0.07 g of a hydride of carbon occupies 56 at S.T.P when vapourised and contains...
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0.07 g of a hydride of carbon occupies 56 at S.T.P when vapourised and contains 14.29% by mass of hydrogen. The formula of the hydrocarbon is (C 12, H = 1)
- A) CH4
- B) C2H2
- C) C2H4
- D) C4H8
Correct Answer: D) C4H8
Explanation
Mass of hydride = 0.07g; Mass of hydrogen = 14.29% \(\times\) 0.07 = 0.01g
Mass of carbon = 0.07g - 0.01g = 0.06g
Mole ratio of H = \(\cfrac{0.01}{1} = 0.01\)
Mole ratio of C = \(\cfrac{0.06}{12} = 0.005\)
Ratio of mixture C = \(\cfrac{0.005}{0.005} = 1\), H = \(\cfrac{0.01}{0.005} = 2\)
Empirical formula = CH\(_2\)
RMM = (12 + 1\(\times\) 2)n = 56
14n = 56 \(\Rightarrow\) n = \(\cfrac{56}{14} = 4\)
Molecular formula = (CH\(_2\))\(_4\) = C\(_4\)H\(_8\)
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