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 3.17P

(a)

Interpretation Introduction

Interpretation:

Total number of ions in 38.1 g of SrF2 is to be calculated.

Concept introduction:

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

For example, the molar mass of NaCl is calculated as follows.

Molar Mass of NaCl=[(number of Na atoms)(atomic massof Na)+(number of  Cl atoms)(atomicmassof Cl )] (1)

A mole is a unit of measurement that has the same number of particles as that in 12.000 g of carbon12. One mole of a chemical substance (atoms/molecules/ions) has 6.022×1023 number of entities.

(a)

Expert Solution
Check Mark

Answer to Problem 3.17P

Total number of ions in 38.1 g of SrF2 is 5.48×1023 ions.

Explanation of Solution

The formula to calculate the molar mass of SrF2. is,

Molar Mass of SrF2=[(number of Sr atoms)(atomic massof Sr)+(number ofF atoms)(atomicmassof F )] (2)

Substitute 1 for number of Sr atoms and 87.62 g/mol for atomic mass of Sr. Similarly, substitute 2 for number of F atoms and 19.00g/mol for atomic mass of F in equation (2).

Molar Mass of SrF2=[(1)(87.62 g/mol)+(2)(19.00g/mol )]=87.62 g/mol+38.00 g/mol=125.62 g/mol

The expression to calculate moles of SrF2 is:

MolesofSrF2=(given mass of SrF2(g))(1 mol SrF2molecular massof SrF2(g)) (3)

Substitute 38.1 g for given mass of SrF2 and 125.62 g/mol for the molecular mass of SrF2 in equation (3).

Moles ofSrF2=(38.1gSrF2)(1mol SrF2125.62gSrF2)=0.303296molSrF2

The expression to calculate moles of ions of SrF2 is:

MolesofionsofSrF2=(mol of SrF2)(mol of ions in SrF21 mol SrF2) (4)

Substitute 0.303296mol for SrF2 and 3 for mol of ions in SrF2 in equation (4).

MolesofionsofSrF2=(0.303296 mol SrF2)(3 mol of ions in SrF21 mol SrF2)=0.909888 molions

The expression to calculate number of ions of SrF2 is:

NumberofionsofSrF2=(mol ions of SrF2)(6.022×1023 ions1 mol ions of SrF2) (5)

Substitute 0.909888mol ions for SrF2 in equation (5).

NumberofionsofSrF2=(0.909888 mol ions)(6.022×1023 ions1 mol ions of SrF2)=5.47935×1023ions5.48×1023 ions

Conclusion

Total number of ions in 38.1 g of SrF2 is 5.48×1023 ions.

(b)

Interpretation Introduction

Interpretation:

Mass (kg) of 3.58 mol of CuCl22H2O is to be calculated.

Concept introduction:

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

For example, the molar mass of NaCl is calculated as follows.

Molar Mass of NaCl=[(number of Na atoms)(atomic massof Na)+(number of  Cl atoms)(atomicmassof Cl )] (1)

The conversion factor for g into kg is given as.

Mass (kg)=(Mass(g))(1 kg103g) (6)

(b)

Expert Solution
Check Mark

Answer to Problem 3.17P

Massof 3.58 mol of CuCl22H2O is 0.610 kg.

Explanation of Solution

The formula to calculate the molar mass of CuCl22H2O is,

Molar Mass of CuCl22H2O=[(number of Cu atoms)(atomic massof Cu)+(number ofCl atoms)(atomicmassof Cl )+(number of H atoms)(atomicmassof H )+(number ofO atoms)(atomicmassof)] (7)

Substitute 1 for number of Cu atoms and 63.55 g/mol for atomic mass of Cu, 2 for number of atoms of Cl and 35.45g/mol for atomic mass of Cl, 4 for number of atoms of H and 1.008g/mol for atomic mass of H, 2 for number of atoms of O and 16.00g/mol for atomic mass of O in equation(7).

Molar Mass of CuCl22H2O=[(1)(63.55 g/mol)+(2)(35.45g/mol )+(4)(1.008g/mol )+(2)(16.00g/mol )]=63.55 g/mol+70.9g/mol+4.032g/mol+32.00g/mol=170.48g/mol

The expression to calculate the mass (g) of CuCl22H2O is,

Mass (g)=(mol of CuCl22H2O)(molecular mass(g) of CuCl22H21 mol CuCl22H2O) (8)

Substitute 3.58 mol for CuCl22H2O and 170.48 g/mol for molecular mass of CuCl22H2O in equation (8).

Mass (g) of CuCl22H2O=(3.58 mol)(170.48 g CuCl22H21 mol CuCl22H2O)=(610.32 g CuCl22H2O)(1 kg103g)=0.61032 kg=0.610 kg of CuCl22H2O

Conclusion

Mass of 3.58 mol of CuCl22H2O is 0.610 kg.

(c)

Interpretation Introduction

Interpretation:

Mass (mg) of 2.88×1022 formula units of Bi(NO3)35H2O is to be calculated.

Concept introduction:

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

For example, the molar mass of NaCl is calculated as follows.

Molar Mass of NaCl=[(number of Na atoms)(atomic massof Na)+(number of  Cl atoms)(atomicmassof Cl )] (1)

The conversion factor for g into mg is given as.

Mass (mg)=(Mass(g))(1 mg103g) (9)

(c)

Expert Solution
Check Mark

Answer to Problem 3.17P

Mass of 2.88×1022 formula units of Bi(NO3)35H2O is 2.32×104 mg.

Explanation of Solution

The formula to calculate the molar mass of Bi(NO3)35H2O. is,

Molar Mass=[(number of Bi atoms)(atomic massof Bi)+(number of N atoms)(atomicmassof N )+(number of H atoms)(atomicmassof H )+(number ofO atoms)(atomicmassof)] (10)

Substitute 1 for number of Bi atoms, 209.0 g/mol for atomic mass of Bi, 3 for number of atoms of N, 14.01g/mol for atomic mass of N, 10 for number of atoms of H, 1.008g/mol for atomic mass of H, 14 for number of atoms of O and 16.00g/mol for atomic mass of O in equation (10).

Molar Mass of Bi(NO3)35H2O=[(1)(209.0 g/mol)+(3)(14.01g/mol )+(10)(1.008g/mol )+(14)(16.00g/mol )]=209.0 g/mol+42.03g/mol+10.08g/mol+224.00g/mol=485.1g/mol

The expression to calculate moles of Bi(NO3)35H2O is.

Molesof Bi(NO3)35H2O=(number of FU)(1 mol6.022×1023 FU) (11)

Substitute 2.88×1022 for formula units of Bi(NO3)35H2O in equation (11).

Molesof Bi(NO3)35H2O=(2.88×1022 FU)(1 mol6.022×1023 FU)=0.047825 mol Bi(NO3)35H2O

The expression to calculate the mass (g) of Bi(NO3)35H2O is:

Mass (g)=(mol Bi(NO3)35H2O)(molecular mass(g) Bi(NO3)35H21 mol CuCl22H2O) (12)

Substitute 0.047825 mol for Bi(NO3)35H2O and 485.1 g/mol for molecular mass of Bi(NO3)35H2O in equation (12).

Mass (g)=(0.047825 mol Bi(NO3)35H2O)(485.1 g Bi(NO3)35H21 mol Bi(NO3)35H2O)=(23.1999 g Bi(NO3)35H2O)(1 mg103g)=23199.9 mg of Bi(NO3)35H2O=2.32×104 mg of Bi(NO3)35H2O

Conclusion

Mass of 2.88×1022 formula units of Bi(NO3)35H2O is 2.32×104 mg.

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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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