QUESTION IMAGE
Question
percent composition and empirical formula worksheet
calculate the percent composition of the following compounds:
- pb(no₃)₂
- fe₂o₃
- a compound containing 1.26 g iron and 0.54 g oxygen
calculate the mass of each of the following:
- the mass of hydrogen in 352 g of ethane that is 79.9% carbon and 21.1% hydrogen.
- the mass of magnesium that can be recovered from 97.4 g if the sample is 41.7% mg.
- the mass of oxygen in 34.6 g of fe₂o₃
Let's solve each problem one by one. We'll start with problem 1: calculating the percent composition of \( \text{Pb(NO}_3\text{)}_2 \).
Problem 1: Percent Composition of \( \text{Pb(NO}_3\text{)}_2 \)
Step 1: Calculate the molar mass of \( \text{Pb(NO}_3\text{)}_2 \)
- Molar mass of \( \text{Pb} \): \( 207.2 \, \text{g/mol} \)
- Molar mass of \( \text{N} \): \( 14.01 \, \text{g/mol} \), and there are 2 \( \text{N} \) atoms: \( 2 \times 14.01 = 28.02 \, \text{g/mol} \)
- Molar mass of \( \text{O} \): \( 16.00 \, \text{g/mol} \), and there are \( 3 \times 2 = 6 \) \( \text{O} \) atoms: \( 6 \times 16.00 = 96.00 \, \text{g/mol} \)
Total molar mass:
\( 207.2 + 28.02 + 96.00 = 331.22 \, \text{g/mol} \)
Step 2: Calculate percent composition of each element
- Percent of \( \text{Pb} \):
\( \frac{207.2}{331.22} \times 100 \approx 62.56\% \)
- Percent of \( \text{N} \):
\( \frac{28.02}{331.22} \times 100 \approx 8.46\% \)
- Percent of \( \text{O} \):
\( \frac{96.00}{331.22} \times 100 \approx 28.98\% \)
Problem 2: Percent Composition of \( \text{Fe}_2\text{O}_3 \)
Step 1: Calculate the molar mass of \( \text{Fe}_2\text{O}_3 \)
- Molar mass of \( \text{Fe} \): \( 55.85 \, \text{g/mol} \), 2 atoms: \( 2 \times 55.85 = 111.70 \, \text{g/mol} \)
- Molar mass of \( \text{O} \): \( 16.00 \, \text{g/mol} \), 3 atoms: \( 3 \times 16.00 = 48.00 \, \text{g/mol} \)
Total molar mass:
\( 111.70 + 48.00 = 159.70 \, \text{g/mol} \)
Step 2: Calculate percent composition of each element
- Percent of \( \text{Fe} \):
\( \frac{111.70}{159.70} \times 100 \approx 69.94\% \)
- Percent of \( \text{O} \):
\( \frac{48.00}{159.70} \times 100 \approx 30.06\% \)
Problem 3: Percent Composition of a Compound with 1.26 g Fe and 0.54 g O
Step 1: Calculate total mass of the compound
Total mass = mass of Fe + mass of O = \( 1.26 + 0.54 = 1.80 \, \text{g} \)
Step 2: Calculate percent composition
- Percent of Fe:
\( \frac{1.26}{1.80} \times 100 = 70.0\% \)
- Percent of O:
\( \frac{0.54}{1.80} \times 100 = 30.0\% \)
Problem 4: Mass of Hydrogen in 352 g of Ethane (79.9% C, 21.1% H)
Step 1: Use percent composition to find mass of H
Mass of H = percent of H × total mass
\( 21.1\% \times 352 \, \text{g} = 0.211 \times 352 \approx 74.3 \, \text{g} \)
Problem 5: Mass of Magnesium Recovered from 97.4 g Sample (41.7% Mg)
Step 1: Use percent composition to find mass of Mg
Mass of Mg = percent of Mg × total mass
\( 41.7\% \times 97.4 \, \text{g} = 0.417 \times 97.4 \approx 40.6 \, \text{g} \)
Problem 6: Mass of Oxygen in 34.6 g of \( \text{Fe}_2\text{O}_3 \)
From Problem 2, we know \( \text{Fe}_2\text{O}_3 \) is 30.06% O (or approximately 30%).
Step 1: Use percent composition to find mass of O
Mass of O = percent of O × total mass
\( 30.06\% \times 34.6 \, \text{g} \approx 0.3006 \times 34.6 \approx 10.4 \, \text{g} \) (or using exact 30%: \( 0.3 \times 34.6 = 10.38 \, \text{g} \), close to 10.4 g)
Final Answers
- Percent composition of \( \text{Pb(NO}_3\text{)}_2 \):
- Pb: ~62.56%, N: ~8.46%, O: ~28.98%
- Percent composition of \( \text{Fe}_2\text{O}_3 \):
- Fe: ~69.94%, O: ~30.06%
- Percent composition of the Fe-O compound:
- Fe: 70.0%, O: 30.0%
- Mass of H in ethane: ~74.3 g
- Mass of Mg recovered: ~40.6 g
- Mass of O in \( \text{Fe}_2\text{O}_3 \): ~10.4 g
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Let's solve each problem one by one. We'll start with problem 1: calculating the percent composition of \( \text{Pb(NO}_3\text{)}_2 \).
Problem 1: Percent Composition of \( \text{Pb(NO}_3\text{)}_2 \)
Step 1: Calculate the molar mass of \( \text{Pb(NO}_3\text{)}_2 \)
- Molar mass of \( \text{Pb} \): \( 207.2 \, \text{g/mol} \)
- Molar mass of \( \text{N} \): \( 14.01 \, \text{g/mol} \), and there are 2 \( \text{N} \) atoms: \( 2 \times 14.01 = 28.02 \, \text{g/mol} \)
- Molar mass of \( \text{O} \): \( 16.00 \, \text{g/mol} \), and there are \( 3 \times 2 = 6 \) \( \text{O} \) atoms: \( 6 \times 16.00 = 96.00 \, \text{g/mol} \)
Total molar mass:
\( 207.2 + 28.02 + 96.00 = 331.22 \, \text{g/mol} \)
Step 2: Calculate percent composition of each element
- Percent of \( \text{Pb} \):
\( \frac{207.2}{331.22} \times 100 \approx 62.56\% \)
- Percent of \( \text{N} \):
\( \frac{28.02}{331.22} \times 100 \approx 8.46\% \)
- Percent of \( \text{O} \):
\( \frac{96.00}{331.22} \times 100 \approx 28.98\% \)
Problem 2: Percent Composition of \( \text{Fe}_2\text{O}_3 \)
Step 1: Calculate the molar mass of \( \text{Fe}_2\text{O}_3 \)
- Molar mass of \( \text{Fe} \): \( 55.85 \, \text{g/mol} \), 2 atoms: \( 2 \times 55.85 = 111.70 \, \text{g/mol} \)
- Molar mass of \( \text{O} \): \( 16.00 \, \text{g/mol} \), 3 atoms: \( 3 \times 16.00 = 48.00 \, \text{g/mol} \)
Total molar mass:
\( 111.70 + 48.00 = 159.70 \, \text{g/mol} \)
Step 2: Calculate percent composition of each element
- Percent of \( \text{Fe} \):
\( \frac{111.70}{159.70} \times 100 \approx 69.94\% \)
- Percent of \( \text{O} \):
\( \frac{48.00}{159.70} \times 100 \approx 30.06\% \)
Problem 3: Percent Composition of a Compound with 1.26 g Fe and 0.54 g O
Step 1: Calculate total mass of the compound
Total mass = mass of Fe + mass of O = \( 1.26 + 0.54 = 1.80 \, \text{g} \)
Step 2: Calculate percent composition
- Percent of Fe:
\( \frac{1.26}{1.80} \times 100 = 70.0\% \)
- Percent of O:
\( \frac{0.54}{1.80} \times 100 = 30.0\% \)
Problem 4: Mass of Hydrogen in 352 g of Ethane (79.9% C, 21.1% H)
Step 1: Use percent composition to find mass of H
Mass of H = percent of H × total mass
\( 21.1\% \times 352 \, \text{g} = 0.211 \times 352 \approx 74.3 \, \text{g} \)
Problem 5: Mass of Magnesium Recovered from 97.4 g Sample (41.7% Mg)
Step 1: Use percent composition to find mass of Mg
Mass of Mg = percent of Mg × total mass
\( 41.7\% \times 97.4 \, \text{g} = 0.417 \times 97.4 \approx 40.6 \, \text{g} \)
Problem 6: Mass of Oxygen in 34.6 g of \( \text{Fe}_2\text{O}_3 \)
From Problem 2, we know \( \text{Fe}_2\text{O}_3 \) is 30.06% O (or approximately 30%).
Step 1: Use percent composition to find mass of O
Mass of O = percent of O × total mass
\( 30.06\% \times 34.6 \, \text{g} \approx 0.3006 \times 34.6 \approx 10.4 \, \text{g} \) (or using exact 30%: \( 0.3 \times 34.6 = 10.38 \, \text{g} \), close to 10.4 g)
Final Answers
- Percent composition of \( \text{Pb(NO}_3\text{)}_2 \):
- Pb: ~62.56%, N: ~8.46%, O: ~28.98%
- Percent composition of \( \text{Fe}_2\text{O}_3 \):
- Fe: ~69.94%, O: ~30.06%
- Percent composition of the Fe-O compound:
- Fe: 70.0%, O: 30.0%
- Mass of H in ethane: ~74.3 g
- Mass of Mg recovered: ~40.6 g
- Mass of O in \( \text{Fe}_2\text{O}_3 \): ~10.4 g