Question

When 6-chloro-2-methyl-3-cyclohexenol was submitted for elemental analysis, the results were C, 51.07%; H, 7.96%; Cl, 21.53%. Are these values within the required ±0.4% of theoretical? If not calculate the apparent empirical/molecular formula and suggest the identity of the contaminant

Answer 1

The molecular formula of the given compound = C₇H₁₁ClO

Theoretical %C = 57.35

Theoretical %H = 7.56

Theoretical %Cl = 24.18

The error in %C = {(57.35 - 51.07)/57.35}*100 = 10.95%

The error in %H = {(7.56-7.96)/7.56}*100 = -5.29%

The error in %Cl = {(24.18 - 21.53)/24.18}*100 = 10.96%

Therefore, the given values are not with in ±0.4% of theoretical values.

The apparent empirical formula can be calculated as follows.

The mass of C = 51.07 g, no. of moles = 51.07/12 = 4.2558 mol

The mass of H = 7.96 g, no. of moles = 7.96/1 = 7.96 mol

The mass of Cl = 21.53 g, no. of moles = 21.53/35.5 = 0.6065 mol

The mass of O = 19.44 g, no. of moles = 19.44/16 = 1.215 mol

Divide all the above no. of moles by smallest one, i.e. 0.6065

Therefore, no. of moles of C = 4.2558/0.6065 ~ 7

No. of moles of H = 7.96/0.6065 ~ 13

No. of moles of Cl = 0.6065/0.6065 = 1

No. of moles of O = 1.215/0.6065 ~ 2

Therefore, the empirical/molecular formula = C₇H₁₃Cl₁O₂

Therefore, the contaminant is H₂O