Explanation Aroclor 1254 with mass of 1 .52 g produces 2 .224 g of CO 2 . Aroclor 1254 with mass of 2 .53 g produces 0 .2530 g of H 2 O . Number of moles of carbon present in 2 .224 g of carbon dioxide is calculated as follows: n (C) = 2.224 g 44.01 g ⋅ mol − 1 × 1 mol C 1 mol CO 2 = 0.0505 mol Mass of carbon present in Aroclor 1254 sample is calculated as given below: m (C) = 0.0505 mol × 12.01 g ⋅ mol − 1 = 0.607 g Number of moles of hydrogen present in 0 .2530 g of water is calculated as follows: n (H) = 0.2530 g 18.02 g ⋅ mol − 1 × 2 mol H 1 mol H 2 O = 0.0281 mol Mass of hydrogen present in Aroclor 1254 sample is calculated as given below: m (H) = 0.0281 mol × 1.008 g ⋅ mol − 1 = 0.0283 g Mass percentage of carbon in 1 .52 g of Aroclor 1254 is calculated as follows: Mass percent of Carbon = Mass of carbon Mass of Aroclor 1254 × 100 = 0

Chemical Principles: The Quest for Insight

7th Edition

ISBN: 9781464183959

Author: Peter Atkins, Loretta Jones, Leroy Laverman

Publisher: W. H. Freeman

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Chapter F, Problem M.13E

Interpretation Introduction

Interpretation:

Number of chlorine atoms present in Aroclor 1254 molecule has to be determined.

Concept Introduction:

Combustion analysis is a technique that is used in order to determine the empirical formula of an organic compound. In this technique the sample is burned in oxygen resulting in formation of carbon dioxide and water. The relative amounts of water and carbon dioxide is calculated from which the empirical formula is deduced.

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Chapter F, Problem M.12E

Chapter F, Problem M.14E

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Chemical Principles: The Quest for Insight

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