1.4 Exam Practice
Exam-style practice questions on Counting Particles (The Mole)
Section B: Data Analysis (Paper 1B Style)
Calculator and Data Booklet permitted. Show all working clearly.
Question 1: Empirical Formula from Combustion Data Determine
5 marksComplete combustion of 2.50 g of an organic compound X produced 3.67 g of CO\(_2\) and 1.50 g of H\(_2\)O. The compound contains only carbon, hydrogen and oxygen.
(a) Calculate the mass of carbon and hydrogen in the sample. [2]
(b) Determine the mass of oxygen in the compound and hence the empirical formula of X. [3]
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(a) Mass of C: \(\frac{12.01}{44.01} \times 3.67 = 1.001\) g [1]
Mass of H: \(\frac{2 \times 1.008}{18.02} \times 1.50 = 0.168\) g [1]
(b) Mass of O: \(2.50 - 1.001 - 0.168 = 1.331\) g [1]
Moles: C = \(\frac{1.001}{12.01}\) = 0.0834, H = \(\frac{0.168}{1.008}\) = 0.167, O = \(\frac{1.331}{16.00}\) = 0.0832 [1]
Ratio C : H : O = 1 : 2 : 1, so empirical formula = CH\(_2\)O [1]
Section C: Structured Questions (Paper 2 Style)
Show all working. State answers with appropriate significant figures and units.
Question 2: Solution Concentration and Dilution Calculate
4 marksA student prepares a stock solution by dissolving 5.85 g of NaCl (M = 58.44 g mol\(^{-1}\)) in water to make 250 cm\(^3\) of solution.
(a) Calculate the concentration of the stock solution in mol dm\(^{-3}\). [2]
(b) The student takes 25.0 cm\(^3\) of the stock solution and dilutes it to 100 cm\(^3\). Calculate the concentration of the diluted solution. [2]
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(a) Moles of NaCl = \(\frac{5.85}{58.44}\) = 0.1001 mol [1]
Concentration = \(\frac{0.1001}{0.250}\) = 0.400 mol dm\(^{-3}\) [1]
(b) Using \(c_1V_1 = c_2V_2\): \(0.400 \times 25.0 = c_2 \times 100\) [1]
\(c_2 =\) 0.100 mol dm\(^{-3}\) [1]
Question 3: Percentage Yield and Atom Economy Calculate
5 marksCalcium carbonate decomposes on heating: CaCO\(_3\)(s) → CaO(s) + CO\(_2\)(g)
A student heated 10.0 g of CaCO\(_3\) and obtained 4.82 g of CaO.
(a) Calculate the theoretical yield of CaO from 10.0 g of CaCO\(_3\). [2]
(b) Calculate the percentage yield. [1]
(c) Calculate the atom economy for the production of CaO in this reaction. [1]
(d) Give one reason why the actual yield is less than the theoretical yield. [1]
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(a) Moles of CaCO\(_3\) = \(\frac{10.0}{100.09}\) = 0.09991 mol [1]
1:1 ratio, so theoretical mass of CaO = 0.09991 × 56.08 = 5.604 g [1]
(b) Percentage yield = \(\frac{4.82}{5.604} \times 100\) = 86.0% [1]
(c) Atom economy = \(\frac{56.08}{100.09} \times 100\) = 56.0% [1]
(d) Any one of: incomplete decomposition / some CaCO\(_3\) did not fully decompose / transfer losses / some CaO absorbed moisture from the air [1]
Question 4: Titration Calculation Determine
4 marks25.0 cm\(^3\) of a solution of sodium hydroxide of unknown concentration was titrated against 0.150 mol dm\(^{-3}\) sulfuric acid. The mean titre was 20.0 cm\(^3\).
The equation for the reaction is: 2NaOH(aq) + H\(_2\)SO\(_4\)(aq) → Na\(_2\)SO\(_4\)(aq) + 2H\(_2\)O(l)
(a) Calculate the number of moles of H\(_2\)SO\(_4\) used. [1]
(b) Determine the number of moles of NaOH in the 25.0 cm\(^3\) sample. [1]
(c) Calculate the concentration of the NaOH solution. [1]
(d) State the colour change observed if methyl orange is used as the indicator. [1]
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(a) \(n = 0.150 \times 0.0200 =\) 3.00 × 10\(^{-3}\) mol [1]
(b) Mole ratio 2:1, so moles NaOH = 2 × 3.00 × 10\(^{-3}\) = 6.00 × 10\(^{-3}\) mol [1]
(c) \(c = \frac{6.00 \times 10^{-3}}{0.0250} =\) 0.240 mol dm\(^{-3}\) [1]
(d) Yellow to orange / red (at the end point, in the flask) [1]