3.782 g of a substance (containing only C, H,
and O) is burned in a combustion analysis apparatus.
The mass of CO2 produced is 4.160 g, and the mass of H2O produced is 1.135 g.
What is the empirical formula of the compound?
In the empirical formula the coefficients must be reduced to the
smallest possible integers.
The convention for entering the empirical formula is to give C first followed by H then by any other elements in alphabetical order.
Coefficients of "1" are not typed.
e.g. C2H2O is OK but the following are NOT: C2H2O1 or C4H4O2 or OC2H2.
Molar mass of CO2 = 44g
Molar mass of C = 12g
CO2 to C conversion factor = 12/44 = 0.2727
Mass of CO2 produced = 4.160g
Mass of C = 4.160g × 0.2727 = 1.1344g
No of mole of C = 1.1344g/12g/mol = 0.09453
Molar mass of H2O = 18g/mol
Molar mass of H = 1g/mol
H2O to H conversion factor = 2/18 = 0.1111
Mass of H2O produced = 1.135g
Mass of H = 0.1111×1.135g = 0.1261g
No of mole of H = 0.1261g/1 = 0.1261
Mass of C + H = 1.1344g + 0.1261g = 1.2605g
Mass of O = 3.782g - 1.2605g = 2.5215g
No of mole of O = 2.5215/16 = 0.1576
(No of mole of C/0.09453)×3 = 3
(No of mole of H/ 0.09453 )×3= 4
(No of mole of O / 0.09453 )×3= 5
So , the empirical formula is
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