MCQ
On complete combustion $0.30 \,g$ of an organic compound gave $0.20\, g$ of carbon dioxide and $0.10\, g$ of water. The percentage of carbon in the given organic compound is $.....$ (Nearest Integer)
- ✓$18$
- B$180$
- C$65$
- D$74$
✓
Answer
Correct option: A.
$18$
a
$C _{ x } HyOz +\left( x +\frac{ y }{4}-\frac{ z }{2}\right) O _{2} \rightarrow xCO _{2}+\frac{ y }{2} H _{2} O$
$C _{ x } HyOz +\left( x +\frac{ y }{4}-\frac{ z }{2}\right) O _{2} \rightarrow xCO _{2}+\frac{ y }{2} H _{2} O$
$0.3g\quad\quad\quad\quad\quad\quad\quad\quad0.2g\quad\quad\quad0.1g$
$\frac{ n _{ CO _{2}}}{ n _{ H _{2} O }}=\frac{ x }{ y / 2}=\frac{0.2 / 44}{.1 / 18}$
$\frac{2 x }{ y }=\frac{36}{44}=\frac{9}{11}$
$x =\frac{9 y }{22}$
$\frac{ n _{ C _{ x } H _{ y } O_{ z } }}{ nCO _{ 2 } }=\frac{1}{ x }$
$\frac{0.3 }{12 x + y +16 z } \times \frac{44}{0.2}=\frac{1}{ x }$
$66 x =12 x + y +16 z$
$54 x = y +16 z$
$\frac{54 \times 9 y }{22}- y =16 z$
$\frac{464 y }{22}=16 z$
$z =\frac{29 y }{22}$
$C _{ x } H _{ y } O _{z}= C _{ x } H _{ y } O _{z}$
$C _{\frac{9 y}{22}} H _{ y } O _{\frac{29 y}{22}}^{22}$
$C _{9} H _{22} O _{29}$
$\,\%$ of $C =\frac{12 \times 9}{(12 \times 9+22+29 \times 16)} \times 100=\frac{108}{594} \times 100$
$18.18 \,\%$
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