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

CHCl₃

Explanation:

We have the following data:

C = 5.03 g

H = 0.42 g

Cl= 44.5 g

First, we divide each mass by the molar mass (MM) of the chemical element to calculate the moles:

MM(C) = 12 g/mol

moles of C = mass/MM(C) = 5.03 g/(12 g/mol) = 0.42 mol C

MM(H) = 1 g/mol

moles of H = mass/MM(H) = 0.42 g/(1 g/mol) = 0.42 mol H

MM(Cl) = 35.4 g/mol

moles of Cl = mass/MM(Cl) = 44.5 g/(35.4 g/mol) = 1.26 mol Cl

Now, we divide the moles by the smallest number of moles (0.42):

0.42 mol C/0.42 = 1 C

0.42 mol H/0.42 = 1 H

1.26 mol Cl/0.42 = 3 Cl

Thus, the C:H:Cl ratio is 1:1:3.

Therefore, the empirical formula is CHCl₃

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