Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

Complete combustion of a sample of hydrocarbon gives a 0.66 g of CO2 and 0.36 g of H2O the empirical formula of the compound is
one year ago

Arun
24497 Points

Molar mass of CO₂ = (12.0 + 16.0×2) g/mol = 44.0 g/mol
Molar mass of H₂O = (1.0×2 + 16.0) g/mol = 18.0 g/mol

n(C in X) = n(C in CO₂) = n(CO₂) = (0.66 g) / (44 g/mol) = 0.015 mol
n(H in X) = n(H in H₂O) = 2 × n(H₂O) = 2 × (0.36 g) / (18.0 g/mol) = 0.04 mol

Mole ratio C : H = 0.015 : 0.04 = 3 : 8
Empirical formula = C₃H₈
one year ago
Think You Can Provide A Better Answer ?

## Other Related Questions on IIT JEE Entrance Exam

View all Questions »

### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution

### Course Features

• 731 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions