Concept explainers
(a)
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
For example, the molar mass of NaCl is calculated as follows.
Molar Mass of NaCl=[(number of Na atoms )(atomic mass of Na)+(number of Cl atoms )(atomic mass of Cl )] (1)
A mole is a unit of measurement that has the same number of particles as that in 12.000 g of carbon−12. One mole of a chemical substance (atoms/molecules/ions) has 6.022×1023 number of entities.
(a)
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 mass of Sr)+(number of F atoms )(atomic mass of 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 19.00 g/mol for atomic mass of F in equation (2).
Molar Mass of SrF2=[(1)(87.62 g/mol)+(2)(19.00 g/mol )]=87.62 g/mol+38.00 g/mol=125.62 g/mol
The expression to calculate moles of SrF2 is:
Moles of SrF2=(given mass of SrF2(g))(1 mol SrF2molecular mass of 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 of SrF2=(38.1 g SrF2)(1mol SrF2125.62 g SrF2)=0.303296 mol SrF2
The expression to calculate moles of ions of SrF2 is:
Moles of ions of SrF2=(mol of SrF2)(mol of ions in SrF21 mol SrF2) (4)
Substitute 0.303296 mol for SrF2 and 3 for mol of ions in SrF2 in equation (4).
Moles of ions of SrF2=(0.303296 mol SrF2)(3 mol of ions in SrF21 mol SrF2)=0.909888 mol ions
The expression to calculate number of ions of SrF2 is:
Number of ions of SrF2=(mol ions of SrF2)(6.022×1023 ions1 mol ions of SrF2) (5)
Substitute 0.909888 mol ions for SrF2 in equation (5).
Number of ions of SrF2=(0.909888 mol ions)(6.022×1023 ions1 mol ions of SrF2)=5.47935×1023 ions≃5.48×1023 ions
Total number of ions in 38.1 g of SrF2 is 5.48×1023 ions.
(b)
Interpretation:
Mass (kg) of 3.58 mol of CuCl2⋅2H2O 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 mass of Na)+(number of Cl atoms )(atomic mass of Cl )] (1)
The conversion factor for g into kg is given as.
Mass (kg)=(Mass (g))(1 kg103 g) (6)
(b)
Answer to Problem 3.17P
Massof 3.58 mol of CuCl2⋅2H2O is 0.610 kg.
Explanation of Solution
The formula to calculate the molar mass of CuCl2⋅2H2O is,
Molar Mass of CuCl2⋅2H2O=[(number of Cu atoms )(atomic mass of Cu)+(number of Cl atoms )(atomic mass of Cl )+(number of H atoms )(atomic mass of H )+(number of O atoms )(atomic mass of O )] (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 35.45 g/mol for atomic mass of Cl, 4 for number of atoms of H and 1.008 g/mol for atomic mass of H, 2 for number of atoms of O and 16.00 g/mol for atomic mass of O in equation(7).
Molar Mass of CuCl2⋅2H2O=[(1 )(63.55 g/mol)+(2 )(35.45 g/mol )+(4 )(1.008 g/mol )+(2 )(16.00 g/mol )]=63.55 g/mol+70.9 g/mol+4.032 g/mol+32.00 g/mol=170.48 g/mol
The expression to calculate the mass (g) of CuCl2⋅2H2O is,
Mass (g)=(mol of CuCl2⋅2H2O)(molecular mass (g) of CuCl2⋅2H2O 1 mol CuCl2⋅2H2O) (8)
Substitute 3.58 mol for CuCl2⋅2H2O and 170.48 g/mol for molecular mass of CuCl2⋅2H2O in equation (8).
Mass (g) of CuCl2⋅2H2O=(3.58 mol)(170.48 g CuCl2⋅2H2O 1 mol CuCl2⋅2H2O)=(610.32 g CuCl2⋅2H2O)(1 kg103 g)=0.61032 kg=0.610 kg of CuCl2⋅2H2O
Mass of 3.58 mol of CuCl2⋅2H2O is 0.610 kg.
(c)
Interpretation:
Mass (mg) of 2.88×1022 formula units of Bi(NO3)3⋅5H2O 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 mass of Na)+(number of Cl atoms )(atomic mass of Cl )] (1)
The conversion factor for g into mg is given as.
Mass (mg)=(Mass (g))(1 mg10−3 g) (9)
(c)
Answer to Problem 3.17P
Mass of 2.88×1022 formula units of Bi(NO3)3⋅5H2O is 2.32×104 mg.
Explanation of Solution
The formula to calculate the molar mass of Bi(NO3)3⋅5H2O. is,
Molar Mass=[(number of Bi atoms )(atomic mass of Bi)+(number of N atoms )(atomic mass of N )+(number of H atoms )(atomic mass of H )+(number of O atoms )(atomic mass of O )] (10)
Substitute 1 for number of Bi atoms, 209.0 g/mol for atomic mass of Bi, 3 for number of atoms of N, 14.01 g/mol for atomic mass of N, 10 for number of atoms of H, 1.008 g/mol for atomic mass of H, 14 for number of atoms of O and 16.00 g/mol for atomic mass of O in equation (10).
Molar Mass of Bi(NO3)3⋅5H2O=[(1)(209.0 g/mol)+(3)(14.01 g/mol )+(10)(1.008 g/mol )+(14)(16.00 g/mol )]=209.0 g/mol+42.03 g/mol+10.08 g/mol+224.00 g/mol=485.1 g/mol
The expression to calculate moles of Bi(NO3)3⋅5H2O is.
Moles of Bi(NO3)3⋅5H2O=(number of FU)(1 mol6.022×1023 FU) (11)
Substitute 2.88×1022 for formula units of Bi(NO3)3⋅5H2O in equation (11).
Moles of Bi(NO3)3⋅5H2O=(2.88×1022 FU)(1 mol6.022×1023 FU)=0.047825 mol Bi(NO3)3⋅5H2O
The expression to calculate the mass (g) of Bi(NO3)3⋅5H2O is:
Mass (g)=(mol Bi(NO3)3⋅5H2O)(molecular mass (g) Bi(NO3)3⋅5H2O 1 mol CuCl2⋅2H2O) (12)
Substitute 0.047825 mol for Bi(NO3)3⋅5H2O and 485.1 g/mol for molecular mass of Bi(NO3)3⋅5H2O in equation (12).
Mass (g)=(0.047825 mol Bi(NO3)3⋅5H2O)(485.1 g Bi(NO3)3⋅5H2O 1 mol Bi(NO3)3⋅5H2O)=(23.1999 g Bi(NO3)3⋅5H2O)(1 mg10−3 g)=23199.9 mg of Bi(NO3)3⋅5H2O=2.32×104 mg of Bi(NO3)3⋅5H2O
Mass of 2.88×1022 formula units of Bi(NO3)3⋅5H2O is 2.32×104 mg.
Want to see more full solutions like this?
Chapter 3 Solutions
Chemistry: The Molecular Nature of Matter and Change
- Nonearrow_forwardHowever, why are intermolecular forces in metallic and ionic compounds not discussed as extensively? Additionally, what specific types of intermolecular attractions exist in metals and ionic compoundsarrow_forwardWhat is the preparation of 1 Liter of 0.1M NH4Cl buffer at pH 9.0 with solid NH4Cl and 0.1M NaOH. How would I calculate the math to describe this preparation? How would I use Henderson-Hasselbach equation?arrow_forward
- C Predict the major products of this organic reaction. Be sure you use wedge and dash bonds when necessary, for example to distinguish between major products with different stereochemistry. : ☐ + x G C RCO₂H Click and drag to start drawing a structure.arrow_forwardFill in the blanks by selecting the appropriate term from below: For a process that is non-spontaneous and that favors products at equilibrium, we know that a) ΔrG∘ΔrG∘ _________, b) ΔunivSΔunivS _________, c) ΔsysSΔsysS _________, and d) ΔrH∘ΔrH∘ _________.arrow_forwardHighest occupied molecular orbital Lowest unoccupied molecular orbital Label all nodes and regions of highest and lowest electron density for both orbitals.arrow_forward
- Relative Intensity Part VI. consider the multi-step reaction below for compounds A, B, and C. These compounds were subjected to mass spectrometric analysis and the following spectra for A, B, and C was obtained. Draw the structure of B and C and match all three compounds to the correct spectra. Relative Intensity Relative Intensity 20 NaоH 0103 Br (B) H2504 → (c) (A) 100- MS-NU-0547 80 40 20 31 10 20 100- MS2016-05353CM 80 60 100 MS-NJ-09-3 80 60 40 20 45 J.L 80 S1 84 M+ absent राग 135 137 S2 62 164 166 11 S3 25 50 75 100 125 150 175 m/zarrow_forwardDon't used hand raiting and don't used Ai solutionarrow_forwardDon't used hand raitingarrow_forward
- Don't used hand raitingarrow_forwardA composite material reinforced with aligned fibers, consisting of 20% by volume of silicon carbide (SiC) fibers and 80% by volume of polycarbonate (PC) matrix. The mechanical characteristics of the 2 materials are in the table. The stress of the matrix when the fiber breaks is 45 MPa. Calculate the longitudinal strength? SiC PC Elastic modulus (GPa) Tensile strength (GPa) 400 2,4 3,9 0,065arrow_forwardQuestion 2 What starting materials or reagents are best used to carry out the following reaction? 2Fe, 3Br2 ○ FeCl3 2Fe, 4Br2 O Heat and Br2 Heat and HBr Brarrow_forward
- ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage LearningChemistryChemistryISBN:9781259911156Author:Raymond Chang Dr., Jason Overby ProfessorPublisher:McGraw-Hill EducationPrinciples of Instrumental AnalysisChemistryISBN:9781305577213Author:Douglas A. Skoog, F. James Holler, Stanley R. CrouchPublisher:Cengage Learning
- Organic ChemistryChemistryISBN:9780078021558Author:Janice Gorzynski Smith Dr.Publisher:McGraw-Hill EducationChemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage LearningElementary Principles of Chemical Processes, Bind...ChemistryISBN:9781118431221Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. BullardPublisher:WILEY