How to solve this combustion analysis question
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How to solve this combustion analysis question

[From: ] [author: ] [Date: 11-11-18] [Hit: ]
(0.0659 x 12G)/1.00g x 100% = 79.Happy Studies.......
Combustion analysis of 1.00g of testosterone yields 2.90g of CO2 and .875g H2O. What are the mass percents of carbon, hydrogen, and oxygen?

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Step 1, write a rough equation.

CxHyOz + O2--> xCO2 and y/2 H2O

So you see that moles of CO2 = moles of C in your compound

Moles of water = 1/2 moles of H in compound


Convert masses to moles of CO2 and H2O by dividing by molar masses ( 44 and 18)

mole CO2 = 0.0659 which contains 0.0659 moles of C
mole H2O =0.0486 which contains 0.0972 moles of H

For percent of C =( mass of C / mass of sample ) x 100%

(0.0659 x 12G)/1.00g x 100% = 79.08% carbon

do the same thing for the hydrogen %

Then for the oxygen - 100% - % carbon -% hydrogen/

Happy Studies.
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