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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: - In the following reaction CHCl₃ + C₆H₅NH₂ + 3NaOH → A + 3B + 3C the product
- Ethylene glycol reacts with excess of PCl₅ to give
- Which is purified by steam distillation
- The oxidation number of oxygen in O₂PtF₆ is
- The normality of 10 volume hydrogen peroxide is
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
- In the following reaction CHCl₃ + C₆H₅NH₂ + 3NaOH → A + 3B + 3C the product
- Ethylene glycol reacts with excess of PCl₅ to give
- Which is purified by steam distillation
- The oxidation number of oxygen in O₂PtF₆ is
- The normality of 10 volume hydrogen peroxide is
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
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