To determine the empirical formula of the compound containing nickel (Ni) and fluorine (F), we need to follow these steps:
1. Determine the number of moles of each element present in the sample:
- The molar mass of nickel (Ni) is 58.6934 g/mol.
- The molar mass of fluorine (F) is 18.998 g/mol.
Given masses:
- Mass of Ni = 9.11 g
- Mass of F = 5.89 g
First, calculate the moles of each element:
[tex]\[
\text{Moles of Ni} = \frac{\text{Mass of Ni}}{\text{Molar mass of Ni}} = \frac{9.11 \, \text{g}}{58.6934 \, \text{g/mol}} \approx 0.155
\][/tex]
[tex]\[
\text{Moles of F} = \frac{\text{Mass of F}}{\text{Molar mass of F}} = \frac{5.89 \, \text{g}}{18.998 \, \text{g/mol}} \approx 0.310
\][/tex]
2. Determine the simplest whole number ratio between the moles of the elements:
Use the smallest number of moles to find the ratio. In this case, the smaller value is the moles of Ni (0.155).
[tex]\[
\text{Ratio of Ni} = \frac{\text{Moles of Ni}}{\text{Smallest number of moles}} = \frac{0.155}{0.155} = 1.0
\][/tex]
[tex]\[
\text{Ratio of F} = \frac{\text{Moles of F}}{\text{Smallest number of moles}} = \frac{0.310}{0.155} \approx 1.997
\][/tex]
3. Round the ratios to the nearest whole number to get the empirical formula:
The ratio of Ni is 1, and the ratio of F is approximately 2. Since the compound ratios round to whole numbers easily:
The empirical formula of the compound is:
[tex]\[
\text{NiF}_2
\][/tex]
Therefore, the empirical formula of the compound is [tex]\(\text{NiF}_2\)[/tex], and the correct choice among the provided options is:
[tex]\[
\boxed{NiF_2}
\][/tex]
