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1. A sample of a chemical compound has 2.96g carbon, 0.414g hydrogen, 0.675g oxygen and 2.96 sulphur.
calculate it's empirical formula.

2.if it's mass was 972g, determine its molecular formula.​

1. The empirical formula => C₆H₁₀OS₂

2. Molecular formula => C₃₆H₆₀O₆S₁₂

Explanation:

1. Determination of the empirical formula.

Carbon (C) = 2.96 g

Hydrogen (H) = 0.414 g

Oxygen (O) = 0.675 g

Sulphur (S) = 2.96 g

Divide by their molar mass

C = 2.96 / 12 = 0.247

H = 0.414 / 1 = 0.414

O = 0.675 / 16 = 0.042

S = 2.96 / 32 = 0.0925

Divide by the smallest

C = 0.247 / 0.042 = 6

H = 0.414 / 0.042 = 10

O = 0.042 / 0.042 = 1

S = 0.0925 /0.042 = 2

Therefore, the empirical formula is

C₆H₁₀OS₂

2. Determination of the molecular formula.

Molar mass of compound = 972 g/mol

Empirical formula => C₆H₁₀OS₂

Molecular formula =>?

Molecular formula = [C₆H₁₀OS₂]ₙ

Molecular formula = molar mass of compound

Thus,

[C₆H₁₀OS₂]ₙ = 972

[(12×6) + (10×1) + 16 + (32×2)]n = 972

[72 + 10 + 16 + 64]n = 32

162n = 972

Divide both side by 162

n = 972 / 162

n = 6

Molecular formula = [C₆H₁₀OS₂]ₙ

Molecular formula = [C₆H₁₀OS₂]₆

Molecular formula = C₃₆H₆₀O₆S₁₂

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.