Mr Cole Chemistry

General Calculation Questions

General Calculations

Amount of Substance Worksheet

Easy Calculation Questions

1. A student reacts 14g of magnesium with excess hydrochloric acid. What volume of gas would the reaction produce if the reaction was conducted at 100kPa and 25oC?
Working Out
Mg + 2HCl → MgCl2 + H2
Step 1: Moles n(Mg) = 14 / 24.3 = 0.576 mol
Ratio 1:1 → n(H2) = 0.576 mol
Step 2: Volume (PV=nRT) V = (0.576 × 8.31 × 298) / 100000 = 0.01426 m3
Answer = 14.3 dm3
2. A student conducts a titration in which average of 26.6cm3 of 0.8moldm-3 sodium hydroxide was required to neutralise 25.0cm3 of sulfuric acid. What is the concentration of the acid?
Working Out
2NaOH + H2SO4 → Na2SO4 + 2H2O
Step 1: Moles of NaOH n = 0.8 × 0.0266 = 0.02128 mol
Step 2: Moles of Acid Ratio 2:1 → n(Acid) = 0.01064 mol
Step 3: Concentration c = 0.01064 / 0.025 = 0.4256
Answer = 0.426 mol dm-3
3. 160.0g of aluminium nitrate dihydrate are heated until constant mass. What will the mass of the anhydrous solid be?
Working Out
Step 1: Molar Masses Mr(Al(NO3)3·2H2O) = 213.0 + 36.0 = 249.0
Mr(Anhydrous) = 213.0
Step 2: Calculation n(Hydrated) = 160.0 / 249.0 = 0.6426 mol
n(Anhydrous) = 0.6426 mol
Mass = 0.6426 × 213.0 = 136.87g
Answer = 137 g
4. How many atoms are there in 100g of dichloroethane?
Working Out
Moles = 100 / 99.0 = 1.01 mol
Molecules = 1.01 × (6.022 × 1023) = 6.08 × 1023 molecules
Each molecule of C2H4Cl2 contains 8 atoms.
Total Atoms = 8 × 6.08 × 1023
Answer = 4.87 × 1024 atoms
5. A substance is analysed and found to contain 28.8g of carbon, 4.20g of hydrogen, 19.2g of oxygen and 8.40g of nitrogen. Mass spec analysis was run and found that the relative formula mass of the substance is 202. What is the molecular formula of the compound?
Working Out
Step 1: Moles C: 28.8/12=2.4 | H: 4.20/1=4.2 | O: 19.2/16=1.2 | N: 8.40/14=0.6
Step 2: Ratio (Divide by 0.6) C: 4 | H: 7 | O: 2 | N: 1
Empirical: C4H7O2N (Mass = 48+7+32+14 = 101)
Step 3: Molecular Formula Mr = 202. Ratio = 202/101 = 2.
Formula = C8H14O4N2

Medium Calculation Questions

1. 16.0g of ammonia are reacted against 50.0g of chloroethane to form 20.2g of ethylamine. What is the percentage yield of the reaction?
Working Out
C2H5Cl + 2NH3 → C2H5NH2 + NH4Cl
Moles of C2H5Cl: 0.775
Moles of NH3: 0.941 (Limiting, need 1.55)
Theoretical Moles Product: 0.941 / 2 = 0.4705
Theoretical Mass: 0.4705 × 45.0 = 21.17g
Yield: (20.2 / 21.17) × 100
Yield = 95.4%
2. An agricultural pH regulator was tested to determine the amount of calcium hydroxide it contains. A 10.0g sample of the regulator was added to 100cm3 of hydrochloric acid. It was then made up to 500cm3 with water and 25cm3 samples of this solution was titrated against 0.5moldm-3 NaOH, finding an average volume of 12.4cm3 was required to neutralise the hydrochloric acid. What was the purity of the pH regulator?
Working Out
Moles of NaOH = 0.5 × 0.0124 = 0.0062
Moles of excess HCl in 25cm3 = 0.0062
Total excess HCl in 500cm3 = 0.0062 × 20 = 0.124
Initial HCl (Assuming 2.0M based on context) = 0.200
Reacted HCl = 0.200 – 0.124 = 0.076
Moles Ca(OH)2 = 0.076 / 2 = 0.038
Mass Ca(OH)2 = 0.038 × 74.1 = 2.82g
Purity = 28.2%
3. A hydrocarbon compound was combusted with excess oxygen, producing 53.1dm3 of carbon dioxide at 298k and 100kPa and 38.6g of water. What was empirical formula of the hydrocarbon?
Working Out
n(CO2) = PV/RT = (100000×0.0531) / (8.31×298) = 2.144 mol
n(C) = 2.144 mol
n(H2O) = 38.6 / 18.0 = 2.144 mol
n(H) = 2.144 × 2 = 4.288 mol
Ratio C:H = 2.144 : 4.288 = 1 : 2
Empirical Formula = CH2
4. Methane can react with iron (III) oxide to product carbon dioxide, water and iron. What mass of iron will you produce from 200g of iron oxide and 60dm3 of CH4 at atmospheric pressure and 400oC if you have an 84% yield.
Working Out
4Fe2O3 + 3CH4 → 8Fe + 3CO2 + 6H2O
n(Fe2O3) = 200 / 159.6 = 1.253 mol
n(CH4) = PV/RT = (101325×0.060) / (8.31×673) = 1.087 mol
Stoichiometry check: 1.253 Fe2O3 needs 0.94 mol CH4. We have excess CH4. Fe2O3 is limiting.
Theoretical Fe = 1.253 × 2 = 2.506 mol
Mass Fe = 2.506 × 55.8 = 139.8g
Actual Mass = 139.8 × 0.84
Answer = 118 g

Hard Calculation Questions

1. Solvay synthesis is used to make sodium hydrogencarbonate. In this reaction, 200g of sodium chloride is dissolved in water, forming a saturated solution of sodium chloride. This solution then has ammonia bubbled through it, during which time, the volume of ammonia decreases from 200dm3 to 120dm3. This is followed by bubbling 100dm3 of carbon dioxide through the mixture. The mixture is cooled, resulting in the formation of sodium hydrogencarbonate precipitate. Ammonium chloride is also formed but remains in solution. What mass of precipitate would you expect to get from the reaction if the percentage yield were 80%?
Working Out
NaCl + NH3 + CO2 + H2O → NaHCO3 + NH4Cl
Step 1: Determine Limiting Reagent n(NaCl) = 200 / 58.5 = 3.42 mol
n(NH3 reacted) = (200 – 120) / 24.0 (assuming RTP) = 3.33 mol
n(CO2) = 100 / 24.0 = 4.17 mol
Ammonia is limiting (3.33 mol).
Step 2: Yield Calculation Theoretical n(NaHCO3) = 3.33 mol
Theoretical Mass = 3.33 × 84.0 = 280 g
Actual Mass = 280 × 0.80
Answer = 224 g
2. Aluminium hydroxide is formed in a multistep process at 100kPa. You start by reacting aluminium with hydrogen chloride gas at a temperature of 700oC. The liquid product is cooled to room temperature where it has water added to it. It exhibits hygroscopic properties, where it absorbs the water molecules into the crystal. Then, once it has stabilised, it is heated back to 700oC, releasing hydrogen chloride gas and water. The aluminium compound has now formed aluminium hydroxide. If you start with 1.00kg of aluminium, what amount of each of the other reactants would you need, and what amount of each of the products and by-products would you create. Give your answer in the most relevant unit (not moles).
Working Out
Reactants Needed:
Moles of HCl needed: 111.1 mol → Vol = 8990 dm3 (at 700C)
Moles of water needed: 111.1 mol → Mass = 2.00 kg
Products Created:
Moles of Al(OH)3: 37.04 mol → Mass = 2.89 kg
Moles of H2: 55.56 mol → Vol = 4500 dm3 (at 700C)
Moles of HCl recovered: 111.1 mol → Vol = 2750 dm3 (at RTP)
3. Ethanoyl chloride can be produced using ethanoic acid. Three potential reactions can be used to form ethanoyl chloride: reacting ethanoic acid with phosphorous (V) chloride, producing ethanoyl chloride and POCl3; reacting ethanoic acid with phosphorous (III) chloride, producing ethanoyl chloride and phosphoric acid; and reacting ethanoic acid with sulfur dichloride oxide, which forms ethanoyl chloride, sulfur dioxide and hydrogen chloride.
a) Which method has the highest atom economy?
Method 2 (using PCl3) has the highest atom economy (74.2%).
Method 1 (PCl5) is 29.2% and Method 3 (SOCl2) is 43.9%.
b) If you had 500g of phosphorous (III) chloride and unlimited ethanoic acid and chlorine, how much less ethanoyl chloride could you make by oxidising the phosphorous to phosphorous (V) chloride before reacting it with the ethanoic acid?
Direct with PCl3: 3 moles product per mole PCl3.
Via PCl5: 1 mole product per mole PCl5 (which comes from 1 mole PCl3).
Mass Diff = 856.4g – 285.5g = 571 g less
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