What are the subscripts associate with C, H and O when a 5.048g sample of a compound undergoes combustion to produce 7.406g of CO2 and 3.027g of H2O ?
The answer I got was C1, H2, and O3
3 answers
I have CH2O
@DrBob222
How did you get C1 H2 O1?
If 5.048g of the substance goes through combustion and produces 7.406g of CO2 and water shouldn't there be 3 oxygens?
How did you get C1 H2 O1?
If 5.048g of the substance goes through combustion and produces 7.406g of CO2 and water shouldn't there be 3 oxygens?
I don't understand your logic. Remember all of this is done in mols and not g.
..................CxHyOz + O2 ---> CO2 + H2O
...................5.048 g...............7.406.....3.027
First we convert 7.048 g CO2 to grams C and 3.027 g H2O to grams H.
7.048 x (atomic mass C/molar mass CO2) = 7.048 x (12/44) = 2.020 g C.
3.027 x (2/18) = 0.336 g H.
grams O = 5.048 - g C - g H = 2.692
Now convert these g to mols C, H, O.
mols C = 2.20/12 = 0.168 C
mols H = 0.336/1 = 0.336 H
mols O = 2.692/16 = 0.168 O
Now find the ratio.
C = 0.168/0.336 = 0.500 for C
H = 0.336/0.336 = 1.00 for H
O = 0.168/0.336 = 0.500 for O
We don't want fractions in the empirical and no subscript less than 1.00 so we multiply those values above to get whole numbers. In this case the multiplier is 2 so C = 1, H = 2, O = 2
..................CxHyOz + O2 ---> CO2 + H2O
...................5.048 g...............7.406.....3.027
First we convert 7.048 g CO2 to grams C and 3.027 g H2O to grams H.
7.048 x (atomic mass C/molar mass CO2) = 7.048 x (12/44) = 2.020 g C.
3.027 x (2/18) = 0.336 g H.
grams O = 5.048 - g C - g H = 2.692
Now convert these g to mols C, H, O.
mols C = 2.20/12 = 0.168 C
mols H = 0.336/1 = 0.336 H
mols O = 2.692/16 = 0.168 O
Now find the ratio.
C = 0.168/0.336 = 0.500 for C
H = 0.336/0.336 = 1.00 for H
O = 0.168/0.336 = 0.500 for O
We don't want fractions in the empirical and no subscript less than 1.00 so we multiply those values above to get whole numbers. In this case the multiplier is 2 so C = 1, H = 2, O = 2