Answer :
To find the empirical formula of the molecule, we follow these steps:
### Step 1: Determine the number of moles of each element.
#### For Nitrogen (N):
1. The mass of nitrogen given is 24.36 grams.
2. The molar mass of nitrogen (N) is 14.01 grams per mole.
[tex]\[ \text{Moles of N} = \frac{\text{mass of N}}{\text{molar mass of N}} = \frac{24.36 \, \text{g}}{14.01 \, \text{g/mol}} \approx 1.738758 \, \text{moles} \][/tex]
#### For Silver (Ag):
1. The mass of silver given is 62.64 grams.
2. The molar mass of silver (Ag) is 107.87 grams per mole.
[tex]\[ \text{Moles of Ag} = \frac{\text{mass of Ag}}{\text{molar mass of Ag}} = \frac{62.64 \, \text{g}}{107.87 \, \text{g/mol}} \approx 0.580699 \, \text{moles} \][/tex]
### Step 2: Determine the simplest mole ratio of the elements.
1. Calculate the mole ratio of nitrogen to silver:
[tex]\[ \text{Ratio of N} = \frac{\text{moles of N}}{\text{moles of Ag}} = \frac{1.738758}{0.580699} \approx 2.994 \][/tex]
This result is very close to 3.
2. Therefore, the ratio of nitrogen to silver is approximately 3:1.
### Step 3: Write the empirical formula.
Since the mole ratio of nitrogen to silver approximates to 3:1, the empirical formula is:
[tex]\[ AgN_3 \][/tex]
Thus, the correct empirical formula for the molecule is AgN₃.
### Step 1: Determine the number of moles of each element.
#### For Nitrogen (N):
1. The mass of nitrogen given is 24.36 grams.
2. The molar mass of nitrogen (N) is 14.01 grams per mole.
[tex]\[ \text{Moles of N} = \frac{\text{mass of N}}{\text{molar mass of N}} = \frac{24.36 \, \text{g}}{14.01 \, \text{g/mol}} \approx 1.738758 \, \text{moles} \][/tex]
#### For Silver (Ag):
1. The mass of silver given is 62.64 grams.
2. The molar mass of silver (Ag) is 107.87 grams per mole.
[tex]\[ \text{Moles of Ag} = \frac{\text{mass of Ag}}{\text{molar mass of Ag}} = \frac{62.64 \, \text{g}}{107.87 \, \text{g/mol}} \approx 0.580699 \, \text{moles} \][/tex]
### Step 2: Determine the simplest mole ratio of the elements.
1. Calculate the mole ratio of nitrogen to silver:
[tex]\[ \text{Ratio of N} = \frac{\text{moles of N}}{\text{moles of Ag}} = \frac{1.738758}{0.580699} \approx 2.994 \][/tex]
This result is very close to 3.
2. Therefore, the ratio of nitrogen to silver is approximately 3:1.
### Step 3: Write the empirical formula.
Since the mole ratio of nitrogen to silver approximates to 3:1, the empirical formula is:
[tex]\[ AgN_3 \][/tex]
Thus, the correct empirical formula for the molecule is AgN₃.
