4.00 g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 152. gimol, is burned completely in excess oxygen, and the mass of the products carefully measured: product mass
carbon dioxide 5.79 g
water 2.84 g
Use this information to find the molecular formula of X.

C₄H₁₀O₄

Explanation:

From the question given above, the following data were obtained:

Mass of compound = 4 g

Molar mass of compound = 152 g/mol

Mass of CO₂ = 5.79 g

Mass of H₂O = 2.84 g

Molecular formula of compound X =?

Next, we shall determine the mass of Carbon, Hydrogen and Oxygen in the compound. This can be obtained as follow:

For Carbon (C):

Mass of CO₂ = 5.79 g

Molar mass of CO₂ = 12 + (2×16) = 44 g/mol

Mass of C = 12/44 × 5.79

Mass of C = 1.58 g

For Hydrogen (H):

Mass of H₂O = 2.84 g

Molar mass of H₂O = (2×1) + 16 = 18 g/mol

Mass of H = 2/18 × 2.84

Mass of H = 0.32 g

For Oxygen (O):

Mass of C = 1.58 g

Mass of H = 0.32 g

Mass of compound = 4 g

Mass of O =?

Mass of O = (Mass of compound) – (Mass of C + Mass of H)

Mass of O = 4 – (1.58 + 0.32)

Mass of O = 4 – 1.9

Mass of O = 2.1 g

Next, we shall determine the empirical for compound X. This can be obtained as follow:

C = 1.58 g

H = 0.32 g

O = 2.1 g

Divide by their molar mass

C = 1.58 / 12 = 0.13

H = 0.32 / 1 = 0.32

O = 2.1 / 16 = 0.13

Divide by the smallest

C = 0.13 / 0.13 = 1

H = 0.32 / 0.13 = 2.46

O = 0.131 / 0.13 = 1

Multiply by 2 to express in whole number

C = 1 × 2 = 2

H = 2.46 × 2 = 5

O = 1 × 2 = 2

Thus, the empirical formula for the compound is C₂H₅O₂

Finally, we shall determine the molecular formula for the compound. This can be obtained as follow:

Molar mass of compound = 152 g/mol

Empirical formula => C₂H₅O₂

Molecular formula =?

Molecular formula => [C₂H₅O₂]ₙ

[C₂H₅O₂]ₙ = 152

[(2×12) + (5×1) + (2×16)]n = 152

[24 + 5 + 32]n = 152

61n = 152

Divide both side by 61

n = 152 / 61

n = 2

Molecular formula => [C₂H₅O₂]₂

Molecular formula => C₄H₁₀O₄

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