Grams to Moles, Moles to Grams Worksheet by Back Pocket Science | TPT - Free Printable

Problem Overview:

The task involves converting between moles and grams for various elements and compounds. This requires using the concept of molar mass, which is the mass of one mole of a substance in grams per mole (g/mol). The general relationship is:

\[
\text{Mass (grams)} = \text{Moles} \times \text{Molar Mass (g/mol)}
\]

Conversely:

\[
\text{Moles} = \frac{\text{Mass (grams)}}{\text{Molar Mass (g/mol)}}
\]

We will solve each part step by step.

---

Part 1: Moles to Grams

#### 1. 3.24 mol of Fe
- Element: Iron (Fe)
- Molar Mass of Fe: 55.85 g/mol
- Calculation:
\[
\text{Mass} = \text{Moles} \times \text{Molar Mass} = 3.24 \, \text{mol} \times 55.85 \, \text{g/mol} = 180.978 \, \text{g}
\]
- Answer: 180.98 g

#### 2. 3.24 mol of NaCl
- Compound: Sodium Chloride (NaCl)
- Molar Mass of NaCl:
- Molar mass of Na = 22.99 g/mol
- Molar mass of Cl = 35.45 g/mol
- Total = 22.99 + 35.45 = 58.44 g/mol
- Calculation:
\[
\text{Mass} = \text{Moles} \times \text{Molar Mass} = 3.24 \, \text{mol} \times 58.44 \, \text{g/mol} = 189.6756 \, \text{g}
\]
- Answer: 189.68 g

#### 3. 0.77 mol of Ca(OH)₂
- Compound: Calcium Hydroxide (Ca(OH)₂)
- Molar Mass of Ca(OH)₂:
- Molar mass of Ca = 40.08 g/mol
- Molar mass of O = 16.00 g/mol (2 atoms)
- Molar mass of H = 1.01 g/mol (2 atoms)
- Total = 40.08 + (2 × 16.00) + (2 × 1.01) = 40.08 + 32.00 + 2.02 = 74.10 g/mol
- Calculation:
\[
\text{Mass} = \text{Moles} \times \text{Molar Mass} = 0.77 \, \text{mol} \times 74.10 \, \text{g/mol} = 57.057 \, \text{g}
\]
- Answer: 57.06 g

#### 4. 4.15 mol of Al₂O₃
- Compound: Aluminum Oxide (Al₂O₃)
- Molar Mass of Al₂O₃:
- Molar mass of Al = 26.98 g/mol (2 atoms)
- Molar mass of O = 16.00 g/mol (3 atoms)
- Total = (2 × 26.98) + (3 × 16.00) = 53.96 + 48.00 = 101.96 g/mol
- Calculation:
\[
\text{Mass} = \text{Moles} \times \text{Molar Mass} = 4.15 \, \text{mol} \times 101.96 \, \text{g/mol} = 422.634 \, \text{g}
\]
- Answer: 422.63 g

#### 5. 0.123 mol of C₆H₁₂O₆
- Compound: Glucose (C₆H₁₂O₆)
- Molar Mass of C₆H₁₂O₆:
- Molar mass of C = 12.01 g/mol (6 atoms)
- Molar mass of H = 1.01 g/mol (12 atoms)
- Molar mass of O = 16.00 g/mol (6 atoms)
- Total = (6 × 12.01) + (12 × 1.01) + (6 × 16.00) = 72.06 + 12.12 + 96.00 = 180.18 g/mol
- Calculation:
\[
\text{Mass} = \text{Moles} \times \text{Molar Mass} = 0.123 \, \text{mol} \times 180.18 \, \text{g/mol} = 22.16154 \, \text{g}
\]
- Answer: 22.16 g

#### 6. 0.63 mol of Na₃PO₄
- Compound: Sodium Phosphate (Na₃PO₄)
- Molar Mass of Na₃PO₄:
- Molar mass of Na = 22.99 g/mol (3 atoms)
- Molar mass of P = 30.97 g/mol (1 atom)
- Molar mass of O = 16.00 g/mol (4 atoms)
- Total = (3 × 22.99) + 30.97 + (4 × 16.00) = 68.97 + 30.97 + 64.00 = 163.94 g/mol
- Calculation:
\[
\text{Mass} = \text{Moles} \times \text{Molar Mass} = 0.63 \, \text{mol} \times 163.94 \, \text{g/mol} = 103.1322 \, \text{g}
\]
- Answer: 103.13 g

#### 7. 0.835 mol of (NH₄)₂SO₄
- Compound: Ammonium Sulfate ((NH₄)₂SO₄)
- Molar Mass of (NH₄)₂SO₄:
- Molar mass of N = 14.01 g/mol (2 atoms)
- Molar mass of H = 1.01 g/mol (8 atoms)
- Molar mass of S = 32.07 g/mol (1 atom)
- Molar mass of O = 16.00 g/mol (4 atoms)
- Total = (2 × 14.01) + (8 × 1.01) + 32.07 + (4 × 16.00) = 28.02 + 8.08 + 32.07 + 64.00 = 132.17 g/mol
- Calculation:
\[
\text{Mass} = \text{Moles} \times \text{Molar Mass} = 0.835 \, \text{mol} \times 132.17 \, \text{g/mol} = 110.46695 \, \text{g}
\]
- Answer: 110.47 g

#### 8. 4.35 mol of CH₄
- Compound: Methane (CH₄)
- Molar Mass of CH₄:
- Molar mass of C = 12.01 g/mol (1 atom)
- Molar mass of H = 1.01 g/mol (4 atoms)
- Total = 12.01 + (4 × 1.01) = 12.01 + 4.04 = 16.05 g/mol
- Calculation:
\[
\text{Mass} = \text{Moles} \times \text{Molar Mass} = 4.35 \, \text{mol} \times 16.05 \, \text{g/mol} = 70.0075 \, \text{g}
\]
- Answer: 70.01 g

---

Part 2: Grams to Moles

#### 1. 3.24 g of S
- Element: Sulfur (S)
- Molar Mass of S: 32.07 g/mol
- Calculation:
\[
\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{3.24 \, \text{g}}{32.07 \, \text{g/mol}} = 0.101026 \, \text{mol}
\]
- Answer: 0.1010 mol

#### 2. 0.55 g of SO₂
- Compound: Sulfur Dioxide (SO₂)
- Molar Mass of SO₂:
- Molar mass of S = 32.07 g/mol
- Molar mass of O = 16.00 g/mol (2 atoms)
- Total = 32.07 + (2 × 16.00) = 32.07 + 32.00 = 64.07 g/mol
- Calculation:
\[
\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{0.55 \, \text{g}}{64.07 \, \text{g/mol}} = 0.008583 \, \text{mol}
\]
- Answer: 0.0086 mol

#### 3. 8.00 g of MgF₂
- Compound: Magnesium Fluoride (MgF₂)
- Molar Mass of MgF₂:
- Molar mass of Mg = 24.31 g/mol
- Molar mass of F = 19.00 g/mol (2 atoms)
- Total = 24.31 + (2 × 19.00) = 24.31 + 38.00 = 62.31 g/mol
- Calculation:
\[
\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{8.00 \, \text{g}}{62.31 \, \text{g/mol}} = 0.12837 \, \text{mol}
\]
- Answer: 0.1284 mol

#### 4. 0.97 g of C₂H₅OH
- Compound: Ethanol (C₂H₅OH)
- Molar Mass of C₂H₅OH:
- Molar mass of C = 12.01 g/mol (2 atoms)
- Molar mass of H = 1.01 g/mol (6 atoms)
- Molar mass of O = 16.00 g/mol (1 atom)
- Total = (2 × 12.01) + (6 × 1.01) + 16.00 = 24.02 + 6.06 + 16.00 = 46.08 g/mol
- Calculation:
\[
\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{0.97 \, \text{g}}{46.08 \, \text{g/mol}} = 0.02105 \, \text{mol}
\]
- Answer: 0.0211 mol

#### 5. 6.00 g of LiOH
- Compound: Lithium Hydroxide (LiOH)
- Molar Mass of LiOH:
- Molar mass of Li = 6.94 g/mol
- Molar mass of O = 16.00 g/mol
- Molar mass of H = 1.01 g/mol
- Total = 6.94 + 16.00 + 1.01 = 23.95 g/mol
- Calculation:
\[
\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{6.00 \, \text{g}}{23.95 \, \text{g/mol}} = 0.2506 \, \text{mol}
\]
- Answer: 0.2506 mol

#### 6. 2.00 g of CO₂
- Compound: Carbon Dioxide (CO₂)
- Molar Mass of CO₂:
- Molar mass of C = 12.01 g/mol
- Molar mass of O = 16.00 g/mol (2 atoms)
- Total = 12.01 + (2 × 16.00) = 12.01 + 32.00 = 44.01 g/mol
- Calculation:
\[
\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{2.00 \, \text{g}}{44.01 \, \text{g/mol}} = 0.04544 \, \text{mol}
\]
- Answer: 0.0454 mol

#### 7. 0.48 g of NH₄Cl
- Compound: Ammonium Chloride (NH₄Cl)
- Molar Mass of NH₄Cl:
- Molar mass of N = 14.01 g/mol
- Molar mass of H = 1.01 g/mol (4 atoms)
- Molar mass of Cl = 35.45 g/mol
- Total = 14.01 + (4 × 1.01) + 35.45 = 14.01 + 4.04 + 35.45 = 53.50 g/mol
- Calculation:
\[
\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{0.48 \, \text{g}}{53.50 \, \text{g/mol}} = 0.008972 \, \text{mol}
\]
- Answer: 0.0090 mol

#### 8. 13.0 g of He
- Element: Helium (He)
- Molar Mass of He: 4.00 g/mol
- Calculation:
\[
\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{13.0 \, \text{g}}{4.00 \, \text{g/mol}} = 3.25 \, \text{mol}
\]
- Answer: 3.25 mol

#### 9. 0.03 g of C₆H₁₂O₆
- Compound: Glucose (C₆H₁₂O₆)
- Molar Mass of C₆H₁₂O₆: 180.18 g/mol (calculated earlier)
- Calculation:
\[
\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{0.03 \, \text{g}}{180.18 \, \text{g/mol}} = 0.0001665 \, \text{mol}
\]
- Answer: 0.000167 mol

#### 10. 0.45 g of C₂H₄O₂
- Compound: Ethyl Acetate (C₂H₄O₂)
- Molar Mass of C₂H₄O₂:
- Molar mass of C = 12.01 g/mol (2 atoms)
- Molar mass of H = 1.01 g/mol (4 atoms)
- Molar mass of O = 16.00 g/mol (2 atoms)
- Total = (2 × 12.01) + (4 × 1.01) + (2 × 16.00) = 24.02 + 4.04 + 32.00 = 60.06 g/mol
- Calculation:
\[
\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{0.45 \, \text{g}}{60.06 \, \text{g/mol}} = 0.007492 \, \text{mol}
\]
- Answer: 0.0075 mol

---

Final Answers:

#### Part 1: Moles to Grams
1. 180.98 g
2. 189.68 g
3. 57.06 g
4. 422.63 g
5. 22.16 g
6. 103.13 g
7. 110.47 g
8. 70.01 g

#### Part 2: Grams to Moles
1. 0.1010 mol
2. 0.0086 mol
3. 0.1284 mol
4. 0.0211 mol
5. 0.2506 mol
6. 0.0454 mol
7. 0.0090 mol
8. 3.25 mol
9. 0.000167 mol
10. 0.0075 mol

\boxed{
\begin{array}{ll}
\text{Part 1:} & 180.98 \, \text{g}, 189.68 \, \text{g}, 57.06 \, \text{g}, 422.63 \, \text{g}, 22.16 \, \text{g}, 103.13 \, \text{g}, 110.47 \, \text{g}, 70.01 \, \text{g} \\
\text{Part 2:} & 0.1010 \, \text{mol}, 0.0086 \, \text{mol}, 0.1284 \, \text{mol}, 0.0211 \, \text{mol}, 0.2506 \, \text{mol}, 0.0454 \, \text{mol}, 0.0090 \, \text{mol}, 3.25 \, \text{mol}, 0.000167 \, \text{mol}, 0.0075 \, \text{mol}
\end{array}
}