JPJC 2026 Mole & Stoichiometry Tutorial Solution

2026 JPJC JC1 H2 Chemistry (9476) 1 Topic 2: The Mole Concept and Stoichiometry Jurong Pioneer Junior College 2026 JC1 H2 Chemistry (9476) Tutorial 2: The Mole Concept and Stoichiometry Level 1 – Self–Check Questions 1 (a) Define the term relative atomic mass. Relative atomic mass (Ar) of an element is defined as the ratio of the weighted average mass of one atom of the element to 1 12 the mass of an atom of 12C. *(b) [2017 P1 Q1] & [2023 P1 Q8] A sample of silicon contains three naturally occurring isotopes, 28Si, 29Si and 30Si. The sample is made up of 92.23% 28Si and the relative atomic mass of silicon in this sample is 28.10. Calculate the percentage of the isotope 29Si in the sample. Let the % of 29Si in the sample be x% and hence, the % of 30Si in the sample would be (7.77–x)%. Ar =  (relative abundance  isotopic mass)                  92.23 7.77-28.10= 28 + 29 + 30100 100 100 xx  % of 29Si in the sample, x = 5.54% (3 sf + units) 2 Calculate the amount (in mol) of the stated particles in each of the following cases. (a) F2 molecules in 10.0 g of fluorine, F2 n(F2) = 10.0 38.0 = 0.263 mol (3 sf + units) (b) CaCO3 units in 250 g of calcium carbonate, CaCO3 n(CaCO3) = 250 100.1 = 2.50 mol (3 sf + units) (c) K+ ions in 9.03  1023 K2SO4 units n(K2SO4) =   23 23 9.03 10 6.02 10 = 1.50 mol K2SO4 : K+ 1 : 2 n(K+) = 2  1.50 = 3.00 mol (3 sf + units) (d) H atoms in 370 cm3 of ammonia, NH3, gas at standard temperature and pressure. n(NH3) at s.t.p. = 370 22700 = 0.0163 mol NH3 : H 1 : 3 n(H) = 3  0.0163 = 0.0489 mol (3 sf + units) [0.263 mol; 2.50 mol ; 3.00 mol; 0.0489 mol] N n L m n M V n Vm r mn= M or M

2026 JPJC JC1 H2 Chemistry (9476) 2 Topic 2: The Mole Concept and Stoichiometry 3 Calculate the number of the stated particles in each of the following cases. (a) O atoms in 320 g of oxygen, O2, gas. {Molar mass, M, of O2 = 32.0 g mol-1  1 mol of O2 contains 32.0 g of O2.} n(O2) = 320 32.0 = 10.0 mol O2 : O 1 : 2 n(O) = 2  10.0 = 20.0 mol No. of O atoms = 20.0  6.02  1023 = 1.20  1025 (3 sf) (b) H atoms in 60 cm3 of methane, CH4, gas at room temperature and pressure. {Molar volume, Vm, of gas at r.t.p. = 24 dm3 mol-1  1 mol of gas occupies 24 dm3.} n(CH4) = 60 24000 = 0.00250 mol CH4 : H 1 : 4 n(H) = 4  0.00250 = 0.0100 mol No. of H atoms = 0.0100  6.02  1023 = 6.02  1021 (3 sf) (c) K+ ions in 27.5 cm3 of 0.120 mol dm–3 potassium sulfate, K2SO4, solution. { [K2SO4] = 0.120 mol dm-3  1 dm3 of solution contains 0.120 mol of K2SO4.} n(K2SO4) = 27.5 0.1201000 = 0.00330 mol K2SO4 : K+ 1 : 2 n(K+) = 2  0.00330 = 0.00660 mol No. of K+ ions = 0.00660  6.02  1023 = 3.97  1021 (3 sf) (d) electrons in 5 g of hydrogen, H2, gas. {Molar mass, M, of H2 = 2.0 g mol-1  1 mol of H2 contains 2.0 g of H2.} n(H2) = 5 2(1.0) = 2.50 mol H2 : e- 1 : 2 n(electron) = 2  2.50 = 5.00 mol Number of electrons = 5.00  6.02  1023 = 3.01  1024 (3 sf) [1.20  1025; 6.02  1021; 3.97  1021; 3.01  1024] m n M V n Vm n c V m n M 3 nC= V in dm r mn= M or M

2026 JPJC JC1 H2 Chemistry (9476) 3 Topic 2: The Mole Concept and Stoichiometry 4 A solution is prepared by dissolving 1.86 g of potassium chloride, KCl, and 8.29 g of potassium carbonate, K2CO3, in enough water to make up the final volume to 500 cm3. Calculate the concentration of the potassium, K+, ions in the solution. [0.290 mol dm–3] n(KCl) = 1.86 74.6 = 0.02493 mol n(K2CO3) = 8.29 138.2 = 0.05999 mol Total amount of K+ in 500 cm3 resulting solution = 0.02493 + 2(0.05999) = 0.1449 mol [K+] =   0.1449 1000 or 0.1449500 5001000 = 0.290 mol dm–3 (3 sf + units) 5 What volume of 0.0500 mol dm –3 potassium dichromate(VI), K2Cr2O7, solution should you transfer into a 250 cm3 volumetric flask in order to dilute the solution to 0.0200 mol dm–3? [100 cm3] = =  Amount of in of solution Amount of in of solution 1000 1000 Upon dilution, the amount will not change. 33 12 12 12 X V cm ORIGINAL X V cm DILUTED cc VV Using, c1V1 = c2V2 0.0500  V1 = 0.0200  250 V1 = 100 cm3 Volume of original solution required, V1 = 100 cm3 (3 sf + units) KCl : K+ 1 : 1 K2CO3 : K+ 1 : 2 3 nC= V in dm r mn= M or M

2026 JPJC JC1 H2 Chemistry (9476) 4 Topic 2: The Mole Concept and Stoichiometry *6 Phosgene, COCl2, was once used as a war gas. It is poisonous because when inhaled, it reacts with water in the lungs to produce carbon dioxide, CO 2, gas and hydrochloric acid, HCl, which causes severe lung damage, leading to death ultimately. (a) Calculate the percentage by mass of chlorine in COCl2. Molar mass of COCl2 = 99.0 g mol-1 % by mass of Cl in COCl2 = ( )  2 35.5 100 % 71.7 % 99.0 = (3 sf + units) (b) Write a balanced equation for the reaction between COCl2 and H2O. COCl2 + H2O → CO2 + 2HCl (c) Calculate the amount of HC l that will be produced by the complete reaction of 0.430 mol of COCl2. COCl2 : HCl 1 : 2 n(HCl) produced = 2  0.430 = 0.860 mol (3 sf + units) (d) Identify the limiting reagent when 0.200 mol of COC l2 is mixed with 6.20 g of H 2O. Hence, calculate the amount of HCl produced at the end of the reaction. n(H2O) used = 6.20 18.0 = 0.3444 mol n(COCl2) used = 0.200 mol COCl2 : H2O 1 : 1 Since 0.3444 mol of H2O will react with 0.3444 mol of COCl2, COCl2 is the limiting reagent. COCl2 : HCl 1 : 2 n(HCl) produced = 2  0.200 = 0.400 mol (3 sf + units) [71.7%; 0.860 mol; 0.400 mol]

2026 JPJC JC1 H2 Chemistry (9476) 5 Topic 2: The Mole Concept and Stoichiometry *7 7.18 g of a sample of hydrated salt, Na2CO3.xH2O, was heated strongly until there was no further decrease in mass. Na2CO3.xH2O(s) → Na2CO3(s) + xH2O(g) If the remaining residue, Na2CO3 has a mass of 2.65 g, calculate the relative formula mass of the hydrated salt. Hence, determine the value of x. [x=10] Na2CO3.xH2O(s) → Na2CO3(s) + xH2O(g) 7.18 g 2.65 g (7.18 - 2.65) g (mass lost) n(Na2CO3) formed = 2.65 106.0 = 0.0250 mol Na2CO3.xH2O : Na2CO3 1 : 1 n(Na2CO3.xH2O) decomposed = 0.0250 mol Mr of Na2CO3.xH2O = 7.18 0.0250 = 287 (no units) 106.0 + 18.0x = 287  x = 10 (nearest whole number) *8 An organic compound R with molar mass 134 g mol–1 has the following composition by mass: C, 35.8 % H, 4.5 % O, 59.7 % Calculate the empirical formula of R and hence deduce its molecular formula. [C4H6O5; C4H6O5] element C H O % mass 35.8 4.5 59.7 amount /mol =35.812.0 2.98 =4.51.0 4.50 =59.716.0 3.73 mole ratio 1 1.5 1.25 (x4) 4 6 5  the empirical formula of R is C4H6O5. Let the molecular formula of R be C4nH6nO5n. n(4x12.0 + 6x1.0 + 5x16.0) = 134  n = 1  the molecular formula of R is C4H6O5. ror mn= M M

2026 JPJC JC1 H2 Chemistry (9476) 6 Topic 2: The Mole Concept and Stoichiometry 9 St