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.
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.