Question

aestion 1 of 3 > A 20.101 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion analysis apparatus, yielding 35.808 mg of carbon dioxide and 14.658 mg of water. In another experiment, 47.029 mg of the compound is reacted with excess oxygen to produce 20.32 mg of sulfur dioxide. Add subscripts to the formula provided to correctly identify the empirical formula of this compound. Do not change the order of the elements. empirical formula: CHSO Publisher University Question Source: MRG - General Chemistry

Answer 1

Combustion of 20.101 mg of compound gives 35.808 mg of CO₂ and 14.658 mg of H₂O

Mass of Carbon present in the compound = Mass of CO₂ * (Molar mass of C / Molar mass of CO₂)

= 35.808 * 12.01 / 44.01 = 9.77 mg

20.101 mg of sample contains 9.77 mg of C

Therefore, 100 mg sample will contain mass of C =  (9.77 / 20.101) * 100 = 48.604 mg

Moles of C contained in 100 mg of sample = 48.604 / 12.01

= 4.047 moles of C

Mass of Hydrogen present in the compound = Mass of H₂O * (2 * Molar mass of H / Molar mass of H₂O)

= 14.658 * (2.016/18.016) = 1.64 mg

20.101 mg of sample contains 1.64 mg of H

Therefore, 100 mg sample will contain mass of H =  (1.64 / 20.101) * 100 = 8.159 mg

Moles of H contained in 100 mg of sample = 8.159 / 1.008

= 8.094 moles of H

Now 47.029 mg of compound is reacted with excess O₂ to form 20.32 mg of SO₂

Therefore, Mass of Sulphur present in the compound = Mass of SO₂ * (Molar mass of S / Molar mass of SO₂)

= 20.32 * (32.065/64.06) = 10.17 mg

47.029 mg of sample contains 10.17 mg of S

Therefore, 100 mg sample will contain mass of S =  (10.17 / 47.029) * 100 = 21.624 mg

Moles of S contained in 100 mg of sample = 21.624 / 32.065

= 0.6744 moles of S

Now mass of O present in 100 mg of sample will be = 100 - Mass of C - Mass of H - Mass of S

= 100 - 48.604 - 8.159 - 21.624

= 21.613 mg

Therefore, Moles of O contained in 100 mg of sample = 21.613 / 16

= 1.3508 moles of O

Now, we will divide the moles of all the elements present by the smallest number of moles present (of Sulphur in this case) to get the ratio of the elements present in compound.

Therefore, Ratio of C = 4.047/0.6744 = 6

Ratio of H = 8.094/0.6744 = 12

Ratio of S = 0.6744/0.6744 = 1

Ratio of O = 1.3508/0.6744 = 2

Therefore, Empirical formula of compound is C₆H₁₂SO₂