Answer:
Molar mass of the gas = Mass of 22.4 L
gas at STP
\[=\frac{11.6}{10}\times 22.4=26\]
Let the molecular
formula of the gas is \[{{C}_{x}}{{H}_{y}}\]. It undergoes combustion as,
\[{{C}_{x}}{{H}_{y}}+\left(
x+\frac{y}{4} \right){{O}_{2}}\to xC{{O}_{2}}(g)+\frac{y}{2}{{H}_{2}}O\]
Number of
moles of carbon = Number of moles of \[C{{O}_{2}}\]
\[=\frac{3.38}{44}=0.0768\]
Number of
moles of hydrogen atom
= Number of
moles of \[{{H}_{2}}O\times 2\]
\[=2\times \frac{0.69}{18}=0.076\]
\[C:H=0.076:0.076\]
\[=1:1\]
\[\therefore \] Empirical
formula = CH
Molecular
formula = \[{{(CH)}_{n}}\]
\[n=\frac{\text{Molar}\,\text{mass}}{\text{Empirical}\,\text{formula}\,\text{mass}}\text{=}\frac{26}{13}=2\]
Molecular formula of
the gas \[={{C}_{2}}{{H}_{2}}\]
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