The ratio of mass percent of C and H of an organic compound (CXHYOZ) is 6 : 1. If one molecule of the above compound (CXHYOZ) contains half as much oxygen as required to burn one molecule of compound CXHY completely to CO2 and H2O. The empirical formula of compound CXHYOZ is :

Engineering

Chemistry

Eudiometry

Question

The ratio of mass percent of C and H of an organic compound (C_XH_YO_Z) is 6 : 1. If one molecule of the above compound (C_XH_YO_Z) contains half as much oxygen as required to burn one molecule of compound C_XH_Y completely to CO₂ and H₂O. The empirical formula of compound C_XH_YO_Z is :

C₃H₄O₂

C₃H₆O₃

C₂H₄O

C₂H₄O₃

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C_XH_YO_Z Given

Ratio of C : H $C_{X} H_{Y} + \frac{1}{2} (2 x + \frac{y}{2}) O_{2} \overset{}{\to} {xCO}_{2} + \frac{y}{2} H_{2} O$

mass 6 : 1 $z = \frac{1}{2} \times \frac{1}{2} (2 x + \frac{y}{2}) \times 2$

moles $\frac{6}{12} : \frac{1}{1}$ $4 z = (2 x + \frac{y}{2}) \times 2$

moles 1 : 2 2Z = 2x + x

y = 2x 2Z = 3x

C_xH_2xO_z

C_xH_2x $O_{\frac{3 x}{2}}$ → C₂H₄O₃