Question

Copper metal of 1.23 g sample is reacted completely with chlorine gas to produce
2.61 g of copper chloride. Determine the empirical formula of the compound?​

Answer 1

Answer: The empirical formula of the compound is $CuCl_2$

Explanation:

Mass of Copper (Cu) = 1.23 g

Mass of Chlorine (Cl)  = Mass of copper chloride - mass of copper = (2.61 - 1.23) g = 1.38 g

Step 1 : convert given masses into moles

Moles of Cu =$\frac{\text{ given mass of Cu}}{\text{ molar mass of Cu}}= \frac{1.23g}{63.5g/mole}=0.019moles$

Moles of Cl =$\frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{1.38g}{35.5g/mole}=0.038moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Cu =$\frac{0.019}{0.019}=1$

For Cl = $\frac{0.038}{0.019}=2$

The ratio of Cu : Cl = 1 : 2

Hence the empirical formula is $CuCl_2$