A 25.0 g sample of a compound is found to contain 6.075 g carbon, 1.025 g hydrogen, and 17.9 g chlorine. Its molar mass is determined to be 99 g/mol. What is the molecular formula for this compound

C₂H₄Cl₂ or CH₃CHCl₂

Explanation:

The following data were obtained from the question:

Carbon (C) = 6.075 g

Hydrogen (H) = 1.025 g

Chlorine (Cl) = 17.9 g

Molar mass of compound = 99 g/mol.

Molecular formula for compound =.?

Next, we shall determine the empirical formula for the compound. This can be obtained as illustrated below:

C = 6.075 g

H = 1.025 g

Cl = 17.9 g

Divide by their molar mass

C = 6.075 / 12 = 0.506

H = 1.025 / 1 = 1.025

Cl = 17.9 /35.5 = 0.504

Divide by the smallest

C = 0.506 / 0.504 = 1

H = 1.025 / 0.504 = 2

Cl = 0.504 / 0.504 = 1

Therefore, the empirical formula for the compound is CH₂Cl.

Finally, we shall determine the molecular formula for the compound as follow:

[CH₂Cl]ₙ = 99

[12 + (2×1) + 35.5]n = 99

[12 + 2 + 35.5]n = 99

49.5n = 99

Divide both side by 49.5

n = 99/49.5

n = 2

Thus,

[CH₂Cl]ₙ => [CH₂Cl]₂ => C₂H₄Cl₂

the molecular formula for the compound C₂H₄Cl₂ or CH₃CHCl₂