World of Chemistry, 3rd edition
World of Chemistry, 3rd edition
3rd Edition
ISBN: 9781133109655
Author: Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste
Publisher: Brooks / Cole / Cengage Learning
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Chapter 6, Problem 16A

(a)

Interpretation Introduction

Interpretation:

The mass of the lithium in grams is to be calculated.

Concept Introduction:

The average atomic mass is numerically equal to the atomic mass which is given in the periodic table.

The formula to calculate the mass of any species for a given number of moles is,

  n=massmolecularmass ……….. (1)

Where,

n represents the number of moles.

(a)

Expert Solution
Check Mark

Answer to Problem 16A

The mass of lithium in 0.251mol is 1.7422g .

Explanation of Solution

The average molecular mass is also equal to molecular mass. The mass of lithium is calculated as follows:

The formula to calculate the number of moles is,

  n=massmolecularmass ……….. (1)

Where,

n represents the number of moles.

Therefore the mass of the species becomes,

  mass=n×molecularmass ………… (2)

Given data:

n = 0.251mol

The molecular mass of lithium = 6.941g/mol

Plugin the values in equation (2)

  mass=0.251mol×6.941g/mol=1.7422g

(b)

Interpretation Introduction

Interpretation:

The mass of the aluminum present in 1.51mol is to be calculated.

Concept Introduction:

The average atomic mass is numerically equal to the atomic mass which is given in the periodic table.

The formula to calculate the mass of any species for a given number of moles is,

  n=massmolecularmass ……….. (1)

Where,

n represents the number of moles.

(b)

Expert Solution
Check Mark

Answer to Problem 16A

The mass of aluminum in 1.51mol is 40.7398g .

Explanation of Solution

The average molecular mass is also equal to molecular mass. The mass of aluminum is calculated as follows:

The formula to calculate the number of moles is,

  n=massmolecularmass ……….. (1)

Where,

n represents the number of moles.

Therefore the mass of the species becomes,

  mass=n×molecularmass ………… (2)

Given data:

n = 1.51mol

The molecular mass of aluminum = 26.98g/mol

Plugin the values in equation (2)

  mass=1.51mol×26.98g/mol=40.7398g

(c)

Interpretation Introduction

Interpretation:

The mass of lead in grams is to be calculated.

Concept Introduction:

The average atomic mass is numerically equal to the atomic mass which is given in the periodic table.

The formula to calculate the mass of any species for a given number of moles is,

  n=massmolecularmass ……….. (1)

Where,

n represents the number of moles.

(c)

Expert Solution
Check Mark

Answer to Problem 16A

The mass of lead in 0.0785mol is 18.13g .

Explanation of Solution

The average molecular mass is also equal to molecular mass. The mass of lead is calculated as follows:

The formula to calculate the number of moles is,

  n=massmolecularmass ……….. (1)

Where,

n represents the number of moles.

Therefore the mass of the species becomes,

  mass=n×molecularmass ………… (2)

Given data:

n = 8.75×102or0.0875mol

The molecular mass of lead = 207.2g/mol

Plugin the values in equation (2)

  mass=0.0875mol×207.2g/mol=18.13g

(d)

Interpretation Introduction

Interpretation:

The mass of chromium present in 125mol is to be calculated.

Concept Introduction:

The average atomic mass is numerically equal to the atomic mass which is given in the periodic table.

The formula to calculate the mass of any species for a given number of moles is,

  n=massmolecularmass ……….. (1)

Where,

n represents the number of moles.

(d)

Expert Solution
Check Mark

Answer to Problem 16A

The mass of chromium in 125mol is 6498.75g .

Explanation of Solution

The average molecular mass is also equal to molecular mass. The mass of chromium is calculated as follows:

The formula to calculate the number of moles is,

  n=massmolecularmass ……….. (1)

Where,

n represents the number of moles.

Therefore the mass of the species becomes,

  mass=n×molecularmass ………… (2)

Given data:

n = 125mol

The molecular mass of chromium = 51.99g/mol

Plugin the values in equation (2)

  mass=125mol×51.99g/mol=6498.75g

(e)

Interpretation Introduction

Interpretation:

The mass of iron present in 4.25×103mol is to be calculated.

Concept Introduction:

The average atomic mass is numerically equal to the atomic mass which is given in the periodic table.

The formula to calculate the mass of any species for a given number of moles is,

  n=massmolecularmass ……….. (1)

Where,

n represents the number of moles.

(e)

Expert Solution
Check Mark

Answer to Problem 16A

The mass of iron in 4.25×103mol is 237341.25g .

Explanation of Solution

The average molecular mass is also equal to molecular mass. The mass of chromium is calculated as follows:

The formula to calculate the number of moles is,

  n=massmolecularmass ……….. (1)

Where, n represents the number of moles.

Therefore the mass of the species becomes,

  mass=n×molecularmass ………… (2)

Given data:

n = 4.25×103or4250mol

The molecular mass of iron = 55.845g/mol

Plugin the values in equation (2)

  mass=4.25×103mol×55.845g/mol=237341.25g

(f)

Interpretation Introduction

Interpretation:

The mass of magnesium is to be calculated.

Concept Introduction:

The average atomic mass is numerically equal to the atomic mass which is given in the periodic table.

The formula to calculate the mass of any species for a given number of moles is,

  n=massmolecularmass ……….. (1)

Where,

n represents the number of moles.

(f)

Expert Solution
Check Mark

Answer to Problem 16A

The mass of magnesium 0.000105mol is 0.00255g .

Explanation of Solution

The average molecular mass is also equal to molecular mass. The mass of chromium is calculated as follows:

The formula to calculate the number of moles is,

  n=massmolecularmass ……….. (1)

Where,

n represents the number of moles.

Therefore the mass of the species becomes,

  mass=n×molecularmass ………… (2)

Given data:

n = 0.000105mol

The molecular mass of magnesium = 24.305g/mol

Plugin the values in equation (2)

  mass=0.000105mol×24.305g/mol=0.00255g

Chapter 6 Solutions

World of Chemistry, 3rd edition

Ch. 6.2 - Prob. 5RQCh. 6.2 - Prob. 6RQCh. 6.3 - Prob. 1RQCh. 6.3 - Prob. 2RQCh. 6.3 - Prob. 3RQCh. 6.3 - Prob. 4RQCh. 6.3 - Prob. 5RQCh. 6.3 - Prob. 6RQCh. 6 - Prob. 1ACh. 6 - Prob. 2ACh. 6 - Prob. 3ACh. 6 - Prob. 4ACh. 6 - Prob. 5ACh. 6 - Prob. 6ACh. 6 - Prob. 7ACh. 6 - Prob. 8ACh. 6 - Prob. 9ACh. 6 - Prob. 10ACh. 6 - Prob. 11ACh. 6 - Prob. 12ACh. 6 - Prob. 13ACh. 6 - Prob. 14ACh. 6 - Prob. 15ACh. 6 - Prob. 16ACh. 6 - Prob. 17ACh. 6 - Prob. 18ACh. 6 - Prob. 19ACh. 6 - Prob. 20ACh. 6 - Prob. 21ACh. 6 - Prob. 22ACh. 6 - Prob. 23ACh. 6 - Prob. 24ACh. 6 - Prob. 25ACh. 6 - Prob. 26ACh. 6 - Prob. 27ACh. 6 - Prob. 28ACh. 6 - Prob. 29ACh. 6 - Prob. 30ACh. 6 - Prob. 31ACh. 6 - Prob. 32ACh. 6 - Prob. 33ACh. 6 - Prob. 34ACh. 6 - Prob. 35ACh. 6 - Prob. 36ACh. 6 - Prob. 37ACh. 6 - Prob. 38ACh. 6 - Prob. 39ACh. 6 - Prob. 40ACh. 6 - Prob. 41ACh. 6 - Prob. 42ACh. 6 - Prob. 43ACh. 6 - Prob. 44ACh. 6 - Prob. 45ACh. 6 - Prob. 46ACh. 6 - Prob. 47ACh. 6 - Prob. 48ACh. 6 - Prob. 49ACh. 6 - Prob. 50ACh. 6 - Prob. 51ACh. 6 - Prob. 52ACh. 6 - Prob. 53ACh. 6 - Prob. 54ACh. 6 - Prob. 55ACh. 6 - Prob. 56ACh. 6 - Prob. 57ACh. 6 - Prob. 58ACh. 6 - Prob. 59ACh. 6 - Prob. 60ACh. 6 - Prob. 61ACh. 6 - Prob. 62ACh. 6 - Prob. 63ACh. 6 - Prob. 64ACh. 6 - Prob. 65ACh. 6 - Prob. 66ACh. 6 - Prob. 67ACh. 6 - Prob. 68ACh. 6 - Prob. 69ACh. 6 - Prob. 70ACh. 6 - Prob. 1STPCh. 6 - Prob. 2STPCh. 6 - Prob. 3STPCh. 6 - Prob. 4STPCh. 6 - Prob. 5STPCh. 6 - Prob. 6STPCh. 6 - Prob. 7STPCh. 6 - Prob. 8STPCh. 6 - Prob. 9STPCh. 6 - Prob. 10STPCh. 6 - Prob. 11STPCh. 6 - Prob. 12STP
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Step by Step Stoichiometry Practice Problems | How to Pass ChemistryMole Conversions Made Easy: How to Convert Between Grams and Moles; Author: Ketzbook;https://www.youtube.com/watch?v=b2raanVWU6c;License: Standard YouTube License, CC-BY