2. Unknown compound /has a melting point of 102-3°C. Carbon/hydrogen analysis of

Question

2. Unknown compound /has a melting point of 102-3°C. Carbon/hydrogen analysis of compound / showed 30.4% carbon and 21% hydrogen Carius sulfur analysis of a 8.45 mg sample of compound 1 yielded 8.32 mg of BaSO4. A 10.50-mg sample of compound / was reacted in a sodium fusion reactor and the resulting solution produced 8.32 mg of AgBr. Based on mass spectral data, the M peak was found to be approximately 235-240. Treatment of compound / with aqueous acid and heat yielded a compound with the formula C&HsBr.; Determine the molecular formula for compound I. (2 points)

Explanation / Answer

Given % Carbon = 30.4%

Hydrogen = 2.1%

a) 8.45 mg of sample yielded 8.32 mg of BaSO4

Molar mass BaSO4 = 233.38 g/mole

Atomic mass of Sulfur = 32.065 g/mole

Weight of sulfur in 8.32 mg of BaSO4 = (atomic mass of Sulfur x Weight of BaSO4)/ molar mass of BaSO4 =(32.065 x 8.32)/ 233.38 = 1.1431 mg

Weight percentage of Sulfur in Compound I = (weight of sulfur in sample x 100)/ weight of sample = (1.1431 x 100)/8.45 = 13.53 %

b) 10.50 mg of sample yielded 8.32 mg of AgBr during sodium fusion

Molar mass of AgBr = 187.77 g/mole

Atomic mass of Br = 79.904 g/mole

So mass of Br in sample = (79.904 x 8.32)/187.77 = 3.5405 mg

weight percentage of Br in Compound = (3.5405 x 100)/10.50 = 33.72%

c) Since after reaction with aqueous acid heat gives C6H5Br, It must be a sulfonic acid group (SO3H) because benzene sulfonic acids undergoes desulfonation after heating in aqueous acid. So other element will be Oxygen and Weight percentage of oxygen can be calculated as

%weight of O = (100 - % weight of C - %weight of H - %weight of Br - %weight of S) = (100 - 30.4 - 2.1 - 13.53 - 33.72) = 20.25 %

d) weight fraction = %weight / atomic mass

weight fraction of C = 30.4/12 = 2.53

weight fraction of H = 2.1/ 1 = 2.1

weight fraction of S = 13.53/32 = 0.42

weight fraction of Br = 33.72/79.9 = 0.42

Weight fraction of O = 20.25/16 = 1.27

We have to make weight fraction to be an integer value so that it can give number atoms in molecule.

We will take least weight fraction element make it as one by multiplying it with (1 /0.42) = 2.38

After multiplying weight fraction each element with 2.38 we get

For Carbon = 2.53 x 2.38 = 6.02 = 6 (approx.)

For Hydrogen = 2.1 x 2.38 = 4.998 = 5 (approx.)

For Bromine = 0.42 x 2.38 = 1 (approx)

For Sulfur = 0.42 x 2.38 = 1 (approx.)

For Oxygen = 1.27 x 2.38 = 3.02 = 3 (approx.)

So emperical formula of compound I will be (C6H5BrSO3)

Emperical formula weight = 238

From mass spectroscopy molecular ion peak is around 235-240 , Our emperical formula is molecular formula compound.

Molecular formula of compound I = C6H5BrSO3

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