Question

Four gram of copper chloride contain 1.890 g of copper and 2.110 g of chlorine. find it's empirical formula

Answer 1

Given,mass of copper in copper chloride = 1.89 g
mass of chlorine in copper chloride = 2.11 g
Percentage of copper in 4g
copper chloride = (1.89/4) x 100 = 47.25%
Percentage of chlorine in 4 g copper chloride = (2.11/4) x 100 = 52.75 %
Therefore, in a 100 g sample:
mass of copper = 47.25g
moles of copper = 47.25/63.5 = 0.74mass of chlorine = 52.75 g
moles of chlorine = 52.75/35.5 = 1.48
Ratio of moles of copper and chlorine = 0.7
4:1.48= 1 : 2
Thus, the empirical formula of the compound = CuCl2

Answer 2

Explanation:

When a chemical formula that represents proportions of elements depicted in a compound is known as an empirical formula.

Since, it is given that in four gram copper chloride there is 1.890 g of copper and 2.110 g of chlorine. So, calculate the number of moles present in copper and chlorine as follows.

       No. of moles in copper = $\frac{mass}{molar mass of copper}$

                                              = $\frac{1.890 g}{63.54 g/mol}$

                                              = 0.0297 g/mol

       No. of moles in chlorine = $\frac{mass}{molar mass of chlorine}$

                                              = $\frac{2.110 g}{35.45 g/mol}$

                                              = 0.0595 g/mol

Therefore, molar ratio will be calculated as follows.

           Molar ratio for copper = $\frac{0.0595 g/mol}{0.0595 g/mol}$

                                                 = 1

          Molar ratio for chlorine = $\frac{0.0595 g/mol}{0.0297 g/mol}$

                                                 = 2

Thus, we can conclude that empirical formula of given compound is $CuCl_{2}$.