Free Printable Mole Conversion Worksheets - Free Printable

To solve this "Moles to Grams" worksheet, we use the formula:

$$
\text{Mass (grams)} = \text{Number of moles} \times \text{Molar mass (g/mol)}
$$

We need to calculate the molar mass of each compound and then multiply by the given number of moles.

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Step-by-step Solutions:

#### 1. 4 moles of Cu(CN)₂
- Molar mass of Cu(CN)₂:
- Cu: 63.55 g/mol
- C: 12.01 × 2 = 24.02
- N: 14.01 × 2 = 28.02
- Total = 63.55 + 24.02 + 28.02 = 115.59 g/mol
- Mass = 4 mol × 115.59 g/mol = 462.36 g

#### 2. 6.6 moles of ZnO
- Zn: 65.38 g/mol
- O: 16.00 g/mol
- Molar mass = 65.38 + 16.00 = 81.38 g/mol
- Mass = 6.6 × 81.38 = 537.11 g

#### 3. 5.6 moles of C₆H₆ (benzene)
- C: 12.01 × 6 = 72.06
- H: 1.008 × 6 = 6.048
- Total = 78.108 g/mol
- Mass = 5.6 × 78.108 ≈ 437.40 g

#### 4. 0.11 moles of Cu
- Atomic mass of Cu = 63.55 g/mol
- Mass = 0.11 × 63.55 = 7.00 g

#### 5. 1.2 moles of (NH₄)₃PO₄
- N: 14.01 × 3 = 42.03
- H: 1.008 × 12 = 12.096
- P: 30.97
- O: 16.00 × 4 = 64.00
- Total = 42.03 + 12.096 + 30.97 + 64.00 = 149.096 g/mol
- Mass = 1.2 × 149.096 ≈ 178.92 g

#### 6. 88.4 moles of NI₃
- N: 14.01
- I: 126.90 × 3 = 380.70
- Total = 14.01 + 380.70 = 394.71 g/mol
- Mass = 88.4 × 394.71 ≈ 34,878.56 g

#### 7. 9.3 × 10⁻³ moles of SmO
- Sm: 150.36 g/mol
- O: 16.00
- Total = 166.36 g/mol
- Mass = 0.0093 × 166.36 ≈ 1.54 g

#### 8. 5.4 moles of K₂SO₄
- K: 39.10 × 2 = 78.20
- S: 32.07
- O: 16.00 × 4 = 64.00
- Total = 174.27 g/mol
- Mass = 5.4 × 174.27 ≈ 939.06 g

#### 9. 0.2 moles of Pb(CH₃COO)₂ (Pb(C₂H₃O₂)₂)
- Pb: 207.2
- C: 12.01 × 4 = 48.04
- H: 1.008 × 6 = 6.048
- O: 16.00 × 4 = 64.00
- Total = 207.2 + 48.04 + 6.048 + 64.00 = 325.288 g/mol
- Mass = 0.2 × 325.288 ≈ 65.06 g

#### 10. 0.31 moles of H₃PO₄
- H: 1.008 × 3 = 3.024
- P: 30.97
- O: 16.00 × 4 = 64.00
- Total = 97.994 g/mol
- Mass = 0.31 × 97.994 ≈ 30.38 g

#### 11. 21.3 moles of BaCO₃
- Ba: 137.33
- C: 12.01
- O: 16.00 × 3 = 48.00
- Total = 197.34 g/mol
- Mass = 21.3 × 197.34 ≈ 4,207.14 g

#### 12. 3.8 moles of Ra(OH)₂
- Ra: 226.03
- O: 16.00 × 2 = 32.00
- H: 1.008 × 2 = 2.016
- Total = 226.03 + 32.00 + 2.016 = 260.046 g/mol
- Mass = 3.8 × 260.046 ≈ 988.17 g

#### 13. 1.31 moles of NaHCO₃
- Na: 22.99
- H: 1.008
- C: 12.01
- O: 16.00 × 3 = 48.00
- Total = 84.008 g/mol
- Mass = 1.31 × 84.008 ≈ 109.95 g

#### 14. 0.28 moles of CO₂
- C: 12.01
- O: 16.00 × 2 = 32.00
- Total = 44.01 g/mol
- Mass = 0.28 × 44.01 ≈ 12.32 g

#### 15. 1.25 moles of HF
- H: 1.008
- F: 19.00
- Total = 20.008 g/mol
- Mass = 1.25 × 20.008 ≈ 25.01 g

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Final Table:

| Moles | Grams |
|-------|--------|
| 4 moles of Cu(CN)₂ | 462.36 g |
| 6.6 moles of ZnO | 537.11 g |
| 5.6 moles of C₆H₆ | 437.40 g |
| 0.11 moles of Cu | 7.00 g |
| 1.2 moles of (NH₄)₃PO₄ | 178.92 g |
| 88.4 moles of NI₃ | 34,878.56 g |
| 9.3 × 10⁻³ moles of SmO | 1.54 g |
| 5.4 moles of K₂SO₄ | 939.06 g |
| 0.2 moles of Pb(CH₃COO)₂ | 65.06 g |
| 0.31 moles of H₃PO₄ | 30.38 g |
| 21.3 moles of BaCO₃ | 4,207.14 g |
| 3.8 moles of Ra(OH)₂ | 988.17 g |
| 1.31 moles of NaHCO₃ | 109.95 g |
| 0.28 moles of CO₂ | 12.32 g |
| 1.25 moles of HF | 25.01 g |

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✔ Explanation Summary:
For each substance, we:
1. Found its molar mass using atomic masses from the periodic table.
2. Multiplied the number of moles by the molar mass to get grams.
3. Rounded appropriately based on significant figures.

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