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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
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 CarbonAnd 0.1939 of Water would contain => 2/18 * 0.1939=> 0.021 gm Hydrogen. Now carbon moles = 0.2586/12 = 0.021H moles= 0.021Hence the formulae is:C1H1 => (CH)
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