Chapter 5.1, 5.3, 5.5, 5.8, 5.9 and 5.10

Homework from the book

1, 11, 35, 37, 39, 41, 43, 45, 45, 47,

67, 69, 75, 77, 79, 83, 85, 87, 89, 91,

97, 103, 107, 109, 115, 119, 121, 127,

129 and 131

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

MoleculesMolecules and Compounds and Compounds5.1

A molecule is an aggregate of two or more atoms in a definite arrangement held together by chemical bonds

H2 H2O NH3 CH4

A molecule is formed when two or more atoms join together chemically. A compound is a molecule that contains at least two different elements. All compounds are molecules but not all molecules are compounds

Which of the following is a molecule and NOT a compound?

Answer:

H2

One way to remember these elements is:

Mr. BrINClHOF

There are 7 elements that occur in nature as a diatomic molecule

A molecular formula shows the exact number of atoms of each element in the smallest unit of a substance

An empirical formula shows the simplest whole-number ratio of the atoms in a substance

H2OH2O

molecular empirical

C6H12O6 CH2O

O3 O

N2H4 NH2

Empirical and Molecular formulas

Ions

Atoms are neutral—meaning that the number of protons is equal to the number of electrons

If an atom loses or gains electrons the atom is No longer neutral but has a charge.

Why do Ions form?

Electrons are transferred from the Cation to the Anion and the charged ions attract each other

Na11 protons11 electrons

Na+11 protons10 electrons

Cations form from metals because they have a low Ionizationenergy and will readily give up electrons to obtain theelectron configuration of a noble gas

+ energy + e-

Anions form from non-metals because they have low Electronaffinity and will readily accept electrons to obtain the electron configuration of a noble gas

Cl 17 protons17 electrons Cl-

17 protons18 electrons

+ e- + energy

Ionic Bonds

Occur between a metal and a non-metal. Ionic bonds form from theAttraction of positive and negative ions that form because of a transferOf electrons.

Lattice Energy

The amount that the potential energy of the system decreasesWhen the ions in one mole of the compound are brought fromA gaseous state to the positions the ions occupy in a crystal of the Compound. This is the reason Ionic Compounds form

Lattice energy (E) increases as Q increases and/or

as r decreases.

cmpd lattice energyMgF2

MgO

LiF

LiCl

2957

3938

1036

853

Q= +2,-1

Q= +2,-2

r F < r Cl

Electrostatic (Lattice) Energy

E = kQ+Q-r

Q+ is the charge on the cation

Q- is the charge on the anionr is the distance between the ions

Lattice energy (E) is the energy required to completely separate one mole of a solid ionic compound into gaseous ions.

The Octet Rule

Atoms tend to gain or lose electrons until they have achieved an outer shell that contains an octet (8) of electrons

Exceptions to the Octet Rule

Atoms that have valence electrons in the d and f orbitalsAnd all atoms before Carbon in the periodic table

Recognizing Ionic Compounds

A compound is ionic if it contains a metal from group 1Or group 2 or one of the polyatomic ions. Binary metal Oxides and sulfides also have ionic character

Fig 2-23Pg 59

Pure water(left) and a solution of sugar(right) do not conduct electricitybecause they contain virtually no ions. Asolution of salt (center) conducts electricity wellbecause it contain mobile cations and anions.Courtesy Ken Karp

When Atoms Combine to make Molecules

Atoms contain both positive and negative charges. When they come Together they arrange themselves so that the attractive forges of oppositeCharges is greater than the repulsive forces of like charges

Bond Length (Bond Distance)

The distance in picometers between the nucleii of twoBound atoms in a molecule

Bond Energy

The strength of the bond between atoms in a molecule. The amount of energy (J) required to break the bond.

Bond Length and Bond Energy

Fig 8-3 Pg 330

The interaction energy of a pair of hydrogen atoms varies with internuclear separation.

Change in electron density as two hydrogen atoms approach each other.

A covalent bond is a chemical bond in which two or more electrons are shared by two atoms.

Why should two atoms share electrons?

F F+

7e- 7e-

F F

8e- 8e-

F F

F F

Lewis structure of F2

lone pairslone pairs

lone pairslone pairs

single covalent bond

single covalent bond

The Octet Rule

Atoms tend to share electrons until they have achieved an outer shell that contains an octet (8) of electrons

Exceptions to the Octet Rule

Atoms that have valence electrons in the d and f orbitalsAnd all atoms before Carbon in the periodic table

8e-

H HO+ + OH H O HHor

2e- 2e-

Lewis structure of water

Double bond – two atoms share two pairs of electrons

single covalent bonds

O C O or O C O

8e- 8e-8e-double bonds double bonds

Triple bond – two atoms share three pairs of electrons

N N8e-8e-

N N

triple bondtriple bond

or

The molar mass (MM ) 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

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

Molar mass (or molecular weight) is the sum ofthe atomic masses (in amu) in a molecule.

SO2

1S 32.07 amu

2O + 2 x 16.00 amu SO2 64.07 amu

For any molecule

molecular mass (amu) = molar mass (grams/mole)

1 molecule SO2 = 64.07 amu

1 mole SO2 = 64.07 g SO2

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.

Interconverting Mass, Moles, and Interconverting Mass, Moles, and Numbers atoms or moleculesNumbers atoms or molecules

How many H atoms are in 72.5 g of C3H8O ?

1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O

1 mol H = 6.022 x 1023 atoms H

5.82 x 1024 atoms H

1 mol C3H8O molecules = 8 mol H atoms

72.5 g C3H8O1 mol C3H8O

60 g C3H8Ox

8 mol H atoms

1 mol C3H8Ox

6.022 x 1023 H atoms

1 mol H atomsx =

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.

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

Percent composition of an element in a compound

n x molar mass of elementmolar mass of compound

x 100%

n is the number of moles of the element in 1 mole of the compound

C2H6O

%C =2 x (12.01 g)

46.07 gx 100% = 52.14%

%H =6 x (1.008 g)

46.07 gx 100% = 13.13%

%O =1 x (16.00 g)

46.07 gx 100% = 34.73%

52.14% + 13.13% + 34.73% = 100.0%

3.5

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.

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.

Worked Example 5.16 (cont.)

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.

Empirical Formula

The listing of the mass of each element present in100 g of a compound (elemental analysis)

What is the empirical formula for a compound with The following elemental analysis?

49.5% C 5.2% H 28.8%N 16.5% O

Empirical formula: contains the smallest set of whole-number subscripts that match the elemental analysis

1. Assume you have a 100 g sample so the mass of eachelement is the same as the percentage

2. Divide by the molar mass of each element to get moles

49.5 g C x 1 mol C = 4.125 mol C 12.00 g

5.20 g H x 1 mol H = 5.2 mol H 1.00 g

28.8 g N x 1 mol N = 2.05 mol N 14.00 g

16.5 g O x 1 mol O = 1.03 mol O 16.00 g

3. Divide the moles of each element by the smallest (O) to get ratios of the atoms in the molecule

C4H5N2O

Finding Empirical Formula from percentages………

÷ 1.03 = 4

÷ 1.03 = 5

÷ 1.03 = 2

÷ 1.03 = 1

Molecular formula: contains the real set of whole-number subscripts that are found in the molecule

Finding Molecular Formula……………

Need: Empirical Formula and Molar Mass of the compound……….

For the preceeding problem where the empirical formula was found to be C4H5N2O with a molar mass of 291 g/ mol

1. Find the molar mass of the empirical formula (97g/ mol)

2. Divide the Molecular Formula MM by theempirical formula MM giving us the “multiplying factor”

291 ÷ 97 = 3 therefore 3 is the multiplying factor

3. Multiply all subscripts by the multiplying factor to get the molecular formula

C12H15N6O3