Explanation Mass of caffeine is given as 0 .376 g . On combustion, 0 .682 g of carbon dioxide, 0 .174 g of water, and 0 .110 g of nitrogen is produced. Number of moles of carbon present in 0 .682 g of carbon dioxide is calculated as follows: n (C) = 0.682 g 44.01 g ⋅ mol − 1 × 1 mol C 1 mol CO 2 = 0.0155 mol Mass of carbon present in caffeine sample is calculated as given below: m (C) = 0.0155 mol × 12.01 g ⋅ mol − 1 = 0.186 g Number of moles of hydrogen present in 0 .174 g of water is calculated as follows: n (H) = 0.174 g 18.02 g ⋅ mol − 1 × 2 mol H 1 mol H 2 O = 0.0193 mol Mass of hydrogen present in caffeine sample is calculated as given below: m (H) = 0.0193 mol × 1.008 g ⋅ mol − 1 = 0.0194 g Number of moles of nitrogen present in 0 .110 g of nitrogen is calculated as follows: n (N) = 0.110 g 28.02 g ⋅ mol − 1 × 2 mol N 1 mol N 2 = 0.00785 mol Mass of nitrogen present in caffeine sample is calculated as given below: m (N) = 0.00785 mol × 14.01 g ⋅ mol − 1 = 0.110 g The total mass of carbon, nitrogen and hydrogen is calculated as shown: 0.186 g + 0.0194 g + 0.110 g = 0.3154 g Mass of oxygen that is present in 0 .376 g of caffeine is calculated as shown: m (O) = 0.376 g − 0.3154 g = 0.061 g Moles of oxygen present in 0

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 1TERM

7th Edition

ISBN: 9781319399849

Author: ATKINS

Publisher: MAC HIGHER

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Chapter F, Problem M.19E

Interpretation Introduction

Interpretation:

Empirical formula and molecular formula of caffeine along with the equation for the combustion 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.18E

Chapter F, Problem M.20E

Chapter F Solutions

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 1TERM

Ch. F - Prob. A.1AST Ch. F - Prob. A.1BST Ch. F - Prob. A.2AST Ch. F - Prob. A.2BST Ch. F - Prob. A.3AST Ch. F - Prob. A.3BST Ch. F - Prob. A.4AST Ch. F - Prob. A.4BST Ch. F - Prob. A.5AST Ch. F - Prob. A.5BST

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Solution Summary: The author explains that the empirical formula and molecular formula of caffeine along with the equation for the combustion has to be written.

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