InfinitySheets1. Convert mass to moles
How many moles are in 9.0 g of water? Use H = 1.0 and O = 16.0.
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Water is H₂O, so its molar mass is 2 × 1.0 + 16.0 = 18.0 g/mol. Divide 9.0 g by 18.0 g/mol to get 0.50 mol.
Free practice · 5 questions
Chemistry mole calculations with worked solutions
Practise molar mass, concentration, reacting quantities and limiting reagents. Use the atomic masses given.
These are original InfinitySheets questions. Try each one first, then open the worked solution to check your method.
2. Calculate concentration
A solution contains 0.075 mol of sodium chloride in 250 cm³ of solution. Find its concentration in mol/dm³.
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Convert the volume first. 250 cm³ is 0.250 dm³. Concentration = amount ÷ volume = 0.075 ÷ 0.250 = 0.30 mol/dm³.
3. Use a balanced equation
Calcium carbonate decomposes as CaCO₃ → CaO + CO₂. What mass of CO₂ forms from 15.0 g of CaCO₃ if it decomposes completely? Use Ca = 40.0, C = 12.0 and O = 16.0.
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CaCO₃ has a molar mass of 100.0 g/mol, so 15.0 g is 0.150 mol. The balanced equation gives one mole of CO₂ for every mole of CaCO₃. CO₂ has a molar mass of 44.0 g/mol, so the mass is 0.150 × 44.0 = 6.60 g.
4. Identify the limiting reagent
React 0.30 mol of hydrogen with 0.10 mol of oxygen according to 2H₂ + O₂ → 2H₂O. Which reagent limits the reaction, and how much water forms?
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The 0.10 mol of oxygen needs 0.20 mol of hydrogen. There is 0.30 mol of hydrogen available, so oxygen is the limiting reagent. One mole of oxygen makes two moles of water, so 0.20 mol of water forms and 0.10 mol of hydrogen remains.
5. Calculate percentage yield
The theoretical yield of a product is 8.40 g. A student collects 6.30 g. Find the percentage yield.
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Divide actual yield by theoretical yield and multiply by 100. The calculation is (6.30 ÷ 8.40) × 100 = 75.0%. The theoretical yield is the denominator because it is the maximum predicted amount.