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Some Basic Concepts of Chemistry

Empirical formula or Molecular formula

The ratio of mass per cent of C and H of an organic compound (

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The ratio of mass per cent of $C$ and $H$ of an organic compound $(C_{x} H_{y} O_{z})$ is $6 : 1$ . If one molecule of the above compound $(C_{x} H_{y} O_{z})$ contains half as much oxygen as required to burn one molecule of compound $C_{x} H_{y}$ completely to $CO_{2}$ and $H_{2} O$ . The empirical formula of compound $C_{x} H_{y} O_{z}$ is (a) $C_{3} H_{6} O_{3}$ (b) $C_{2} H_{4} O$ (c) $C_{3} H_{4} O_{2}$ (d) $C_{2} H_{4} O_{3}$

Views: 5,835 students

Updated on: Jun 1, 2023

Book:

Arihant Chemistry JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2018

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The ratio of mass per cent of

C

and

H

of an organic compound

(C_{x} H_{y} O_{z})

6 : 1

. If one molecule of the above compound

(C_{x} H_{y} O_{z})

contains half as much oxygen as required to burn one molecule of compound

C_{x} H_{y}

completely to

CO_{2}

and

H_{2} O

. The empirical formula of compound

C_{x} H_{y} O_{z}

is (a)

C_{3} H_{6} O_{3}

(b)

C_{2} H_{4} O

(c)

C_{3} H_{4} O_{2}

(d)

C_{2} H_{4} O_{3}

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Question Text	The ratio of mass per cent of $C$ and $H$ of an organic compound $(C_{x} H_{y} O_{z})$ is $6 : 1$ . If one molecule of the above compound $(C_{x} H_{y} O_{z})$ contains half as much oxygen as required to burn one molecule of compound $C_{x} H_{y}$ completely to $CO_{2}$ and $H_{2} O$ . The empirical formula of compound $C_{x} H_{y} O_{z}$ is (a) $C_{3} H_{6} O_{3}$ (b) $C_{2} H_{4} O$ (c) $C_{3} H_{4} O_{2}$ (d) $C_{2} H_{4} O_{3}$
Updated On	Jun 1, 2023
Topic	Some Basic Concepts of Chemistry
Subject	Chemistry
Class	Class 11
Answer Type	Text solution:1 Video solution: 35
Upvotes	3399
Avg. Video Duration	7 min

Text solutionVerified

Filo tutor solutions (35)

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