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`

Step-by-step 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}}= (1.23g)/(63.5g/mole)=0.019moles`

Moles of Cl =
`\frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= (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 =
`(0.019)/(0.019)=1`

For Cl =
`(0.038)/(0.019)=2`

The ratio of Cu : Cl = 1 : 2

Hence the empirical formula is
`CuCl_2`