Find empirical formula of compound containing 40% C, 6.7% H, 53.3% O.

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Given: C = 40%, H = 6.7%, O = 53.3% Assume 100 g compound - C=40 g, H=6.7 g, O=53.3 g Moles: C: 40 / 12 = 3.33 H: 6.7 / 1 = 6.7 O: 53.3 / 16 = 3.33 Divide by smallest (3.33): C: 3.33/3.33 = 1 H: 6.7/3.33 = 2 O: 3.33/3.33 = 1 Empirical formula = CH O Answer: CH O

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Asked 34 days ago|2/1/2026 10:36:34 AM

Updated 33 days ago|2/2/2026 5:54:07 AM

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cruz30

Given:
C = 40%, H = 6.7%, O = 53.3%
Assume 100 g compound - C=40 g, H=6.7 g, O=53.3 g
Moles:
C: 40 / 12 = 3.33
H: 6.7 / 1 = 6.7
O: 53.3 / 16 = 3.33
Divide by smallest (3.33):
C: 3.33/3.33 = 1
H: 6.7/3.33 = 2
O: 3.33/3.33 = 1
Empirical formula = CH O
Answer: CH O

Added 33 days ago|2/2/2026 5:53:01 AM

This answer has been confirmed as correct and helpful.

Salas1987

Convert mass to moles: - Moles of C = 40g / 12.01 g/mol = 3.33 mol - Moles of H = 6.7g / 1.008 g/mol = 6.65 mol - Moles of O = 53.3g / 16.00 g/mol = 3.33 mol2.

(3.33): - C: 3.33 / 3.33 = 1 - H: 6.65 / 3.33 = 2 - O: 3.33 / 3.33 = 1
The empirical formula is CH2O.

Added 33 days ago|2/2/2026 5:54:07 AM

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Deleted by Salas1987 [2/2/2026 5:50:04 AM]

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