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Mole Concept and Stoichiometry Worksheet

General chemistryGrades 9–12 · GCSE / A-level · AP · IB · CBSE Class 1118 items · answer key included

This mole concept and stoichiometry worksheet with answers covers every core calculation in one printable PDF: molar mass and percentage composition, converting between mass, moles and number of particles with Avogadro's constant, limiting reagent and excess reagent problems, theoretical and percent yield, empirical and molecular formula, and solution concentration (molarity) and dilution. Eighteen graded questions with fully worked answers make it ideal for GCSE, A-level, AP Chemistry, IB and CBSE Class 11 Some Basic Concepts of Chemistry revision. All numbers use standard molar masses (H 1.008, C 12.01, O 16.00…) and are computer-checked.

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Mole Concept and Stoichiometry Worksheet

Name Class Date

Part A — Molar mass and percentage composition

Use Ar: H 1.008, C 12.01, N 14.01, O 16.00, Na 22.99, Mg 24.31, Al 26.98, S 32.06, Cl 35.45, Ca 40.08, Cu 63.55. NA = 6.022 × 1023 mol−1.

1.
Calculate the molar mass of sulfuric acid, H2SO4.
2(1.008) + 32.06 + 4(16.00) = 98.08 g mol−1
2.
Calculate the molar mass of calcium carbonate, CaCO3.
40.08 + 12.01 + 3(16.00) = 100.09 g mol−1
3.
Calculate the molar mass of copper(II) sulfate pentahydrate, CuSO4·5H2O.
63.55 + 32.06 + 4(16.00) + 5(18.016) = 159.61 + 90.08 = 249.69 g mol−1
4.
Calculate the percentage by mass of nitrogen in ammonium nitrate, NH4NO3.
M = 2(14.01) + 4(1.008) + 3(16.00) = 80.05 g mol−1
%N = 28.02 ÷ 80.05 × 100 = 35.0 %

Part B — Mass, moles and particles

Give answers to 3 significant figures.

5.
How many moles are in 4.50 g of water?
M(H2O) = 18.016 g mol−1
n = 4.50 ÷ 18.016 = 0.250 mol
6.
What is the mass of 0.350 mol of sodium chloride?
M(NaCl) = 22.99 + 35.45 = 58.44 g mol−1
m = 0.350 × 58.44 = 20.5 g
7.
How many molecules of CO2 are in 11.0 g of carbon dioxide? How many atoms in total?
n = 11.0 ÷ 44.01 = 0.2499 mol
Molecules = 0.2499 × 6.022 × 1023 = 1.51 × 1023
Atoms = 3 × 1.505 × 1023 = 4.52 × 1023
8.
What is the mass of 3.011 × 1023 atoms of magnesium?
n = 3.011 × 1023 ÷ 6.022 × 1023 = 0.5000 mol
m = 0.5000 × 24.31 = 12.2 g

Part C — Limiting reagent and yield

Show all working: moles → ratio → moles of product → mass.

9.
Aluminium reacts with chlorine: 2Al + 3Cl2 → 2AlCl3. 5.40 g of Al reacts with 14.2 g of Cl2. Identify the limiting reagent and calculate the theoretical yield of AlCl3.
n(Al) = 5.40 ÷ 26.98 = 0.2001 mol; n(Cl2) = 14.2 ÷ 70.90 = 0.2003 mol
Al needs 1.5 × 0.2001 = 0.3002 mol Cl2 — only 0.2003 mol present, so Cl2 is limiting.
n(AlCl3) = ⅔ × 0.2003 = 0.1335 mol; M(AlCl3) = 133.33 g mol−1
Theoretical yield = 0.1335 × 133.33 = 17.8 g
10.
In the reaction in C1, only 15.0 g of AlCl3 was collected. Calculate the percentage yield.
% yield = 15.0 ÷ 17.80 × 100 = 84.3 %
11.
N2 + 3H2 → 2NH3. 28.0 g of N2 is mixed with 8.00 g of H2. Which reagent is limiting? What mass of NH3 forms, and what mass of the excess reagent remains?
n(N2) = 28.0 ÷ 28.02 = 0.9993 mol; n(H2) = 8.00 ÷ 2.016 = 3.968 mol
N2 needs 3 × 0.9993 = 2.998 mol H2 < 3.968 → N2 is limiting
n(NH3) = 2 × 0.9993 = 1.999 mol × 17.03 = 34.0 g NH3
H2 left = 3.968 − 2.998 = 0.970 mol × 2.016 = 1.96 g
12.
25.0 g of CaCO3 is heated strongly: CaCO3 → CaO + CO2. Calculate the theoretical yield of CaO. If 12.5 g of CaO is obtained, what is the percentage yield?
n(CaCO3) = 25.0 ÷ 100.09 = 0.2498 mol = n(CaO)
Theoretical yield = 0.2498 × 56.08 = 14.0 g
% yield = 12.5 ÷ 14.01 × 100 = 89.2 %

Part D — Formulae, concentration and dilution

Show all working.

13.
A compound contains 40.0 % C, 6.7 % H and 53.3 % O by mass. Its relative molecular mass is 180. Find its empirical and molecular formulae.
C: 40.0 ÷ 12.01 = 3.33; H: 6.7 ÷ 1.008 = 6.65; O: 53.3 ÷ 16.00 = 3.33
Divide by 3.33 → 1 : 2 : 1 → empirical formula CH2O (M = 30.03)
180 ÷ 30.03 ≈ 6 → molecular formula C6H12O6
14.
2.43 g of magnesium burns in air to form 4.03 g of magnesium oxide. Find the empirical formula of the oxide.
Mass O = 4.03 − 2.43 = 1.60 g
n(Mg) = 2.43 ÷ 24.31 = 0.1000; n(O) = 1.60 ÷ 16.00 = 0.1000
Ratio 1 : 1 → MgO
15.
5.85 g of NaCl is dissolved in water and made up to 250 cm3. Calculate the concentration in mol dm−3.
n = 5.85 ÷ 58.44 = 0.1001 mol
c = 0.1001 ÷ 0.250 = 0.400 mol dm−3
16.
25.0 cm3 of 2.00 mol dm−3 HCl is diluted with water to 500 cm3. What is the new concentration?
c1V1 = c2V2: c2 = 2.00 × 25.0 ÷ 500 = 0.100 mol dm−3
17.
What volume of 12.0 mol dm−3 HCl is needed to make 250 cm3 of 0.500 mol dm−3 HCl?
V1 = 0.500 × 250 ÷ 12.0 = 10.4 cm3 (dilute to 250 cm3 in a volumetric flask)
18.
Concentrated sulfuric acid is 98.0 % H2SO4 by mass and has a density of 1.84 g cm−3. Calculate its molarity.
Mass of 1 dm3 = 1000 × 1.84 = 1840 g; mass H2SO4 = 0.980 × 1840 = 1803 g
n = 1803 ÷ 98.08 = 18.4 mol dm−3

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Answer key: Mole Concept and Stoichiometry Worksheet

Part A — Molar mass and percentage composition

1.
2(1.008) + 32.06 + 4(16.00) = 98.08 g mol−1
2.
40.08 + 12.01 + 3(16.00) = 100.09 g mol−1
3.
63.55 + 32.06 + 4(16.00) + 5(18.016) = 159.61 + 90.08 = 249.69 g mol−1
4.
M = 2(14.01) + 4(1.008) + 3(16.00) = 80.05 g mol−1
%N = 28.02 ÷ 80.05 × 100 = 35.0 %

Part B — Mass, moles and particles

5.
M(H2O) = 18.016 g mol−1
n = 4.50 ÷ 18.016 = 0.250 mol
6.
M(NaCl) = 22.99 + 35.45 = 58.44 g mol−1
m = 0.350 × 58.44 = 20.5 g
7.
n = 11.0 ÷ 44.01 = 0.2499 mol
Molecules = 0.2499 × 6.022 × 1023 = 1.51 × 1023
Atoms = 3 × 1.505 × 1023 = 4.52 × 1023
8.
n = 3.011 × 1023 ÷ 6.022 × 1023 = 0.5000 mol
m = 0.5000 × 24.31 = 12.2 g

Part C — Limiting reagent and yield

9.
n(Al) = 5.40 ÷ 26.98 = 0.2001 mol; n(Cl2) = 14.2 ÷ 70.90 = 0.2003 mol
Al needs 1.5 × 0.2001 = 0.3002 mol Cl2 — only 0.2003 mol present, so Cl2 is limiting.
n(AlCl3) = ⅔ × 0.2003 = 0.1335 mol; M(AlCl3) = 133.33 g mol−1
Theoretical yield = 0.1335 × 133.33 = 17.8 g
10.
% yield = 15.0 ÷ 17.80 × 100 = 84.3 %
11.
n(N2) = 28.0 ÷ 28.02 = 0.9993 mol; n(H2) = 8.00 ÷ 2.016 = 3.968 mol
N2 needs 3 × 0.9993 = 2.998 mol H2 < 3.968 → N2 is limiting
n(NH3) = 2 × 0.9993 = 1.999 mol × 17.03 = 34.0 g NH3
H2 left = 3.968 − 2.998 = 0.970 mol × 2.016 = 1.96 g
12.
n(CaCO3) = 25.0 ÷ 100.09 = 0.2498 mol = n(CaO)
Theoretical yield = 0.2498 × 56.08 = 14.0 g
% yield = 12.5 ÷ 14.01 × 100 = 89.2 %

Part D — Formulae, concentration and dilution

13.
C: 40.0 ÷ 12.01 = 3.33; H: 6.7 ÷ 1.008 = 6.65; O: 53.3 ÷ 16.00 = 3.33
Divide by 3.33 → 1 : 2 : 1 → empirical formula CH2O (M = 30.03)
180 ÷ 30.03 ≈ 6 → molecular formula C6H12O6
14.
Mass O = 4.03 − 2.43 = 1.60 g
n(Mg) = 2.43 ÷ 24.31 = 0.1000; n(O) = 1.60 ÷ 16.00 = 0.1000
Ratio 1 : 1 → MgO
15.
n = 5.85 ÷ 58.44 = 0.1001 mol
c = 0.1001 ÷ 0.250 = 0.400 mol dm−3
16.
c1V1 = c2V2: c2 = 2.00 × 25.0 ÷ 500 = 0.100 mol dm−3
17.
V1 = 0.500 × 250 ÷ 12.0 = 10.4 cm3 (dilute to 250 cm3 in a volumetric flask)
18.
Mass of 1 dm3 = 1000 × 1.84 = 1840 g; mass H2SO4 = 0.980 × 1840 = 1803 g
n = 1803 ÷ 98.08 = 18.4 mol dm−3

How to use this worksheet

  1. Parts A–B are foundation (molar mass, mass ↔ moles ↔ particles); Parts C–D are higher tier/A-level (limiting reagent, yield, formulae, dilution).
  2. Provide a periodic table — the printable periodic table worksheet on MolDraw uses the same atomic masses as this answer key.
  3. Insist that students carry unrounded values through multi-step answers and round only at the end to 3 significant figures.
  4. The answer key shows each step on a separate line so students can find exactly where their method went wrong.
  5. Set the title, student lines and answer option on the left, then press Download PDF and choose “Save as PDF” (or print directly). Your choices are remembered on this device.

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Frequently asked questions

What is the mole concept?

One mole is 6.022 × 10²³ particles (Avogadro's constant). The mass of one mole of a substance in grams equals its relative formula mass, so moles = mass ÷ molar mass and particles = moles × 6.022 × 10²³.

How do you find the limiting reagent?

Convert each reactant to moles, divide by its coefficient in the balanced equation (or work out how much of the other reactant each needs). The reactant that runs out first is limiting and fixes the theoretical yield.

How do you calculate percent yield?

Percent yield = actual mass obtained ÷ theoretical (maximum) mass × 100. The theoretical yield is calculated from moles of the limiting reagent using the balanced equation.

Which atomic masses are used in the answers?

IUPAC standard atomic weights to four significant figures, e.g. H 1.008, C 12.01, N 14.01, O 16.00, Na 22.99, Cl 35.45, with Avogadro's constant 6.022 × 10²³ mol⁻¹.

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