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two cpds have the same composition; 85.62 persent C. 14.38 persent of H. A/ OPTAIN THE EMPERICAL formula correspondig to this composition. B/one of cpds has amolecular mass of 28.03 amu; the other, of 56.06 amu . obtain the molecular formulas of both cpds.

User Peck
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The empirical formula of the compounds is CH₂. Dividing the given molecular masses by the empirical formula mass gives factors of 2 and 4, leading to molecular formulas of C₂H₄ for the first compound and C₄H₈ for the second.

To find the empirical formula of a compound with 85.62% carbon (C) and 14.38% hydrogen (H), we first assume a 100 g sample to simplify calculations. This means we have 85.62 g of C and 14.38 g of H. To convert these into moles we use atomic masses: C (12.01 g/mol) and H (1.008 g/mol).

Moles of C = 85.62 g ÷ 12.01 g/mol = approximately 7.13 mol

Moles of H = 14.38 g ÷ 1.008 g/mol = approximately 14.26 mol

The mole ratio of C to H is roughly 1:2, so the most simplified ratio of the two elements is CH₂, which is the empirical formula.

Next, to find the molecular formulas, we divide the given molecular masses by the empirical formula mass (C: 12.01 amu, H: 1.008 amu). The empirical formula mass of CH₂ is 14.026 amu.

Molecular Mass of Compound 1 = 28.03 amu

Molecular Mass of Compound 2 = 56.06 amu

Divide these by the empirical formula mass:

28.03 amu ÷ 14.026 amu = 2

56.06 amu ÷ 14.026 amu = 4

These are the factors by which we multiply the empirical formula. Thus, the molecular formula for the first compound is C₂H₄ (ethene) and for the second it's C₄H₈ (butene). These compounds are different isomers of each other, sharing the same empirical formula but having distinct molecular structures and properties.

User Kamaya
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