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

Question

Unknown compound / has a melting point of 102-3°c. Carbon/hydrogen analysis of com 304% carbon and 21% hydrogen. A Carius sulfur analysis of a 8.45-mg sample of 8.32 mg of Ba resulting solution produced 8.32 mg of AgB approximately 235-240. Treatment of compound I with aqueous acid and the formula C&HsBr.; Determine the molecular formula for compound I. SO4. A 10.50-mg sample of compound I was reacted in a sodium fusion reactor compound I showed e of compound I yielded r. Based on mass spectral data, the M' peak was found to be. heat yielded a compound with (2 points)

Explanation / Answer

Given: Melting Point of Unknown (Compound I): 102 -103 0C   

Conclusion: It is a pure compound as the the meting range is very narrow and since it melts at 102 deg C, it is a solid.

Given: Compound I contains 30.4% Carbon nad 2.1% H   

Conclusion: For every 30.4g of C, moles of C = 30.4 g/12.00 g/mol = 2.533 moles of C

For every 2.1g of H, moles of H = 2.1 g/1.01 g/mol = 2.080 moles of H

Given: Carious sulfur analysis of 8.45 mg of compound I yielded 8.32 mg of BaSO4

Conclusion: In carius sulfur analysis,

C in the organic compound is oxidised to CO2 (C + 2O-----> CO2)

H in the the organic compound is oxidised to H2O (2H + O-----> H2O)

S in the the organic compound is oxidised to H2SO4 (S + H2O + 3O -----> H2SO4)

The H2SO4 thus produced reacts with BaCl2 to produce BaSO4 (BaCl2 + H2SO4 ----------> BaSO4 + 2 HCl )

It indicates that 1 mole of S would produce 1 mole of BaSO4 and 1 mole of BaSO4 contains 32.00 g of Sulfur (molar mass of S = 32.00 g/mol)

Total mass of organic compound reacted = 8.45 mg

Mass of BaSO4 formed = 8.32 mg

1 mole of BaSO4 = 233.4 g/mol = 32.00 g of Sulfur

So g of S in 8.32 mg of BaSO4 = (8.32 x 10-3 g / 233.4 g/mol) * 32.00 g/mol = 1.139 x10-3 g or 1.139 mg S

So % Sulfur = (mass of S/ mass of organic compd)* 100% = (1.139 mg S/ 8.45 mg org. compd) * 100% = 13.4 % S

mole ratio of S = 13.4 g /32.00 g/mol = 0.419 moles of S

Given: Carius halide analysis of 10.50 mg of compound I yielded 8.32 mg of AgBr

Conclusion: The reaction involved here would be:

This is similar to sulfur analysis done above. Br from the organic compound reacts with AgNO3 to produce AgBr.Which indicates that 1 mole of Br would produce 1 mole of AgBr and 1 mole of BaSO4 contains 79.90 g of Br.

Total mass of organic compound reacted = 10.50 mg

Mass of AgBrformed = 8.32 mg

1 mole of AgBr = 187.8 g/mol = 79.90 g of Br

So g of Br in 8.32 mg of AgBr= (8.32 x 10-3 g / 187.8 g/mol) * 79.90 g/mol = 3.540 x10-3 g or 3.540 mg Br

So % Br = (mass of Br/ mass of organic compd)* 100% = (3.540 mg S/ 10.50 mg org. compd) * 100% = 33.7 % Br

mole ratio of Br = 33.7 g / 79.90 g/mol = 0.422 moles of S

The rest of the % mass should be from O,

So % O = 100 - (30.4 + 2.1 + 13.4 + 33.7) = 20.4 g of O

mole ratio of O = 20.4 g of O/ 15.99 g/mol O = 1.276 moles O

We now divide all mole ratios of C, H, S and Br with the least number of moles to get the number of relative atoms.

So the % composition of each of the components would be:

C--------2.53/0.419 = 6.04

H--------2.08/0.419 = 4.96

S--------0.419/0.419 = 1.00

Br--------0.421/0.419 = 1.00

O--------1.276/0.419 = 3.04

So the emperical formula would be C6H5O3SBr

Molar mass of C6H5O3SBr = {(6x12.00) + (5*1.01) + (3*15.99) + (1*32.00) + (1*79.90)}

= 72.00 + 5.05 + 47.97 + 32.00 + 79.90 = 236.92 g/mol, which is in the range of 235 - 240 as given in the question.

So Compound I is C6H5O3SBr

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