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WHAT IS THE EMPIRICAL FORMULAE OF A COMPOUND 0.2801 GRAM OF WHICH GAVE ON COMPLETE COMBUSTION 0.9482 GRAM OF CARBON DIOXIDE AND 0.1939 GRAM OF WATER

WHAT IS THE EMPIRICAL FORMULAE OF A COMPOUND 0.2801 GRAM OF WHICH GAVE ON COMPLETE COMBUSTION 0.9482 GRAM OF CARBON DIOXIDE AND 0.1939 GRAM OF WATER

Grade:11

1 Answers

Vikas TU
14149 Points
7 years ago
From unitary method,
If 44g of CO2 has 12 g carbon then 0.9482 gm Carbon Dioxide would contain 
=> 12/44   * 0.9482 
=> 0.2586 gms of Carbon
And 0.1939 of Water would contain 
=> 2/18  * 0.1939
=> 0.021 gm Hydrogen.
 
Now carbon moles = 0.2586/12  = 0.021
H moles= 0.021
Hence the formulae is:
C1H1 => (CH)

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