A bromoalkane contains 35% carbon and 6.57% hydrogen by mass. Calculate the empirical

⇦

⇨

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:

1. The reactive species in the nitration of benzene is
2. The flame colours of metals ions are due to
3. Which of the following is dihydric alcohol
4. The equivalent weight of phosphoric acid(H₃PO₄) in the reaction
5. which of the following non-metal possesses the atomicity double than that

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