A combustion devise was used to determine the empirical formula for a compound t

Question

A combustion devise was used to determine the empirical formula for a compound that only contains C, H, and O. A 0.6349 g sample of the unknown produced 1.603 g of CO2 and 0.2810 g H2O. The grams of carbon in 1.603 g CO2 are _____.

Based on the information from the previous question, determine the grams of hydrogen in 0.2810 g of H2O.

Based on the information and your answers to the previous two questions, the amount of grams of oxygen in the original compound is ____ g.

Based on the information and your answers for the last three questions, the corresponding mol of C, H, and O are _____, _____, and _____, respectively.

Base on the information provided in the last 4 questions and your answers, the empirical formula for the compound is C7H6O2.

Explanation / Answer

mass of CO2 = 1.603g or
0.036moles of CO2
or0.036moles of C or
0.44g of C

mass of H2O =0.281g or
0.016moles of H2Oor
0.031moles of H or
0.031g of H

mass of compound burnt =0.6349g

mass of C+H =0.469g
mass of O =0.166g
moles of O =0.01039


molar ratio of C : H :O = 0.0364:0.0312:0.0104

or after dividing by the smallest = 3.506:3.005:1.000
multiply by 2 to get whole numbers = 7.012:6.010:2.000

empirical formula is C7H6O2

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