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Bell Work

• Put your name, class period, and today’s date on a sheet of paper.

• Use dimensional analysis to convert 3.000×10 1 mi/hr to km/s. There are

exactly

2.54 centimeters in one inch and one mile.

exactly

5280 feet in  You must show the setup.

 Use rules for significant figures.

 No group work.

 Calculator allowed.

Bell Work

• Convert 3.000×10 1  There are

exactly

 There are

exactly

mi/hr to km/s.

2.54 centimeters in one inch.

5280 feet in one mile.

3.000

 10 1 mi hr 5280ft 12in mi ft 2.54cm

in m km 100cm 1000m hr 60min min 60s  0.0134112

km s  0.01341

km s

Chapter 11: The Mole

• In this chapter, you will learn how to use the mole and related quantities to make conversions and calculations necessary to perform typical laboratory work, including making determinations about quantities of chemicals.

• As usual, understanding this part of your book is essential to understanding the chapters thereafter.

Measuring Matter

• Just as normal people use known quantities to express useful amounts of a certain item – roses and eggs come in dozens, paper comes in reams – chemists pretend to be normal and do the same.

• We need a reasonable way to keep track of the number of atoms, molecules, ions, etc., when we’re doing chemistry.

Measuring Matter

• But normal people have it easy because counting flowers and eggs is a simple process – it’s not so easy to count atoms.

• In fact, these particles are so small that there’s no way to actually

count

them, so we have a unique way to count using our own counting unit: the mole.

Measuring Matter

•

A mole is defined as the number of carbon atoms within exactly 12 grams of pure carbon-12.

• We’ve experimentally determined that

this number is 6.02 × 10 23 .

•

This is known as Avogadro’s number.

 It is sometimes symbolized

A .

• So there are 6.02 × 10 23 atoms of carbon-12 ( 12 C) in one mole of 12 C, which is 12.000 grams of 12 C.

Measuring Matter

•

The mole (abbrev: mol) is the SI base unit for measuring the amount of a substance.

• Remember that the mole is simply a counting unit that can be applied to anything – just as a dozen or a gross can be applied to flowers or eggs or pencils.

Measuring Matter

• Just as we can have a mole of carbon (6.02 × 10 23 atoms of carbon), we could have:  a mole of carbon dioxide (6.02 × 10 23 molecules of CO 2 )  a mole of desks (6.02 × 10 23 desks)  a mole of moles (6.02 × 10 23 furry vermin).

• (So we

can

use the mole with anything, but unless we’re working with the physical sciences, we probably won’t.)

Measuring Matter

• Suppose we have one mole of sodium. How many atoms of sodium do we have?

1 mol 6.02

 10 23 atoms  6.02

 10 23 atoms 1 mol

Measuring Matter

• We can also convert in the opposite direction: • Suppose we have 2.00×10 20 atoms of oxygen. How many moles of O are present?

2.00

 10 20 atoms O mol O 6.02

 10 23 atoms O  3 .

32  10 4 mol O

Measuring Matter

• Suppose we have one mole of water. How many molecules of water do we have?

1 mol 6.02

 10 23 molecules  6.02

 10 23 molecules mol

Measuring Matter

• Suppose we have 2.00×10 20 molecules of ozone, O 3 . How many moles of O 3 are present?

2.00

 10 20 molecules O 3 6.02

mol O 3  10 23 molecules O 3  3 .

32  10 4 mol O 3

Practice Problems

1. How many atoms are in 2.50 mol of Zn?

2. How many molecules are in 11.5 mol of H 2 O?

3. 5.75

×10 24 of Al?

atoms of Al is how many moles 4. 3.58

×10 23 formula units of ZnCl 2 many moles of ZnCl 2 ?

is how

Measuring Matter

• Let’s use some analogies to help us understand why the mole is useful:  Twelve 300-lb football players weigh 3600lbs.

 Twelve 100-lb cheerleaders weight 1200lbs.

• Even though the count of each sample is the same (a dozen), the mass of the samples differ due to the different masses of the pieces within the samples.

Measuring Matter

• Just as there must be some number of 12 oz marbles in 12 tons of marbles, there must be some number of 12.000-amu 12 C atoms in 12 grams of 12 C.

 The latter number is Avogadro’s number.

× ×

Measuring Matter

• That number is the number of atomic mass units in one gram.

 The number of 12amu pieces in 12g is the same as the number of 1amu pieces in 1g.

 This must be the number of

amu pieces in

grams: it is a conversion factor.

× ×

Measuring Matter

• Some of you may be helped by thinking of it this way:

1 gram



6 .

02



10 1 amu 6 .

02



10 amu



1 gram 1 6 .

02



10 gram



1 amu

Mass and the Mole

• Recall that the atomic weights listed on the periodic table are the weighted averages of the isotopes of that element.

• In other words, if we gather a large number of oxygen atoms, the

average

mass of one atom will be 15.9994 amu.

• Now apply the knowledge that we can use Avogadro’s number:

Mass and the Mole

• If it takes 6.02×10 23 items that each weigh 12.000 amu to total 12.000 grams in mass, what will be the mass of 6.02

×10 23 items that each weigh (an average of) 15.9994 amu?

 15.9994g

• Thus,

one mole of any substance will have a mass in grams numerically equal to the mass in amu of one particle of that substance.

Mass and the Mole

• • Using the logic described above, we can conclude that one mole of a random sampling of sulfur atoms will have a mass of 32.065g because the average individual atom has a mass of 32.065amu.

The mass in grams of one mole of any substance is called its molar mass.

 Because this is the number of grams in one mole, the unit for a molar mass will be g/mol.

Mass and the Mole

• Avogadro’s number is especially useful because it is effectively a conversion factor that relates the number of realistically uncountable particles to a measurable quantity: mass.

• By using Avogadro’s number, we can work with realistic amounts of substances using units of moles.

Calculations with Moles

• Now that we know the relationship between the number of particles, the mass of those particles, and the number of moles of those particles, we can do calculations with these quantities that will allow us to know how much of a substance we have when we measure a sample with less than-state-of-the-art equipment like balances instead of scanning-tunneling electron microscopes.

Calculations with Moles

• Using dimensional analysis will help you set up your calculation. Use the molar mass to cancel the units you do not want.

• What mass of chromium is in 0.0450mol Cr?

0.0450

mol Cr 52.00

g Cr



2.34 g Cr 1 mol Cr

Calculations with Moles

• We can also convert in the other direction. • How many moles of calcium are in 525g of calcium?

525 g Ca mol Ca 40.078

g Ca



13.1 mol Ca

Calculations with Moles

• We can use these concepts along with Avogadro’s number to relate mass to number of representative particles, and vice versa.

• 5.50×10 22 atoms of He has what mass?

5.50

 10 22 atoms He 1 mol He 6.02

 10 23 atoms He 4.003g

He  0 .

366 g He 1 mol He

Moles of Compounds

• Suppose we have one mole of water. How many molecules of water do we have?

1 mol 6.02

 10 23 molecules  6.02

 10 23 molecules mol

Moles of Compounds

• How many atoms of hydrogen are in that mole of water?

1 mol H 2 O 6.02

 10 23 molecules H 2 O mol H 2 O 2 atoms H 1 molecule H 2 O  1 .

20  10 24 atoms H

Moles of Compounds

• We can also convert in the opposite direction: • Suppose we have 2.00×10 20 molecules of ozone, O 3 . How many moles of O 3 are present?

2.00

 10 20 molecules O 3 6.02

mol O 3  10 23 molecules O 3  3 .

32  10 4 mol O 3

Moles of Compounds

• How many moles of oxygen atoms are present in this sample of ozone?

2.00

 10 20 molecules O 3 6.02

mol O 3  10 23 molecules O 3 3 mol O  9 .

97  10 4 mol O 3 mol O 3

Moles of Compounds

• A mole of hydrogen

atoms

has a mass of 1.0079g, but a mole of

hydrogen gas

has a different molar mass. Why?

 There are two atoms of hydrogen in one molecule of hydrogen gas.

 What is the molar mass of hydrogen gas, H 2 ?

 1.0079g × 2 = 2.0158g

• or you could reason that each molecule weighs 1.0079amu

×2 = 2.0158amu, so a mole is 2.0158g.

Moles of Compounds

• Because a water molecule is composed of two atoms of hydrogen and one of oxygen, a mole of water will have a mass of (2 ×1.0079g + 1×15.9994g or) 18.0152g.

 Each water molecule has a mass of 2 ×1.0079amu + 1×15.9994amu or 18.0152amu.

 A mole of these 18.0152amu molecules must weight 18.0152 grams.

Moles of Compounds

• What is the molar mass of potassium chromate?

K 2 CrO 4 2K × 39.10g = 78.20g

1Cr × 52.00g = 52.00g

4O × 16.00g = 64.00g

Total: 194.20g

Moles of Compounds

• Find the molar mass of each of the following compouds: 1. NaCl 2. NaOH 3. potassium chloride 4. sulfuric acid H 2 SO 4 5. ammonium phosphate KCl (NH 4 ) 3 PO 4

Moles of Compounds

• Suppose we have 5.50 moles of freon, CCl 2 F 2 .

• How many molecules of freon are in the sample?

5.50

mol CCl 2 F 2 6.02

 10 23 molecules CCl 2 F 2 mol CCl 2 F 2  3.31

 10 24 molecules CCl 2 F 2

Moles of Compounds

• In our 5.50mol freon, we have how many:  carbon atoms?

• There is one carbon atom per molecule, so we have 5.50mol carbon.

 chlorine atoms?

• There are two chlorine atoms per molecule, so we have 5.50mol

×2 or 11.0mol chlorine.

 fluorine atoms?

• There are two fluorine atoms per molecule, so we have 5.50mol

×2 or 11.0mol fluorine.

Moles of Compounds

• What mass of freon is in our 5.50 moles?

 First, we must find the

molar mass

• 1mol C @ 12.011g: 12.011g

• 2mol Cl @ 35.453g: 70.906g

• 2mol F @ 18.998g: 37.996g

• Total: 120.913g CCl 2 F 2 per mole CCl 2 F 2 of freon, CCl 2 F 2 : 5.50

mol CCl 2 F 2 120.913g

CCl 2 F 2 1 mol CCl 2 F 2  665 g CCl 2 F 2

Moles of Compounds

• Once we have molar masses, we can do the same types of conversions for compounds as we can do for atoms.

• 2.75 moles of potassium chromate is how much mass?

2.75

mol K 2 CrO 4 194.20

g K 2 CrO 4 1 mol K 2 CrO 4  5 .

34 g K 2 CrO 4

Moles of Compounds

• Or we can go in the other direction.

• Or we can convert from moles to representative particles.

 Or vice versa.

• Or from representative particles to mass.

 Or vice versa.

Moles of Compounds

• Determine the number of moles present in each of the following.

1. 22.6g AgNO 3 2. 6.50g ZnSO 4 3. 35.0g HCl 4. 25.0g Fe 2 O 3 5. 254g PbCl 4

Moles of Compounds

• On the same sheet of paper as your bell work, work practice problems 31-35 on p326.

molar mass × ×

Empirical and Molecular Formulas

• When a synthetic chemist produces a new compound, the task of verifying the identity of that compound falls upon an analytical chemist.

• This can be done by finding the percent composition, the empirical formula, and the molecular formula of the compound.

Percent Composition

• We calculated percent composition (“percent by mass”) of compounds in an earlier chapter.

• In this chapter, we will take the very same idea further so that we can determine empirical and molecular formulas.

Percent Composition

•

The percent by mass of each element in a compound is called the percent composition of that compound.

 This can also be called

percent by mass mass percent

or Mass % of

= mass of

in sample total mass of sample  100

Percent Composition

• Because we are calculating a percentage, standard math rules apply: find the ratio of the part to the whole and convert to percent:

part whole



100%



percentage

Percent Composition

• If you know the mass breakdown of a compound, then you can calculate the percent composition by comparing the mass of each element in the compound to the mass of the entire compound.

Percent Composition

• For example, if element X makes up 55g of a 100g sample, and element Y makes up 45g of the sample, then compound’s percent composition is:

55 g element X 100 g compound



100 %



55 % element X 45 g element X 100 g compound



100 %



45 % element X

Percent Composition

• If you know the chemical formula of a compound, its percent composition can be determined from the formula and the molar masses.

mass of element in 1 mol of compound  100% molar mass of compound

Percent Composition

• The molar mass of water is: 2(

H )+

O = 2(1.008g)+16.00g = 18.02g/mol • There are two atoms of H per molecule, and one atom of O per molecule: 2 (

H )  100 %

m H

 2 ( 1 .

01 g )  100 % 18.02g

 2 .

02 g 18.02g

 100 %  11 .

2 % H

m H

 100 %  16 .

00 g 18.02g

 100 %  88 .

80 % O

Percent Composition Practice

• Find the percent composition of iron and oxygen in magnetite, Fe 3 O 4 .

3 (

m Fe

)  100 %

m Fe

4  3 ( 55 .

85 g )  100 % 231.55g

 72 .

36 % Fe 4 (

m O

)  100 %

m Fe

4  4 ( 16 .

00 g )  100 % 231.55g

 27 .

64 % O

Percent Composition Practice

• Find the mass percent of: 1. calcium chloride 2. sodium sulfate 3. phosphoric acid

Empirical Formulas

• • Once the percent composition of a compound is found, it can be used to find the simplest-form ratio of the number of elements in the compound.

The empirical formula for a compound is the formula with the smallest whole number mole ratio of the elements therein.

Empirical Formulas

•

An empirical formula might or might not be the same as the chemical formula.

• For example, the chemical formula for hydrogen peroxide is H 2 O 2 , but the empirical formula is HO (implied H 1 O 1 ) because a 2:2 ratio reduces to 1:1.

• The empirical formula for water, H 2 O, is H 2 O because the ratio is already in simplest form, 2:1.

Empirical Formulas

• The empirical formula for glucose, C 6 H 12 O 6 , is CH 2 O (6:12:6 reduces to 1:2:1).

• What is the empirical formula for hydroquinone, C 6 H 6 O 2 ?

C 3 H 3 O

Empirical Formulas

• To find an empirical formula from a percent composition, start by considering the case where your sample is 100g.

• Suppose your sample contains 40.05% S and 59.95% O.

• Use the molar masses to find the number of moles of each element in the theoretical 100g sample.

Empirical Formulas

40 .

05 g S mol S 32.07g

S



1 .

249 mol S 59 .

95 g O mol O 16.00g

O



3 .

747 mol O

Empirical Formulas

• Notice that these are not whole numbers, so they cannot be the subscripts of the formula.

• However, we can express them as a ratio in terms of the smaller one: 1 .

249 mol S  1 mol S 1 .

249 3 .

747 mol O  3 mol O 1 .

249

Empirical Formulas

• Now we can see that the ratio of O atoms to S atoms is 3:1, so the empirical formula must be SO 3 .

Empirical Formulas

• Find the empirical formulas of: 1. calcium chloride 2. sodium sulfate 3. phosphoric acid

Empirical Formulas

• However, not all mole ratios will simplify to whole numbers when compared to the smallest as above.

• Consider methyl acetate, which has a percent composition of: 48.64% C 8.16% H 43.20% O

Empirical Formulas

• To find the empirical formula, use the molar masses to find a ratio of number of atoms: 48 .

64 g C mol C 12.01

g C  4 .

050 mol C 8 .

16 g H mol H 1.008

g H  8 .

10 mol H 43 .

20 g O mol O 16.00

g O  2 .

700 mol O

Empirical Formulas

• Divide each number of moles by the smallest value: 4 .

050 mol C  1 .

500 mol C 2 .

700 8 .

10 mol H  3 .

00 mol H 2 .

700 2 .

700 mol O  1 .

000 mol O 2 .

700

Empirical Formulas

• Multiply these numbers by the smallest number that produces only whole-number values.

Since our only fractional value is 3/

, multiply each number by two: 2 ×1.5 mol C = 3 mol C 2 ×3 mol H = 6 mol H 2 ×1 mol O = 2 mol O • And we conclude that the empirical formula of methyl acetate is C 3 H 6 O 2 .

Empirical Formulas

• Find the empirical formula of each of the following samples: 1. a blue solid is 36.84% N and 63.13% O 2. a cooking fuel is 81.82% C and 18.18% H 3. aspirin contains 60.00% C, 4.44%H, and 35.56% O

Warm Up

• Find the empirical formula of a compound with the following percent composition:  62.1% C  5.21% H  12.1% N  20.7% O

Molecular Formulas

• • Now that we know how to find empirical formulas, we need to know how to differentiate between different compounds with the same empirical formula, like acetylene (C 2 H 2 ) and benzene (C 6 H 6 ).

The molecular formula of a compound specifies the actual number of atoms of each element in one molecule or formula unit of a compound.

Molecular Formula

• We can do this using the empirical formula and the molar mass, which can be determined experimentally.

• The molar mass of acetylene is 26.04g/mol.

• The mass of the empirical formula of acetylene (CH) is 13.02g/mol.

• Simply divide the molar mass of the compound by the molar mass of the empirical formula: 26 .

04 g/mol  2 .

000 13 .

02 g/mol

Molecular Formulas

•

The molecular formula is the empirical formula multiplied by the ratio of the molar mass of the compound to the molar mass of the empirical formula.

 2(CH) → C 2 H 2 • We can do the same for benzene:

benzene

emp.formul

a  78 .

12 g/mol 13 .

02 g/mol  6 .

000  The molecular formula for benzene is therefore 6(CH) or C 6 H 6 .

Determining Formulas

• In general, we can follow this procedure to determine empirical and molecular formulas from percent composition.

1. Express percent by mass in grams.

2. Find the number of moles of each element.

3. Examine the mole ratio: divide by the lowest number and convert to whole numbers.

4. Write the empirical formula.

5. Determine the integer relationship between the empirical formula and the molecular formula,

6. Multiply the subscripts of the empirical formula by

7. Write the molecular formula.

Molecular Formulas Practice

• Determine the molecular formula of a compound composed of 40.68% carbon, 5.08% hydrogen, and 54.24% oxygen with a molar mass of 118.1 g/mol.

Molecular Formulas

• Determine molecular formulas for each of the following: 1. 65.45% C, 5.45% H, 29.09% O, 110.0 g/mol 2. 49.98g C, 10.47g H, 58.12 g/mol 3. 46.68% N, 53.32% O, 60.01 g/mol

End §11.4

Hydrates

• • Sometimes water molecules stick to the ions in an ionic compound. When this happens, a

hydrate

is formed.

A hydrate is a compound that has a specific number of water molecules bound to one unit of that substance.

Hydrates

• We signify a hydrate by writing the formula of the unit and an indicator of the number of bound water molecules separated by a dot.

•

The general form is AB•xH 2 O, where x is the number of bound water molecules.

• For example, if a hydrate of sodium fluoride has two water molecules bound to the salt, its formula would be NaF•2H 2 O.

Hydrates

•

We name these compounds by naming the ionic compound, then adding a prefix indicating the number of bound water molecules to the word “hydrate

.” • The previous example, NaF•2H 2 O, would be named

sodium fluoride dihydrate

• A list of prefixes is on p338 in your textbook.

prefix

mono di tri tetra penta hexa hepta octa nona deca-

mol H 2 O

1 2 3 4 5 6 7 8 9 10

formula

(NH 4 ) 2 C 2 O 4 •H 2 O CaCl 2 •2H 2 O NaC 2 H 3 O 2 •3H 2 O FePO 4 •4H 2 O CuSO 4 •5H 2 O CoCl 2 •6H 2 O MgSO 4 •7H 2 O Ba(OH) 2 •8H 2 O Cr(NO 3 ) 3 •9H 2 O Na 2 CO 3 •10H 2 O

name

… mono hydrate … di hydrate … tri hydrate … tetra hydrate … penta hydrate … hexa hydrate … hepta hydrate … octa hydrate … nona hydrate … deca hydrate

Hydrates

• How would we name SrCl 2 •6H 2 O?

 strontium chloride hexahydrate • What is the name of MgSO 4 •7H 2 O?

 magnesium sulfate heptahydrate • What is the formula of barium hydroxide octahydrate?

 Ba(OH) 2 •8H 2 O

Determining the Formulas of Hydrates

• To determine the formula of a hydrate, we heat the hydrate to evaporate the water from the compound-water complex.

• The mass that is lost from the sample must be the mass of the water that was in the original sample. • We then use composition data to find the formula.

Determining the Formulas of Hydrates

• Suppose we have 2.50g of CoCl 2 •

H 2 O.

• We heat the sample and are left with 1.36g of anhydrous (waterless) CoCl 2 .

• 2.50g - 1.36g or 1.14g of the original sample was water.

• We then convert each mass to moles and use the mole ratio to find the formula.

Determining the Formulas of Hydrates

1 .

36 g CoCl 2 1 .

14 g H 2 O 1 mol CoCl 18.02g

H 2 O 2  0 .

0105 mol CoCl 2 129.8g

CoCl 2 1 mol H 2 O  .

0633 mol H 2 O x  mol H 2 O mol CoCl 2  .

0633 mol H 2 O 0 .

0105 mol CoCl 2  6 .

03  6 • Thus the formula of the hydrate is CoCl 2 • 6 H 2 O.

Determining the Formulas of Hydrates

1. A hydrate is found to be 48.8% MgSO 4 and 51.2% H 2 O. What is the formula of this hydrate? What is the name of this hydrate?

2. If 11.75g of a cobalt(II) chloride hydrate is heated and 9.25g of anhydrous salt remains, what are the formula and name of this hydrate?