Question

What is the empirical formula of a compound containing 5.03 grams carbon, 0.42 grams hydrogen, and 44.5 grams chlorine?

Answer 1

The empirical formula of the compound is CHCl₃.

Calculation:

Given,

Mass of carbon = 5.03 g

Mass of hydrogen = 0.42 g

Mass of chlorine = 44.5 g

Molecular weight of carbon = 12 g

Molecular weight of hydrogen = 1 g

Molecular weight of chlorine = 35.4 g

First, calculate the moles of each element,

                      Moles = given mass/ molecular weight

Moles of carbon = 5.03/12 = 0.42

Moles of hydrogen = 0.42/1 = 0.42

Moles of chlorine = 44.5/ 35.4 = 1.26

Divide the moles of each element by the smallest number of moles,

0.42 mol of C/ 0.42 = 1 C

0.42 mol of H/ 0.42 = 1 H

1.26 mol of Cl/0.42 = 3 Cl

The ratio of elements is 1:1:3

Therefore the empirical formula of the compound will be CHCl₃.

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