Pre-AP Chemistry worksheet focusing on mole calculations and percent composition with 23 practice problems.

Let’s solve each problem step by step. We’ll use the periodic table to find atomic masses and Avogadro’s number (6.022 × 10²³) for converting between moles, atoms, molecules, and mass.

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1. How many moles of Na are in 42 g of Na?

Atomic mass of Na = 23.0 g/mol
Moles = mass / molar mass = 42 g / 23.0 g/mol ≈ 1.826 mol

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2. How many moles of O are in 8.25 g of O?

Atomic mass of O = 16.0 g/mol
Moles = 8.25 g / 16.0 g/mol = 0.5156 mol

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3. How much does 2.18 mol of Cu weigh?

Atomic mass of Cu = 63.5 g/mol
Mass = moles × molar mass = 2.18 mol × 63.5 g/mol = 138.43 g

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4. What is the mass of 0.28 mol of iron?

Atomic mass of Fe = 55.8 g/mol
Mass = 0.28 mol × 55.8 g/mol = 15.624 g

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5. How many atoms are in 7.2 mol of chlorine?

Chlorine here means Cl atoms (not Cl₂ unless specified).
Atoms = moles × Avogadro’s number = 7.2 × 6.022×10²³ = 4.336×10²⁴ atoms

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6. How many atoms are in 36 g of bromine?

Bromine is Br₂ as a molecule, but question says “atoms”, so we need to find how many Br atoms.

First, molar mass of Br₂ = 2 × 79.9 = 159.8 g/mol
Moles of Br₂ = 36 g / 159.8 g/mol ≈ 0.2253 mol
Each mole of Br₂ has 2 moles of Br atoms → 0.2253 × 2 = 0.4506 mol Br atoms
Atoms = 0.4506 × 6.022×10²³ ≈ 2.714×10²³ atoms

Wait — let me double-check: The question says “bromine” — if it’s elemental bromine, it’s diatomic (Br₂), but when asking for “atoms”, we count individual atoms. So yes, above is correct.

Alternatively, some might interpret “bromine” as Br atoms — but standardly, elemental bromine is Br₂. However, since the question says “atoms”, we must convert to atoms regardless.

But actually — let’s check common practice: In such worksheets, sometimes they treat “bromine” as Br atoms unless specified as Br₂. But to be safe, let’s assume it’s Br atoms because the question says “atoms”.

Actually, no — if you have 36 g of bromine element, it’s Br₂. So we should calculate based on Br₂, then multiply by 2 for atoms.

So:

Molar mass Br₂ = 159.8 g/mol
Moles Br₂ = 36 / 159.8 ≈ 0.2253 mol
Number of Br₂ molecules = 0.2253 × 6.022e23 ≈ 1.357e23
Each molecule has 2 atoms → total atoms = 2.714e23

✔ Final answer: 2.71 × 10²³ atoms (rounded)

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7. How many moles are in 1.0 × 10⁹ atoms?

Moles = atoms / Avogadro’s number = 1.0e9 / 6.022e23 ≈ 1.66 × 10⁻¹⁵ mol

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8. What is the mass of 1.20 × 10²⁵ atoms of sulfur?

Sulfur atomic mass = 32.1 g/mol
First, moles = atoms / Avogadro’s number = 1.20e25 / 6.022e23 ≈ 19.927 mol
Mass = moles × molar mass = 19.927 × 32.1 ≈ 639.6 g

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9. How many moles of CO molecules are in 52 g of CO?

Molar mass CO = 12.0 + 16.0 = 28.0 g/mol
Moles = 52 g / 28.0 g/mol ≈ 1.857 mol

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10. How many moles of C₂H₆ are in 124 g?

Molar mass C₂H₆ = 2×12.0 + 6×1.0 = 24.0 + 6.0 = 30.0 g/mol
Moles = 124 g / 30.0 g/mol ≈ 4.133 mol

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11. How many moles of CCl₄ are there in 56 g?

Molar mass CCl₄ = 12.0 + 4×35.5 = 12.0 + 142.0 = 154.0 g/mol
Moles = 56 g / 154.0 g/mol ≈ 0.3636 mol

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12. How much does 2.50 mol of H₂SO₄ weigh?

Molar mass H₂SO₄ = 2×1.0 + 32.1 + 4×16.0 = 2.0 + 32.1 + 64.0 = 98.1 g/mol
Mass = 2.50 mol × 98.1 g/mol = 245.25 g

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13. How much does 0.25 mol of Fe₂O₃ weigh?

Molar mass Fe₂O₃ = 2×55.8 + 3×16.0 = 111.6 + 48.0 = 159.6 g/mol
Mass = 0.25 mol × 159.6 g/mol = 39.9 g

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14. How many molecules are there in 52 g of CO?

From #9, moles of CO in 52 g = 52 / 28.0 ≈ 1.857 mol
Molecules = moles × Avogadro’s number = 1.857 × 6.022e23 ≈ 1.118 × 10²⁴ molecules

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15. How many formula units are in 22.4 g SnO₂?

Formula units = same as molecules for ionic compounds.

Molar mass SnO₂ = 118.7 + 2×16.0 = 118.7 + 32.0 = 150.7 g/mol
Moles = 22.4 g / 150.7 g/mol ≈ 0.1486 mol
Formula units = 0.1486 × 6.022e23 ≈ 8.95 × 10²²

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16. How many molecules are in 116 g CCl₄?

From #11, molar mass CCl₄ = 154.0 g/mol
Moles = 116 g / 154.0 g/mol ≈ 0.7532 mol
Molecules = 0.7532 × 6.022e23 ≈ 4.536 × 10²³

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17. What is the mass of 3.01 × 10²³ formula units of Fe₂O₃?

First, moles = formula units / Avogadro’s number = 3.01e23 / 6.022e23 ≈ 0.500 mol
Molar mass Fe₂O₃ = 159.6 g/mol (from #13)
Mass = 0.500 mol × 159.6 g/mol = 79.8 g

Note: 3.01e23 is half of Avogadro’s number → exactly 0.5 mol → nice round number.

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18. What is the mass of 1.2 × 10²⁵ molecules of CO?

Moles = 1.2e25 / 6.022e23 ≈ 19.927 mol
Molar mass CO = 28.0 g/mol
Mass = 19.927 × 28.0 ≈ 557.96 g

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19. How many O atoms are in 1.25 mol of SO₂?

Each SO₂ molecule has 2 O atoms.
So, moles of O atoms = 1.25 mol × 2 = 2.50 mol
Atoms = 2.50 × 6.022e23 = 1.5055 × 10²⁴ atoms

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20. How many moles of O atoms do you have when you have 1.20 × 10²⁵ N₂O₅ molecules?

Each N₂O₅ molecule has 5 O atoms.
First, moles of N₂O₅ = 1.20e25 / 6.022e23 ≈ 19.927 mol
Moles of O atoms = 19.927 × 5 = 99.635 mol

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21. How many formula units are in 5.33 mol of CuCl₂?

Formula units = moles × Avogadro’s number = 5.33 × 6.022e23 ≈ 3.210 × 10²⁴

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22. How many copper atoms are in 5.33 mol of CuCl₂?

Each formula unit of CuCl₂ has 1 Cu atom.
So, moles of Cu atoms = 5.33 mol
Atoms = 5.33 × 6.022e23 ≈ 3.210 × 10²⁴ atoms

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23. How many moles of Cl atoms are in 5.33 mol of CuCl₂?

Each CuCl₂ has 2 Cl atoms.
Moles of Cl atoms = 5.33 mol × 2 = 10.66 mol

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Final Answer:
1. 1.83 mol
2. 0.516 mol
3. 138 g
4. 15.6 g
5. 4.34 × 10²⁴ atoms
6. 2.71 × 10²³ atoms
7. 1.66 × 10⁻¹⁵ mol
8. 640 g
9. 1.86 mol
10. 4.13 mol
11. 0.364 mol
12. 245 g
13. 39.9 g
14. 1.12 × 10²⁴ molecules
15. 8.95 × 10²² formula units
16. 4.54 × 10²³ molecules
17. 79.8 g
18. 558 g
19. 1.51 × 10²⁴ atoms
20. 99.6 mol
21. 3.21 × 10²⁴ formula units
22. 3.21 × 10²⁴ atoms
23. 10.7 mol