Final answer:
The percent compositions of the elements in Mg(NO₃)₂ are as follows: Mg (magnesium) contributes approximately 16.87%, N (nitrogen) contributes around 36.86%, and O (oxygen) contributes about 46.27%.
Step-by-step explanation:
The percent composition of a compound is the percentage by mass of each element in the compound. To find the percent composition of Mg(NO₃)₂, we need to determine the molar mass of each element and the molar mass of the entire compound.
Firstly, calculate the molar mass of Mg(NO₃)₂ by adding the individual molar masses. The molar mass of Mg is approximately 24.31 g/mol, N is about 14.01 g/mol, and O is around 16.00 g/mol. Since there are two nitrate ions (NO₃) in the formula, multiply the molar mass of NO₃ by 2. Then, add the molar masses of Mg and the two NO₃ groups to find the total molar mass of Mg(NO₃)₂.
Next, find the percent composition of each element. Divide the molar mass of each element by the total molar mass of the compound and multiply by 100 to get the percentage.
- Percent Mg: (24.31 g/mol / total molar mass) * 100 ≈ 16.87%
- Percent N: (2 * 14.01 g/mol / total molar mass) * 100 ≈ 36.86%
- Percent O: (6 * 16.00 g/mol / total molar mass) * 100 ≈ 46.27%
Therefore, the percent composition of Mg(NO₃)₂ is 16.87% Mg, 36.86% N, and 46.27% O. This information is crucial for understanding the elemental makeup of compounds, aiding in various scientific and chemical applications.