Answer:
C₂H₄Cl₂ or CH₃CHCl₂
Step-by-step 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