Connect 2-Year Access Card for Chemistry: The Molecular Nature of Matter and Change
Connect 2-Year Access Card for Chemistry: The Molecular Nature of Matter and Change
7th Edition
ISBN: 9780078129865
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Chapter 3, Problem 3.15P

(a)

Interpretation Introduction

Interpretation:

The mass of 4.6×1021 molecules of NO2 in kg 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 number of molecules is given.

Step 1: Determine the amount of substance 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 moleswith the molar mass of the chemical substance to obtain the mass of chemical substance in grams. The mass can be converted to kilogram by using the relation between gram and kilogram. The formula to calculate the mass of a substance in kilogram is as follows:

Mass (kg )=(Amount (mol))(No. of grams1 mol)(103 kg1 g)

(a)

Expert Solution
Check Mark

Answer to Problem 3.15P

The mass of 4.6×1021 molecules of NO2 in kg is 3.5×104 g NO2.

Explanation of Solution

The expression to calculate the molar mass of NO2 is as follows:

Molar mass of NO2=(1)(M of N)+(2)(M of O)

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

Molar mass of NO2=(1)(14.01 g/mol)+(2)(16.00 g/mol)=14.01 g/mol+32.00 g/mol=46.01 g/mol

Therefore, the molar mass of NO2 is 46.01 g/mol.

One mole of NO2 contains 6.022×1023 NO2 molecules. To calculate the moles of NO2, divide 4.6×1021 molecules of NO2 by the Avogadro’s number as follows:

Amount of NO2(mol)=(4.6×1021 molecules NO2)(1 mol NO26.022×1023 NO2 molecules)=7.63866×103 mol NO2

Multiply 7.63866×103 mol NO2 with the molar mass to calculate the mass of NO2 in gram as follows:

Mass of NO2(g)=(7.63866×103 mol NO2)(46.01 g NO21 mol NO2)=3.514547×101 g NO2

The relation between gram and kilogram is as follows:

1 g=103 kg

Therefore, the conversion factor is (103 kg1 g).

Multiply 3.514547×101 g NO2 with the conversion factor to calculate the mass of NO2 in kilogram as follows:

Mass of NO2(kg)=(3.514547×101 g NO2)(103 kg1 g)=3.514547×104 g NO23.5×104 g NO2

Conclusion

The mass of 4.6×1021 molecules of NO2 in kg is 3.5×104 g NO2.

(b)

Interpretation Introduction

Interpretation:

The amount of chlorine atoms in 0.0615 g of C2H4Cl2 in mol 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.

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

Following are the steps to calculate the amount of an element in a compound.

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

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 an element is as follows:

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

(b)

Expert Solution
Check Mark

Answer to Problem 3.15P

The amount of chlorine atoms in 0.0615 g of C2H4Cl2 in mol is 1.2431×103 mol.

Explanation of Solution

The expression to calculate the molar mass of C2H4Cl2 is as follows:

Molar mass of C2H4Cl2=(2)(M of C)+(4)(M of H)+(2)(M of Cl)

Substitute 12.01 g/mol for M of C, 1.008 g/mol for M of H and 35.45 g/mol for M of Cl in the above equation as follows:

Molar mass of C2H4Cl2=(2)(12.01 g/mol)+(4)(1.008 g/mol)+(2)(35.45 g/mol)=24.02 g/mol+4.032 g/mol+70.9 g/mol=98.95 g/mol

Therefore, the molar mass of C2H4Cl2 is 98.95 g/mol. Hence, the mass of one mole of C2H4Cl2 is 98.95 g. Multiply 0.0615 g of C2H4Cl2 with the molar mass of C2H4Cl2 to calculate the moles as follows:

Amount of C2H4Cl2(mol)=(0.0615 g C2H4Cl2)(1 mol C2H4Cl298.95 g C2H4Cl2)=6.21526×104 mol C2H4Cl2

From the molecular formula of C2H4Cl2, it is concluded that 2 mol of chlorine atoms are present in 1 mol of C2H4Cl2. Calculate the moles of chlorine atoms in 6.21526×104 mol C2H4Cl2 as follows:

Amount of Cl atoms(mol)=(6.21526×104 mol C2H4Cl2)(2 mol Cl atoms1 mol C2H4Cl2)=1.2431×103 mol Cl atoms

Conclusion

The amount of chlorine atoms in 0.0615 g of C2H4Cl2 in mol is 1.2431×103 mol.

(c)

Interpretation Introduction

Interpretation:

The number of H ions in 5.82 g of SrH2 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 ions 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 ions of an element from the molecular formula of the compound. The formula to calculate the moles of ions is as follows:

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

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

Number of ions=(Amount (mol))(6.022×1023 ions1 mol)

(c)

Expert Solution
Check Mark

Answer to Problem 3.15P

In 5.82 g of SrH2, 7.82×1022 H ions are present.

Explanation of Solution

The expression to calculate the molar mass of SrH2 is as follows:

Molar mass of SrH2=(1)(M of Sr)+(2)(M of H)

Substitute 87.62 g/mol for M of Sr and 1.008 g/mol for M of H in the above equation as follows:

Molar mass of SrH2=(1)(87.62 g/mol)+(2)(1.008 g/mol)=87.62 g/mol+2.016 g/mol+70.9 g/mol=89.636 g/mol89.64 g/mol

Therefore, the molar mass of SrH2 is 89.64 g/mol. Hence, the mass of one mole of SrH2 is 89.64 g. Divide 5.82 g of SrH2 with the molar mass of SrH2 to calculate the moles of SrH2 as follows:

Amount of SrH2(mol)=(5.82 g SrH2)(1 mol SrH289.64 g SrH2)=0.064926372 mol SrH20.0649264 mol SrH2

From the molecular formula of SrH2, it is concluded that 2 mol of hydride (H) ions are present in 1 mol of SrH2. Calculate the moles of hydride ion in 0.0649264 mol SrH2 as follows:

Amount of H ions(mol)=(0.0649264 mol SrH2)(2 mol H ions1 mol SrH2)=0.1298528 mol H ions

One mole of hydride ion has 6.022×1023 H ions. Multiply 0.1298528 mol H ions with the Avogadro’s number to obtain the number of H ions. Calculate the number of H ions as follows:

Number of H ions=(0.1298528 mol H ions)(6.022×1023 H ions1 mol H ions)=7.8187×1022 H ions7.82×1022 H ions

Conclusion

In 5.82 g of SrH2, 7.82×1022 H ions are present.

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Chapter 3 Solutions

Connect 2-Year Access Card 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 - Elemental phosphorus occurs as tetratomic...Ch. 3 - Prob. 3.75PCh. 3 - Solid iodine trichloride is prepared in two steps:...Ch. 3 - Prob. 3.77PCh. 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 - When 20.5 g of methane and 45.0 g of chlorine gas...Ch. 3 - Prob. 3.91PCh. 3 - Prob. 3.92PCh. 3 - Prob. 3.93PCh. 3 - Prob. 3.94PCh. 3 - Prob. 3.95PCh. 3 - Sodium borohydride (NaBH4) is used industrially in...Ch. 3 - Prob. 3.97PCh. 3 - The first sulfur-nitrogen compound was prepared in...Ch. 3 - Prob. 3.99PCh. 3 - Prob. 3.100PCh. 3 - Prob. 3.101PCh. 3 - Serotonin () transmits nerve impulses between...Ch. 3 - In 1961, scientists agreed that the atomic mass...Ch. 3 - Prob. 3.104PCh. 3 - Isobutylene is a hydrocarbon used in the...Ch. 3 - The multistep smelting of ferric oxide to form...Ch. 3 - Prob. 3.107PCh. 3 - Prob. 3.108PCh. 3 - Prob. 3.109PCh. 3 - Prob. 3.110PCh. 3 - Prob. 3.111PCh. 3 - Prob. 3.112PCh. 3 - Prob. 3.113PCh. 3 - Prob. 3.114PCh. 3 - Prob. 3.115PCh. 3 - For the reaction between solid tetraphosphorus...Ch. 3 - Prob. 3.117PCh. 3 - Prob. 3.118PCh. 3 - Prob. 3.119PCh. 3 - Prob. 3.120PCh. 3 - Prob. 3.121PCh. 3 - Prob. 3.122PCh. 3 - Prob. 3.123PCh. 3 - Ferrocene, synthesized in 1951, was the first...Ch. 3 - Prob. 3.125PCh. 3 - Prob. 3.126PCh. 3 - Citric acid (below) is concentrated in citrus...Ch. 3 - Prob. 3.128PCh. 3 - Nitrogen monoxide reacts with elemental oxygen to...Ch. 3 - Prob. 3.130PCh. 3 - Prob. 3.131PCh. 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. 3.135PCh. 3 - Prob. 3.136PCh. 3 - Prob. 3.137PCh. 3 - When powdered zinc is heated with sulfur, a...Ch. 3 - Cocaine (C17H21O4N) is a natural substance found...Ch. 3 - Prob. 3.140P
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