Question
Lead (II) acetate (Pb(C2H3O2)2) exista as a crystalline hydrate that is 14.25% water by mass. What is the molecular formula
Answers
GPT-4o mini
To find the molecular formula of lead (II) acetate hydrate, we first need to determine the formula of the hydrate based on the percentage of water by mass.
1. **Calculate the molar mass of lead (II) acetate (Pb(C2H3O2)2)**:
\[
\text{Molar mass of Pb} = 207.2 \, \text{g/mol}
\]
\[
\text{Molar mass of C} = 12.01 \, \text{g/mol} \quad (2 \text{ C in 2 Acetate})
\]
\[
\text{Molar mass of H} = 1.008 \, \text{g/mol} \quad (3 \text{ H in each Acetate})
\]
\[
\text{Molar mass of O} = 16.00 \, \text{g/mol} \quad (2 \text{ O in each Acetate})
\]
\[
\text{Molar mass of Pb(C2H3O2)2} = 207.2 + (2 \times (12.01 \times 2 + 1.008 \times 3 + 16.00 \times 2))
\]
\[
= 207.2 + 2 \times (24.02 + 3.024 + 32.00)
\]
\[
= 207.2 + 2 \times 59.044
\]
\[
= 207.2 + 118.088 = 325.288 \, \text{g/mol}
\]
2. **Calculate the mass of water in the hydrate**:
Let \( x \) be the number of water molecules in the hydrate, then the mass of water is:
\[
\text{Mass of water} = x \times 18.015 \, \text{g/mol}
\]
3. **Set up the equation based on percentage by mass**:
The total mass of the hydrate is the sum of the mass of lead (II) acetate and the mass of water:
\[
\text{Total mass} = 325.288 + x \times 18.015
\]
The percentage of water in the hydrate is given as 14.25%:
\[
\frac{x \times 18.015}{325.288 + x \times 18.015} = 0.1425
\]
4. **Solve the equation**:
Cross-multiplying gives:
\[
x \times 18.015 = 0.1425 (325.288 + x \times 18.015)
\]
This expands to:
\[
x \times 18.015 = 46.31307 + 0.1425x \times 18.015
\]
Rearranging gives:
\[
x \times 18.015 - 0.1425x \times 18.015 = 46.31307
\]
Factoring out \( x \):
\[
x (1 - 0.1425) \times 18.015 = 46.31307
\]
\[
x \times 0.8575 \times 18.015 = 46.31307
\]
\[
x \times 15.4194 = 46.31307
\]
\[
x = \frac{46.31307}{15.4194} \approx 3.01
\]
Since \( x \) must be a whole number, we round \( x \) to 3.
5. **Final formula**:
Thus, the molecular formula of lead (II) acetate hydrate is:
\[
\text{Pb(C2H3O2)2} \cdot 3 \text{H2O}
\]
Which is commonly written as:
\[
\text{Pb(C2H3O2)2} \cdot 3\text{H2O}
\]
1. **Calculate the molar mass of lead (II) acetate (Pb(C2H3O2)2)**:
\[
\text{Molar mass of Pb} = 207.2 \, \text{g/mol}
\]
\[
\text{Molar mass of C} = 12.01 \, \text{g/mol} \quad (2 \text{ C in 2 Acetate})
\]
\[
\text{Molar mass of H} = 1.008 \, \text{g/mol} \quad (3 \text{ H in each Acetate})
\]
\[
\text{Molar mass of O} = 16.00 \, \text{g/mol} \quad (2 \text{ O in each Acetate})
\]
\[
\text{Molar mass of Pb(C2H3O2)2} = 207.2 + (2 \times (12.01 \times 2 + 1.008 \times 3 + 16.00 \times 2))
\]
\[
= 207.2 + 2 \times (24.02 + 3.024 + 32.00)
\]
\[
= 207.2 + 2 \times 59.044
\]
\[
= 207.2 + 118.088 = 325.288 \, \text{g/mol}
\]
2. **Calculate the mass of water in the hydrate**:
Let \( x \) be the number of water molecules in the hydrate, then the mass of water is:
\[
\text{Mass of water} = x \times 18.015 \, \text{g/mol}
\]
3. **Set up the equation based on percentage by mass**:
The total mass of the hydrate is the sum of the mass of lead (II) acetate and the mass of water:
\[
\text{Total mass} = 325.288 + x \times 18.015
\]
The percentage of water in the hydrate is given as 14.25%:
\[
\frac{x \times 18.015}{325.288 + x \times 18.015} = 0.1425
\]
4. **Solve the equation**:
Cross-multiplying gives:
\[
x \times 18.015 = 0.1425 (325.288 + x \times 18.015)
\]
This expands to:
\[
x \times 18.015 = 46.31307 + 0.1425x \times 18.015
\]
Rearranging gives:
\[
x \times 18.015 - 0.1425x \times 18.015 = 46.31307
\]
Factoring out \( x \):
\[
x (1 - 0.1425) \times 18.015 = 46.31307
\]
\[
x \times 0.8575 \times 18.015 = 46.31307
\]
\[
x \times 15.4194 = 46.31307
\]
\[
x = \frac{46.31307}{15.4194} \approx 3.01
\]
Since \( x \) must be a whole number, we round \( x \) to 3.
5. **Final formula**:
Thus, the molecular formula of lead (II) acetate hydrate is:
\[
\text{Pb(C2H3O2)2} \cdot 3 \text{H2O}
\]
Which is commonly written as:
\[
\text{Pb(C2H3O2)2} \cdot 3\text{H2O}
\]