Chemistry High School

I really need help on this question! Thank you so much!

Answers

Answer 1

Answer:

A. Empirical formula => C₂H₄O

B. Molecular formula => C₄H₈O₂

Explanation:

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

Mass of C = 0.273 g

Mass of H = 0.046 g

Mass of O = 0.182 g

Molar mass of compound = 88 g/mol

A. Determination of the empirical formula of the compound.

C = 0.273 g

H = 0.046 g

O = 0.182 g

Divide by their molar mass

C = 0.273 /12 = 0.023

H = 0.046 /1 = 0.046

O = 0.182 /16 = 0.011

Divide by the smallest

C = 0.023 / 0.011 = 2

H = 0.046 / 0.011 = 4

O = 0.011 / 0.011 = 1

Thus, the empirical formula of the compound is C₂H₄O

B. Determination of the molecular formula of the compound.

Empirical formula = C₂H₄O

Molar mass of compound = 88 g/mol

Molecular formula =?

Molecular formula = [C₂H₄O]ₙ

[C₂H₄O]ₙ = 88

[(12×2) + (4×1) + 16]n = 88

[24 + 4 + 16]n = 88

44n = 88

Divide both side by 44

n = 88 / 44

n = 2

Molecular formula = [C₂H₄O]ₙ

Molecular formula = [C₂H₄O]₂

Molecular formula = C₄H₈O₂

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Explanation:

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Answer:

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Answer:

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