An unknown compound contains only carbon, hydrogen, and oxygen (). Combustion of 7.50 of this compound produced 11.0 of carbon dioxide and 4.50 of water.

1) How many moles of carbon, C, were in the original sample?
An unknown compound contains only carbon, hydrogen, and oxygen (). Combustion of 7.50 of this compound produced 11.0 of carbon dioxide and 4.50 of water.

1) How many moles of carbon, C, were in the original sample?
@Chemistry

Question

An unknown compound contains only carbon, hydrogen, and oxygen (). Combustion of 7.50  of this compound produced 11.0  of carbon dioxide and 4.50  of water.

1) How many moles of carbon, C, were in the original sample?
 An unknown compound contains only carbon, hydrogen, and oxygen (). Combustion of 7.50  of this compound produced 11.0  of carbon dioxide and 4.50  of water.

1) How many moles of carbon, C, were in the original sample?
 @Chemistry

anonymous · Answer

n(C)=m/M    m(CO2)=11.0 M(CO2)=44.01
n=11/44.01  =0.249943194moles = moles of Carbon
If you wanted to find the empirical formula though,
n(H)=4.5/18.016  =0.249777975x2= 0.49955595moles of H
m(H)= 0.499556x1.008=0.50355     m(C)= 0.2499x12.01= 3.0018
7.50-0.50355-3.0018 = 3.99g = 0.24966moles of O
C  =0.2499/0.2497   =1
H  = 0.4996/0.2497  =2
0  = 0.2497/0.2497  =1
therefore the empirical formula=CH2O