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 3.16P

(a)

Interpretation Introduction

Interpretation:

The mass of 6.44×102 mol of MnSO4 in grams 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.

Molar mass is used to convert the mass of a substance to moles. The expression to calculate the mass of a chemical substance in grams is as follows:

Mass (g)=(amount(mol))(no. of grams1 mol)

(a)

Expert Solution
Check Mark

Answer to Problem 3.16P

The mass of 6.44×102 mol of MnSO4 in grams is 9.72 g.

Explanation of Solution

The expression to calculate the molar mass of MnSO4 is as follows:

Molar mass of MnSO4=(1)(M of Mn)+(1)(M of S)+(4)(M of O)

Substitute 54.94 g/mol for M of Mn, 32.06 g/mol for M of S, and 16.00 g/mol for M of O in the above equation as follows:

Molar mass of MnSO4=(1)(54.94 g/mol)+(1)(32.06 g/mol)+(4)(16.00 g/mol)=54.94 g/mol+32.06 g/mol+64.00 g/mol=151.00 g/mol

Therefore, the molar mass of MnSO4 is 151.00 g/mol.

Calculate the mass of 6.44×102 mol of MnSO4 as follows:

Mass of MnSO4(g)=(6.44×102 mol MnSO4)( 151.00 g MnSO41 mol MnSO4)=9.7244 g MnSO49.72 g MnSO4

Conclusion

The mass of 6.44×102 mol of MnSO4 in grams is 9.72 g.

(b)

Interpretation Introduction

Interpretation:

The amount of 15.8 kg of Fe(ClO4)3 in moles 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.

Molar mass is used to calculate the mass of a chemical substance in grams. The expression to calculate the mass of a chemical substance in grams is as follows:

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

(b)

Expert Solution
Check Mark

Answer to Problem 3.16P

The amount of 15.8 kg of Fe(ClO4)3 is 4.6 mol.

Explanation of Solution

The expression to calculate the molar mass of Fe(ClO4)3 is as follows:

Molar mass of Fe(ClO4)3=(1)(M of Fe)+(3)(M of Cl)+(12)(M of O)

Substitute 55.85 g/mol for M of Fe, 35.45 g/mol for M of Cl, and 16.00 g/mol for M of O in the above equation as follows:

Molar mass of Fe(ClO4)3=(1)(55.85 g/mol)+(3)(35.45 g/mol)+(12)(16.00 g/mol)=55.85 g/mol+106.35 g/mol+192.00 g/mol=354.20 g/mol

Therefore, the molar mass of Fe(ClO4)3 is 354.20 g/mol.

The relation between kg and g is as follows:

1 kg=103 g

Therefore, the conversion factor is (103 g1 kg).

Calculate the amount of 15.8 kg of Fe(ClO4)3 as follows:

Amount of Fe(ClO4)3(mol)=(15.8 kg Fe(ClO4)3)(103 g1 kg)( 1 mol Fe(ClO4)3354.20 g Fe(ClO4)3)=44.60756635 mol Fe(ClO4)34.6 mol Fe(ClO4)3

Conclusion

The amount of 15.8 kg of Fe(ClO4)3 is 4.6 mol.

(c)

Interpretation Introduction

Interpretation:

The number of N atoms in 92.6 mg of NH4NO2 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.

The molecular formula of a compound tells the number of atoms/ions of each element present in the compound.

Following are the steps to calculate the number of atoms in a compound.

Step 1: Convert the mass of the compound to moles by using the molar mass of that compound as follows:

Amount(mol)=(Given mass(g))(1 molNo. of grams)

Step 2: Determine the number of moles of atoms of an element from the molecular formula of the compound. The formula to calculate the moles of atoms is as follows:

Amount of atoms(mol)=(amount ofcompound(mol))(moles of atoms inmolecular formula1 mol compound)

Step 3: Determine the number of atoms by multiplying the moles of atoms with the Avogadro’s number. The expression to calculate the number of atoms of an element is as follows:

Number of atoms=(Amount (mol))(6.022×1023 atoms1 mol)

(c)

Expert Solution
Check Mark

Answer to Problem 3.16P

In 92.6 mg of NH4NO2, 1.74×1021 N atoms are present.

Explanation of Solution

The expression to calculate the molar mass of NH4NO2 is as follows:

Molar mass of NH4NO2=(2)(M of N)+(4)(M of H)+(2)(M of O)

Substitute 14.01 g/mol for M of N , 16.00 g/mol for M of O , and 1.008 g/mol for M of H  in the above equation as follows:

Molar mass of NH4NO2=(2)(14.01 g/mol)+(4)(1.008 g/mol)+(2)(16.00 g/mol)=28.02 g/mol+4.032 g/mol+32.00 g/mol=64.052 g/mol64.05 g/mol

Therefore, the molar mass of NH4NO2 is 64.05 g/mol. Hence, the mass of one mole of NH4NO2 is 64.05 g.

The relation between mg and g is as follows:

1 mg=103 g

Therefore, the conversion factor is (103 g1 mg).

Multiply 92.6 mg with the conversion factor and then divide with the molar mass of NH4NO2 to calculate the moles of NH4NO2 as follows:

Amount of NH4NO2(mol)=(92.6 mg NH4NO2)(103 g1 mg)(1 mol NH4NO264.05 g NH4NO2)=1.44575×103 mol NH4NO2

From the molecular formula of NH4NO2, it is concluded that 2 mol of nitrogen (N) atoms are present in 1 mol of NH4NO2. Calculate the moles of nitrogen atom in 1.44575×103 mol NH4NO2 as follows:

Amount of N atoms (mol)=(1.44575×103 mol NH4NO2)(2 mol N atoms1 mol NH4NO2)=2.8915×103 mol N atoms

One mole of nitrogen atom contains 6.022×1023 N atoms. Multiply 2.8915×103 mol N atoms with the Avogadro’s number to obtain the number of N atoms. Calculate the number of N atoms as follows:

Number of N atoms=(2.8915×103 mol N atoms)(6.022×1023 N atoms1 mol N atoms)=1.74×1021 N atoms

Conclusion

In 92.6 mg of NH4NO2, 1.74×1021 N atoms are present.

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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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