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Author Question: Determine the molecular formula of a compound that is 49.48% carbon, 5.19% hydrogen, 28.85% ... (Read 49 times)

Capo

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on: Nov 18, 2019

Question 1

Combustion analysis of an unknown hydrocarbon produced the following results: C 59.96%, H 13.42%, and O 26.62%. Determine the empirical formula.
◦ C6H16O4
◦ C8H8O2
◦ C3H8O
◦ C5H8O4
◦ C9H16O2

Question 2

Determine the molecular formula of a compound that is 49.48% carbon, 5.19% hydrogen, 28.85% nitrogen, and 16.48% oxygen. The molecular weight is 194.19 g mol-1.
◦ C8H10N4O2
◦ C4H5N2O
◦ C8H10N2O
◦ C8H12N4O2
Chemistry: A Molecular Approach (Canadian) ¦ Tro, Fridgen, Shaw ¦ 2nd Edition
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Marked as best answer by Capo on Nov 18, 2019

sarah_brady415

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Reply #1 on: Nov 18, 2019
Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.
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Capo

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Reply 2 on: Nov 18, 2019
Wow, this really help

dreamfighter72

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Reply 3 on: Yesterday
Excellent

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