An organic substance ( $0.2115 g m$ ) on complete combustion gave $0.4655 g m$ of carbon dioxide and $0.2533 g m$ of water. Determine the percentage composition of the compound.

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Solution

$M a s s o f C O_{2} f o r m e d = 0.4655 g m$
$M a s s o f H_{2} O f o r m e d = 0.2533 g m$
$M a s s o f t h e o r g a n i c s u b s t a n c e t a k e n = 0.2115 g m$
$P e r c e n t a g e o f c a r b o n = \frac{12}{44} \times \frac{M a s s o f C O_{2}}{M a s s o f c o m p o u n d} \times 100$
$= \frac{12}{44} \times \frac{0.4655}{0.2115} \times 100 = 60.03$
$P e r c e n t a g e o f h y d r o g e n = \frac{2}{18} \times \frac{M a s s o f H_{2} O}{M a s s o f c o m p o u n d} \times 100$
$= \frac{2}{18} \times \frac{0.2533}{0.2115} \times 100 = 13.30$
$P e r c e n t a g e o f o x y g e n = 100 - (60.03 + 13.30) = 26.67$
$C o m p o s i t i o n o f t h e s u b s t a n c e := C = 60.03 %; H = 13.30 %; a n d O = 26.67 %$

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0.2115

0.4655

0.2533

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