Chemistry (9-1) 0971 | Paper 4 Mock 01 | Theory (Extended)
Question 1 Report
The table gives the melting points and boiling points of four substances, P, Q, R and S, together with the way each substance melts.
| Substance | Melting point / °C | Boiling point / °C | Melting behaviour |
|---|---|---|---|
| P | 0 | 100 | melts sharply at one temperature |
| Q | -7 to 3 | 95 to 103 | melts over a range of temperatures |
| R | 801 | 1413 | melts sharply at one temperature |
| S | -39 | 357 | melts sharply at one temperature |
(a) State what is meant by a pure substance. [1]
(b) Using the data, deduce which substance is impure (a mixture) and explain how the data shows this. [2]
(c) Determine the physical state of substance R and of substance S at 25 °C. [2]
(d) Substance R and substance S are both elements. Explain why the melting point data alone cannot tell you whether a pure substance is an element or a compound. [2]
(e) Describe how the melting point of pure ice (substance P) changes when salt is dissolved in it, and state how the melting behaviour changes. [2]
(f) State how a chemist could use a melting point measurement to check whether a solid is pure. [1]
(g) Deduce which substance has the highest boiling point and state its physical state at 1000 °C. [2]
This question tests how melting and boiling data distinguish a pure substance from a mixture, and how to read physical state from the data.
(a) A pure substance is a single substance with nothing else mixed in (made of only one type of particle, with a fixed composition) [1].
(b) The impure substance is Q [1]. A pure substance melts sharply at one fixed temperature, but Q melts over a range (-7 to 3 °C), which shows it is impure, a mixture [1].
(c) Compare each fixed point with 25 °C:
- R melts at 801 °C, which is above 25 °C, so at 25 °C R is a solid [1];
- S melts at -39 °C (below 25 °C) and boils at 357 °C (above 25 °C), so at 25 °C S is a liquid [1].
(d) Both elements and pure compounds melt sharply at a single fixed temperature [1]. A sharp melting point therefore only tells you the substance is pure; it does not reveal whether the substance is one element or two or more elements joined as a compound [1].
(e) Dissolving salt in ice lowers the melting point below 0 °C [1], and the mixture then melts over a range of temperatures instead of sharply [1]. This is exactly why salt is spread on icy roads.
(f) Measure the melting point: a pure solid melts sharply at its known melting point, so if the sample melts over a range or at a different temperature it is impure [1].
(g) The highest boiling point belongs to R (1413 °C) [1]. At 1000 °C, R is above its melting point (801 °C) but below its boiling point (1413 °C), so R is a liquid [1].
Answer Details
This question tests how melting and boiling data distinguish a pure substance from a mixture, and how to read physical state from the data.
(a) A pure substance is a single substance with nothing else mixed in (made of only one type of particle, with a fixed composition) [1].
(b) The impure substance is Q [1]. A pure substance melts sharply at one fixed temperature, but Q melts over a range (-7 to 3 °C), which shows it is impure, a mixture [1].
(c) Compare each fixed point with 25 °C:
- R melts at 801 °C, which is above 25 °C, so at 25 °C R is a solid [1];
- S melts at -39 °C (below 25 °C) and boils at 357 °C (above 25 °C), so at 25 °C S is a liquid [1].
(d) Both elements and pure compounds melt sharply at a single fixed temperature [1]. A sharp melting point therefore only tells you the substance is pure; it does not reveal whether the substance is one element or two or more elements joined as a compound [1].
(e) Dissolving salt in ice lowers the melting point below 0 °C [1], and the mixture then melts over a range of temperatures instead of sharply [1]. This is exactly why salt is spread on icy roads.
(f) Measure the melting point: a pure solid melts sharply at its known melting point, so if the sample melts over a range or at a different temperature it is impure [1].
(g) The highest boiling point belongs to R (1413 °C) [1]. At 1000 °C, R is above its melting point (801 °C) but below its boiling point (1413 °C), so R is a liquid [1].
Question 2 Report
Long-chain alkanes from crude oil can be broken down in the laboratory. Fig. 6.1 shows apparatus a student uses to do this.
Fig. 6.1
(a) Name the process shown in Fig. 6.1 and state what happens to the large hydrocarbon molecules during it. [2]
(b) State two conditions needed for this process to take place. [2]
(c) State the role of the broken pottery in the boiling tube. [1]
(d) One large alkane breaks down as shown. Complete the equation by giving the formula of the missing product:
C10H22 → C8H18 + ......... [2]
(e) Explain why this process is carried out on a large scale in the petrochemical industry. [2]
(f) The gas collected includes an alkene. Describe a test, with its result, that would show an alkene is present. [2]
What this tests: cracking long-chain alkanes, the conditions used, and testing for an alkene.
(a) [2] The process is cracking [1]. During it the large, long-chain hydrocarbon molecules are broken down into smaller molecules [1].
(b) Two conditions [2] Any two of: a high temperature (strong heat); a catalyst (broken pottery, aluminium oxide or silica); the vapour passed over the hot catalyst [1 + 1].
(c) Role of the broken pottery [1] It acts as a catalyst, speeding up the breakdown so it happens at a lower temperature; it is not used up.
(d) Missing product [2] Balance each element in \(C_{10}H_{22} \rightarrow C_8H_{18} + ?\): carbon \(10 - 8 = 2\), hydrogen \(22 - 18 = 4\), so the missing product is C2H4 (ethene).
\[ C_{10}H_{22} \rightarrow C_8H_{18} + C_2H_4 \]1 mark for the carbon balance (C2), 1 mark for the hydrogen balance (H4).
(e) Why cracking matters [2] It converts less useful long-chain alkanes into smaller molecules in high demand as fuels such as petrol [1]; and it makes alkenes such as ethene that are needed to manufacture plastics and other chemicals [1].
(f) Test for an alkene [2] Add bromine water and shake [1]; with an alkene the orange bromine water is decolourised (turns colourless) [1]. An alkane would leave it orange.
Answer Details
What this tests: cracking long-chain alkanes, the conditions used, and testing for an alkene.
(a) [2] The process is cracking [1]. During it the large, long-chain hydrocarbon molecules are broken down into smaller molecules [1].
(b) Two conditions [2] Any two of: a high temperature (strong heat); a catalyst (broken pottery, aluminium oxide or silica); the vapour passed over the hot catalyst [1 + 1].
(c) Role of the broken pottery [1] It acts as a catalyst, speeding up the breakdown so it happens at a lower temperature; it is not used up.
(d) Missing product [2] Balance each element in \(C_{10}H_{22} \rightarrow C_8H_{18} + ?\): carbon \(10 - 8 = 2\), hydrogen \(22 - 18 = 4\), so the missing product is C2H4 (ethene).
\[ C_{10}H_{22} \rightarrow C_8H_{18} + C_2H_4 \]1 mark for the carbon balance (C2), 1 mark for the hydrogen balance (H4).
(e) Why cracking matters [2] It converts less useful long-chain alkanes into smaller molecules in high demand as fuels such as petrol [1]; and it makes alkenes such as ethene that are needed to manufacture plastics and other chemicals [1].
(f) Test for an alkene [2] Add bromine water and shake [1]; with an alkene the orange bromine water is decolourised (turns colourless) [1]. An alkane would leave it orange.
Question 3 Report
A small brewery ferments a solution that contains 360 g of glucose, C6H12O6. The equation for the fermentation is:
C6H12O6 → 2C2H5OH + 2CO2
Use the relative atomic masses Ar: H = 1, C = 12, O = 16.
(a) Calculate the relative formula mass (Mr) of glucose. [1]
(b) Calculate the number of moles of glucose that are fermented. [1]
(c) Calculate the maximum mass of ethanol that could be produced. The Mr of ethanol is 46. [3]
(d) The brewery actually collected 147.2 g of ethanol. Calculate the percentage yield of ethanol. [2]
(e) Calculate the volume of carbon dioxide produced, measured at room temperature and pressure, assuming the glucose reacts completely. The molar gas volume at r.t.p. is 24 dm3/mol. [2]
(f) Suggest one reason why the actual yield of ethanol is less than the maximum. [1]
What this tests: the mole map (mass to moles, mole ratio, then back to mass or gas volume) and percentage yield for fermentation of glucose.
(a) The relative formula mass is the sum of the relative atomic masses of every atom in C6H12O6: \(M_r = (6\times12)+(12\times1)+(6\times16) = 72+12+96 = 180\). [1]
(b) Moles = mass divided by Mr, so \(n(\text{glucose}) = \dfrac{360}{180} = 2\ \text{mol}\). [1] Convert mass to moles first, because the balancing numbers in the equation compare moles, not grams.
(c) The equation shows 1 mol of glucose gives 2 mol of ethanol, so \(n(\text{ethanol}) = 2\times2 = 4\ \text{mol}\) [1]. Mass = moles times Mr = \(4\times46\) [1] = 184 g [1]. This is the maximum (theoretical) mass, assuming every glucose molecule reacts exactly as the equation predicts.
(d) Percentage yield compares the mass actually made with that maximum: \(\text{yield} = \dfrac{147.2}{184}\times100\) [1] = 80% [1].
(e) From the equation 1 mol of glucose also gives 2 mol of CO2, so \(n(CO_2)=2\times2 = 4\ \text{mol}\) [1]. At r.t.p. each mole of any gas occupies 24 dm3, so volume = \(4\times24 = 96\ \text{dm}^3\) [1].
(f) Any one reason the real yield falls below 184 g: some glucose remained unreacted, some ethanol evaporated or was lost during handling, side reactions consumed some glucose, or the yeast was killed (for example by the rising ethanol concentration) before all the glucose had fermented. [1]
Exam tip: keep the chain mass to moles to mole ratio to answer, and never apply a mole ratio directly to masses.
Answer Details
What this tests: the mole map (mass to moles, mole ratio, then back to mass or gas volume) and percentage yield for fermentation of glucose.
(a) The relative formula mass is the sum of the relative atomic masses of every atom in C6H12O6: \(M_r = (6\times12)+(12\times1)+(6\times16) = 72+12+96 = 180\). [1]
(b) Moles = mass divided by Mr, so \(n(\text{glucose}) = \dfrac{360}{180} = 2\ \text{mol}\). [1] Convert mass to moles first, because the balancing numbers in the equation compare moles, not grams.
(c) The equation shows 1 mol of glucose gives 2 mol of ethanol, so \(n(\text{ethanol}) = 2\times2 = 4\ \text{mol}\) [1]. Mass = moles times Mr = \(4\times46\) [1] = 184 g [1]. This is the maximum (theoretical) mass, assuming every glucose molecule reacts exactly as the equation predicts.
(d) Percentage yield compares the mass actually made with that maximum: \(\text{yield} = \dfrac{147.2}{184}\times100\) [1] = 80% [1].
(e) From the equation 1 mol of glucose also gives 2 mol of CO2, so \(n(CO_2)=2\times2 = 4\ \text{mol}\) [1]. At r.t.p. each mole of any gas occupies 24 dm3, so volume = \(4\times24 = 96\ \text{dm}^3\) [1].
(f) Any one reason the real yield falls below 184 g: some glucose remained unreacted, some ethanol evaporated or was lost during handling, side reactions consumed some glucose, or the yeast was killed (for example by the rising ethanol concentration) before all the glucose had fermented. [1]
Exam tip: keep the chain mass to moles to mole ratio to answer, and never apply a mole ratio directly to masses.
Question 4 Report
Fig. 5.1 shows the electron arrangements of atoms of two different elements, L and M. The letters are not the chemical symbols of the elements.
(a) State the feature of the electron arrangements that L and M have in common, and state which part of the periodic table (group or period) is fixed by this shared feature. [2]
(b) Use the diagrams to state which of L and M is in the earlier (lower-numbered) period, and explain how the diagrams show this. [2]
(c) Explain, in terms of electronic structure, why M is more reactive than L. [3]
(d) State the charge of the ions formed by L and M. [1]
(e) Write a balanced symbol equation for the reaction of M with water. Use the symbol M in your equation and assume M behaves like a typical Group I metal. [2]
(f) State the trend in melting point on going from L to M and predict how the melting point of the next element in the same group would compare with that of M. [2]
What this tests: using shell diagrams to place Group I metals and explaining the reactivity trend down a group.
(a) Shared feature [2] Both L and M have one electron in their outer shell [1]; the number of outer electrons fixes the group, so both are in Group I [1].
(b) Earlier period [2] L is in the earlier (lower-numbered) period [1]; L has 2 occupied electron shells while M has 3, and the period number equals the number of occupied shells [1].
(c) Why M is more reactive [3] Reactivity increases down Group I. M has more electron shells, so its single outer electron is further from the nucleus [1]; there is also more shielding by the inner shells [1]; so the nucleus holds the outer electron less strongly and it is lost more easily, making M more reactive than L [1].
(d) Ion charges [1] Both form ions with a charge of 1+, since each loses its single outer electron.
(e) Reaction of M with water [2]
\[ 2M + 2H_2O \rightarrow 2MOH + H_2 \]1 mark for the correct products (the hydroxide MOH and hydrogen gas), 1 mark for balancing.
(f) Melting point trend [2] The melting point decreases from L to M [1]; the next element down the same group would have an even lower melting point than M [1].
Answer Details
What this tests: using shell diagrams to place Group I metals and explaining the reactivity trend down a group.
(a) Shared feature [2] Both L and M have one electron in their outer shell [1]; the number of outer electrons fixes the group, so both are in Group I [1].
(b) Earlier period [2] L is in the earlier (lower-numbered) period [1]; L has 2 occupied electron shells while M has 3, and the period number equals the number of occupied shells [1].
(c) Why M is more reactive [3] Reactivity increases down Group I. M has more electron shells, so its single outer electron is further from the nucleus [1]; there is also more shielding by the inner shells [1]; so the nucleus holds the outer electron less strongly and it is lost more easily, making M more reactive than L [1].
(d) Ion charges [1] Both form ions with a charge of 1+, since each loses its single outer electron.
(e) Reaction of M with water [2]
\[ 2M + 2H_2O \rightarrow 2MOH + H_2 \]1 mark for the correct products (the hydroxide MOH and hydrogen gas), 1 mark for balancing.
(f) Melting point trend [2] The melting point decreases from L to M [1]; the next element down the same group would have an even lower melting point than M [1].
Question 5 Report
Ethene reacts readily with bromine at room temperature. Fig. 4.1 shows the displayed formula of ethene, the molecule that reacts.
Fig. 4.1
(a) Bromine can be added as bromine water. Describe what you would see when ethene is bubbled through bromine water. [2]
(b) State the type of reaction that occurs between ethene and bromine. [1]
(c) Name the product formed when ethene reacts with bromine. [1]
(d) Give the molecular formula of this product. [1]
(e) Draw the displayed formula of the product, showing every atom and every bond. [2]
(f) State whether the product is saturated or unsaturated, and explain your answer. [2]
(g) State what the fast reaction of ethene with bromine tells you about the ethene molecule. [1]
This question is about the addition reaction between ethene and bromine, testing the observation, reaction type, product, its formula and displayed structure, and what the reaction reveals about ethene.
(a) The orange (yellow-brown) colour of the bromine water [1] disappears; the mixture turns colourless, that is the bromine water is decolourised [1]. This colour change is the standard test for an unsaturated compound.
(b) The reaction is an addition reaction [1]: the bromine adds across the double bond and nothing is given off.
(c) The product is 1,2-dibromoethane (dibromoethane) [1].
(d) Its molecular formula is \( C_2H_4Br_2 \) [1]: the two bromine atoms simply add to the ethene, \( C_2H_4 \).
(e) The displayed formula shows the C=C double bond opened to a single bond, with one bromine atom added to each carbon and every bond drawn in full:
One mark for the correct single C-C bond with a bromine on each carbon, one mark for the correct four hydrogen atoms with all bonds shown [2].
(f) The product is saturated [1]: the C=C double bond has opened, leaving only single bonds, so no more atoms can add on [1].
(g) The fast reaction tells you that ethene is unsaturated, that is it contains a reactive carbon-carbon double bond that readily takes part in addition [1].
Answer Details
This question is about the addition reaction between ethene and bromine, testing the observation, reaction type, product, its formula and displayed structure, and what the reaction reveals about ethene.
(a) The orange (yellow-brown) colour of the bromine water [1] disappears; the mixture turns colourless, that is the bromine water is decolourised [1]. This colour change is the standard test for an unsaturated compound.
(b) The reaction is an addition reaction [1]: the bromine adds across the double bond and nothing is given off.
(c) The product is 1,2-dibromoethane (dibromoethane) [1].
(d) Its molecular formula is \( C_2H_4Br_2 \) [1]: the two bromine atoms simply add to the ethene, \( C_2H_4 \).
(e) The displayed formula shows the C=C double bond opened to a single bond, with one bromine atom added to each carbon and every bond drawn in full:
One mark for the correct single C-C bond with a bromine on each carbon, one mark for the correct four hydrogen atoms with all bonds shown [2].
(f) The product is saturated [1]: the C=C double bond has opened, leaving only single bonds, so no more atoms can add on [1].
(g) The fast reaction tells you that ethene is unsaturated, that is it contains a reactive carbon-carbon double bond that readily takes part in addition [1].
Question 6 Report
This question is about copper and about the arrangement of elements in the Periodic Table.
Table 18.1 gives information about the two isotopes of copper.
| Isotope of copper | Nucleon number | Number of protons | Number of neutrons | Abundance / % |
|---|---|---|---|---|
| copper-63 | 63 | 29 | … | 70 |
| copper-65 | 65 | 29 | … | 30 |
(a) Complete the shaded 'Number of neutrons' column of Table 18.1. [2]
(b) Explain, in terms of subatomic particles, why the two isotopes are both atoms of copper yet have different masses. [2]
(c) Calculate the relative atomic mass of copper from the abundance data. Show all your working. [3]
(d) Copper is a transition element. State two properties of copper, other than its density, that are typical of a transition metal but not of a Group I metal. [2]
(e) Fig. 18.1 shows the electron arrangement of a different element, Z, in the same period as copper.
(i) Deduce the proton number, group and period of element Z. [3]
(ii) Element Z reacts vigorously with cold water. Write a balanced symbol equation for the reaction, giving the hydroxide of Z (formula ZOH) and a gas. [2]
(iii) Explain why element Z is more reactive than the element directly above it in the same group. [2]
(f) Copper forms ions Cu+ and Cu2+. State the term for this behaviour and give the formulae of the two oxides these ions form with O2−. [3]
(g) State, giving a reason, whether copper or element Z is the better conductor of heat, or whether they would be similar. [1]
(h) Both copper and element Z are metals. State one feature of their structure that they share. [1]
(a) Neutrons = nucleon number \(-\) protons. Copper-63: \(63 - 29 = 34\) neutrons [1]; copper-65: \(65 - 29 = 36\) neutrons [1].
(b) Both isotopes have the same number of protons, 29, so both are copper [1]; they differ in neutrons (34 and 36), giving different nucleon numbers and therefore different masses [1]. Chemical identity is fixed by proton number, not by mass.
(c) Relative atomic mass is the weighted mean [method 1]:
\[ A_r = \frac{(63\times 70) + (65\times 30)}{100} = \frac{4410 + 1950}{100} = \frac{6360}{100} \quad [1] \] \[ = 63.6 \quad [1] \](d) Any two transition-metal properties, other than density: forms coloured compounds; shows variable valency (\(\text{Cu}^{+}\) and \(\text{Cu}^{2+}\)); acts as a catalyst; has a high melting point [2]. A Group I metal shows none of these (white compounds, a single 1+ charge, low melting point).
(e) Element Z has \(2 + 8 + 8 + 1 = 19\) electrons.
(i) Proton number 19 [1]; Group I [1] (1 outer electron); Period 4 [1] (4 occupied shells).
(ii) A Group I metal reacts with cold water to give its hydroxide and hydrogen gas [products 1; balanced 1]:
\[ 2\text{Z} + 2\text{H}_2\text{O} \rightarrow 2\text{ZOH} + \text{H}_2 \](iii) Z is more reactive than the element directly above it because it has one more electron shell, so its outer electron is further from the nucleus and more shielded [1]; the weaker nuclear pull means that outer electron is lost more easily, and losing that electron is how a Group I metal reacts [1].
(f) Forming both \(\text{Cu}^{+}\) and \(\text{Cu}^{2+}\) is variable valency (variable oxidation state) [1]. With the oxide ion \(\text{O}^{2-}\) the charges balance to give \(\text{Cu}_2\text{O}\) [1] (two 1+ ions per oxide) and \(\text{CuO}\) [1] (one 2+ ion per oxide).
(g) They would be similar; both are good conductors of heat [1], because both are metals with delocalised electrons free to move and transfer thermal energy.
(h) They share a giant metallic lattice of positive ions in a sea of delocalised electrons [1].
Answer Details
(a) Neutrons = nucleon number \(-\) protons. Copper-63: \(63 - 29 = 34\) neutrons [1]; copper-65: \(65 - 29 = 36\) neutrons [1].
(b) Both isotopes have the same number of protons, 29, so both are copper [1]; they differ in neutrons (34 and 36), giving different nucleon numbers and therefore different masses [1]. Chemical identity is fixed by proton number, not by mass.
(c) Relative atomic mass is the weighted mean [method 1]:
\[ A_r = \frac{(63\times 70) + (65\times 30)}{100} = \frac{4410 + 1950}{100} = \frac{6360}{100} \quad [1] \] \[ = 63.6 \quad [1] \](d) Any two transition-metal properties, other than density: forms coloured compounds; shows variable valency (\(\text{Cu}^{+}\) and \(\text{Cu}^{2+}\)); acts as a catalyst; has a high melting point [2]. A Group I metal shows none of these (white compounds, a single 1+ charge, low melting point).
(e) Element Z has \(2 + 8 + 8 + 1 = 19\) electrons.
(i) Proton number 19 [1]; Group I [1] (1 outer electron); Period 4 [1] (4 occupied shells).
(ii) A Group I metal reacts with cold water to give its hydroxide and hydrogen gas [products 1; balanced 1]:
\[ 2\text{Z} + 2\text{H}_2\text{O} \rightarrow 2\text{ZOH} + \text{H}_2 \](iii) Z is more reactive than the element directly above it because it has one more electron shell, so its outer electron is further from the nucleus and more shielded [1]; the weaker nuclear pull means that outer electron is lost more easily, and losing that electron is how a Group I metal reacts [1].
(f) Forming both \(\text{Cu}^{+}\) and \(\text{Cu}^{2+}\) is variable valency (variable oxidation state) [1]. With the oxide ion \(\text{O}^{2-}\) the charges balance to give \(\text{Cu}_2\text{O}\) [1] (two 1+ ions per oxide) and \(\text{CuO}\) [1] (one 2+ ion per oxide).
(g) They would be similar; both are good conductors of heat [1], because both are metals with delocalised electrons free to move and transfer thermal energy.
(h) They share a giant metallic lattice of positive ions in a sea of delocalised electrons [1].
