| ⇦ | 
| ⇨ |
A bromoalkane contains 35% carbon and 6.57% hydrogen by mass. Calculate the empirical formula of this bromoalkane.
Options
(a) CH₂Br
(b) C₂H₂Br₂
(c) C₄H₄Br
(d) C₄H₉Br
Correct Answer:
C₄H₉Br
Explanation:
1) Assume 100 g of the compound is available:
C ⇒ 35 g
H ⇒ 6.57 g
Br ⇒ 58.43 g (from 100 minus 41.57)
2) Determine moles:
C ⇒ 35 g / 12 gmol = 2.917
H ⇒ 6.57 g / 1 g/mol = 6.57
Br ⇒ 58.43 g / 80 g/mol = 0.730375
3) Divide by smallest to seek lowest whole-number ratio:
C ⇒ 2.917 / 0.730375 = 4
H ⇒ 6.57 / 0.730375 = 9
Br ⇒ 0.730375 / 0.730375 = 1
C₄H₉Br
Related Questions: - The half-life of a reaction is inversely proportional to the square of the
- Which of the following is more acidic in nature
- What is the ratio of mass of an electron to the mass of a proton?
- Non – lanthanide element is
- Molecular mass of a compound having empirical formula C₂H₅O is 90.
Question Type: Memory
(964)
Difficulty Level: Easy
(1008)
Topics: Basic Concepts of Chemistry
(94)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The half-life of a reaction is inversely proportional to the square of the
- Which of the following is more acidic in nature
- What is the ratio of mass of an electron to the mass of a proton?
- Non – lanthanide element is
- Molecular mass of a compound having empirical formula C₂H₅O is 90.
Question Type: Memory (964)
Difficulty Level: Easy (1008)
Topics: Basic Concepts of Chemistry (94)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply