chemistry: atoms first julia burdge & jason overby chapter 5 ionic and covalent compounds kent...

Chemistry: Atoms FirstJulia Burdge & Jason Overby

Chapter 5

Ionic and Covalent Compounds

Kent L. McCorkle

Cosumnes River College

Sacramento, CA

Ionic and Covalent CompoundsIonic and Covalent Compounds

5.1 Compounds5.2 Lewis Dot Symbols5.3 Ionic Compounds and Bonding5.4 Naming Ions and Ionic Compounds

Formulas of Ionic CompoundsNaming Ionic Compounds

5.5 Covalent Bonding and MoleculesMoleculesMolecular FormulasEmpirical Formulas

5.6 Naming Molecular CompoundsSpecifying Numbers of AtomsCompounds Containing HydrogenOrganic Compounds

5

5.7 Covalent Bonding in Ionic SpeciesPolyatomic IonsOxoacidsHydratesFamiliar Inorganic Compounds

5.8 Molecular and Formula Masses5.9 Percent Composition of Compounds5.10 Molar Mass

Interconverting Mass, Moles, and Number of ParticlesDetermination of Empirical Formula and Molecular Formula from Percent Composition

Ionic and Covalent CompoundsIonic and Covalent Compounds5

A compound is a substance composed of two or more elements combined in a specific ratio and held together by chemical bonds.

Familiar examples of compounds are water and salt (sodium chloride).

CompoundsCompounds5.1

When atoms form compounds, it is their valence electrons that actually interact.

A Lewis dot symbol consists of the element’s symbol surround by dots.

Each dot represents a valence electron.

Boron 1s22s22p1

3 valence electrons

B•

• •

Lewis dot symbol for boron

B•

•• B•

•• B•

•

•

other reasonable Lewis dot symbols for boron

Lewis Dot SymbolsLewis Dot Symbols5.2

Atoms combine in order to achieve a more stable electron configuration.

Maximum stability results when a chemical species is isoelectronic with a noble gas.

Na: 1s22s22p63s1 Na+: 1s22s22p6

10 electrons total, isoelectronic with Ne

Cl: 1s22s22p63s23p5 Cl‒: 1s22s22p63s23p6

18 electrons total, isoelectronic with Ar

Lewis Dot SymbolsLewis Dot Symbols


Lewis dot symbols of the main group elements.


Dots are not paired until absolutely necessary.

Na •

O••

• •••

For main group metals such as Na, the number of dots is the number of electrons that are lost.

For nonmetals in the second period, the number of unpaired dots is the number of bonds the atom can form.

B•

• •

1s22s22p1

C•

• ••

1s22s22p2

N••

• ••

1s22s22p3

5 valence electrons; first pair formed in the Lewis dot symbol


Ions may also be represented by Lewis dot symbols.

Na•

Na 1s22s22p63s1 Na+ 1s22s22p6

Na+

O••

• •••

O 1s22s22p4 O2‒ 1s22s22p6

O••

••••••

2‒

Core electrons not represented in the Lewis dot symbol

Valence electron lost in the formation of the Na+ ion.

Remember the charge

Worked Example 5.1

Strategy Starting with the Lewis dot symbols for each element, add dots (for anions) or remove dots (for cations) as needed to achieve the correct charge on each ion. Don’t forget to include the appropriate charge on the Lewis dot symbol.

Write Lewis dot symbols for (a) fluoride ion (F-), (b) potassium ion (K+), and (c) sulfide ion (S2-).

Solution (a)

(b) K+

(c)

F

S2-

Think About It For ions that are isoelectronic with noble gases, cations should have no dots remaining around the element symbol, whereas anions should have eight dots around the element symbol. Note, too, that for anions, we put square brackets around the Lewis dot symbol and place the negative charge outside the brackets. Because the symbol for a common cation such as the potassium ion has no remaining dots, square brackets are not necessary.

Ionic bonding refers to the electrostatic attraction that holds oppositely charged ions together in an ionic compound.

Na • Na+ + e−

+••

••••••

−

Cl••

•••••Cl e−

Na • Na+ ++••

••••••

−

Cl••

•••••Cl

The attraction between the cation and anion draws them together to form NaCl

Ionic Compounds and BondingIonic Compounds and Bonding5.3

Ionic Compounds and BondingIonic Compounds and Bonding

The resulting electrically neutral compound, sodium chloride, is represented with the chemical formula NaCl.

The chemical formula, or simply formula, of an ionic compound denotes the constituent elements and the ratio in which they combine.


A three-dimensional array of oppositely-charged ions is called a lattice. Lattice energy is the amount of energy required to convert a mole of ionic solid to its constituent ions in the gas phase.

−+

−

+

− +

−

+−

+

−

+

−

+

NaCl(s) Na+(g) + Cl−(g) Hlattice = +788 kJ/mol


The magnitude of lattice energy is a measure of an ionic compound’s stability.

Lattice energy depends on the magnitudes of the charge and on the distance between them.

d

Q1 Q2

221

d

QQF

Q = amount of charge

d = distance of separation


The magnitude of lattice energy is a measure of an ionic compound’s stability.

Lattice energy depends on the magnitudes of the charge and on the distance between them.

Ionic Compounds and Bonding Ionic Compounds and Bonding

Worked Example 5.2

Strategy Consider the charges on the ions and the distances between them. Apply Coulomb’s law to determine the relative lattice energies. All three compounds contain O2- and all three cations are +2. Recalling that lattice energy increases as the distance between ions decreases, we need only consider the radii of the cations as all three contain the same anion. From Figure 4.13, the ionic radii are 0.72 Å (Mg2+), 1.00 Å (Ca2+), and 1.18 Å (Sr2+).

Arrange MgO, CaO, and SrO in order of increasing lattice energy.

Solution MgO has the smallest distance between ions, whereas SrO has the largest distance between ions. Therefore, in order of increase lattice energy: SrO < CaO < MgO.

Think About It Mg, Ca, and Sr are all Group 2A metals, so we could have predicted this result without knowing their radii. Recall that ionic radii increase as we move down a column in the periodic table, and charges that are farther apart are more easily separated (meaning the lattice energy will be smaller.) The lattice energies of SrO, CaO, and MgO are 3217, 3414, and 3890 kJ/mol, respectively.

A monatomic ion is named by changing the ending of the element’s name to –ide.

Cl– is chloride O2– is oxide

Some metals can form cations of more than one possible charge.

Fe2+ : ferrous ion [Fe(II)]Fe3+ : ferric ion [Fe(III)]

Mn2+ : manganese(II) ionMn3+ : manganese(III) ionMn4+ : manganese(IV) ion

Naming Ions and Ionic CompoundsNaming Ions and Ionic Compounds5.4

Naming Ions and Ionic CompoundsNaming Ions and Ionic Compounds


Formulas for ionic compounds are generally empirical formulas.

Ionic compounds are electronically neutral.

Al3+ O2–

Al2O3

In order for ionic compounds to be electronically neutral, the sum of the charges on the cation and anion in each formula must be zero.

Aluminum oxide:

Sum of charges: 2(+3) + 3(–2) = 0

Formulas of Ionic CompoundsFormulas of Ionic Compounds

Formulas of Ionic CompoundsFormulas of Ionic Compounds


To name ionic compounds:

1) Name the cation omit the word ionuse a Roman numeral if the cation can have more than one charge

2) Name the anionomit the word ion

Examples:NaCN sodium cyanideFeCl2 iron(II) chlorideFeCl3 iron(III) chloride

Worked Example 5.3

Strategy Begin by identifying the cation and anion in each compound, and then combine the names for each, eliminating the word ion.

Name the following ionic compounds: (a) CaO, (b) Mg3N2, and (c) Fe2S3.

Solution (a) CaO is calcium oxide.

(b) Mg3N2 is magnesium nitride.

(c) Fe2S3 is iron(III) sulfide.

Think About It Be careful not to confuse the subscript in the formula with the charge in the metal ion. In part (c), for example, the subscript on Fe is 2, but this is an iron(III) compound.

Worked Example 5.4

Strategy Identify the ions in each compound, and determine their ratios of combination using the charges on the cation and anion in each.

Deduce the formulas of the following ionic compounds: (a) mercury(II) chloride, (b) lead(II) bromide, and (c) potassium nitride.

Solution (a) Mercury(II) chloride is a combination of Hg2+ and Cl-. To produce a neutral compound, these two ions must combine in a 1:2 ratio – HgCl2.

(b) Lead(II) bromide is a combination of Pb2+ and Cl-. These ions combine in a 1:2 ratio to give PbBr2.

(c) Potassium nitride is a combination of K+ and N3-. These ions combine in a 3:1 ratio to give K3N.

Think About It Make sure that the charges sum to zero in each compound formula. In part (a), for example, Hg2+ + 2Cl- = (+2) + 2(-1) = 0; in part (b), (+2) + 2(-1) = 0; and in part (c), 3(+1) + (-3) = 0.

When compounds form between elements with similar properties, electrons are not transferred from one element to another but instead are shared in order to give each atom a noble gas configuration.

This approach is known as the Lewis theory of bonding, named for it’s proponent, Gilbert Lewis.

Lewis theory depicts bond formation in H2 as

H∙ + ∙H → H:H

This type of arrangement, where two atoms share a pair of electrons, is known as covalent bonding, and the shared pair of electrons constitutes a covalent bond.

Covalent Bonding and MoleculesCovalent Bonding and Molecules5.5

A molecule is a combination of at least two atoms in a specific arrangement held together by chemical forces (chemical bonds).

A molecule may be an element or a compound.

Different samples of a given compound always contain the same elements in the same ratio. This is known as the law of definite proportions.

Covalent Bonding and MoleculesCovalent Bonding and Molecules

Sample Mass of O (g) Mass of C (g) Ratio (g O:g C)

123 g carbon dioxide 89.4 33.6 2.66:1

50.5 g carbon dioxide 36.7 13.8 2.66:1

88.6 g carbon dioxide 64.4 24.2 2.66:2

If two elements can for two or more different compounds, the law of multiple proportions tells us that the ratio of masses of one element that combine with a fixed mass of the other element can be expressed in small whole numbers.

In addition to carbon dioxide, carbon also combines with oxygen to form carbon monoxide.


Sample Mass of O (g) Mass of C (g) Ratio (g O:g C)

16.3 g carbon monoxide 9.31 6.99 1.33:1



The mass ratio of oxygen to carbon in carbon dioxide is 2.66:1, and the ratio of oxygen to carbon in carbon monoxide is 1.33:1.

The ratio of two such mass ratios can be expressed as small whole numbers.


mass ratio of O to C in carbon dioxidemass ratio of O to C in carbon monoxide =

2.661.33 = 2:1

Diatomic molecules contain two atoms and may be either heteronuclear or homonuclear.

Polyatomic molecules contain more than two atoms.


A chemical formula denotes the composition of the substance.

A molecular formula shows the exact number of atoms of each element in a molecule.

Some elements have two or more distinct forms known as allotropes.For example, oxygen (O2) and ozone (O3) are allotropes of oxygen.

A structural formula shows not only the elemental composition, but also the general arrangements.


Worked Example 5.5

Strategy Refer to the labels on the atoms (or see Table 5.3). There are two carbon atoms, six hydrogen atoms, and one oxygen atom, so the subscript on C will be 2 and the subscript on H will be 6, and there will be no subscript on O.

Write the molecular formula of ethanol based on its ball-and-stick model, shown here.

Solution C2H6O

Think About It Often the molecular formula for a compound such as ethanol (consisting of carbon, hydrogen, and oxygen) is written so that the formula more closely resembles the actual arrangement of atoms in the molecule. Thus, the molecular formula for ethanol is commonly written as C2H5OH.

Molecular substances can also be represented using empirical formulas, the whole-number ratio of elements.

While, the molecular formulas tell us the actual number of atoms (the true formula), the empirical formula gives the simplest formula.

Molecular formula: N2H4

Empirical formula: NH2

The molecular and empirical formulas are often the same.


Worked Example 5.6

Strategy To write the empirical formula, the subscripts in the molecular formula must be reduced to the smallest possible whole numbers (without altering the relative numbers of atoms).

Write the empirical formulas for the following molecules: (a) glucose (C6H12O6),

a substance known as blood sugar; (b) adenine (C5H5N5), also known as vitamin

B4; and (c) nitrous oxide (N2O), a gas that is used as an anesthetic (“laughing

gas”) and as an aerosol propellant for whipped cream.

Worked Example 5.6 (cont.)

Solution (a) Dividing each of the subscripts in the molecular formula for glucose by 6, we obtain the empirical formula CH2O. If we had divided the subscripts by 2 or 3, we would have obtained the formulas C3H6O3 and C2H4O2, respectively. Although the ratio of carbon to hydrogen to oxygen atoms in each of these formulas is correct (1:2:1), neither is the simplest formula because the subscripts are not in the smallest possible whole-number ratio.

(b) Dividing each subscript by 5, we get the empirical formula CHN.

(c) Because the subscripts in the formula for nitrous oxide are already the smallest possible whole numbers, its empirical formula is the same as its molecular formula N2O.

Think About It Make sure that the ratio in each empirical formula is the same as that in the corresponding molecular formula and that the subscripts are the smallest possible whole numbers. In part (a), for example, the ratio of C:H:O in the molecular formula is 6:12:6, which is equal to 1:2:1, the ratio expressed in the empirical formula.

Remember that binary molecular compounds are substances that consist of just two different elements.

Nomenclature:1) Name the first element that appears in the formula.

2) Name the second element that appears in the formula, changing its ending to –ide.

Examples:HCl hydrogen chloride

HI hydrogen iodide

Naming Molecular CompoundsNaming Molecular Compounds5.6

Greek prefixes are used to denote the number of atoms of each element present.

Naming Molecular CompoundsNaming Molecular Compounds

The prefix mono- is generally omitted for the first element.

For ease of pronunciation, we usually eliminate the last letter of a prefix that ends in “o” or “a” when naming an oxide.

Example: N2O5 is dinitrogen pentoxide not dinitrogen pentaoxide

Naming Molecular CompoundsNaming Molecular Compounds

Worked Example 5.7

Strategy Each compound will be named using the systematic nomenclature including, where necessary, appropriate Greek prefixes.

Name the following binary molecular compounds: (a) NF3 and (b) N2O4.

Solution (a) nitrogen trifluoride

(b) dinitrogen tetroxide

Think About It Make sure that the prefixes match the subscripts in the molecular formulas and that the word oxide is not preceded immediately by an “a” or an “o”.

Worked Example 5.8

Strategy The formula for each compound will be deduced using the systematic nomenclature guidelines.

Write the chemical formulas for the following binary molecular compounds: (a) sulfur tetrafluoride and (b) tetraphosphorus decasulfide.

Solution (a) SF4

(b) P4S10

Think About It Double-check that the subscripts in the formulas match the prefixes in the compound names: (a) 4 = tetra and (b) 4 = tetra and 10 = deca.

The names of molecular compounds containing hydrogen do not usually conform to the systematic nomenclature guidelines.

Many are called by the common, nonsystematic names or by names that do not indicate explicitly the number of H atoms present.

Examples:B2H6 DiboraneSiH4 SilaneNH3 AmmoniaPH3 PhosphineH2O WaterH2S Hydrogen sulfide

Compounds Containing HydrogenCompounds Containing Hydrogen


One definition of an acid is a substance that produces hydrogen ions (H+) when dissolved in water.

HCl is an example of a binary compound that is an acid when dissolved in water.

To name these types of acids:

1) remove the –gen ending from hydrogen

2) change the –ide ending on the second element to –ic.

hydrogen chloride → hydrochloric acid


A compound must contain at least one ionizable hydrogen atom to be an acid upon dissolving.

Our nomenclature discussion so far has focused on inorganic compounds, generally defined as those without carbon.

Organic compounds contain carbon and hydrogen, sometimes in combination with other atoms.

Hydrocarbons contain only carbon and hydrogen.

The simplest hydrocarbons are called alkanes.

Organic CompoundsOrganic Compounds


Many organic compounds contain groups of atoms known as functional groups, which often determine a molecule’s reactivity.

Polyatomic ions consist of a combination of two or more atoms.

Formulas are determined following the same rule as for ionic compounds containing only monatomic ions: ions must combine in a ratio that give a neutral formula overall.

Calcium phosphate:

Covalent Bonding in Ionic SpeciesCovalent Bonding in Ionic Species5.7

Ca2+ PO4

3–

Ca3(PO4

)2Sum of charges: 3(+2) + 2(–3) = 0

Covalent Bonding in Ionic SpeciesCovalent Bonding in Ionic Species

Worked Example 5.9

Strategy Begin by identifying the cation and anion in each compound, and then combine the names for each, eliminating the word ion.

Name the following ionic compounds: (a) Fe2(SO4)3, (b) Al(OH)3, and (c) Hg2O.

Solution (a) Fe2(SO4)3 is iron(III) sulfate.

(b) Al(OH)3 is aluminum hydroxide.

(c) Hg2O is mercury(I) oxide.

Think About It Be careful not to confuse the subscript in the formula with the charge in the metal ion. In part (a), for example, the subscript on Fe is 2, but this is an iron(III) compound.


Oxoanions are polyatomic anions that contain one or more oxygen atoms and one atom (the “central atom”) of another element.

Starting with the oxoanions that end in –ate, we can name these ions as follows:1)The ion with one more O atom than the –ate ion is called the per…ate ion. Thus, ClO3

- is the chlorate ion, so ClO4- is the

perchlorate ion.2)The ion with one less O atom than the –ate ion is called the –ite ion. Thus, ClO2

- is the chlorite ion.3)The ion with two fewer O atom than the –ate ion is called the hypo…ite ion. Thus, ClO- is the hypochlorite ion.

At minimum, memorize the oxoanions that end in –ate so you can apply these guidelines when necessary.


perchlorate ClO4-

chlorate ClO3-

chlorite ClO2-

hypochlorite ClO-

nitrate NO3-

nitrite NO2-

phosphate PO43-

phosphite PO33-

sulfate SO42-

sulfite SO32-

Worked Example 5.10

Strategy Each species is either an oxoanion or an oxoacid. Identify the “reference oxidation” (the one with the –ate ending) for each, and apply the rules to determine appropriate names.

Name the following species: (a) BrO4-, (b) HCO3

-, and (c) H2CO3.

Solution (a) BrO4- has one more O atom than the bromate ion (BrO3

-), so BrO4-

is the perbromate ion.

(b) CO32- is the carbonate ion. Because HCO3

- has one ionizable hydrogen atom, it is called the hydrogen carbonate ion.

(c) With two ionizable hydrogen atoms and no charge on the compound, H2CO3 is carbonic acid.

Think About It Make sure that the charges sum to zero in each compound formula. In part (a), for example, Hg2+ + 2Cl- = (+2) + 2(-1) = 0; in part (b), (+2) + 2(-1) = 0; and in part (c), 3(+1) + (-3) = 0.

Think About It Remembering all these names and formulas is greatly facilitated by memorizing the common ions that end in –ate.

chlorate ClO3- nitrate NO3

-

iodate IO3- carbonate CO3

2-

bromate BrO3- oxalate C2O4

2-

sulfate SO42- chromate CrO4

2-

phosphate PO43- permanganate MnO4

-

Worked Example 5.11

Strategy The –ous ending in the name of an acid indicates that the acid is derived from an oxoanion ending in –ite. The oxoanion must be sulfite, SO3

2-, so add enough hydrogen ions to make a neutral formula.

Determine the formula of sulfurous acid.

Solution The formula of sulfurous acid is H2SO3.

Think About It Remembering all these names and formulas is greatly facilitated by memorizing the common ions that end in -ate.

HydratesHydrates

A hydrate is a compound that has a specific number of water molecules within its solid structure.

For example, in its normal state, copper(II) sulfate has five water molecules associated with it.

Systematic name: copper(II) sulfate pentahydrateFormula: Cu(SO)4 ∙ 5H2O

Some other hydrates areBaCl2 ∙ 2H2OLiCl ∙ H2OMgSO4 ∙ 7H2OSr(NO3)2 ∙ 4H2O

HydratesHydrates

When the water molecules are driven off by heating, the resulting compound, Cu(SO)4, is sometimes called anhydrous copper(II) sulfate.

Anhydrous means the compound no longer has water molecules associated with it.

Familiar Inorganic CompoundsFamiliar Inorganic Compounds

The molecular mass is the mass in atomic mass units (amu) of an individual molecule.

To calculate molecular mass, multiply the atomic mass for each element in a molecule by the number of atoms of that element and then total the masses

Molecular mass of H2O = 2(atomic mass of H) + atomic mass of O = 2(1.008 amu) + 16.00 amu = 18.02 amu

Because the atomic masses on the periodic table are average atomic masses, the result of such a determination is an average molecular mass, sometimes referred to as the molecular weight.

Molecular and Formula MassMolecular and Formula Mass5.8

Although an ionic compound does not have a molecular mass, we can use its empirical formula to determine its formula mass (the mass of a “formula unit”), sometimes called the formula weight.

Molecular and Formula MassMolecular and Formula Mass

Worked Example 5.12

Strategy Determine the molecular mass (for each molecular compound) or formula mass (for each ionic compound) by summing all the atomic masses.

Calculate the molecular mass or the formula mass, as appropriate, for each of the

following corresponds: (a) propane, C3H8, (b) lithium hydroxide, LiOH, and (c)

barium acetate, Ba(C2H3O2)2.

Solution For each compound, multiply the number of atoms by the atomic mass of each element and then sum the calculated values.

(a) The molecular mass of propane is 3(12.01 amu) + 8(1.008 amu) = 44.09 amu

(b) The formula mass of lithium hydroxide is 6.941 amu + 16.00 amu + 1.008 amu = 23.95 amu.

(c) The formula mass of barium acetate is 137.3 amu + 4(12.01 amu) + 6(1.008 amu) + 4(16.00 amu) = 255.4 amu.

Think About It Double-check that you have counted the number of atoms correctly for each compound and that you have used the proper atomic masses from the periodic table.

A list of the percent by mass of each element in a compound is known as the compound’s percent composition by mass.

where n is the number of atoms of the element in a molecule or formula unit of the compound

atomic mass of elementpercent mass of an element = 100%

molecular or formula mass of compound

n

Percent Composition of CompoundsPercent Composition of Compounds5.9

For a molecule of H2O2:

2 2

2 1.008 amu H%H = 100% = 5.926%

34.02 amu H O

2 2

2 16.00 amu O%O = 100% = 94.06%

34.02 amu H O

Percent Composition of CompoundsPercent Composition of Compounds

We could also have used the empirical formula of hydrogen peroxide (HO) for the calculation.

In this case, we could have used the empirical formula mass, the mass in amu of one empirical formula, in place of the molecular formula.

The empirical formula mass of H2O2 (the mass of HO) is 17.01 amu.

%H = × 100% = 5.926%

%O = × 100% = 94.06%

Percent Composition of CompoundsPercent Composition of Compounds

1.008 amu H17.01 amu H2O2

16.00 amu O17.01 amu H2O2

Worked Example 5.13

Strategy Use Equation 5.1 to determine the percent by mass contributed by each element in the compound.

Lithium carbonate, Li2CO3, was the first “mood-stabilizing” drug approved by

the FDA for the treatment of mania and manic-depressive illness, also known as bipolar disorder. Calculate the percent composition by mass of lithium carbonate.

Solution For each element, multiply the number of atoms by the atomic mass, divide by the formula mass, and multiply by 100 percent.

%Li = ×100% = 18.79%

%C = ×100% = 16.25%

%O = ×100% = 64.96%

2×6.941 amu Li73.89 amu Li2CO3

12.01 amu C73.89 amu Li2CO3

3×16.00 amu O73.89 amu Li2CO3

Think About It Make sure that the percent composition results for a compound sum to approximately 100. (In this case, the results sum to exactly 100 percent––18.79% + 16.25% + 64.96% = 100.00%––but remember that because of rounding, the percentages may sum to very slightly more or very slightly less.

The molar mass (M ) of a substance is the mass in grams of one mole of the substance.

The molar mass of an element is numerically equal to its atomic mass.

1 mol C = 12.01 g 1 C atom = 12.01 amu

The molar mass of a compound is the sum of molar masses of the elements it contains.

1 mol H2O = 2 ×1.008 g + 16.00 g = 18.02 g

1 mol NaCl = 22.99 g + 35.45 g = 58.44 g

Molar MassMolar Mass5.10

When expressing the molar mass of elements such as oxygen and hydrogen, we have to be careful to specify which form of the element we mean.

For instance, the element oxygen exists predominantly as O2. But we might also mean atomic oxygen (O). We must consider the context to tell which form of the element, O2 or O, is intended.

Molar MassMolar Mass

Context Oxygen means Molar mass

How many moles of oxygen react with 2 moles of hydrogen to produce water? O2 32.00 g

How many moles of oxygen are there in 1 mole of water? O 16.00 g

Air is approximately 21 percent oxygen. O2 32.00 g

Many organic compounds contain oxygen. O 16.00 g

Interconverting Mass, Moles, and Numbers of ParticlesInterconverting Mass, Moles, and Numbers of Particles

Worked Example 5.14

Strategy Use molar mass to convert from mass to moles and to convert from moles to mass. The molar mass of carbon dioxide (CO2) is 44.01 g/mol and the molar mass of sodium chloride (NaCl) is 58.44 g/mol.

Determine (a) the number of moles of CO2 in 10.00 g of carbon dioxide and

(b) the mass of 0.905 mole of sodium chloride.

Solution (a) 10.00 g CO2 × = 0.2272 mol CO2

(b) 0.905 mol NaCl × = 52.9 g NaCl

1 mol CO2

44.01 g CO2

58.44 g NaCl1 mol NaCl

Think About It Always double-check unit cancellations in problems such as these–errors are common when molar mass is used as a conversion factor. Also make sure that the results make sense. In both cases, a mass smaller than the molar mass corresponds to less than a mole of substance.

Worked Example 5.15

Strategy Use molar mass and Avogadro’s number to convert from mass to molecules, and vice versa. Use the molecular formula of water to determine the numbers of H and O atoms.

(a) Determine the number of water molecules and the numbers of H and O atoms in 3.26 g of water.

(b) Determine the mass of 7.92×1019 carbon dioxide molecules.

Solution (a) 3.26 g H2O× ×

Using the molecular formula, we can determine the number of H and O atoms in 3.26 g of H2O as follows:

1.09×1023 H2O molecules ×

1.09×1023 H2O molecules ×

1 mol H2O18.02 g H2O

6.022×1023 H2O molecules1 mol H2O

2 H atoms1 H2O molecule

1 O atom1 H2O molecule

= 2.18×1023 H atoms

= 1.09×1023 H atoms

= 1.09×1023 H2O molecules


Solution (b) 7.92×1019 CO2 molecules × ×

44.01 g CO2

1 mol CO2

1 mol CO2

6.022×1023 CO2 molecules

= 5.79×10-3 g CO2

Think About It Again, check the cancellation of units carefully and make sure that the magnitudes of your results are reasonable.

Using the concepts of the mole and molar mass, we can now use an experimentally determined percent composition to determine the empirical and/or molecular formula.

The empirical formula gives only the ratio of atoms in a molecule, so there may be multiple compounds with the same empirical formula.

If we know the approximate molar mass of the compound, we can determine the molecular formula by dividing the molar mass by the empirical formula mass. We then multiply the empirical formula by this number to obtain the molecular formula.

Determination of Empirical Formula and Molecular Determination of Empirical Formula and Molecular Formula from Percent CompositionFormula from Percent Composition

Worked Example 5.16

Strategy Assume a 100-g sample so that the mass percentages of nitrogen and oxygen given in the problem statement correspond to the masses of N and O in the compound. Then, using the appropriate molar masses, convert the grams of each element to moles. Use the resulting numbers as subscripts in the empirical formula, reducing them to the lowest possible whole numbers for the final answer. To calculate the molecular formula, first divide the molar mass given in the problem statement by the empirical formula mass. Then, multiply the subscripts in the empirical formula by the resulting number to obtain the subscripts in the molecular formula.

The molar masses of N and O are 14.01 and 16.00 g/mol, respectively. One hundred grams of a compound that is 30.45 percent nitrogen and 69.35 percent oxygen by contains 30.45 g N and 69.55 g O.

Determine the empirical formula of a compound that is 30.45 percent nitrogen and 69.55 percent oxygen by mass. Given that the molar mass of the compound is approximately 92 g/mol, determine the molecular formula of the compound.


Solution 30.45 g N ×

69.55 g O ×

The gives a formula of N2.173O4.347. Dividing both subscripts by the smaller of the two to get the smallest possible whole numbers (2.173/2.173 = 1, 4.347/2.173 ≈ 2) gives an empirical formula of NO2.

Finally, dividing the approximate molar mass (92 g/mol) by the empirical formula mass [14.01 g/mol + 2(16.00 g/mol) = 46.01 g/mol] gives 92/46.01 ≈ 2. Then, multiplying both subscripts in the empirical formula by 2 gives the molecular formula, N2O4.

1 mol N14.01 g N1 mol O

16.00 g O

= 2.173 mol N

= 4.347 mol O

Think About It Use the method described in Worked Example 5.13 to calculate the percent composition of the molecular formula N2O4 and verify that it is the same as that given in this problem.

Chapter Summary: Key PointsChapter Summary: Key Points5

CompoundsLewis Dot SymbolsIonic Compounds and Ionic BondingChemical FormulasLattice EnergyIonic Compound NomenclatureLewis Theory of BondingCovalent BondingLaws of Definite Proportions and Multiple ProportionsMolecular, Structural, and Empirical FormulasMolecular Compound NomenclatureInorganic and Organic CompoundsPolyatomic IonsOxyanionsOxoacidsHydratesMolecular Mass and Formula MassEmpirical Formula MassMolar Mass

• Identify the type(s) of bonds (ionic or covalent) holding ALL atoms together in the following compounds:

• MgCl2 • NCl3• MgCO3

• Cl2• Na2SO4

Group Quiz #6:

• Write neutral ionic formulas using the following pairs of ions :– Ba and Cl– Co+2 and N– Fe+3 and N – K and P– Zn and O

Group Quiz #7:

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• Write formulas for the following names:– aluminum bromide– titanium(IV) chloride– chromium(III) sulfide

• Write names for the following formulas:– Ba3N2

– K2S– MnF2

Binary Ionic Compounds: Group Quiz #8

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• Write names for formulas and formulas for names:

• XeF6

• ammonium sulfite• Cr3N2

• K3PO4

• hydrochloric acid

Group Quiz #9

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chemistry: atoms first julia burdge & jason overby chapter 5 ionic and covalent compounds kent...

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