# 0.22g sample of volatile compound,containing $C,H$ and $Cl$ only on combustion in $O_2$ gave 0.195g $CO_2$ and 0.0804g $H_2O$.If 0.120g of the compound occupied a volume of 37.24ml at $105^{\large\circ}$ and 768mm of pressure,calculate molecular formula of compound.

$(a)\;CH_3Cl\qquad(b)\;C_2H_4Cl_2\qquad(c)\;C_3H_{12}Cl_5\qquad(d)\;CH_3Cl_2$

## 1 Answer

Step 1:
$PV=\large\frac{w}{m}$$RT \large\frac{768}{760}$$\times \large\frac{37.24}{1000}=\large\frac{0.120}{m}$$\times 0.0821\times 378 m=99 Let formula of compound be C_aH_6Cl_2 C_aH_bCl_2+(a+\large\frac{b}{4})$$O_2\rightarrow aCO_2+\large\frac{b}{2}$$H_2O+\large\frac{z}{2}$$Cl_2$
$99g\Rightarrow 44ag\;CO_2$
$6.22g\Rightarrow \large\frac{44a\times 0.22}{99}$
$\therefore \large\frac{44a\times 0.22}{99}$$=0.195 a=2 Step 2: Similarly 0.22g compound gives =\large\frac{18\times b\times 0.22}{2\times 99}g H_2O \large\frac{18\times b\times 0.22}{2\times 99}$$=0.0804$
$b=4$
Now $C_aH_bCl_2$
$12\times a+b+35.5z=99$
$24+4+3.5z=99$
$z=2$
$\therefore C_2H_4Cl_2$
Hence (b) is the correct answer.
answered Nov 6, 2013

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