(a) A student adds excess large pieces of magnesium carbonate, MgCO3, to dilute hydrochloric acid, HCl, and measures the volume of carbon dioxide gas, CO2, given off.
Add the missing state symbols to the chemical equation for the reaction.
MgCO3 ….. + 2HCl….. → MgCl2(aq) + H2O ….. + CO2 …..
(b) Complete the dot-and-cross diagram to show the electron arrangement of the ions in magnesium chloride.
The inner shells have been drawn.
Give the charges on the ions.
(c) Complete the dot-and-cross diagram to show the electron arrangement in a molecule of carbon dioxide.
Show outer shell electrons only.
(d) The graph shows how the volume of carbon dioxide gas changes with time.
(i) Describe how the graph shows that the rate of this reaction decreases as time increases.
(ii) Explain, in terms of particles, why the rate of this reaction decreases as time increases.
(iii) The student repeats the experiment using powdered MgCO3 instead of large pieces.
All other conditions stay the same.
On the grid, draw the line expected when powdered MgCO3 is used instead of large pieces.
(e) Determine the volume of CO2 gas given off when excess MgCO3 is added to 25.0cm3 of 0.400mol/dm3 HCl at room temperature and pressure.
MgCO3 + 2HCl → MgCl2 + H2O + CO2
Use the following steps.
- Calculate the number of moles of HCl in 25.0cm3 of 0.400mol/dm3 of acid.
- Determine the number of moles of CO2 gas given off.
- Calculate the volume of CO2 gas given off in cm3.
▶️ Answer/Explanation
(a) Ans: MgCO3(s) + 2HCl(aq) → MgCl2(aq) + H2O(l) + CO2(g)
Magnesium carbonate is a solid (s), hydrochloric acid is aqueous (aq), water is liquid (l), and carbon dioxide is a gas (g).
(b) Ans: Eight crosses in the second shell of Mg, 7 dots and 1 cross in the third shell of both Cl, with 2+ charge on Mg and – charge on both Cl ions.
Magnesium loses 2 electrons to form Mg2+, and each chlorine gains 1 electron to form Cl–.
(c) Ans: C atom double bonded to 2 O atoms, with 4 non-bonding electrons on each O and no non-bonding electrons on C.
Carbon dioxide has two double bonds, with each oxygen completing its octet.
(d)(i) Ans: The gradient of the line decreases over time.
The slope becomes less steep, indicating a slower rate of reaction as time progresses.
(d)(ii) Ans: The concentration of acid particles decreases, leading to fewer successful collisions.
As reactants are used up, collisions between particles become less frequent, reducing the reaction rate.
(d)(iii) Ans: A steeper line starting at the origin and leveling off earlier, but at the same final volume.
Powdered MgCO3 increases the surface area, speeding up the reaction initially.
(e) Ans: 120 cm3
Moles of HCl = 0.400 × 0.025 = 0.01 mol. Moles of CO2 = 0.01 / 2 = 0.005 mol. Volume of CO2 = 0.005 × 24000 = 120 cm3.
Many organic compounds contain carbon, hydrogen and oxygen only.
(a) An organic compound V has the following composition by mass.
C, 48.65%; H, 8.11%; O, 43.24%
Calculate the empirical formula of compound V.
(b) Compound W has the empirical formula CH4O and a relative molecular mass of 32.
Calculate the molecular formula of compound W.
(c) Compounds X and Y have the same general formula.
X and Y are both carboxylic acids.
Compound X has the molecular formula \(C_2H_4O_2\).
Compound Y has the molecular formula \(C_4H_8O_2\).
(i) Deduce the general formula of compounds X and Y.
(ii) Draw the structure of compound Y. Show all of the atoms and all of the bonds.
Name compound Y.
(iii) Give the name used to describe a ‘family’ of similar compounds with the same general
formula, similar chemical properties and the same functional group.
(d) Propene is an unsaturated hydrocarbon. The formula of propene is shown.
\(CH_3CH=CH_2\)
(i) State the colour change observed when propene is added to aqueous bromine.
(ii) Propene can be produced by cracking long chain alkanes.
Pentadecane, \(C_{15}H_{32}\), is cracked to produce an alkane and propene in a 1:2 molar ratio.
Complete the chemical equation for this reaction.
\(C_{15}H_{32}\) → ………………………………… + …………………………………
(iii) Propene can be converted into poly(propene).
Name the type of polymerisation that occurs when propene is converted into poly(propene).
(iv) Complete the diagram to show a section of poly(propene).
▶️ Answer/Explanation
(a) Ans: \(C_3H_6O_2\)
Convert mass percentages to moles: C = 48.65/12 ≈ 4.05, H = 8.11/1 ≈ 8.11, O = 43.24/16 ≈ 2.70.
Divide by smallest (2.70): C ≈ 1.5, H ≈ 3, O ≈ 1.
Multiply by 2 to get whole numbers: C = 3, H = 6, O = 2.
Thus, empirical formula is \(C_3H_6O_2\).
(b) Ans: \(CH_4O\)
Empirical formula mass = 12 + 4 + 16 = 32.
Since relative molecular mass is 32, molecular formula = empirical formula = \(CH_4O\).
(c)
(i) General formula: \(C_nH_{2n}O_2\) or \(C_nH_{2n+1}COOH\).
(ii) Name: butanoic acid. Structure: 
(iii) Homologous series.
(d)
(i) Brown to colourless.
(ii) \(C_{15}H_{32} → C_9H_{20} + 2C_3H_6\).
(iii) Addition polymerisation.
(iv) Structure: 