2: Atoms, Molecules, and Ions

 

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. 
2. 
3. 
4. 

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, NH_3, 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:

1. 
2. 
3. 
4. 

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 H₂O 
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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