Question

A 1.24g sample of a hydrocarbon, when completely burned in an excess of O2 yields 4.04g Co2 and 1.24g H20. Draw plausible structure for the hydrocarbon molecule

Answer 1

Plausible structure has been given below

Step-by-step explanation:

• Molar mass of
  `CO_(2)` is 44 g/mol and molar mass of
  `H_(2)O` is 18 g/mol
• Number of mole = (mass/molar mass)

4.04 g of
`CO_(2)` =
`(4.04)/(44)moles`
`CO_(2)` = 0.0918 moles of
`CO_(2)`

1 mol of
`CO_(2)` contains 1 mol of C atom

So, 0.0918 moles of
`CO_(2)` contains 0.0918 moles of C atom

1.24 g of
`H_(2)O` =
`(1.24)/(18)moles`
`H_(2)O` = 0.0689 moles of
`H_(2)O`

1 mol of
`H_(2)O` contain 2 moles of H atom

So, 0.0689 moles of
`H_(2)O` contain
`(2* 0.0689)moles` of
`H_(2)O` or 0.138 moles of
`H_(2)O`

Moles of C : moles of H = 0.0918 : 0.138 = 2 : 3

Empirical formula of hydrocarbon is
`C_(2)H_(3)`

So, molecular formula of one of it's analog is
`C_(4)H_(6)`

Plausible structure of
`C_(4)H_(6)` has been given below.