Calculate the empirical formula of a compound with 70% Fe and 30% O by mass.

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Given: Fe = 70%, O = 30% Moles Fe = 70 / 55.85 = 1.253 Moles O = 30 / 16 = 1.875 Divide by smallest: Fe: 1.253/1.253 = 1, O: 1.875/1.253 = 1.495 = 1.5 Multiply by 2 to get whole numbers: Fe2O3 Empirical formula: Fe2O3

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Asked 38 days ago|1/28/2026 8:22:14 AM

Updated 38 days ago|1/28/2026 9:15:24 AM

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cruz30

Given: Fe = 70%, O = 30%
Moles Fe = 70 / 55.85 = 1.253
Moles O = 30 / 16 = 1.875
Divide by smallest: Fe: 1.253/1.253 = 1, O: 1.875/1.253 = 1.495 = 1.5
Multiply by 2 to get whole numbers: Fe2O3
Empirical formula: Fe2O3

Added 38 days ago|1/28/2026 8:51:42 AM

This answer has been confirmed as correct and helpful.

Flora1988

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Added 38 days ago|1/28/2026 9:15:24 AM

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Deleted by Flora1988 [1/28/2026 9:18:42 AM]

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