A 30.141 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion analysis apparatus, yielding 53.694 mg of carbon dioxide and 21.980 mg of water. In another experiment, 30.199 mg of the compound is reacted with excess oxygen to produce 13.05 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

The empirical formula of the chemical compound is C₆H₁₂SO₂

Explanation:

To determine the empirical formula, the mass of each element in the compound is first determined from their respective compounds formed after combustion analysis.

Mass of element = mass of compound × molar mass of element / molar mass of compound

Carbon: 53.694 g x (12g/mol / 44.0 g/mol) = 14.643 mg of C in 30.141 mg sample Hydrogen: 21.980 mg x ( 2.00 / 18.00) = 2.442 mg of H in 30.141 mg sample

Sulfur: 13.05 mg x (32.00/ 64.00) = 6.525 mg of S in 30.199 mg in sample

The percentage mass composition of the elements in the compound is then determined:

Carbon: 14.643 mg / 30.141 mg = 48.58%

Hydrogen: 2.442 mg / 30.141 mg = 8.10%

Sulfur: 6.525 mg / 30.199 mg = 21.60 %

Oxygen: 100% - (48.58% + 8.10% +21.60%) = 21.72%

Number of moles of each element is then determined:

Number of moles = percentage mass/molar mass

Carbon: 48.58/12 = 4.07

Hydrogen: 8.10/1 = 8.10

Sulfur: 21.60/32 = 0.68

Oxygen: 21.72/16 = 1.36

Dividing with the smallest amount to obtain a whole number mole ratio

Carbon: 4.07 ÷ 0.68 = 6

Hydrogen: 8.10 ÷ 0.68 = 12

Sulfur: 0.68 ÷ 0.68 = 1

Oxygen: 1.36 ÷ 0.68 = 2

Therefore, the empirical formula of the chemical compound is C₆H₁₂SO₂

- lead + phosphoric acid = hydrogen + lead(ii) phosphate

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