3.782 g of a substance (containing only C, H, and O) is burned in a combustion a

Question

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.

Explanation / Answer

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

C3H4O5

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