A 36.165 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion analysis apparatus, yielding 64.425 mg of carbon dioxide and 26.373 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.

C6H12SO2

Explanation:

Step 1:

Data obtained from the question.

First experiment:

Mass of the compound = 36.165 mg

Mass of CO2 = 64.425 mg

Mass of H2O = 26.373 mg

Second experiment:

Mass of compound = 47.029 mg

Mass of SO2 = 20.32 mg

Step 2:

Determination of the mass of C, H and S. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C in CO2 = 12/44 x 64.425 Mass of C = 17.57 mg

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H in H2O = 2/18 x 26.373

Mass of H = 2.93 mg

Molar Mass of SO2 = 32 + (16x2) = 64g/mol

Mass of S in SO2 = 32/64 x 20.32

Mass of S = 10.16 mg

Step 3:

Determination of the the percentage composition of C, H, S and O. This is illustrated below:

% of C = 17.57/36.165 x 100 = 48.58%

% of H = 2.93/36.165 x 100 = 8.10%

% of S = 10.16/47.029 x 100 = 21.60%

% of O = 100 - (48.58 + 8.1 + 21.6)

% of O = 21.72%

Step 4:

Divide by their molar mass

C = 48.58x10^-3/12 = 4.0483

H = 8.10/1 = 8.1

S = 21.60/32 = 0.675

O = 21.72/16 = 1.3575

Step 5:

Divide by the smallest:

C = 4.0483/0.675 = 6

H = 8.1/0.675 = 12

S = 0.675/0.675 = 1

O = 1.3575/0.675 = 2

Therefore, the empirical formula is C6H12SO2

d: the voltage of the secondary coil is less than that of the primary coil.

im not suree sorry if wrong

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