Chemistry: The Molecular Nature of Matter and Change
Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781259631757
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Chapter 3, Problem 3.19P

(a)

Interpretation Introduction

Interpretation:

The mass (g) of 8.42 mol of chromium(III)sulfate decahydrate is to be calculated.

Concept introduction:

Molar mass is defined as the mass of 1 mol of a chemical substance in grams. The molar mass of a compound is calculated by the addition of the molar mass of each element multiplied by its number of atoms present in the chemical formula.

(a)

Expert Solution
Check Mark

Answer to Problem 3.19P

The mass (g) of 8.42 mol chromium(III)sulfate decahydrate is 4.82×103 g.

Explanation of Solution

Chromium(III)sulfate decahydrate is a chemical compound with the chemical formula Cr2(SO4)310H2O.

The formula to calculate the molar mass of Cr2(SO4)310H2O is,

Molar Mass=[(number of Cr atoms)(atomic massof Cr)+(number ofS atoms)(atomicmassof S)+(number ofH atoms)(atomicmassofH)+(number ofO atoms)(atomicmassofO)] (1)

Substitute 2 for number of Cr atoms, 52.00 g/mol for the atomic mass of Cr, 3 for number of atoms of S, 32.06g/mol for the atomic mass of S, 20 for number of atoms of H, 1.008g/mol for the atomic mass of H, 22 for number of atoms of O and 16.00g/mol for the atomic mass of O in equation (1).

Molar Mass of Cu2CO3=[(2)(52.00 g/mol)+(3)(32.06g/mol)+(20)(1.008g/mol)+(22)(16.00g/mol)]=104.00 g/mol+96.18g/mol+20.16g/mol+352.00g/mol=572.34g/mol

The expression to calculate the mass (g) of Cr2(SO4)310H2O is:

Mass (g)=[(mol of Cr2(SO4)310H2O)(molecular mass(g) Cr2(SO4)310H21 mol Cr2(SO4)310H2O)] (2)

Substitute 8.42 mol for Cr2(SO4)310H2O and 572.34g/mol for the molecular mass of Cr2(SO4)310H2O in equation (2).

Mass (g) of Cr2(SO4)310H2O=(8.42 mol)(572.34 g Cr2(SO4)310H21 mol Cr2(SO4)310H2O)=4819.103 g=4.82×103 g Cr2(SO4)310H2

Conclusion

The mass (g) of 8.42 molchromium(III)sulfate decahydrate is 4.82×103 g.

(b)

Interpretation Introduction

Interpretation:

The mass (g) of 1.83×1024 molecules of dichlorine heptoxide is to be calculated.

Concept introduction:

One mole is defined as the amount of substance that contains the same number of entities such as molecules, ions, atoms as the number of atoms in 12 g of C-12. This number is called Avogadro’s number. The value of Avogadro’s number is 6.022×1023 (entitites/mol).

Molar mass is defined as the mass of 1 mol of a chemical substance in grams. The molar mass of a compound is calculated by the addition of the molar mass of each element multiplied by its number of atoms present in the chemical formula.

Following are the steps to calculate the mass of a chemical substance when a number of molecules are given.

Step 1: Determine the amount of substance in moles by using Avogadro’s number. The expression to calculate the moles of a chemical substance is as follows:

Amount (mol )=(Given molecules)(1 mol6.022×1023 molecules)

Step 2: Multiply the moles with the molar mass of the chemical substance to obtain the mass of chemical substance in grams. The formula to calculate the mass of a substance in grams is as follows:

Mass ()=(Amount (mol))(No. of grams1 mol)

(b)

Expert Solution
Check Mark

Answer to Problem 3.19P

The mass (g) of 1.83×1024 molecules of dichlorine heptoxide is 5.56×102 g.

Explanation of Solution

Dichlorine heptoxide is an inorganic compound with the chemical formula Cl2O7.

The formula to calculate the molar mass of Cl2O7 is,

Molar Mass of Cl2O7=[(number of Cl atoms)(atomic massof Cl)+(number ofO atoms)(atomicmassofO)] (3)

Substitute 2 for number of Cl atoms, 35.45 g/mol for the atomic mass of Cl, 7 for number of atoms of O and 16.00g/mol for the atomic mass of O in equation (3).

Molar Mass of Cl2O7=[(2)(35.45 g/mol)+(7)(16.00g/mol )]=70.9 g/mol+112.00g/mol=182.9g/mol

The expression to calculate moles of Cl2O7 is:

MolesofCl2O7=(molecules of Cl2O7)(1 mol Cl2O76.022×1023of Cl2O7) (4)

Substitute 1.83×1024 molecules for Cl2O7 in equation (4).

MolesofCl2O7=(1.83×1024molecules of Cl2O7)(1 mol Cl2O76.022×1023of Cl2O7)=3.038858 mol Cl2O7

The expression to calculate the mass (g) of Cl2O7 is:

Mass (g)=(mol of Cl2O7)(molecular mass(g) of Cl2O7 1 mol Cl2O7) (5)

Substitute 3.038858 mol for mol of Cl2O7 and 182.9g/mol for the molecular mass of Cl2O7 in equation (5).

Mass (g) of Cl2O7=(3.038858 mol)( 182.9 g Cl2O7 1 mol Cl2O7)=555.807 g=5.56×102 g Cl2O7

Conclusion

The mass (g) of 1.83×1024 molecules of dichlorine heptoxide is 5.56×102 g.

(c)

Interpretation Introduction

Interpretation:

The number of moles and formula units in 6.2 g of lithium sulfate is to be calculated.

Concept introduction:

One mole is defined as the amount of substance that contains the same number of entities such as molecules, ions, atoms as the number of atoms in 12 g of C-12. This number is called Avogadro’s number. The value of Avogadro’s number is 6.022×1023 (entitites/mol).

Molar mass is defined as the mass of 1 mol of a chemical substance in grams. The molar mass of a compound is calculated by the addition of the molar mass of each element multiplied by its number of atoms present in the chemical formula.

A formula unit is used for the ionic compound to represent their empirical formula. The steps to determine the formula unit of an ionic compound from the given mass are as follows:

Step 1: Divide the given mass of the ionic compound with the molar mass to calculate the moles. The expression to calculate the moles of an ionic compound when the mass is given is as follows:

Amount(mol)=(Mass (g) )(1 molNo. of grams)

Step 2: Multiply the calculated moles with the Avogadro’s number to determine the formula units of an ionic compound. The expression to determine the formula unit is as follows:

Number of formula units=(Amount (mol))(6.022×1023 formula units1 mol)

(c)

Expert Solution
Check Mark

Answer to Problem 3.19P

The number of moles and formula unit in 6.2 g of lithium sulfate is 0.056 mol and 3.4×1022 FU respectively.

Explanation of Solution

Lithium sulfate is an ionic compound with the chemical formula Li2SO4.

The formula to calculate the molar mass of Li2SO4 is,

Molar Mass of Li2SO4=[(number of Li atoms)(atomic massof Li)+(number ofS atoms)(atomicmassof S)+(number ofO atoms)(atomicmassofO)] (6)

Substitute 2 for number of Li atoms, 6.941 g/mol for the atomic mass of Li, 1 for number of atoms of S, 32.06g/mol for the atomic mass of S, 4 for number of atoms of O and 16.00g/mol for the atomic mass of O in equation (2).

Molar Mass of Li2SO4=[(2)(6.941 g/mol)+(1)(32.06g/mol )+(4)(16.00g/mol )]=13.882 g/mol+32.06g/mol+64.00g/mol=109.94g/mol

The expression to calculate moles of Li2SO4 is:

MolesofLi2SO4=(given mass of Li2SO4(g))(1 mol Li2SO4molecular mass(g)) (7)

Substitute 6.2 g for given mass of Li2SO4 and 109.94g/mol for the molecular mass of Li2SO4 in equation (7).

Moles ofLi2SO4=(6.2 gLi2SO4)(1mol Li2SO4109.94 g Li2SO4)=0.056394mol=0.056 mol Li2SO4

The expression to calculate formula units (FU) of Li2SO4 is:

FU of Li2SO4=(mol of Li2SO4)(6.022×1023 FU Li2SO41mol Li2SO4) (8)

Substitute 0.056394mol for Li2SO4 in equation (8).

FU of Li2SO4=(0.056394 mol Li2SO4)(6.022×1023 FU Li2SO41 mol Li2SO4)=3.3960×1022 FU=3.4×1022 FU Li2SO4

Conclusion

The number of moles and formula unit in 6.2 g of lithium sulfate is 0.056 mol and 3.4×1022 FU respectively.

(d)

Interpretation Introduction

Interpretation:

The number of lithium ions, sulfate ions, sulfur atoms and oxygen atoms in the mass of the compound Li2SO4 is to be calculated.

Concept introduction:

A formula unit is used for the ionic compound to represent their empirical formula. The molecular formula of a compound tells the number of atoms/ions of each element present in the compound.

A number of ions in a chemical compound is directly linked to the formula unit of the compound.

(d)

Expert Solution
Check Mark

Answer to Problem 3.19P

The number of lithium ions is 6.8×1022, sulfate ions are 3.4×1022, sulfur atoms are 3.4×1022 and oxygen atoms are 1.4×1023 in the mass of the compound Li2SO4.

Explanation of Solution

The expression to calculate the number of ions/atoms in Li2SO4 is:

Number of ions=(FU of Li2SO4)(Number of ions/atoms1 FUofLi2SO4) (9)

Substitute 3.3960×1022 FU for Li2SO4 and 2 for Li+ ion in equation (9).

Number of  Li+ ions=(3.3960×1022 FU)(1 Li+ ion1 FUofLi2SO4)=6.7920×1022=6.8×1022 Li+ ions

Substitute 3.3960×1022 FU for Li2SO4 and 1 for SO42 ion in equation (9).

Number of SO42 ions=(3.3960×1022 FU)(1 SO42 ion1 FUofLi2SO4)=3.3960×1022=3.4×1022 SO42 ions

Substitute 3.3960×1022 FU for Li2SO4 and 1 for S atom in equation (9).

Number of  S atoms=(3.3960×1022 FU)(1 S atom1 FUofLi2SO4)=3.3960×1022=3.4×1022 S atoms

Substitute 3.3960×1022 FU for Li2SO4 and 4 for O atom in equation (9).

Number of  O atoms=(3.3960×1022 FU)(4 O atom1 FUofLi2SO4)=1.3584×1023=1.4×1023 O atoms

Conclusion

The number of lithium ions is 6.8×1022, sulfate ions are 3.4×1022, sulfur atoms are 3.4×1022 and oxygen atoms are 1.4×1023 in the mass of the compound Li2SO4.

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Chapter 3 Solutions

Chemistry: The Molecular Nature of Matter and Change

Ch. 3.1 - Prob. 3.6AFPCh. 3.1 - For many years, compounds known as...Ch. 3.1 - Use the information in Follow-up Problem 3.6A to...Ch. 3.1 - Prob. 3.7BFPCh. 3.2 - Prob. 3.8AFPCh. 3.2 - A sample of an unknown compound contains 6.80 mol...Ch. 3.2 - A sample of an unknown compound is found to...Ch. 3.2 - Prob. 3.9BFPCh. 3.2 - Prob. 3.10AFPCh. 3.2 - Prob. 3.10BFPCh. 3.2 - A dry-cleaning solvent (ℳ = 146.99 g/mol) that...Ch. 3.2 - Prob. 3.11BFPCh. 3.3 - Prob. 3.12AFPCh. 3.3 - Prob. 3.12BFPCh. 3.3 - Prob. 3.13AFPCh. 3.3 - Prob. 3.13BFPCh. 3.4 - Prob. 3.14AFPCh. 3.4 - The tarnish that forms on objects made of silver...Ch. 3.4 - Prob. 3.15AFPCh. 3.4 - In the reaction that removes silver tarnish (see...Ch. 3.4 - Prob. 3.16AFPCh. 3.4 - Prob. 3.16BFPCh. 3.4 - Prob. 3.17AFPCh. 3.4 - Prob. 3.17BFPCh. 3.4 - Prob. 3.18AFPCh. 3.4 - Prob. 3.18BFPCh. 3.4 - In the reaction in Follow-up Problem 3.18A, how...Ch. 3.4 - Prob. 3.19BFPCh. 3.4 - Prob. 3.20AFPCh. 3.4 - Prob. 3.20BFPCh. 3.4 - Marble (calcium carbonate) reacts with...Ch. 3.4 - Prob. 3.21BFPCh. 3 - Prob. 3.1PCh. 3 - Prob. 3.2PCh. 3 - Why might the expression “1 mol of chlorine” be...Ch. 3 - Prob. 3.4PCh. 3 - Prob. 3.5PCh. 3 - Prob. 3.6PCh. 3 - Prob. 3.7PCh. 3 - Prob. 3.8PCh. 3 - Calculate the molar mass of each of the...Ch. 3 - Prob. 3.10PCh. 3 - Prob. 3.11PCh. 3 - Calculate each of the following quantities: Mass...Ch. 3 - Calculate each of the following quantities: Amount...Ch. 3 - Prob. 3.14PCh. 3 - Prob. 3.15PCh. 3 - Prob. 3.16PCh. 3 - Prob. 3.17PCh. 3 - Prob. 3.18PCh. 3 - Prob. 3.19PCh. 3 - Calculate each of the following: Mass % of H in...Ch. 3 - Calculate each of the following: Mass % of I in...Ch. 3 - Calculate each of the following: Mass fraction of...Ch. 3 - Calculate each of the following: Mass fraction of...Ch. 3 - Oxygen is required for the metabolic combustion of...Ch. 3 - Cisplatin (right), or Platinol, is used in the...Ch. 3 - Allyl sulfide (below) gives garlic its...Ch. 3 - Iron reacts slowly with oxygen and water to form a...Ch. 3 - Prob. 3.28PCh. 3 - Prob. 3.29PCh. 3 - The mineral galena is composed of lead(II) sulfide...Ch. 3 - Prob. 3.31PCh. 3 - Prob. 3.32PCh. 3 - List three ways compositional data may be given in...Ch. 3 - Prob. 3.34PCh. 3 - Prob. 3.35PCh. 3 - Prob. 3.36PCh. 3 - Prob. 3.37PCh. 3 - Prob. 3.38PCh. 3 - Prob. 3.39PCh. 3 - What is the molecular formula of each...Ch. 3 - Prob. 3.41PCh. 3 - Prob. 3.42PCh. 3 - Find the empirical formula of each of the...Ch. 3 - An oxide of nitrogen contains 30.45 mass % N. (a)...Ch. 3 - Prob. 3.45PCh. 3 - A sample of 0.600 mol of a metal M reacts...Ch. 3 - Prob. 3.47PCh. 3 - Prob. 3.48PCh. 3 - Prob. 3.49PCh. 3 - Prob. 3.50PCh. 3 - Prob. 3.51PCh. 3 - Prob. 3.52PCh. 3 - Prob. 3.53PCh. 3 - Prob. 3.54PCh. 3 - Prob. 3.55PCh. 3 - Prob. 3.56PCh. 3 - Prob. 3.57PCh. 3 - Prob. 3.58PCh. 3 - Prob. 3.59PCh. 3 - Prob. 3.60PCh. 3 - Prob. 3.61PCh. 3 - Prob. 3.62PCh. 3 - Prob. 3.63PCh. 3 - Prob. 3.64PCh. 3 - Prob. 3.65PCh. 3 - Prob. 3.66PCh. 3 - Prob. 3.67PCh. 3 - Prob. 3.68PCh. 3 - Prob. 3.69PCh. 3 - Prob. 3.70PCh. 3 - Prob. 3.71PCh. 3 - Prob. 3.72PCh. 3 - Prob. 3.73PCh. 3 - Prob. 3.74PCh. 3 - Elemental phosphorus occurs as tetratomic...Ch. 3 - Prob. 3.76PCh. 3 - Solid iodine trichloride is prepared in two steps:...Ch. 3 - Prob. 3.78PCh. 3 - Prob. 3.79PCh. 3 - Prob. 3.80PCh. 3 - Prob. 3.81PCh. 3 - Prob. 3.82PCh. 3 - Prob. 3.83PCh. 3 - Prob. 3.84PCh. 3 - Prob. 3.85PCh. 3 - Prob. 3.86PCh. 3 - Prob. 3.87PCh. 3 - Prob. 3.88PCh. 3 - Prob. 3.89PCh. 3 - Prob. 3.90PCh. 3 - Prob. 3.91PCh. 3 - Prob. 3.92PCh. 3 - Prob. 3.93PCh. 3 - Prob. 3.94PCh. 3 - When 20.5 g of methane and 45.0 g of chlorine gas...Ch. 3 - Prob. 3.96PCh. 3 - Prob. 3.97PCh. 3 - Prob. 3.98PCh. 3 - Prob. 3.99PCh. 3 - Prob. 3.100PCh. 3 - Sodium borohydride (NaBH4) is used industrially in...Ch. 3 - Prob. 3.102PCh. 3 - The first sulfur-nitrogen compound was prepared in...Ch. 3 - Prob. 3.104PCh. 3 - Prob. 3.105PCh. 3 - Prob. 3.106PCh. 3 - Serotonin () transmits nerve impulses between...Ch. 3 - In 1961, scientists agreed that the atomic mass...Ch. 3 - Prob. 3.109PCh. 3 - Isobutylene is a hydrocarbon used in the...Ch. 3 - The multistep smelting of ferric oxide to form...Ch. 3 - Prob. 3.112PCh. 3 - Prob. 3.113PCh. 3 - Prob. 3.114PCh. 3 - Prob. 3.115PCh. 3 - Prob. 3.116PCh. 3 - Prob. 3.117PCh. 3 - Prob. 3.118PCh. 3 - Prob. 3.119PCh. 3 - Prob. 3.120PCh. 3 - For the reaction between solid tetraphosphorus...Ch. 3 - Prob. 3.122PCh. 3 - Prob. 3.123PCh. 3 - Prob. 3.124PCh. 3 - Prob. 3.125PCh. 3 - Prob. 3.126PCh. 3 - Prob. 3.127PCh. 3 - Prob. 3.128PCh. 3 - Ferrocene, synthesized in 1951, was the first...Ch. 3 - Prob. 3.130PCh. 3 - Prob. 3.131PCh. 3 - Citric acid (below) is concentrated in citrus...Ch. 3 - Prob. 3.133PCh. 3 - Nitrogen monoxide reacts with elemental oxygen to...Ch. 3 - Prob. 3.135PCh. 3 - Prob. 3.136PCh. 3 - Manganese is a key component of extremely hard...Ch. 3 - The human body excretes nitrogen in the form of...Ch. 3 - Aspirin (acetylsalicylic acid, C9H8O4) is made by...Ch. 3 - Prob. 3.140PCh. 3 - Prob. 3.141PCh. 3 - Prob. 3.142PCh. 3 - When powdered zinc is heated with sulfur, a...Ch. 3 - Cocaine (C17H21O4N) is a natural substance found...Ch. 3 - Prob. 3.145P
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