Water of Crystallisation
Worksheet
Easy Water of Crystallisation Questions
1. A 12.0g crystal of hydrated magnesium sulfate is heated until a constant mass. The final mass is 5.86g. What is the formula of the hydrated magnesium sulfate?
Working Out
Mass H2O = 12.0 – 5.86 = 6.14 g
Moles MgSO4 = 5.86 / 120.4 = 0.0487 mol
Moles H2O = 6.14 / 18.0 = 0.341 mol
Ratio = 0.341 / 0.0487 = 7.00
Formula = MgSO4 • 7H2O
Moles MgSO4 = 5.86 / 120.4 = 0.0487 mol
Moles H2O = 6.14 / 18.0 = 0.341 mol
Ratio = 0.341 / 0.0487 = 7.00
Formula = MgSO4 • 7H2O
2. A 15.0g crystal of hydrated copper sulfate is heated until a constant mass. The final mass is 9.591g. What is the formula of the hydrated copper sulfate?
Working Out
Mass H2O = 15.0 – 9.591 = 5.409 g
Moles CuSO4 = 9.591 / 159.6 = 0.0601 mol
Moles H2O = 5.409 / 18.0 = 0.3005 mol
Ratio = 0.3005 / 0.0601 = 5.00
Formula = CuSO4 • 5H2O
Moles CuSO4 = 9.591 / 159.6 = 0.0601 mol
Moles H2O = 5.409 / 18.0 = 0.3005 mol
Ratio = 0.3005 / 0.0601 = 5.00
Formula = CuSO4 • 5H2O
3. A 32.0g crystal of hydrated titanium nitrate is heated until a constant mass. It has a formula of Ti(NO3)4•6H2O. What mass did it reach?
Working Out
Mr (Hydrated) = 47.9 + 4(62) + 6(18) = 403.9
Mr (Anhydrous) = 47.9 + 4(62) = 295.9
Moles = 32.0 / 403.9 = 0.0792 mol
Mass Anhydrous = 0.0792 × 295.9 = 23.44 g
(32.0g has 3 s.f.)
Answer = 23.4 g
Mr (Anhydrous) = 47.9 + 4(62) = 295.9
Moles = 32.0 / 403.9 = 0.0792 mol
Mass Anhydrous = 0.0792 × 295.9 = 23.44 g
(32.0g has 3 s.f.)
Answer = 23.4 g
4. 100g of anhydrous Cobalt chloride was dissolved in excess water. The water was then removed by crystallisation, resulting in a solid with a mass of 183.2g. What is the formula of the hydrated salt?
Working Out
Mass H2O = 183.2 – 100 = 83.2 g
Moles CoCl2 = 100 / 129.9 = 0.770 mol
Moles H2O = 83.2 / 18.0 = 4.62 mol
Ratio = 4.62 / 0.770 = 6.00
Formula = CoCl2 • 6H2O
Moles CoCl2 = 100 / 129.9 = 0.770 mol
Moles H2O = 83.2 / 18.0 = 4.62 mol
Ratio = 4.62 / 0.770 = 6.00
Formula = CoCl2 • 6H2O
5. 20.0g of aluminium nitrate monohydrate are heated until constant mass. What will the mass of the anhydrous solid be?
Working Out
(Assuming “monohydrate” implies 1H2O as stated)
Mr Al(NO3)3•H2O = 213.0 + 18.0 = 231.0
Moles = 20.0 / 231.0 = 0.0866 mol
Mass Anhydrous = 0.0866 × 213.0 = 18.44 g
Answer = 18.4 g
Mr Al(NO3)3•H2O = 213.0 + 18.0 = 231.0
Moles = 20.0 / 231.0 = 0.0866 mol
Mass Anhydrous = 0.0866 × 213.0 = 18.44 g
Answer = 18.4 g
Medium Difficulty Water of Crystallisation Questions
1. 100cm3 of 0.50moldm-3 sulfuric acid was reacted with excess magnesium. The solution was filtered to remove the excess magnesium and the salt was crystallised. The crystals that formed had a mass of 11.4g. What was the formula of the hydrated crystal?
Working Out
Moles H2SO4 = 0.100 × 0.50 = 0.050 mol
Moles MgSO4 formed = 0.050 mol
Mass of 0.050 mol hydrated salt = 11.4 g
Mr (Hydrated) = 11.4 / 0.050 = 228
Mr (Anhydrous MgSO4) = 120.4
Mass H2O = 228 – 120.4 = 107.6
Moles H2O per mol = 107.6 / 18.0 = 5.98
Formula = MgSO4 • 6H2O
Moles MgSO4 formed = 0.050 mol
Mass of 0.050 mol hydrated salt = 11.4 g
Mr (Hydrated) = 11.4 / 0.050 = 228
Mr (Anhydrous MgSO4) = 120.4
Mass H2O = 228 – 120.4 = 107.6
Moles H2O per mol = 107.6 / 18.0 = 5.98
Formula = MgSO4 • 6H2O
2. A 5.20g of calcium carbonate reacted with sulfuric acid. The solution was filtered to remove the excess calcium carbonate and after drying, the excess calcium carbonate weighed 1.20g. The salt solution was crystallised resulting in a crystal mass of 8.32g; what was the formula of the hydrated salt?
Working Out
Reacted CaCO3 = 5.20 – 1.20 = 4.00 g
Moles CaCO3 = 4.00 / 100.1 = 0.0400 mol
Moles CaSO4 formed = 0.0400 mol
Mr (Hydrated) = 8.32 / 0.0400 = 208
Mr (Anhydrous CaSO4) = 136.2
Mass H2O in Mr = 208 – 136.2 = 71.8
Moles H2O = 71.8 / 18.0 = 3.99
Formula = CaSO4 • 4H2O
Moles CaCO3 = 4.00 / 100.1 = 0.0400 mol
Moles CaSO4 formed = 0.0400 mol
Mr (Hydrated) = 8.32 / 0.0400 = 208
Mr (Anhydrous CaSO4) = 136.2
Mass H2O in Mr = 208 – 136.2 = 71.8
Moles H2O = 71.8 / 18.0 = 3.99
Formula = CaSO4 • 4H2O
Hard Water of Crystallisation Questions
1. A 10.1g mass of hydrated iron (II) nitrate was dissolved in 100cm3 of water and then heated up to 120°C… The iron oxide was filtered… Its mass was 2.8g. What is the formula of the hydrated iron nitrate?
Working Out
Step 1: Moles of Iron in Product (Fe2O3)
Mr Fe2O3 = (55.8×2) + (16×3) = 159.6
Moles Fe2O3 = 2.8 / 159.6 = 0.01754 mol
Moles Fe = 0.01754 × 2 = 0.03509 mol
Step 2: Moles of Reactant
Reaction conserves Iron atoms. Moles Fe(NO3)2 = Moles Fe = 0.03509 mol
Step 3: Molar Mass of Hydrated Salt
Mr (Hydrated) = Mass / Moles = 10.1 / 0.03509 = 287.8
Step 4: Calculate Water
Mr (Anhydrous Fe(NO3)2) = 55.8 + 2(14+48) = 179.8
Mass Water = 287.8 – 179.8 = 108.0
Moles Water = 108.0 / 18.0 = 6.00
Formula = Fe(NO3)2 • 6H2O
Mr Fe2O3 = (55.8×2) + (16×3) = 159.6
Moles Fe2O3 = 2.8 / 159.6 = 0.01754 mol
Moles Fe = 0.01754 × 2 = 0.03509 mol
Step 2: Moles of Reactant
Reaction conserves Iron atoms. Moles Fe(NO3)2 = Moles Fe = 0.03509 mol
Step 3: Molar Mass of Hydrated Salt
Mr (Hydrated) = Mass / Moles = 10.1 / 0.03509 = 287.8
Step 4: Calculate Water
Mr (Anhydrous Fe(NO3)2) = 55.8 + 2(14+48) = 179.8
Mass Water = 287.8 – 179.8 = 108.0
Moles Water = 108.0 / 18.0 = 6.00
Formula = Fe(NO3)2 • 6H2O
2. 4.00g of a hydrated zinc sulfate crystal reacted with 55cm3 of 0.8moldm-3 barium nitrate solution, producing barium sulfate, which, when dehydrated, had a mass of 3.25g. What is the formula of hydrated zinc sulfate?
Working Out
Moles BaSO4 = 3.25 / 233.4 = 0.0139 mol
Moles ZnSO4 (in 4.00g) = 0.0139 mol (1:1 sulfate ratio)
Mr (Hydrated) = 4.00 / 0.0139 = 287.8
Mr (Anhydrous) = 161.5
Mass Water = 287.8 – 161.5 = 126.3
Moles Water = 126.3 / 18.0 = 7.01
Formula = ZnSO4 • 7H2O
Moles ZnSO4 (in 4.00g) = 0.0139 mol (1:1 sulfate ratio)
Mr (Hydrated) = 4.00 / 0.0139 = 287.8
Mr (Anhydrous) = 161.5
Mass Water = 287.8 – 161.5 = 126.3
Moles Water = 126.3 / 18.0 = 7.01
Formula = ZnSO4 • 7H2O
3. 6.00g of hydrated iron (II) sulfate was dissolved in excess 0.8moldm-3 HNO3… On analysis, it is discovered that 200cm3 of gas is produced at 100kPa and 20°C. What is the formula of the hydrated iron (II) sulfate?
Working Out
Step 1: Moles of Gas (pV=nRT)
p = 100,000 Pa
V = 200 cm3 = 0.0002 m3
T = 20°C = 293 K
n = (100,000 × 0.0002) / (8.31 × 293) = 0.008214 mol
Step 2: Stoichiometry
Ionic Eq: 3Fe2+ + 4H+ + NO3– → 3Fe3+ + NO + 2H2O
Ratio Fe2+ : NO is 3 : 1.
Moles FeSO4 = 3 × 0.008214 = 0.02464 mol
Step 3: Find Formula
Mr (Hydrated) = Mass / Moles = 6.00 / 0.02464 = 243.5
Mr (Anhydrous FeSO4) = 55.8 + 32.1 + 64.0 = 151.9
Mass Water part = 243.5 – 151.9 = 91.6
Moles Water = 91.6 / 18.0 = 5.09
Formula = FeSO4 • 5H2O
p = 100,000 Pa
V = 200 cm3 = 0.0002 m3
T = 20°C = 293 K
n = (100,000 × 0.0002) / (8.31 × 293) = 0.008214 mol
Step 2: Stoichiometry
Ionic Eq: 3Fe2+ + 4H+ + NO3– → 3Fe3+ + NO + 2H2O
Ratio Fe2+ : NO is 3 : 1.
Moles FeSO4 = 3 × 0.008214 = 0.02464 mol
Step 3: Find Formula
Mr (Hydrated) = Mass / Moles = 6.00 / 0.02464 = 243.5
Mr (Anhydrous FeSO4) = 55.8 + 32.1 + 64.0 = 151.9
Mass Water part = 243.5 – 151.9 = 91.6
Moles Water = 91.6 / 18.0 = 5.09
Formula = FeSO4 • 5H2O
4. 100cm3 of 0.6moldm-3 hydrochloric acid and 5.00 grams of cobalt (II) oxide are reacted together to form a salt. The salt was crystalised, dried and weighed. It was found to have a mass of 4.98g. What is the formula of the hydrated salt?
Working Out
Moles HCl = 0.100 × 0.6 = 0.060 mol
Moles CoO = 5.00 / 74.9 = 0.0667 mol
Ratio CoO : HCl = 1 : 2. HCl is Limiting.
Moles CoCl2 formed = 0.060 / 2 = 0.030 mol
Mr (Hydrated) = 4.98 / 0.030 = 166
Mr (Anhydrous CoCl2) = 129.9
Mass H2O = 166 – 129.9 = 36.1
Moles H2O = 36.1 / 18.0 = 2.00
Formula = CoCl2 • 2H2O
Moles CoO = 5.00 / 74.9 = 0.0667 mol
Ratio CoO : HCl = 1 : 2. HCl is Limiting.
Moles CoCl2 formed = 0.060 / 2 = 0.030 mol
Mr (Hydrated) = 4.98 / 0.030 = 166
Mr (Anhydrous CoCl2) = 129.9
Mass H2O = 166 – 129.9 = 36.1
Moles H2O = 36.1 / 18.0 = 2.00
Formula = CoCl2 • 2H2O