Question

1. a compound consists of 72.2% magnesium and 27.8% nitrogen by mass. what is the empirical formula? element mass (g) atomic mass mole mole ratio empirical formula:

Explanation:

Step1: Assume 100g of the compound

Since the percentages are given by mass, assuming 100g of the compound means there are 72.2g of magnesium (Mg) and 27.8g of nitrogen (N).

Step2: Calculate the moles of each element

The molar mass of Mg is approximately 24.31 g/mol, and for N is approximately 14.01 g/mol.
The number of moles of Mg, $n_{Mg}=\frac{72.2g}{24.31g/mol}\approx2.97mol$.
The number of moles of N, $n_{N}=\frac{27.8g}{14.01g/mol}\approx1.98mol$.

Step3: Find the mole - ratio

Divide each number of moles by the smaller number of moles. Here, 1.98mol is smaller.
The mole - ratio of Mg:$ \frac{2.97mol}{1.98mol}\approx1.5$.
The mole - ratio of N: $\frac{1.98mol}{1.98mol} = 1$.
Since we need whole - number ratios, multiply both by 2 to get a ratio of Mg:N = 3:2.

Answer:

The empirical formula is $Mg_3N_2$