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2: Atoms, Molecules, and Ions

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    163846
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    Protons, Neutrons, and Electrons 

    Exercise \(\PageIndex{1.a}\)

    How are electrons and protons similar? How are they different?

    Answer

    Both are subatomic particles. Protons are positively charged, and reside in an atom’s nucleus. Whereas, electrons are negatively charged and reside outside an atom’s nucleus.

    Exercise \(\PageIndex{1.b}\)

    How are protons and neutrons similar? How are they different?

    Answer

    Both are subatomic particles that reside in an atom’s nucleus. Both have approximately the same mass. Protons are positively charged, whereas neutrons are uncharged.

     

    Dalton's Theory

    Exercise \(\PageIndex{2.a}\)

    The existence of isotopes violates one of the original ideas of Dalton’s atomic theory. Which one?

    Answer

    Dalton originally thought that all atoms of a particular element had identical properties, including mass. Thus, the concept of isotopes, in which an element has different masses, was a violation of the original idea. To account for the existence of isotopes, the second postulate of his atomic theory was modified to state that atoms of the same element must have identical chemical properties. 

    (See 2.1.2.2)

     

    Atomic Number, Mass Number, and Atomic Mass Units

    Exercise \(\PageIndex{3.a}\)

    Write the symbol for each of the following ions:

    1. the ion with a 1+ charge, atomic number 55, and mass number 133
    2. the ion with 54 electrons, 53 protons, and 74 neutrons
    3. the ion with atomic number 15, mass number 31, and a 3− charge
    4. the ion with 24 electrons, 30 neutrons, and a 3+ charge

    (see 

    Answer a

    133Cs+

    Answer b

    127I

    Answer c

    31P3+

    Answer d

    57Co3+

    Exercise \(\PageIndex{3.b}\)

    Average atomic masses listed by IUPAC are based on a study of experimental results. Bromine has two isotopes 79Br and 81Br, whose masses (78.9183 and 80.9163 amu) and abundances (50.69% and 49.31%) were determined in earlier experiments. Calculate the average atomic mass of bromine based on these experiments.

    Answer
    Isotope Mass Percent Abundance Fraction
    79Br 78.9183amu 50.69% 0.5069
    81Br 80.9163 amu 49.31% 0.4931


    \[{X_A}{(m_A)} +{X_B}{(m_B)} = {m_{ave}}\]

    \[{X_{79Br}}{(m_{79Br})} +{X_{81Br}}{(m_{81Br})} = {m_{ave}}\]

    \[{0.5069}{(78.9183amu)} +{0.4931}{(80.9163 amu)} = {m_{ave}}\]

    \[{m_{ave}}=79.904 amu\]

     

    Exercise \(\PageIndex{3.c}\)

    Average atomic masses listed by IUPAC are based on a study of experimental results. The average atomic mass of Copper is 63.546 amu. Copper has two isotopes 63Cu and 65Cu. The mass of 63Cu was determined in an earlier experiment and found to be 62.930 amu. The abundance of 63Cu was found to be 69.17%. Calculate the mass and abundance of 65Cu.

    Answer
    Isotope Mass Percent Abundance Fraction
    63Cu 62.930 amu 69.17% 0.6917
    65Cu \(m_B\nonumber\) \(%_B\nonumber\) \(X_B\nonumber\)

    \[X_A + X_B=1\]

    \[X_B=1-X_A \]

    \[X_B=1-0.6917=0.3083=30.83% \]

    \[{X_A}{(m_A)} +{X_B}{(m_B)} = {m_{ave}}\]

    \[m_B=\frac{m_{ave}-X_A(m_A)}{X_B}\]

    \[m_B=\frac{63.546amu-0.6917(62.930 amu)}{0.3083}\]

    \[m_B= 64.928amu\]

     

    Exercise \(\PageIndex{3.d}\)

    Average atomic masses listed by IUPAC are based on a study of experimental results. The average atomic mass of Gallium is 69.723 amu. Gallium has two isotopes 69Ga and 71Ga, whose masses (68.9257 and 70.9249 amu) were determined in earlier experiments. What is the natural abundance of each isotope?

    Answer
    Isotope Mass Fraction
    69Ga 68.9257 amu \(X_A\nonumber\)
    71Ga 70.9249 amu \(X_B\nonumber\)

    \[{X_A}{(m_A)} +{X_B}{(m_B)} = {m_{ave}}\]

    \[X_A + X_B=1\]

    \[X_B=1-X_A \]

    \[{X_A}{(m_A)} +{1-X_A}{(m_B)} = {m_{ave}}\]

    \[{X_A}=\frac{m_{ave}-m_B}{m_A-m_B}\]

    \[{X_A}=\frac{69.723 amu-70.9249amu}{68.9257amu-70.9249amu}\]

    \[{X_A}=0.6012\]

    \[X_B=1-X_A \]

    \[X_B=1-0.6012=0.3988 \]

    The natural abundance of 69Ga is 60.12% and of 71Ga is 39.88%

     

    Exercise \(\PageIndex{3.e}\)

    In an alternate universe on the planet Htrae, the element Minebro has two isotopes 79Mr and 81Mr, whose masses (78.9183 and 80.9163 amu) and abundances (49.31% and 50.69%) were determined in earlier experiments. Calculate the average atomic mass of Minebro based on these experiments.

    Answer
    Isotope Mass Percent Abundance Fraction
    79Br 78.9183amu 49.31% 0.4931
    81Br 80.9163 amu 50.69% 0.5069


    \[{X_A}{(m_A)} +{X_B}{(m_B)} = {m_{ave}}\]

    \[{X_{79Br}}{(m_{79Br})} +{X_{81Br}}{(m_{81Br})} = {m_{ave}}\]

    \[{0.4931}{(78.9183amu)} +{0.5069}{(80.9163 amu)} = {m_{ave}}\]

    \[{m_{ave}}= 79.931amu\]

     

    Exercise \(\PageIndex{3.f}\)

    In an alternate universe on the planet Htrae, the element Robber has an average atomic mass of 64.494 amu. Robber has two isotopes 63Ru and 65Ru. The mass of 63Ru was determined in an earlier experiment and found to be 62.930 amu. The abundance of 63Ru was found to be 21.72%. Calculate the mass and abundance of 65Ru.

    Answer
    Isotope Mass Percent Abundance Fraction
    63Cu 62.930 amu 21.72% 0.2172
    65Cu \(m_B\nonumber\) \(%_B\nonumber\) \(X_B\nonumber\)

    \[X_A + X_B=1\]

    \[X_B=1-X_A \]

    \[X_B=1-0.2172=0.7828=78.28% \]

    \[{X_A}{(m_A)} +{X_B}{(m_B)} = {m_{ave}}\]

    \[m_B=\frac{m_{ave}-X_A(m_A)}{X_B}\]

    \[m_B=\frac{64.494amu-0.2172(62.930 amu)}{0.7828}\]

    \[m_B= 64.928\]

    Identify Isotopes

    Exercise \(\PageIndex{4.a}\)

    The following are properties of isotopes of two elements that are essential in our diet. Determine the number of protons, neutrons and electrons in each and name them.

    1. atomic number 26, mass number 58, charge of 2+
    2. atomic number 53, mass number 127, charge of 1−
    Answer a

    Iron, 26 protons, 24 electrons, and 32 neutrons

    Answer b

    Iodine, 53 protons, 54 electrons, and 74 neutrons

     

    Isotope Notation

    Exercise \(\PageIndex{5.a}\)

    Give the number of protons, electrons, and neutrons in neutral atoms of each of the following isotopes:

    1. \(_{3}^{7}\textrm{Li}\)
    2. \(_{52}^{125}\textrm{Te}\)
    3. \(_{47}^{109}\textrm{Ag}\)
    4. \(_{7}^{15}\textrm{N}\)
    5. \(_{15}^{31}\textrm{P}\)
    Answer a

    3 protons, 3 electrons, 4 neutrons 

    Answer b

    52 protons, 52 electrons, 73 neutrons 

    Answer c

    47 protons, 47 electrons, 62 neutrons 

    Answer d

    7 protons, 7 electrons, 8 neutrons 

    Answer e

    15 protons, 15 electrons, 16 neutrons

    Exercise \(\PageIndex{5.b}\)

    Write a symbol for each of the following neutral isotopes. Include the atomic number and mass number for each.

    1. the alkali metal with 11 protons and a mass number of 23
    2. the noble gas element with and 75 neutrons in its nucleus and 54 electrons in the neutral atom
    3. the isotope with 33 protons and 40 neutrons in its nucleus
    4. the alkaline earth metal with 88 electrons and 138 neutrons
    Answer a

    \(_{11}^{23}\textrm{Na}\)

    Answer b

    \(_{54}^{129}\textrm{Xe}\)

    Answer c

    \(_{33}^{73}\textrm{As}\)

    Answer d

    \(_{88}^{226}\textrm{Ra}\)

     

    Isotope Abundance and Average Atomic Mass

    Exercise \(\PageIndex{6.a}\)

    Average atomic masses listed by IUPAC are based on a study of experimental results. Bromine has two isotopes 79Br and 81Br, whose masses (78.9183 and 80.9163 amu) and abundances (50.69% and 49.31%) were determined in earlier experiments. Calculate the average atomic mass of bromine based on these experiments.

    NOTE: This problem is also in 2.2: Atomic Number, Mass Number, and Atomic Mass Unit

    Answer

    79.904 amu

    Exercise \(\PageIndex{6.b}\)

    The average atomic masses of some elements may vary, depending upon the sources of their ores. Naturally occurring boron consists of two isotopes with accurately known masses (10B, 10.0129 amu and 11B, 11.0931 amu). The actual atomic mass of boron can vary from 10.807 to 10.819, depending on whether the mineral source is from Turkey or the United States. Calculate the percent abundances leading to the two values of the average atomic masses of boron from these two countries.

    Answer

    Turkey source: 0.2649 (of 10.0129 amu isotope)

    US source: 0.2537 (of 10.0129 amu isotope)

     

    Metal or Nonmetal

    Exercise \(\PageIndex{7.a}\)

    Using the periodic table, classify each of the following elements as a metal or a nonmetal, and then further classify each as a main-group (representative) element, transition metal, or inner transition metal:

    1. uranium
    2. bromine
    3. strontium
    4. neon
    5. gold
    6. americium
    7. rhodium
    8. sulfur
    9. carbon
    10. potassium
    Answer a

    metal, inner transition metal

    Answer b

    nonmetal, representative element

    Answer c

    metal, representative element

    Answer d

    nonmetal, representative element

    Answer e

    metal, transition metal

    Answer f

    metal, inner transition metal

    Answer g

    metal, transition metal

    Answer h

    nonmetal, representative element

    Answer i

    nonmetal, representative element

    Answer j

    metal, representative element

     

    Periodic Table: Groups

    Exercise \(\PageIndex{8.a}\)

    Using the periodic table, identify the lightest and heaviest member of each of the following groups:

    1. noble gases
    2. alkaline earth metals
    3. alkali metals
    4. chalcogens
    Answer a

    He, Og

    Answer b

    Be, Ra

    Answer c

    Li, Fr

    Answer d

    O, Lv

     

     Periodic Table: Periods

    Exercise \(\PageIndex{9.a}\)

    Use the periodic table to give the name and symbol for each of the following elements:

    1. the noble gas in the same period as germanium
    2. the alkaline earth metal in the same period as selenium
    3. the halogen in the same period as lithium
    4. the chalcogen in the same period as cadmium
    Answer a

    krypton, Kr

    Answer b

    calcium, Ca

    Answer c

    fluorine, F

    Answer d

    tellurium, Te

     

    Diatomic Elements 

    Exercise \(\PageIndex{10.a}\)

    Explain why the symbol for an atom of the element oxygen and the formula for a molecule of oxygen differ.

    Answer

    The symbol for the element oxygen, O, represents both the element and one atom of oxygen. A molecule of oxygen, O2, contains two oxygen atoms; the subscript 2 in the formula must be used to distinguish the diatomic molecule from two single oxygen atoms.

     

    Isomers

    Exercise \(\PageIndex{11.a}\)

    Draw or build a molecule with two carbons, six hydrogens, and one oxygen.

    1. Draw the structural formula of this molecule and state its name.
    2. Can you arrange these atoms to make a different molecule? If so, draw its structural formula and state its name.
    3. How are the molecules drawn in (a) and (b) the same? How do they differ? What are they called (the type of relationship between these molecules, not their names).
    Answer a

    Ethanol 

    Answer b

    Methoxymethane, more commonly known as dimethly ether

    Answer c

    These molecules has the same chemical composistion (types and number of atoms) but have different chemical structures. They are structural isomers

     

    Ionic or Covalent Compounds

    Exercise \(\PageIndex{12.a}\)

    Using the periodic table, predict whether the following chlorides are ionic or covalent: KCl, NCl3, ICl, MgCl2, PCl5, and CCl4.

    Answer

    Ionic: KCl, MgCl2

    Covalent: NCl3, ICl, PCl5, CCl4

    Exercise \(\PageIndex{12.b}\)

    For each of the following compounds, state whether it is ionic or covalent. If it is ionic, write the symbols for the ions involved:

    1. NF3
    2. BaO
    3. (NH4)2CO3
    4. Sr(H2PO4)2
    5. IBr
    6. Na2O
    Answer a

    covalent

    Answer b

    ionic, Ba2+, O−2

    Answer c

    ionic, NH4+, CO3−2

    Answer d

    ionic, Sr2+, H2PO4-

    Answer e

    covalent

    Answer f

    ionic, Na+, O−2

     

    Ions to Compounds

    Exercise \(\PageIndex{13.a}\)

    For each of the following pairs of ions, write the symbol for the formula of the compound they will form:

    1. Ca2+, S2−
    2. \(NH_{4}^{+}\), \(SO_{4}^{2-}\)
    3. Al3+, Br
    4. Na+, \(HPO_{4}^{2-}\)
    5. Mg2+, \(PO_{4}^{3-}\)
    Answer a

    CaS

    Answer b

    (NH4)2SO4

    Answer c

    AlBr3

    Answer d

    Na2HPO4

    Answer e

    Mg3(PO4)2

     

    Name from Molecular Formula

    Exercise \(\PageIndex{14.a}\)

    Name the following compounds:

    1. CsCl
    2. BaO
    3. K2S
    4. BeCl2
    5. HBr
    6. AlF3
    Answer a

    cesium chloride

    Answer b

    barium oxide

    Answer c

    potassium sulfide

    Answer d

    beryllium chloride

    Answer e

    hydrogen bromide

    Answer f

    aluminum fluoride

    Exercise \(\PageIndex{14.b}\)

    Each of the following compounds contains a metal that can exhibit more than one ionic charge. Name these compounds:

    1. Cr2O3
    2. FeCl2
    3. CrO3
    4. TiCl4
    5. CoO
    6. MoS2
    Answer a

    chromium(III) oxide

    Answer b

    iron(II) chloride

    Answer c

    chromium(VI) oxide

    Answer d

    titanium(IV) chloride

    Answer e

    cobalt(II) oxide

    Answer f

    molybdenum(IV) sulfide

    Molecular Formula from Name

    Exercise \(\PageIndex{15.a}\)

    Write the formulas of the following compounds:

    1. rubidium bromide
    2. magnesium selenide
    3. sodium oxide
    4. calcium chloride
    5. hydrogen fluoride
    6. gallium phosphide
    7. aluminum bromide
    8. ammonium sulfate
    Answer a
    RbBr
    Answer b
    MgSe
    Answer c
    Na2O
    Answer d
    CaCl2
    Answer e
    HF
    Answer f
    GaP
    Answer g
    AlBr3
    Answer h
    (NH4)2SO4

    Exercise \(\PageIndex{15.b}\)

    Write the formulas of the following compounds:

    1. chlorine dioxide
    2. dinitrogen tetraoxide
    3. potassium phosphide
    4. silver(I) sulfide
    5. aluminum nitride
    6. silicon dioxide
    Answer a
    ClO2
    Answer b
    N2O4
    Answer c
    K3P
    Answer d
    Ag2S
    Answer e
    AlN
    Answer f
    SiO2

    Exercise \(\PageIndex{15.c}\)

    The following ionic compounds are found in common household products. Write the formulas for each compound:

    1. potassium phosphate
    2. copper(II) sulfate
    3. calcium chloride
    4. titanium dioxide
    5. ammonium nitrate
    6. sodium bisulfate (the common name for sodium hydrogen sulfate)
    Answer a
    K3PO4
    Answer b
    CuSO4
    Answer c
    CaCl2
    Answer d
    TiO2
    Answer e
    NH4NO3
    Answer f
    NaHSO4

     

    Common Names to IUPAC Names

    Exercise \(\PageIndex{16.a}\)

    What are the IUPAC names of the following compounds?

    1. manganese dioxide
    2. mercurous chloride (Hg2Cl2)
    3. ferric nitrate [Fe(NO3)3]
    4. titanium tetrachloride
    5. cupric bromide (CuBr2)
    Answer a
    manganese(IV) oxide
    Answer b
    mercury(I) chloride
    Answer c
    iron(III) nitrate
    Answer d
    titanium(IV) chloride
    Answer e
    copper(II) bromide
     

     

    Atoms and the Mole

    Exercise \(\PageIndex{17.a}\)

    What is the total mass (amu) of carbon in each of the following molecules?

    1. CH4
    2. CHCl3
    3. C12H10O6
    4. CH3CH2CH2CH2CH3
    Answer a

    \(12.01\;amu\)

    Answer b

    \(12.01\;amu\)

    Answer c

    \(12(12.01)\;=\;144.12\;amu\)

    Answer d

    \(5(12.01)\;=\;60.05\;amu\)

    Exercise \(\PageIndex{17.b}\)

    Calculate the molecular or formula mass of each of the following:

    1. CH4
    2. CHCl3
    3. C12H10O6
    4. CH3CH2CH2CH2CH3
    Answer a

    \(12.01 \;+\;4(1.007)\;=\;16.04\;amu\)

    Answer b

    \(12.01 \;+\;1.007\;+\;3(35.45)\;=\;119.37\;amu\)

    Answer c

    \(12(12.01)\;+\;10(1.007)\;+\;6(16.00)\;=\;250.19\;amu\)

    Answer d

    \(5(12.01)\;+\;12(1.007)\;=\;72.13\;amu\)

    Exercise \(\PageIndex{17.c}\)

    Calculate the molecular or formula mass of each of the following:

    1. P4
    2. H2O
    3. Ca(NO3)2
    4. CH3CO2H (acetic acid)
    5. C12H22O11 (sucrose, cane sugar)
    Answer a

    \(4(30.974)\;=\;123.896\;amu\)

    Answer b

    \(2(1.007)\;+\;15.999\;=\;18.015\;amu\)

    Answer c

    \(40.078\;+\;2(14.007\;+\;3(15.99))\;=\;164.086\;amu\)

    Answer d

    \(2(12.011)\;+\;2(15.999)\;+\;4(1.007)\;=\;60.052\;amu\)

    Answer e

    \(12(12.011)\;+\;22(1.007)\;+\;11(15.999)\;=\;342.297\;amu\)

    Exercise \(\PageIndex{17.d}\)

    Determine the molecular mass of the following compounds:

    1. A structure is shown. Two C atoms form double bonds with each other. The C atom on the left forms a single bond with two H atoms each. The C atom on the right forms a single bond with an H atom and with a C H subscript 2 C H subscript 3 group.
    2. A structure is shown. There is a C atom which forms single bonds with three H atoms each. This C atom is bonded to another C atom. This second C atom forms a triple bond with another C atom which forms a single bond with a fourth C atom. The fourth C atom forms single bonds with three H atoms each.
    3. A structure is shown. An S i atom forms a single bond with a C l atom, a single bond with a C l atom, a single bond with an H atom, and a single bond with another S i atom. The second S i atom froms a single bond with a C l atom, a single bond with a C l atom, and a single bond with an H atom.
    4. A structure is shown. A P atom forms a double bond with an O atom. It also forms a single bond with an O atom which forms a single bond with an H atom. It also forms a single bond with another O atom which forms a single bond with an H atom. It also forms a single bond with another O atom which forms a single bond with an H atom.
    Answer a

    \(4(12.011)\;+\;8(1.007)\;=\;56.107\;amu\)

    Answer b

    \(4(12.011)\;+\;6(1.007)\;=\;54.091\;amu\)

    Answer c

    \(2(28.085)\;+\;4(35.45)\;+\;2(1.007)\;=\;199.998\;amu\)

    Answer d

    \(30.974\;+\;4(15.99)\;3(1.007)\;=\;97.995\;amu\)

     

    Mass, Moles, and Number of Particles

    Exercise \(\PageIndex{18.a}\)

    Which contains the greatest mass of oxygen: 0.75 mol of ethanol (C2H5OH), 0.60 mol of formic acid (HCO2H), or 1.0 mol of water (H2O)? Explain why.

    Answer

    Formic acid. Its formula has twice as many oxygen atoms as the other two compounds (one each). Therefore, 0.60 mol of formic acid would be equivalent to 1.20 mol of a compound containing a single oxygen atom.

    Exercise \(\PageIndex{18.b}\)

    Calculate the molar mass of each of the following:

    1. S8
    2. C5H12
    3. Sc2(SO4)3
    4. CH3COCH3 (acetone)
    5. C6H12O6 (glucose)
    Answer a

    \(8(32.070)\;=\;256.560\;g/mol\)

    Answer b

    \(5(12.011)\;+\;12(1.008)\;=\;72.150\;g/mol\)

    Answer c

    \(2(44.956)\;+\;3(32.07+4(15.999)\;=\;378.110\;g/mol\)

    Answer d

    \(3(12.011)\;+\;6(1.008)\;+\;15.999\;=\;58.080\;g/mol\)

    Answer e

    \(6(12.011)\;+\;12(1.008)\;+\;6(15.999)\;=\;180.156\;g/mol\)

    Exercise \(\PageIndex{18.c}\)

    Calculate the molar mass of each of the following:

    1. the anesthetic halothane, C2HBrClF3
    2. the herbicide paraquat, C12H14N2Cl2
    3. caffeine, C8H10N4O2
    4. urea, CO(NH2)2
    5. a typical soap, C17H35CO2Na
    Answer a

    \(2(12.011)\;+\;1.008\;+\;79.900\;+\;35.45\;+\;3(18.998)\;=\;197.382\;g/mol\)

    Answer b

    \(12(12.011)\;+\;14(1.008)\;+\;2(14.007)\;2(35.45)\;=\;257.158 g/mol\)

    Answer c

    \(8(12.011)\;+\;10(1.008)\;4(14.007)\;+\;2(15.999)\;=\;194.194 g/mol\)

    Answer d

    \(12.011\;+\;15.999\;+\;2(14.007\;+\;2(1.008))\;=\;60.056 g/mol\)

    Answer e

    \(18(12.011)\;+\;35(1.008)\;+\;2(15.999)\;+\;22.990\;=\;306.466 g/mol\)

    Exercise \(\PageIndex{18.d}\)

    Determine the mass of each of the following:

    1. 0.0146 mol KOH
    2. 10.2 mol ethane, C2H6
    3. 1.6 × 10⁻³mol Na2SO4
    4. 6.854 × 10³ mol glucose, C6H12O6
    5. 2.86 mol Co(NH3)6Cl3
    Answer a

    \(38.098\;+\;15.999\;+\;1.008\;=\;55.105\;g/mol\\55.105\;g/mol\;\times\;.0146\;mol\;=\;0.805\;g\)

    Answer b

    \(2(12.011\;+\;6(1.008)\;=\;30.070\;g/mol\\30.070\;g/mol\;\times\;10.2\;mol\;=306.714\;g\)

    Answer c

    \(2(22.990)\;+\;32.07\;+\;4(15.999)\;=\;142.046\;g/mol\\142.046\;g/mol\;\times\;1.6\;\times10^{-3}\;mol\;=\;.227\;g\)

    Answer d

    \(6(12.011)\;+\;12(1.008)\;+\;6(15.999)\;=\;180.156\;g/mol\\180.156\;g/mol\;\times\;6.854\;\times\;10^{3}\;mol\;=\;1.235*10^6\;g\)

    Answer e

    \(58.933\;+\;6(14.007\;+\;3(1.007))\;+\;3(35.45)\;=\;267.451\;g/mol\\267.451\;g/mol\;\times\;2.86mol\;=\;764.909\;g\)

    Exercise \(\PageIndex{18.e}\)

    Determine the mass of each of the following:

    1. (a) 2.345 mol LiCl
    2. (b) 0.0872 mol acetylene, C2H2
    3. (c) 3.3 × 10−2 mol Na2 CO3
    4. (d) 1.23 × 103 mol fructose, C6 H12 O6
    5. (e) 0.5758 mol FeSO4(H2O)7
    Answer a

    \(7.0\;+35.45\;=\;42.45\;g/mol\\42.45\;g/mol\;\times\;2.345\;mol\;=99.545\;g\)

    Answer b

    \(2(12.011)\;+\;2(1.008)\;=\;26.038\;g/mol\\26.038\;g/mol\;\times\;0.0872\;mol\;=2.271\;g\)

    Answer c

    \(2(22.99)\;+\;12.011\;+\;3(15.999)\;=\;105.988\;g/mol\\105.988\;g/mol\;\times\;3.3\;\times\;10^{-2}\;mol\;=\;3.498\;g\)

    Answer d

    \(6(12.011)\;+\;12(1.008)\;+\;6(15.999)\;=\;180.156\;g/mol\\180.156\;g/mol\;\times\;1.23\;\times\;10^3\;mol\;=2.216\;\times\;10^5\;g\)

    Answer e

    \(55.84\;+\;32.07\;+\;4(15.999)\;+\;7(2(1.008)\;+\;15.999)\;=\;278.011\;g/mol\\278.011\;g/mol\;\times\;0.5758\;mol\;=\;160.079\;g\)

    Exercise \(\PageIndex{18.f}\)

    Determine the mass in grams of each of the following:

    1. 0.600 mol of oxygen atoms
    2. 0.600 mol of oxygen molecules, O2
    3. 0.600 mol of ozone molecules, O3
    Answer a

    \(15.999\;g/mol\;\times\;0.600\;mol\;=\;9.599\;g\)

    Answer b

    \(2(15.999)\;g/mol\;\times\;0.600\;mol\;=\;19.199\;g\)

    Answer c

    \(3(15.999)\;g/mol\;\times\;0.600\;mol\;=\;28.798\;g\)

    Exercise \(\PageIndex{18.g}\)

    Determine which of the following contains the greatest mass of aluminum: 122 g of AlPO4, 266 g of Al2Cl6, or 225 g of Al2S3.

    Answer

    \(26.982\;+\;30.974\;+\;4(15.999)\;=\;121.952\;g/mol\\ \frac{122\;g}{121.952\;g/mol}\;=\;1\;mol\;AlPO_4\\1\;mol\;Al\;in\;AlPO_4\\1.000\;\times\;1\;=\;1\;mol\;Al\)

     

    \(2(26.982)\;+\;6(35.45)\;=\;266.664\;g/mol\\ \frac{266\;g}{266.664\;g/mol}\;=\;0.998\;mol\;Al_2Cl_6\\2\;mol\;Al\;in\;Al_2Cl_6\\0.998\;\times\;2\;=\;1.996\;mol\;Al\)

     

    \(2(26.982)\;+\;3(32.07)\;=\;150.174\;g/mol\\ \frac{225\;g}{150.174\;g/mol}\;=\;1.498\;mol\;Al_2S_3\\2\;mol\;Al\;in\;Al_2S_3\\1.498\;\times\;2\;=\;2.996\;mol\;Al\)

    Exercise \(\PageIndex{18.h}\)

    The Cullinan diamond was the largest natural diamond ever found (January 25, 1905). It weighed 3104 carats (1 carat = 200 mg). How many carbon atoms were present in the stone?

    Answer

    \(3104\;carats\;\times\;200\;mg\;=\;\frac{620800\;mg}{1000\;g}\;=\;620.8\;g\\\frac{620.8\;g}{12.011}\;=\;51.686\;mol\;C\\51.686\;times\;6.022\;\times\;10^{23}\;=\;3.113\;\times\;10^{25}\)

    Exercise \(\PageIndex{18.i}\)

    A certain nut crunch cereal contains 11.0 grams of sugar (sucrose, C12H22O11) per serving size of 60.0 grams. How many servings of this cereal must be eaten to consume 0.0278 moles of sugar?

    Answer

    0.865 servings, or about 1 serving

    Exercise \(\PageIndex{18.j}\)

    Which of the following represents the least number of molecules?

    1. 20.0 g of H2O (18.02 g/mol)
    2. 77.0 g of CH4 (16.06 g/mol)
    3. 68.0 g of CaH2 (42.09 g/mol)
    4. 100.0 g of N2O (44.02 g/mol)
    5. 84.0 g of HF (20.01 g/mol)
    Answer

    a. 20.0 g H2O represents the least number of molecules since it has the least number of moles.

     

    Percent Composition

    Exercise \(\PageIndex{19.a}\)

    Calculate the following to four significant figures:

    1. the percent composition of ammonia, NH3
    2. the percent composition of photographic “hypo,” Na2S2O3
    3. the percent of calcium ion in Ca3(PO4)2
    Answer

    a.

    \(M_N: \;1\;\times\;14.007 \;=\;14.007  \; \\M_H:\;3\;\times\;1.008\;=\;3.024 \\M_T\;= \; 1(14.007)+ 3(1.008)=17.031\\\; \\ \%\; N\;=\;\frac{M_N}{M_T}\;=\;82.24\;\%\\\%\;H\;=\;\frac{M_H}{M_T}\;=\;17.76\;\%\)\

     

    b.

    \(M_{Na}: \;2\;\times\;22.99\;=\;45.98\\M_{S}:\;2\;\times\;32.07\;=\;64.14\\M_O:\;3\;\times\;15.999\;=\;47.997\\M_T\;=\;2(22.99)\;+\;2(32.07)\;=\;158.117\\\%\; Na\;=\;\frac{M_{Na}}{M_T}\;=\;29.07\;\%\\\%\; S\;=\;\frac{M_{S}}{M_T}\;=\;40.56\;\%\\\%\; O\;=\;\frac{M_{O}}{M_T}\;=\;30.36\;\%\)

     

    c.

    \(M_{Ca^{+2}}: \;3\;\times\;40.08\;=\;120.24\\P:\;2\;\times\;30.974\;=\;61.948\\O:\;8\;\times\;15.999\;=\;127.992\\M_T\;=\;3(40.08)\;+\;2(30.974)\;+8(15.999)\;=\;310.18\\\%\; Ca^{+2}\;=\;\frac{M_{Ca^{+2}}}{M_T}\;=\;38.76\;\%\)

    Exercise \(\PageIndex{19.b}\)

    Determine the percent ammonia, NH3, in Co(NH3)6Cl3, to three significant figures.

    Answer

    \(NH_3: \;6\;\times\;(14.007\;+\;3(1.008))\;=\;102.186\\Co:\;1\;\times\;58.933\;=\;58.933\\Cl:\;3\;\times\;35.45\;=\;106.35\\M_T\;=267.469\\M_{NH_{3}}\;=\;6(14.007\;+\;3(1.008))\\ \%\; NH_{3}\;=\;\frac{M_{NH_{3}}}{M_T}\;=\;38.20\;\%\)

    Exercise \(\PageIndex{19.c}\)

    Determine the empirical formulas for compounds with the following percent compositions:

    1. 15.8% carbon and 84.2% sulfur
    2. 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen
    Answer

    \(C\;100\;\times\;.158\;=\;15.8/12.011\;=\;1.315/1.315\;=\;1\\ S\;100\;\times\;.842\;=\;84.2/32.07\;=\;2.62/1.315\;=\;2\\CS_2\)

     

     

    \(C\;100\;\times\;.400\;=\;40.0/12.011\;=\;3.33/3.33\;=\;1\\ H\;100\;\times\;.067\;=\;6.7/1.008\;=\;6.64/3.33\;=\;2\\O\;100\;\times\;.533\;=\;55.3/15.999\;=\;3.46/3.33\;=\;1\\CH_2O\)

    Exercise \(\PageIndex{19.d}\)

    A compound of carbon and hydrogen contains 92.3% C and has a molar mass of 78.1 g/mol. What is its molecular formula?

    Answer

    \(78.1\;\times\;.923\;=\;72.08/12.011\;=\;6\\78.1\;-\;72.08\;=\;6.02/1.008\;=\;6\\C_6H_6\)

    Exercise \(\PageIndex{19.e}\)

    Determine the empirical and molecular formula for chrysotile asbestos. Chrysotile has the following percent composition: 28.03% Mg, 21.60% Si, 1.16% H, and 49.21% O. The molar mass for chrysotile is 520.8 g/mol.

    Answer

    Mg3Si2H3O8 (empirical formula),

    Mg6Si4H6O16 (molecular formula)

    Empirical Formula

    \(Mg\;100\;\times\;.2803\;=\;28.03/24.305\;=\;1.15/.77\;=\;1.5\;\times\;2\;=\;3\\ Si\;100\;\times\;.2160\;=\;21.60/28.085\;=\;.77/.77\;=\;1\;\times\;2\;=\;2\\H\;100\;\times\;.0116\;=\;1.16/1.008\;=\;1.15/.77\;=\;1.5\;\times\;2\;=\;3\\O\;100\;\times\;.4921\;=\;49.21/15.999\;=\;3.08/.77\;=\;4\;\times\;2\;=\;8\\Mg_3Si_2H_3O_{16}\)

    Molecular Formula

    \(Mg\;520.8\;\times\;.2803\;=\;145.98/24.305\;=6\\ Si\;520.8\;\times\;.2160\;=\;112.49/28.085\;=4\\H\;520.8\;\times\;.0116\;=\;6.04/1.008\;=6\\O\;520.8\;\times\;.4921\;=\;256.28/15.999\;=16\\Mg_6Si_4H_6O_{16}\)

    Exercise \(\PageIndex{19.f}\)

    A major textile dye manufacturer developed a new yellow dye. The dye has a percent composition of 75.95% C, 17.72% N, and 6.33% H by mass with a molar mass of about 240 g/mol. Determine the molecular formula of the dye.

    Answer

    \(C\;240\;\times\;.7595\;=\;182.28/12.011\;=15\\ H\;240\;\times\;.0633\;=\;15.19/1.008\;=15\\N\;240\;\times\;.1772\;=\;42.53/14.007\;=4\\C_{15}H_{15}N_{3}\)

     

    Empirical Formulas

    Exercise \(\PageIndex{20.a}\)

    Determine the empirical formulas for the following compounds:

    1. caffeine, C8H10N4O2
    2. fructose, C12H22O11
    3. hydrogen peroxide, H2O2
    4. glucose, C6H12O6
    5. ascorbic acid (vitamin C), C6H8O6
    Answer
    1. C4H5N2O

    2. C12H22O11

    3. HO

    4. CH2O

    5. C3H4O3

     

    Molecular and Empirical Formulas from Structures

    Exercise \(\PageIndex{21.a}\)

    Write the molecular and empirical formulas of the following compounds:

    Answer
    1. molecular CO2, empirical CO2
    2. molecular C2H2, empirical CH
    3. molecular C2H4, empirical CH2
    4. molecular H2SO4, empirical H2SO4

     

    Hydrates

    Exercise \(\PageIndex{22.a}\)

    Name the following compounds:

    1. FeCl3 * 6 H2O
    2. CuSO4 * 5 H2O
    3. Na2SO4 * 10 H2O
    Answer
    1. Iron (III) chloride hexahydrate
    2. Copper (II) sulfate pentahydrate
    3. Sodium sulfate decahydrate

    Exercise \(\PageIndex{22.b}\)

    Write the formulas for the following compounds:

    1. Barium chloride dihydrate
    2. Magnesium sulfate heptahydrate
    3. Sodium carbonate decahydrate
    Answer
    1. BaCl2 * 2 H2O
    2. MgSO4 * 7 H2
    3. Na2CO3 * 10 H2O

    Exercise \(\PageIndex{22.c}\)

    What is the percent composition of water in the compounds in 2b?

    Answer

    126.0 / 246.4 * 100 = 51.14%

    Exercise \(\PageIndex{22.d}\)

    If 125 grams of magnesium sulfate heptahydrate is completely dehydrated, how many grams of anhydrous magnesium sulfate will remain?

    Answer

    100 – 51.14 = 48.86% magnesium sulfate

    0.4886 * 125 = 61.1 grams

     

     

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