Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781260151749
Author: Silberberg Dr., Martin; Amateis Professor, Patricia
Publisher: McGraw-Hill Education
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Chapter 3, Problem 3.18P

(a)

Interpretation Introduction

Interpretation:

The mass (g) of 8.35 mol of copper(I)carbonate is to be calculated.

Concept introduction:

Molar mass of a substance is defined as the mass of one mole of a chemical entity. It is calculated as the summation of product of number of atoms of each element and the atomic mass of the respective element.

(a)

Expert Solution
Check Mark

Answer to Problem 3.18P

The mass (g) of 8.35 mol of copper(I)carbonate is 1.56×103 g.

Explanation of Solution

Copper(I)carbonate is an ionic compound with chemical formula Cu2CO3.

The formula ot calculate the molar mass of Cu2CO3 is,

Molar Mass of Cu2CO3=[(number of Cu atoms)(atomic massof Cu)+(number ofC atoms)(atomicmassof C)+(number ofO atoms)(atomicmassofO)] (1)

Substitute 2 for number of Cu atoms, 63.55 g/mol for the atomic mass of Cu, 1 for number of C atoms, 12.01g/mol for the atomic mass of C, 3 for number of O atoms and 16.00g/mol for the atomic mass of O in equation (1).

Molar Mass of Cu2CO3=[(2)(63.55 g/mol)+(1)(12.01g/mol)+(3)(16.00g/mol)]=127.1 g/mol+12.01g/mol+48.00g/mol=187.11g/mol

The expression to calculate mass (g) of Cu2CO3 is,

Mass (g)=(mol of Cu2CO3)(molecular mass(g) of Cu2CO3 1 mol Cu2CO3) (2)

Substitute 8.35 mol for mol of Cu2CO3 and 187.11g/mol for molecular mass of Cu2CO3 in equation (2).

Mass (g) of Cu2CO3=(8.35 mol)(187.11 g Cu2CO3 1 mol Cu2CO3)=156.24 g=1.56×103 g Cu2CO3

Conclusion

The mass (g) of 8.35 mol of copper(I)carbonate is 1.56×103 g.

(b)

Interpretation Introduction

Interpretation:

The mass (g) of 4.04×1020 molecules of dinitrogen pentoxide 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 3.18P

The mass (g) of 4.04×1020 molecules of dinitrogen pentoxide is 0.0725 g.

Explanation of Solution

Dinitrogen pentoxide is an inorganic compound with chemical formula N2O5.

The formula to calculate the molar mass of N2O5 is,

Molar Mass of N2O5=[(number of N atoms)(atomic massof N)+(number ofO atoms)(atomicmassofO)] (3)

Substitute 2 for number of N atoms, 14.01 g/mol for atomic mass of N, 5 for number of O atoms, 16.00g/mol for atomic mass of O in equation (3).

Molar Mass of N2O5=[(2)(14.01 g/mol)+(5)(16.00g/mol )]=28.02 g/mol+80.00g/mol=108.02g/mol

One mole of N2O5 contains 6.022×1023 N2O5 molecules. Divide 4.04×1020 molecules of N2O5 by the Avogadro’s number to calculate the moles of N2O5 as follows:

Amount of N2O5(mol)=(4.04×1020 N2O5 molecules)(1 mol N2O56.022×1023 N2O5 molecules)=6.7087×104 mol N2O5

The expression to calculate mass (g) of N2O5 is,

Mass (g)=(mol of N2O5)(molecular mass(g) of N2O5 1 mol N2O5) (4)

Substitute 6.7087×104 mol for mol of N2O5 and 108.02g/mol for molecular mass of N2O5 in equation (4).

Mass (g) of N2O5=(6.7087×104 mol)( 108.02 g N2O5 1 mol N2O5)=0.072467 g=0.0725 g N2O5

Conclusion

The mass (g) of 4.04×1020 molecules of dinitrogen pentoxide is 0.0725 g.

(c)

Interpretation Introduction

Interpretation:

The number of moles and formula unit in 78.9 g of sodium perchlorate 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 3.18P

The number of moles and formula unit in 78.9 g of sodium perchlorate is 0.644 mol and 3.88×1023 FU respectively.

Explanation of Solution

Sodium perchlorate is an ionic compound with chemical formula NaClO4.

The formula to calculate the molar mass of NaClO4 is,

Molar Mass of NaClO4=[(number of Na atoms)(atomic massof Na)+(number ofCl atoms)(atomicmassof Cl)+(number ofO atoms)(atomicmassofO)] (5)

Substitute 1 for number of Na atoms, 22.99 g/mol for the atomic mass of Na, 1 for number of Cl atoms, 35.45g/mol for the atomic mass of Cl, 4 for number of O atoms and 16.00g/mol for the atomic mass of O in equation (5).

Molar Mass of NaClO4=[(1)(22.99 g/mol)+(1)(35.45g/mol )+(4)(16.00g/mol )]=22.99 g/mol+35.45g/mol+64.00g/mol=122.44g/mol

The expression to calculate moles of NaClO4 is:

MolesofNaClO4=(given mass of NaClO4(g))(1 mol NaClO4molecular mass(g)) (6)

Substitute 78.9 g for given mass of NaClO4 and 122.44g/mol for the molecular mass of NaClO4 in equation (6).

Moles ofNaClO4=(78.9 gNaClO4)(1mol NaClO4122.44 g NaClO4)=0.644397mol=0.644 mol NaClO4

The expression to calculate formula units (FU) of NaClO4 is:

FU of NaClO4=(mol of NaClO4)(6.022×1023 FU NaClO41mol NaClO4) (7)

Substitute 0.644397mol for mol of NaClO4 in equation (7).

FU of NaClO4=(0.644397mol NaClO4)(6.022×1023 FU NaClO41mol NaClO4)=3.88056×1023 FU=3.88×1023 FU NaClO4

Conclusion

The number of moles and formula unit in 78.9 g of sodium perchlorate is 0.644 mol and 3.88×1023 FU respectively.

(d)

Interpretation Introduction

Interpretation:

The number of sodium ions, perchlorate ions, chlorine atoms and oxygen atoms in the mass of compound NaClO4 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.

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 3.18P

The number of sodium ions, perchlorate ions, chlorine atoms are 3.88×1023 and oxygen atoms are 1.55×1024 in the mass of compound NaClO4.

Explanation of Solution

From the formula of NaClO4, it is concluded that 1 sodium ion, one perchlorate ion, one chlorine atom and four oxygen atoms are present in one formula unit of NaClO4.

The expression to calculate the number of ions/atoms in NaClO4 is:

Number of ions=(FU of NaClO4)(Number of ions/atoms1FUofNaClO4) (8)

Substitute 3.88056×1023 FU for NaClO4 and 1 for number of Na+ ion in equation (8).

Number of  Na+ ions=(3.88056×1023 FU)(1 Na+ ion1FUofNaClO4)=3.88×1023 Na+ ions

Substitute 3.88056×1023 FU for NaClO4 and 1 for number of ClO4 ion in equation (8).

Number of  ClO4 ions=(3.88056×1023 FU)(1 ClO4 ion1FUofNaClO4)=3.88×1023 ClO4 ions

Substitute 3.88056×1023 FU for NaClO4 and 1 for number of Cl atom in equation (8).

Number of  Cl atoms=(3.88056×1023 FU)(1 Cl atom1FUofNaClO4)=3.88×1023 Cl atoms

Substitute 3.88056×1023 FU for NaClO4 and 4 for number of O atom in equation (8).

Number of  O atoms=(3.88056×1023 FU)( 4 O atom1FUofNaClO4)=1.55×1024 O atoms

Conclusion

The number of sodium ions, perchlorate ions, chlorine atoms are 3.88×1023 and oxygen atoms are 1.55×1024 in the mass of compound NaClO4.

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Chapter 3 Solutions

Loose Leaf for 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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