Answer:
[tex]\text{CaCl}_2[/tex]
Explanation:
Here, we want to determine the empirical formula of the compound
We start by dividing the masses by the atomic mass of the individual elements
[tex]\begin{gathered} Cl\text{ = }\frac{2.128}{35.5}\text{ = 0.0599437} \\ \\ Ca\text{ = }\frac{1.203}{40}\text{ = 0.030075} \end{gathered}[/tex]
The next thing to do is to divide each of the values by the smallest value from above
Mathematically, we have that as:
[tex]\begin{gathered} Cl\text{ = }\frac{0.0599437}{0.030075\text{ = 2}} \\ \\ Ca\text{ = }\frac{0.030075}{0.03375}\text{ = 1} \end{gathered}[/tex]
The empirical formula of the compound is thus:
[tex]\text{CaCl}_2[/tex]
