Question

1- 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

2- Elemental analysis of a compound gave the results C, 65.44%; H, 5.49%; N, 8.58%. Calculate the empirical formula. Determination of the molecular weight by "wet" methods indicated that the molecular weight is between 150 and 180 g/mole. Determine the molecular formula, the molecular weight, and the index of hydrogen deficiency. Suggest a possible structure for the compound.

3- Elemental analysis of a compound gave the results C, 73.60%; H, 8.03%; N, 8.58%. The molecular weight of the compound is 326. Calculate the molecular formula and the index of hydrogen deficiency.

4-. Elemental analysis of a compound gave the results C, 69.72%; H, 11.7%, and the mass spectrum gave the highest amu ion at m/z 71. Calculate the empirical formula, the molecular formula, and the true molecular weight.

Answer 1

1.

From the given data,

moles C = 51.07 g/12 g/mol = 4.25 moles

moles H = 7.96 g/1 g/mol = 7.96 moles

moles Cl = 21.53 g/35.5 g/mol = 0.61 moles

moles O = (100 - (51.07+7.96+21.53)) g/16 g/mol = 1.215 moles

divide by smallest factor

C = 4.25/0.61 = 7

H = 7.96/0.61 = 13

O = 0.61/0.61 = 1

Cl = 1.215/0.61 = 2

Empirical formula = C7H13ClO2

formula for 6-chloro-2-methyl-3-cyclohexenol = C7H11ClO

remainder = H2O

so the identity of the conaminant = H2O

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2.

From the given data,

moles C = 65.44 g/12 g/mol = 5.45 moles

moles H = 5.49 g/1 g/mol = 5.49 moles

moles N = 8.58 g/14 g/mol = 0.61 moles

moles O = (100 - (65.44 + 5.49 + 8.58) g/16 g/mol = 1.28 moles

Divide by smallest factor

moles C = 5.45/0.61 = 9

moles H = 5.49/0.61 = 9

moles N = 0.61/0.61 = 1

moles O = 1.28/0.61 = 2

Empirical formula = C9H9NO2

Hydrogen deficiency = 6

Possible structure

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3. From the given data

moles C = 73.60 g/12 g/mol = 6.13 moles

moles H = 8.03 g/1 g/mol = 8.03 moles

moles N = 8.58 g/14 g/mol = 0.61 moles

moles O = (100 - (73.60 + 8.03 + 8.58))/16 = 0.61 moles

divide by smallest factor,

C = 6.13/0.61 = 10

H = 8.03/0.61 = 13

N = 0.61/0.61 = 1

O = 0.61/0.61 = 1

empirical formula = C10H13NO

empirical formula mass = 10 x 12 + 13 x 1 + 1 x 14 + 1 x 16 = 163 g/moles

molar mass/empircial formula mass = 326/163 = 2

Molecular formula = C20H26N2O2

Degree of hydrogen defiiciency = 9

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4. From given data

moles C = 69.72 g/12 g/mol = 5.81 moles

moles H = 11.7 g/1 g/mol = 11.7 moles

moles O = (100 - (69.72 + 11.7)) g/16 g/mol = 1.16 moles

Divide by smallest factor

C = 5

H = 10

O = 1

empirical formula = C5H10O

Degree of unsaturation = 1

molecular formula = C5H10O

True molecular weight = 5 x 12 + 10 x 1 + 1 x 16 = 86 g/mol

Structure,

CH3-CH2-CH2-CO-CH3