Explanation Given that 2 .20 g of CO 2 and 0 .901 g of H 2 O is produced from 1 .35 g of compound and 0 .130 g of nitrogen is produced from 0 .500 g of compound. Number of moles of carbon present in 2 .20 g of carbon dioxide is calculated as follows: n (C) = 2.20 g 44.01 g ⋅ mol − 1 × 1 mol C 1 mol CO 2 = 0.050 mol Mass of carbon present in 2 .20 g of carbon dioxide is calculated as follows: m (C) = 0.050 mol × 12.01 g ⋅ mol − 1 = 0.60 g Number of moles of hydrogen present in 0 .901 g of water is calculated as follows: n (H) = 0.901 g 18.02 g ⋅ mol − 1 × 2 mol H 1 mol H 2 O = 0.10 mol Mass of hydrogen present in 0 .901 g of water is calculated as follows: m (H) = 0.10 mol × 1.008 g ⋅ mol − 1 = 0.10 g Number of moles of nitrogen present in 0 .130 g of nitrogen is calculated as follows: n (N) = 0.130 g 28.02 g ⋅ mol − 1 × 2 mol N 1 mol N 2 = 0.0093 mol Mass of nitrogen present in 0 .130 g of nitrogen is calculated as follows: m (N) = 0.0093 mol × 14.01 g ⋅ mol − 1 = 0.130 g Mass percentage of elements: Mass percentage of carbon is determined as follows: Mass percentage of carbon = 0

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ISBN: 9781319421175

Author: ATKINS

Publisher: MAC HIGHER

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Chapter F, Problem M.21E

Interpretation Introduction

Interpretation:

Empirical formula for the compound has to be written.

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.20E

Chapter F, Problem M.22E

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Solution Summary: The author explains combustion analysis as a technique used to determine the empirical formula of an organic compound.

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