Leaded gasoline contains an additive to prevent engine knocking. On analysis, the additive compound is found to contain carbon, hydrogen, and lead (Pb) (hence, leaded gasoline). When 51.36 g of this compound is burned in a combustion analysis apparatus, 55.90 g of CO2 and 28.61 g of H2O are produced. Determine the empirical formula of the gasoline additive.

Question
Asked Oct 25, 2019

Leaded gasoline contains an additive to prevent engine knocking. On analysis, the additive compound is found to contain carbon, hydrogen, and lead (Pb) (hence, leaded gasoline). When 51.36 g of this compound is burned in a combustion analysis apparatus, 55.90 g of CO2 and 28.61 g of H2O are produced. Determine the empirical formula of the gasoline additive.

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Step 1

Given,

51.36 g sample produced 55.90 g of CO2 and 28.61 g of H2O

Molar mass of carbon = 12.0107

Molar mass of carbon dioxide = 44.01 g/mol

Mass of carbon in 55.90 g carbon dioxide
12.01
x55.90 g
44.01
Mass of Carbon=
Mass of Carbon=15.25g
Molar mass of hydrogen= 1.008. Molar mass of water = 18.0 g/mol
Mass of hydrogen in 28.61 g water
2.01
x 28.61 gg
18
Mass of hydrogen
Mass of hydrogen = 3.17g
Therefore
Mass of oxygen 51.36g - (15.25
3.17)g
Mass of oxygen 32.94 g
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Mass of carbon in 55.90 g carbon dioxide 12.01 x55.90 g 44.01 Mass of Carbon= Mass of Carbon=15.25g Molar mass of hydrogen= 1.008. Molar mass of water = 18.0 g/mol Mass of hydrogen in 28.61 g water 2.01 x 28.61 gg 18 Mass of hydrogen Mass of hydrogen = 3.17g Therefore Mass of oxygen 51.36g - (15.25 3.17)g Mass of oxygen 32.94 g

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Step 2

Mass of the carbon, hydrogen an...

Divide through by respective atomic mass:
15.25
12.01
C 1.27
3.17g
H
1.008
H 3.14g
32.94 g
=2.05
15.999
O 2.05
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Divide through by respective atomic mass: 15.25 12.01 C 1.27 3.17g H 1.008 H 3.14g 32.94 g =2.05 15.999 O 2.05

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