Mol calculations
How many moles are in 12.0 g of Mg? (Ar: Mg=24)
Moles in 12.0 g Mg
n = m / M = 12.0 / 24 = 0.500 mol
Calcium reacts with oxygen to form calcium oxide in the equation Ca + O → CaO.
Calculate the mass of calcium oxide formed when 4 g of calcium reacts completely with excess oxygen.
(Ca = 40, O = 16)
Ca + O -> CaO
Moles of Ca = 4 / 40 = 0.10
Ratio Ca : CaO = 1 : 1 so moles of CaO = 0.10
Mr of CaO = 40 + 16 = 56
Mass of CaO = 0.10 x 56 = 5.6 g
ANSWER: 5.6 g
Calculate the concentration (in mol/dm³) of a solution containing 0.2 moles of sodium hydroxide dissolved in 1 dm³ of water.
Concentration = moles / volume in dm3
Concentration = 0.2 / 1 = 0.20 mol per dm3
ANSWER: 0.20 mol per dm3
Calculate the volume of oxygen gas produced when 0.5 moles of oxygen gas are formed.
Volume = moles x 24
Volume = 0.5 x 24 = 12 dm3
ANSWER: 12 dm3
Calculate the Mr of Cu(NO₃)₂. (Ar: Cu=63.5, N=14, O=16)
3) Mr of Cu(NO₃)₂
NO₃ = 14 + (3 × 16) = 14 + 48 = 62
Mr = 63.5 + (2 × 62) = 63.5 + 124 = 187.5
Magnesium reacts with sulfur to form magnesium sulfide.
Calculate the mass of magnesium sulfide formed when 12 g of magnesium reacts with excess sulfur.
(Mg = 24, S = 32)
Mg + S -> MgS
Moles of Mg = 12 / 24 = 0.50
Ratio Mg : MgS = 1 : 1 so moles of MgS = 0.50
Mr of MgS = 24 + 32 = 56
Mass of MgS = 0.50 x 56 = 28 g
ANSWER: 28 g
Calculate the concentration (in mol/dm³) of a solution made by dissolving 5.85 g of sodium chloride in 500 cm³ of water.
(Na = 23, Cl = 35.5)
Mr of NaCl = 23 + 35.5 = 58.5
Moles of NaCl = 5.85 / 58.5 = 0.10
Volume = 500 cm3 = 0.500 dm3
Concentration = 0.10 / 0.500 = 0.20 mol per dm3
ANSWER: 0.20 mol per dm3
What volume of hydrogen gas is produced when 0.25 moles of hydrogen gas are formed?
Volume = 0.25 x 24 = 6 dm3
ANSWER: 6 dm3
Calculate the percentage by mass of oxygen in Mg(OH)₂. If a sample has a mass of 87.0 g, calculate the mass of oxygen in the sample. (Ar: Mg=24, O=16, H=1)
5) %O in Mg(OH)₂ and mass of O in 87.0 g
Mr(Mg(OH)₂) = 24 + 2(16 + 1) = 24 + 34 = 58
Mass of O in 1 mole = 2 × 16 = 32
%O = (32 / 58) × 100 = 55.2%
Mass of O in 87.0 g = 0.552 × 87.0 = 48.0 g
Zinc reacts with hydrochloric acid according to the equation:
Zn + 2HCl → ZnCl₂ + H₂
Calculate the mass of zinc chloride formed when 6.5 g of zinc reacts completely.
(Zn = 65, Cl = 35.5)
Moles of Zn = 6.5 / 65 = 0.10
Ratio Zn : ZnCl2 = 1 : 1 so moles of ZnCl2 = 0.10
Mr of ZnCl2 = 65 + (2 x 35.5) = 136
Mass of ZnCl2 = 0.10 x 136 = 13.6 g
ANSWER: 13.6 g
What mass of potassium nitrate is needed to prepare 250 cm³ of a 0.4 mol/dm³ solution?
(K = 39, N = 14, O = 16)
Volume = 250 cm3 = 0.250 dm3
Moles needed = concentration x volume = 0.4 x 0.250 = 0.10
Mr of KNO3 = 39 + 14 + (3 x 16) = 101
Mass needed = moles x Mr = 0.10 x 101 = 10.1 g
ANSWER: 10.1 g
Magnesium reacts with hydrochloric acid:
Mg + 2HCl → MgCl₂ + H₂
Calculate the volume of hydrogen gas produced when 6 g of magnesium reacts completely.
(Mg = 24)
Moles of Mg = 6 / 24 = 0.25
Ratio Mg : H2 = 1 : 1 so moles of H2 = 0.25
Volume of H2 = 0.25 x 24 = 6 dm3
ANSWER: 6 dm3
A compound contains 4.2 g N and 9.6 g O. Find the empirical formula, then use Mr = 92 to determine the molecular formula. (Ar: N=14, O=16)
n(N) = 4.2 / 14 = 0.300
n(O) = 9.6 / 16 = 0.600
Ratio 1 : 2 ⇒ empirical NO₂ (Mᵉ = 46)
Molecular: factor = 92 / 46 = 2 ⇒ N₂O₄
Calcium carbonate decomposes when heated:
CaCO₃ → CaO + CO₂
Calculate the mass of calcium oxide formed when 10 g of calcium carbonate decomposes completely.
(Ca = 40, C = 12, O = 16)
Mr of CaCO3 = 40 + 12 + (3 x 16) = 100
Moles of CaCO3 = 10 / 100 = 0.10
Ratio CaCO3 : CaO = 1 : 1 so moles of CaO = 0.10
Mr of CaO = 40 + 16 = 56
Mass of CaO = 0.10 x 56 = 5.6 g
ANSWER: 5.6 g
A student uses 25.0 cm3 of sodium hydroxide solution of concentration 0.20 mol per dm3 to neutralise hydrochloric acid of unknown concentration. The average volume of hydrochloric acid used is 18.0 cm3.
Balanced equation:
HCl + NaOH -> NaCl + H2O
Calculate the concentration of the hydrochloric acid in mol per dm3.
Moles of NaOH = concentration x volume in dm3
Volume of NaOH = 25.0 cm3 = 0.0250 dm3
Moles of NaOH = 0.20 x 0.0250 = 0.00500
From the equation, HCl : NaOH is 1 : 1
So moles of HCl = 0.00500
Volume of HCl = 18.0 cm3 = 0.0180 dm3
Concentration of HCl = moles / volume
Concentration of HCl = 0.00500 / 0.0180 = 0.277...
Round suitably to 3 significant figures
ANSWER: 0.278 mol per dm3
Calcium carbonate reacts with hydrochloric acid:
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
Calculate the volume of carbon dioxide produced when 5 g of calcium carbonate reacts completely.
(Ca = 40, C = 12, O = 16)
Mr of CaCO3 = 100
Moles of CaCO3 = 5 / 100 = 0.05
Ratio CaCO3 : CO2 = 1 : 1 so moles of CO2 = 0.05
Volume of CO2 = 0.05 x 24 = 1.2 dm3
ANSWER: 1.2 dm3
A compound contains 40.0% C, 6.7% H, 53.3% O by mass. Determine the empirical formula. (Ar: C=12, H=1, O=16)
16) Empirical formula from 40.0% C, 6.7% H, 53.3% O
Assume 100 g:
n(C) = 40.0 / 12 = 3.33
n(H) = 6.7 / 1 = 6.7
n(O) = 53.3 / 16 = 3.33
Divide by smallest (≈3.33): C 1.00, H 2.01, O 1.00 ⇒ CH₂O
Iron reacts with sulfur to form iron(II) sulfide:
Fe + S → FeS
If 14 g of iron reacts with 8 g of sulfur, calculate the mass of iron(II) sulfide formed and identify the limiting reagent.
(Fe = 56, S = 32)
Moles of Fe = 14 / 56 = 0.25
Moles of S = 8 / 32 = 0.25
The ratio is 1 : 1 and the moles are equal, so neither is limiting
Moles of FeS formed = 0.25
Mr of FeS = 56 + 32 = 88
Mass of FeS = 0.25 x 88 = 22 g
ANSWER: 22 g, limiting reagent: none
25.0 cm3 of sulfuric acid of concentration 0.10 mol per dm3 is neutralised by potassium hydroxide solution of unknown concentration. The average volume of potassium hydroxide used is 30.0 cm3.
Balanced equation:
H2SO4 + 2KOH -> K2SO4 + 2H2O
Calculate the concentration of the potassium hydroxide in mol per dm3.
Volume of H2SO4 = 25.0 cm3 = 0.0250 dm3
Moles of H2SO4 = 0.10 x 0.0250 = 0.00250
From the equation, 1 mole H2SO4 reacts with 2 moles KOH
So moles of KOH = 2 x 0.00250 = 0.00500
Volume of KOH = 30.0 cm3 = 0.0300 dm3
Concentration of KOH = moles / volume
Concentration of KOH = 0.00500 / 0.0300 = 0.1666...
Round suitably to 3 significant figures
ANSWER: 0.167 mol per dm3
Hydrogen reacts with oxygen according to the equation:
2H₂ + O₂ → 2H₂O
If 48 dm³ of hydrogen gas reacts completely, calculate the volume of oxygen required.
Ratio H2 : O2 = 2 : 1
If hydrogen volume is 48 dm3, oxygen volume is half
Oxygen volume = 48 / 2 = 24 dm3
ANSWER: 24 dm3