CxHyOz+(x+4y−2z)O2→xCO2+2yH2O0.3g0.2g0.1g
H2OnCO2n=2yx=180.1440.2=119
x=229y
Now, CO2nCxHyO2n=x1
⇒12x+y+16z0.3×0.244=x1
66x=12x+y+16z54x=y+16z
2254×9y–y=16z
z=2229y
CxHyOz=C229yHyO2229y=C9H22O29
On complete combustion 0.30g of an organic compound gave 0.20g of carbon dioxide and 0.10g of water. The percentage of carbon in the given organic compound is____(Nearest Integer)
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