the water condenses and collects as a liquid. [1] Without cooling, the water would stay as vapour and pass straight through, so none could be collected or tested.
(b) A chemical test for water, worth two marks: add the liquid to anhydrous copper(II) sulfate (or use anhydrous cobalt(II) chloride paper). [1] With water present the anhydrous copper(II) sulfate turns from white to blue (cobalt(II) chloride paper turns from blue to pink). [1] These colour changes are caused by water becoming chemically combined into the crystal structure.
(c) In tube B the limewater turns milky / cloudy. [1] This is the standard positive result for carbon dioxide (a white precipitate of calcium carbonate forms).
(d) The gas causing this change is carbon dioxide. [1]
(e) A hydrocarbon contains only two elements: carbon and hydrogen. [1] That is the definition of a hydrocarbon, and it is why complete combustion gives only CO2 (from the carbon) and H2O (from the hydrogen).
(f) The word equation for complete combustion, worth two marks:
hydrocarbon + oxygen → carbon dioxide + water [1] for the correct reactants and [1] for the correct products. Complete combustion needs a plentiful supply of oxygen; if oxygen is limited you would also get carbon monoxide and soot.
Exam tip: the water test and the carbon dioxide (limewater) test together confirm a hydrocarbon has burned completely. Learn both positive results word for word.
Question 8 Report
The table gives information about some fractions obtained from petroleum.
| fraction | boiling range / °C | number of carbon atoms per molecule | use |
|---|---|---|---|
| refinery gas | below 25 | 1 to 4 | bottled fuel gas |
| gasoline | 25 to 75 | 5 to 10 | ...................... |
| kerosene | 150 to 250 | 11 to 15 | aircraft fuel |
| diesel oil | 220 to 350 | 16 to 20 | fuel for lorries |
| bitumen | above 350 | more than 50 | road surfacing |
(a) Use the table. State which fraction has the lowest boiling range. [1]
(b) Complete the table by giving one use of the gasoline fraction. [1]
(c) Describe how the number of carbon atoms per molecule changes as the boiling range increases. [1]
(d) Name the raw material from which all these fractions are obtained. [1]
(e) Explain why petroleum is described as a mixture. [2]
(f) State what is meant by the term hydrocarbon. [1]
(g) Petroleum is a finite resource. State what is meant by finite. [2]
This question is about reading a data table on petroleum fractions and the key definitions linked to crude oil. As the chain length rises, both boiling range and molecule size increase together, which is the pattern the table is designed to show.
(a) The fraction with the lowest boiling range is refinery gas (below 25°C). [1] It has the shortest molecules (1 to 4 carbon atoms), so it boils at the lowest temperature.
(b) A use of the gasoline fraction is as fuel for cars (petrol for vehicles). [1]
(c) As the boiling range increases, the number of carbon atoms per molecule increases. [1] Longer chains have stronger intermolecular forces, so more energy is needed to boil them, giving a higher boiling range.
(d) All the fractions come from petroleum (crude oil). [1]
(e) Petroleum is a mixture because it contains many different substances (different hydrocarbons) [1] that are not chemically combined, so they can be separated by a physical method such as fractional distillation. [1] In a compound the elements would be chemically bonded in fixed proportions and could not be separated this way.
(f) A hydrocarbon is a compound containing hydrogen and carbon only. [1] The word "only" is essential for the mark.
(g) "Finite" means the resource is being used up / will eventually run out [1] and cannot be replaced quickly (it takes millions of years to form, so it is non-renewable). [1]
Exam tip: for "mixture" answers you must say the components are not chemically combined; simply saying "it has many substances" is not enough for full marks.
Answer Details
This question is about reading a data table on petroleum fractions and the key definitions linked to crude oil. As the chain length rises, both boiling range and molecule size increase together, which is the pattern the table is designed to show.
(a) The fraction with the lowest boiling range is refinery gas (below 25°C). [1] It has the shortest molecules (1 to 4 carbon atoms), so it boils at the lowest temperature.
(b) A use of the gasoline fraction is as fuel for cars (petrol for vehicles). [1]
(c) As the boiling range increases, the number of carbon atoms per molecule increases. [1] Longer chains have stronger intermolecular forces, so more energy is needed to boil them, giving a higher boiling range.
(d) All the fractions come from petroleum (crude oil). [1]
(e) Petroleum is a mixture because it contains many different substances (different hydrocarbons) [1] that are not chemically combined, so they can be separated by a physical method such as fractional distillation. [1] In a compound the elements would be chemically bonded in fixed proportions and could not be separated this way.
(f) A hydrocarbon is a compound containing hydrogen and carbon only. [1] The word "only" is essential for the mark.
(g) "Finite" means the resource is being used up / will eventually run out [1] and cannot be replaced quickly (it takes millions of years to form, so it is non-renewable). [1]
Exam tip: for "mixture" answers you must say the components are not chemically combined; simply saying "it has many substances" is not enough for full marks.
