The empirical formula of the compound is CHCl₃.
<h3>Calculation:</h3>
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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