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

Answer:

Plausible structure has been given below

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 $CO_{2}$ = $\frac{4.04}{44}moles$ $CO_{2}$ = 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 $H_{2}O$ = $\frac{1.24}{18}moles$ $H_{2}O$ = 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 $(2\times 0.0689)moles$ 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.