Please help!! A compound with a molar mass of 544.0 g/mol is made up of 26.5 grams Carbon, 2.94 grams Hydrogen, and 70.6 grams Oxygen. What is its empirical and molecular formula?

1. Empirical formula = C₃H₄O₆

2. Molecular formula = C₁₂H₁₆O₂₄

Explanation:

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

Molar mass of compound = 544 g/mol

Mass of Carbon (C) = 26.5 g

Mass of Hydrogen (H) = 2.94 g

Mass of oxygen (O) = 70.6 g

Empirical formula =?

Molecular formula =?

1. Determination of the empirical formula of the compound.

C = 26.5 g

H = 2.94 g

O = 70.6 g

Divide by their molar mass

C = 26.5 / 12 = 2.208

H = 2.94 / 1 = 2.94

O = 70.6 / 16 = 4.4125

Divide by the smallest

C = 2.208 / 2.208 = 1

H = 2.94 / 2.208 = 1.33

O = 4.4125 / 2.208 = 2

Muitiply through by 3 to express in whole number.

C = 1 × 3 = 3

H = 1.33 × 3 = 4

O = 2 × 3 = 6

Empirical formula = C₃H₄O₆

2. Determination of the molecular formula of the compound.

Molar mass of compound = 544 g/mol

Empirical formula = C₃H₄O₆

Molecular formula = [C₃H₄O₆]ₙ

[C₃H₄O₆]ₙ = 544

[(12×3) + (4×1) + (16×6)]n = 544

[36 + 4 + 96]n = 544

136n = 544

Divide both side by 136

n = 544 / 136

n = 4

Molecular formula = [C₃H₄O₆]ₙ

Molecular formula = [C₃H₄O₆]₄

Molecular formula = C₁₂H₁₆O₂₄

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