Calculate the percent composition by mass of magnesium in Mg(NO3)2.
The Correct Answer and Explanation is:
To calculate the percent composition by mass of magnesium in magnesium nitrate, Mg(NO3)2\text{Mg(NO}_3)_2Mg(NO3)2, follow these steps:
Step 1: Find the molar mass of Mg(NO3)2\text{Mg(NO}_3)_2Mg(NO3)2
Start by calculating the molar mass of magnesium nitrate. This involves adding the molar masses of each element in the compound. The atomic masses are approximately:
- Magnesium (Mg) = 24.31 g/mol
- Nitrogen (N) = 14.01 g/mol
- Oxygen (O) = 16.00 g/mol
Now, break down the compound Mg(NO3)2\text{Mg(NO}_3)_2Mg(NO3)2:
- One magnesium atom: 1×24.31=24.31 g/mol1 \times 24.31 = 24.31 \, \text{g/mol}1×24.31=24.31g/mol
- Two nitrate ions, NO3\text{NO}_3NO3: Each nitrate consists of one nitrogen atom and three oxygen atoms. Therefore:
- Nitrogen: 2×14.01=28.02 g/mol2 \times 14.01 = 28.02 \, \text{g/mol}2×14.01=28.02g/mol
- Oxygen: 6×16.00=96.00 g/mol6 \times 16.00 = 96.00 \, \text{g/mol}6×16.00=96.00g/mol
Now add all these together:Molar mass of Mg(NO3)2=24.31+28.02+96.00=148.33 g/mol\text{Molar mass of Mg(NO}_3)_2 = 24.31 + 28.02 + 96.00 = 148.33 \, \text{g/mol}Molar mass of Mg(NO3)2=24.31+28.02+96.00=148.33g/mol
Step 2: Calculate the mass percent of magnesium
Next, calculate the mass percent of magnesium in magnesium nitrate by dividing the mass of magnesium by the total molar mass of the compound and then multiplying by 100:Percent of Mg=(24.31148.33)×100=16.4%\text{Percent of Mg} = \left( \frac{24.31}{148.33} \right) \times 100 = 16.4\%Percent of Mg=(148.3324.31)×100=16.4%
Conclusion:
The percent composition by mass of magnesium in Mg(NO3)2\text{Mg(NO}_3)_2Mg(NO3)2 is approximately 16.4%. This means that 16.4% of the mass of magnesium nitrate is contributed by magnesium.
