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Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

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BuyFindarrow_forward

Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 5, Problem 86E
Textbook Problem
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A sample of urea contains 1.121 g N, 0.161 g H, 0.480 g C, and 0.640 g O. What is the empirical formula of urea?

Interpretation Introduction

Interpretation: The mass (in grams) of nitrogen, hydrogen, carbon and oxygen of urea is given. Using these values, the empirical formula of urea is to be calculated.

Concept introduction: The empirical formula is a formula which gives elemental composition of a compound. It is the smallest whole number ratio of atoms of each element.

To determine: The empirical formula of urea.

Explanation of Solution

Given

The mass of oxygen (O) is 0.640g .

The mass of nitrogen (N) is 1.121g .

The mass of hydrogen (H) is 0.0.161g .

The mass of carbon (C) is 0.480g .

The atomic mass of nitrogen (N) is 14.006g .

The atomic mass of oxygen (O) is 15.999g .

The atomic mass of hydrogen (H) is 1.008g .

The atomic mass of carbon (C) is 12.01g .

Formula

The number of moles in each element is calculated by using the formula,

Molesofatom=GivenmassofatomAtomicmass                                                          (1)

Substitute the values of mass and atomic mass of nitrogen in above equation.

MolesofN=GivenmassofNAtomicmass=(1.12114.006)mol=0.0800mol

Substitute the values of mass and atomic mass of oxygen in equation (1).

MolesofO=GivenmassofOAtomicmass=(0

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Chapter 5 Solutions

Chemistry: An Atoms First Approach
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