. A 5.0 g sample of calcium nitrate (Ca(NO3)2, M = 164 ) contaminated with silica (SiO₂, M = 60.1 ) is found to contain 1.0 g calcium. What is the mass percent purity of calcium nitrate in the sample?
a. 20%
b. 24%
c. 73%
d. 82%
The Correct Answer and Explanation is :
The problem asks to find the mass percent purity of calcium nitrate (Ca(NO₃)₂) in a sample that contains both calcium nitrate and silica (SiO₂), where the total sample mass is 5.0 g, and the mass of calcium in the sample is 1.0 g.
Step 1: Determine the amount of calcium in calcium nitrate
From the chemical formula of calcium nitrate (Ca(NO₃)₂), we know that one mole of Ca(NO₃)₂ contains one mole of calcium (Ca). The molar mass of Ca(NO₃)₂ is 164 g/mol, and the molar mass of calcium is 40.1 g/mol.
Thus, the mass of calcium in one mole of Ca(NO₃)₂ is 40.1 g.
Step 2: Calculate the amount of calcium nitrate corresponding to 1.0 g of calcium
Using the molar mass of calcium nitrate and the molar mass of calcium, we can calculate the amount of calcium nitrate that contains 1.0 g of calcium.
[
\text{mass of Ca(NO₃)₂} = \frac{\text{mass of Ca}}{\text{mass of Ca in Ca(NO₃)₂}} \times \text{molar mass of Ca(NO₃)₂}
]
[
\text{mass of Ca(NO₃)₂} = \frac{1.0 \, \text{g}}{40.1 \, \text{g/mol}} \times 164 \, \text{g/mol}
]
[
\text{mass of Ca(NO₃)₂} = 4.09 \, \text{g}
]
Step 3: Calculate the mass percent purity of calcium nitrate
Now, we can calculate the mass percent purity of calcium nitrate in the sample:
[
\text{mass percent purity} = \frac{\text{mass of Ca(NO₃)₂}}{\text{total sample mass}} \times 100
]
[
\text{mass percent purity} = \frac{4.09 \, \text{g}}{5.0 \, \text{g}} \times 100 = 81.8\%
]
Conclusion
The mass percent purity of calcium nitrate in the sample is approximately 82%, so the correct answer is d. 82%.