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When 3.62 g of a compound containing carbon, hydrogen, and oxygen were burned...
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Let, the empirical formula of the hydrocarbon be CxHyOz.
The burning equation shall be,
CxHyOz + (x+y/4-z/2) O2 = x CO2 + y/2 H2O
Taking one mole of the Hydrocarbon (HC), (12x+y+16z) g HC upon burning produces 44x g CO2
So, 3.62 g HC upon burning produces 44x*3.62/ (12x+y+16z) g CO2. By condition this is equal to 5.19g. Thus, the first equation is generated as: 97x-5.19y-83.04z = 0 ------(i)
Similarly, considering the amount of water produced from the HC, one gets the second equation as:
-33.96x+29.75y-45.28z = 0 ------(ii)
Again, by considering the gain in weight of the products, relative to the amount of the HC, the third equation is obtained as: 31.52x+12.28y-64.16z = 0 ------(iii)
Solving i) & ii), one obtains, y:z = 8:3
Putting these values in iii), one gets, x:y = 3:8
Solving, x:y:z = 3:8:3
Hence, the empirical formula of the hydrocarbon is C3H8O3.
Posted by llltkl on April 14, 2013 at 3:45 PM (Answer #1)
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