0.15 g of a hydrocarbon upon combustion gave 0.44 g of carbon dioxide and 0.27 g of water. If the vapour density of the hydrocarbon is 15, find out its molecular formula.

C_{2} H_{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C_{3} H_{8}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C_{2} H_{6}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

C H_{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $C_{2} H_{6}$
Moles of Carbon $= \frac{0.44}{44} = 0.01$
Moles of Water $= \frac{0.27}{18} = 0.015$
Moles of Hydrogen $= \frac{0.27}{18} \times 2 = 0.03$
Empirical formula $C H_{3}$
n $= \frac{2 \times V . D .}{E m p i r i c a l m a s s} = \frac{30}{15} = 2$
Molecular formula $= (C H_{3})_{n} = C_{2} H_{6}$

Suggest Corrections

Similar questions

4.4 g

C O_{2}

2.7 g

H_{2} O

30

8.8 g C O_{2}

{5.4}_{g} H_{2} O

C O_{2}

0.44 g

1.8 g

0.88 g

0.72 g

3.08 g

C O_{2}

Join BYJU'S Learning Program