THE MOLE Chapter 11

I. Introduction

A. Measuring Matter 1. Many items require exact counting units

a. One dozen = 12 items

b. One baker’s dozen = 13 items

c. A pair = 2 items

d. One gross = 12 dozen = 144 items

e. One ream = 500 items

2. Chemists also have a unit that is used to count atoms or molecules!

a. A mole is the SI unit that describes the amount of a substance

b. One mole = 6.02 x 1023 objects

c. “Objects” can represent number of atoms (ex: Ca), molecules (ex: CO2), formula units (ex: KCl), or number of any type of particle. We refer to any of these as “particles”

d. 1 mole = 6.02 x 1023 atoms (molecules) – this is called Avogadro’s Number

B. Avogadro’s Number 1. The mole is a HUGE unit – we use such a large number because

atoms are tiny

2. The number 6.02 x 1023 specifically is a standard chosen by scientists: carbon!

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

4. The number of atoms contained in exactly 12 grams of carbon- 12 was determined to be 6.02214 x 1023 atoms… rounds to 6.02 x 1023!

5. So, this means that…

a. 1 mole of C atoms = 6.02 x 1023 atoms

b. 1 mole of O atoms = 6.02 x 1023 atoms

c. 1 mole of Cl atoms = 6.02 x 1023 atoms

d. 1 mole of anything = 6.02 x 1023 anything

6. Remember, a mole is NOT a weight (mass)! It is a COUNT

C. Converting Between Moles and Particles 1. To find the number of particles (atoms, ions, molecules, etc.) in a

given number of moles, use the following conversion factor & T-chart setup:

2. To find the number of moles in a given number of particles (atoms, ions, molecules, etc.), use the following conversion factor & T-chart setup:

3. In these problems, the compound or element does not matter – it’s just about the numbers!

# moles given 6.02 x 1023 particles

1 mole

# particles given 1 mole

6.02 x 1023 particles

4. Examples:

a. Determine the number of particles in 2.33 mol of hydrogen gas (H2).

# particles H2 = 1.40 x 1024 particles

b. Find the number of moles contained in 5.30 x 1025 carbon monoxide

(CO) molecules.

# moles CO = 88.0 mol

2.33 mol 6.02 x 1023 particles

1 mol

5.30 x 1025 molecules 1 mole

6.02 x 1023 molecules

Homework

• p. 311 #1-3

• p. 312 #4-10

II. Mass & the Mole

A. We cannot count individual atoms, so we use their

masses to find the exact number of atoms in a sample

1. Example: A candy shop keeper knows that 10 gumballs have a

mass of 21.4 g. How can we measure out 200 gumballs without

actually counting the individual gumballs?

2. We can use this same approach to determine the number of

particles in a sample

B. Molar mass: mass in grams of one mole of a pure

substance; measured in grams/mole (g/mol)

1. For any element, atomic mass (in amu) = molar mass (in grams)

2. Found in upper left-hand corner of the element symbol on your

periodic table (use 3 sig figs if <100, 4 sig figs if > 100)

C. Conversions

1. Given grams, need moles:

a. Example: Convert 153 g of silver to moles.

# grams 1 mol

molar mass (in grams)

2. Given moles, need grams:

a. Example: Convert 7.32 mol of gold to grams.

# moles molar mass (in grams)

1 mol

3. Converting from grams to particles or particles to grams requires

two steps, converting to moles in between.

a. Example: Find the number of atoms contained in 34.2 g of pure nickel.

b. Example: Find the mass of 1.54 x 1024 atoms of titanium.

34.2 g Ni 1 mol Ni 6.02 x 1023 atoms

58.7 g Ni 1 mol Ni

1.54 x 1024 atoms Ti 1 mol Ti 47.9 g Ti

6.02 x 1023 atoms 1 mol Ti

Homework

• p. 316 #11-12

• p. 318 #13-14

• p. 319 #15-19

• p. 876 #1-14 (under Chapter 11)

III. Moles of Compounds

A. Chemical Formulas & the Mole

1. In a compound, the number of moles of an element in the

compound is equal to the subscript on that element

a. Example: find the number of moles of oxygen atoms in 3.25 moles of

CO2 gas.

b. This also applies to ions of a compound in aqueous solution

c. Example: find the number of moles of hydrogen ions in 1.72 mol of

phosphoric acid (H3PO4)

3.25 mol CO2 2 mol O

1 mol CO2

1.72 mol H3PO4 3 mol H+

1 mol H3PO4

2. Molar mass of compounds

a. Also referred to as molecular

mass or formula mass

b. The molar mass of the compound

is the sum of the molar masses of

each element times the number of

atoms of that element in the

compound

i. Example: calculate the molar mass

of CaCl2

1 Ca: 1 x 40.1 g = 40.1 g

2 Cl: 2 x 35.5 g = 71.0 g

1 mol = 111.1 g

ii. Example 2: find the molar mass of C6H12O6

6 C: 6 x 12.0 g = 72.0 g

12 H: 12 x 1.01 g = 12.1 g

6 O: 6 x 16.0 g = 96.0 g

180.1 g = 1 mol

iii. Example 3: find the molar mass of Na2SO4

2 Na: 2 x 23.0 g = 46.0 g

1 S: 1 x 32.1 g = 32.1 g

4 O: 4 x 16.0 g = 64.0 g

142.1 g = 1 mol Na2SO4

iv. Example 4: find the molar mass of Al(NO3)3

1 Al: 1 x 27.0 g = 27.0 g

3 N: 3 x 14.0 g = 42.0 g

9 O: 9 x 16.0 g = 144 g

213 g = 1 mol Al(NO3)3

3. Converting moles of a compound to mass

a. To find the mass of a certain number of moles of a compound:

i. Calculate the molar mass of the compound

ii. Multiply the number of moles by the molar mass of the compound in a T-

chart

b. Example: convert 12.5 moles of water to grams

i. Molar mass of water = (2 x 1.01 g) + 16.0 g = 18.02 g = 18.0 g

ii.

c. Example 2: convert 9.48 moles of Sn(CO3)2 to grams

i. Molar mass = 118.7 g + (2 x 12.0 g) + (6 x 16.0 g) = 238.7 g

ii.

12.5 mol H2O 18.0 g

1 mol H2O

9.48 mol Sn(CO3)2 238.7 g

1 mol Sn(CO3)2

Homework

• p. 321 #20-24

• p. 322 #25-26

• p. 323 #27-29

4. Converting mass of a compound to moles

a. To find the number of moles of a certain mass of a compound:

i. Calculate the molar mass of the compound

ii. Multiply the mass by the reciprocal of the molar mass in a T-chart

b. Example: find the number of moles in 132 g of Ca(OH)2 powder.

i. Molar mass of Ca(OH)2 = 40.1 g + (2 x 1.01 g) + (2 x 16.0 g) = 74.12 g = 74. 1 g

ii.

132 g Ca(OH)2 1 mol Ca(OH)2

74.1 g

5. Converting mass of a compound to particles

a. To find the number of particles of a certain mass of a compound:

i. Calculate the molar mass of the compound

ii. In a T-chart, multiply the mass given by the reciprocal of the molar mass, then

multiply by Avogadro’s number

b. Example: find the number of molecules in a 4.00 g packet of sucrose

(C12H22O11)

i. Molar mass of C12H22O11 = (12 x 12.01 g) + (22 x 1.01 g) + (11 x 16.0 g)

= 342.2 g

= 7.04 x 1021 molecules

4 g C12H22O11 1 mol C12H22O11 6.02 x 1023 molecules

342 g C12H22O11 1 mol

c. Example 2: using the answer to example 1, find the number of atoms of

carbon in one packet of sucrose

i. We know that 1 packet of sucrose is 7.04 x 1021 molecules, and that there are 12

C atoms in each molecule of sucrose

ii. So we can determine # of C atoms by:

= 8.45 x 1022 atoms C

d. Example 3: find the number of calcium & chlorine ions in solution when

15.2 g of CaCl2 is dissolved in water.

i. Step 1: Molar mass of CaCl2

ii. Step 2: Determine # of formula units in 15.2 g of CaCl2 (see example 1)

iii. Step 3: Determine # of Ca2+ & Cl- ions by multiplying the # of formula units by the

subscripts of each ion

7.04 x 1021 molecules C12H22O11 12 atoms C

1 molecule C12H22O11

Molar mass CaCl2: 40.1 g + (2 x 35.5 g) = 111.1 g = 111 g

= 8.25 x 1022 formula units CaCl2

# of Ca2+ ions:

= 8.25 x 1022 ions Ca2+

# of Cl- ions:

= 1.65 x 1023 ions Cl-

15.2 g CaCl2 1 mol CaCl2 6.02 x 1023 molecules

111 g CaCl2 1 mol

8.25 x 1022 formula units CaCl2 1 atom Ca

1 formula unit CaCl2

8.25 x 1022 formula units CaCl2 2 atoms Cl

1 formula unit CaCl2

Homework

• p. 324 #30

• p. 326 #31-35

• p. 327 #36-41

**Quiz on block day over everything we’ve covered so far**

IV. Empirical & Molecular Formulas

A. Percent Composition

1. Definition: percent by mass of each element in a compound

2. Often used to identify the formula of compounds

3. Always adds up to 100%! (may be .01 - .1 off depending on sig figs)

4. Calculation to determine an element’s percent by mass:

5. Can also be used to determine a polyatomic ion’s percent by mass:

mass of element x 100% = percent by mass

mass of compound

mass of polyatomic ion x 100% = percent by mass

mass of compound

6. Examples:

a. A compound has a total mass of 200 g. 110 g are of element X, and 90

are of element Y. Calculate each element’s percent by mass.

b. Calculate the percent by mass of each element in water.

i. To get total mass, calculate molar mass of compound to 4 sig figs:

H: 2 x 1.008 = 2.016 O: 16.00 2.016 + 16.00 = 18.016 = 18.02 g

ii. Use molar mass values to plug into the equation:

c. Calculate the percent by mass of each element in Ce2(CO3)3

110 g X x 100% = 55% X

200 g

90 g Y x 100% = 45% Y

200 g

2.016 g H x 100% = 11.18% H

18.02 g H2O

16.0 g O x 100% = 88.79% O

18.02 g H2O

B. Empirical Formula

1. Percent composition can be used to find the empirical formula of

a compound

a. Definition: formula with the smallest whole number ratio of the number

of moles of the element

b. The mole ratio provides the subscripts for the empirical formula

c. Sometimes the empirical formula is the same as the actual formula

(ex: H2O), but sometimes they differ (ex: H2O2’s empirical formula is

HO)

2. Steps to determine a compound’s empirical formula:

a. If given percent composition, assume the compound is 100g. If given

grams, skip this step.

b. Convert grams to moles for each element using molar mass to 4 sig

figs

c. Using the #s of moles calculated above, find the mole ratio of one

element to the other elements.

i. To do so, divide all #s of moles by the lowest number of moles & round to

whole numbers.

ii. If these do not round nicely to whole #s (see example 2 below), multiply all

the values by the smallest # that will make all values whole #s.

iii. These numbers will become the subscripts on each element

d. Write the empirical formula -- you did it!

3. Examples:

a. Find the empirical formula for a compound containing 40.05% sulfur

and 59.95% oxygen.

40.05 g S 1 mol

32.07 g S

59.95 g O 1 mol

16.00 g O

= 1.25 mol S

1.25 mol

= 3.75 mol O

1.25 mol

= 1

= 3

empirical formula: SO3

b. Find the empirical formula for a compound containing 48.64% carbon,

8.16% hydrogen, and 43.20% oxygen.

48.64 g C 1 mol

12.01 g C

8.16 g H 1 mol

1.008 g H

= 4.05 mol C

2.70 mol

= 8.10 mol H

2.70 mol

= 1.5

= 3

empirical formula: C3H6O2

43.20 g O 1 mol

16.00 g O = 2.70 mol O

2.70 mol

= 1 x 2 = 2

x 2 = 6

x 2 = 3

b. Find the empirical formula for a compound containing 40.0% carbon,

6.70% hydrogen, and 53.3% oxygen.

40.0 g C 1 mol

12.01 g C

6.70 g H 1 mol

1.008 g H

= 3.33 mol C

3.33 mol

= 6.65 mol H

3.33 mol

= 1

= 2

empirical formula: CH2O

53.3 g O 1 mol

16.00 g O = 3.33 mol O

3.33 mol

= 1

Homework

• p. 331 #42-45

• p. 333 #46-50

• Do Pre-Lab on a separate sheet of paper!

• These are the reactions that will occur in lab:

• KCl (aq) + AgNO3 (aq) → KNO3 (aq) + AgCl (s)

• NaCl (aq) + AgNO3 (aq) → NaNO3 (aq) + AgCl (s)

• BaCl2 (aq) + AgNO3 (aq) → Ba(NO3) 2 (aq) + AgCl (s)

• AlCl3 (aq) + AgNO3 (aq) → Al(NO3)3 (aq) + AgCl (s)

• In every reaction, AgCl will precipitate

• Dichlorofluorescein turns pink when all of the chloride has reacted – but we will use sodium chromate, which turns a gross pink/red/orange color (looks like nasty tomato soup)

• The # of drops that it takes for all of the chloride to react will help determine the cation to anion ratio

• Theoretically, the ratio for KCl will be 1:1

• The same volume of BaCl2 should take twice as many drops for all of the chloride to react since there is twice as many chloride ions – that would give it a cation:anion ratio of 1:2. (this will help you with Hypothesis #2)

• Hypothesis #1 is asking you about how many drops of AgNO3 you think it will take for the chloride to completely react. Which solution do you think will need the most AgNO3? The least? Say that in your hypothesis.

• Ms. Sparks’ numbers, using 1 mL of each solution & 3-4

drops indicator:

• What numbers should have been:

Sample T1 Drops T2 Drops Ave. Drops Ratio

KCl 29 28 28.5 1:1

NaCl 38 33 35.5 1:1.25 (4:6)

BaCl2 55 47 51 1:1.8 (5:9)

AlCl3 57 64 60.5 1:2.1 (10:21)

Sample T1 Drops T2 Drops Ave. Drops Ratio

KCl 21 19 20 1:1

NaCl 19 21 20 1:1

BaCl2 38 40 39 1:2

AlCl3 59 61 60 1:3

• Possible causes for error:

• Every group should list the indicator as a cause for error – it

worked, but inconsistently, possibly because the salt that I made it

from was very old

• Type of pipette used – could affect drop sized & number of drops

needle-tip pipette transfer pipette

• Did you measure exactly 1 mL of each chloride solution?

• Did the same person do the dropping every time?

• Any possible contamination?

• Was the pipette held vertically above the test tube? This could

affect size of drops

C. Molecular Formula

1. The empirical formula is not always the same as the molecular

formula! Compounds with the same empirical formula but

different molecular formulas can have very different properties

2. Definition: actual number of atoms of each element in one

molecule or formula unit of a substance

3. Molecular formula can be determined using a compound’s

empirical formula & molar mass

4. Steps to find molecular formula:

a. Find mass of empirical formula (use values for molar mass of each

element)

b. Find the number of times the empirical formula mass will divide into

the molar mass & round to nearest whole number

c. Multiply the subscripts in the empirical formula by the number found

above

d. Write molecular formula!

5. Examples:

a. Find the molecular formula for a compound with the empirical formula

CH2O and a molar mass of 180.1 g/mol.

• Mass of empirical formula:

1 C + 2 H + 1 O = 12.01 + (2 x 1.008) + 16.00 = 30.03 g

• # of times empirical mass goes into molar mass:

180.1 g / 30.03 g = 5.997 = 6

• Multiply subscripts in the empirical formula by number found above

1 C x 6 = 6

2 H x 6 = 12

1 O x 6 = 6

• Molecular formula: C6H12O6

b. Succinic acid is 40.68% carbon, 5.08% hydrogen, and 54.24%

oxygen with a molar mass of 118.1 g/mol. What are the empirical and

molecular formulas?

• Find empirical formula

40.68 g C 1 mol

12.01 g C

5.08 g H 1 mol

1.008 g H

/ 3.387 = 1

/ 3.387 = 1.5

empirical formula: C2H3O2

54.24 g O 1 mol

16.00 g O = 3.390 mol O / 3.387 = 1

= 5.040 mol H

= 3.387 mol C x 2 = 2

x 2 = 3

x 2 = 2

• Mass of empirical formula:

2 C + 3 H + 2 O = (2 x 12.01) + (3 x 1.008) + (2 x 16.00) = 59.04 g

• # of times empirical mass goes into molar mass:

118.1 g / 59.04 g = 2.0003 = 2

• Multiply subscripts in the empirical formula by number found above

2 C x 2 = 4

3 H x 2 = 6

2 O x 2 = 4

• Molecular formula: C4H6O4

c. Determine the empirical & molecular formulas for a compound that is

26.7% P, 12.1% N, and 61.2% Cl with a molar mass of 695.3 g/mol.

• Find empirical formula

• Mass of empirical formula

• # of times empirical mass goes into molar mass

• Multiply subscripts in the empirical formula by number found above

• Molecular formula

Homework

• p. 337 #58-62

• Worksheet: side 1

V. Hydrates

A. Definition: solid crystals that have a specific number of

water molecules bound to its atoms 1. Anhydrous/anhydrate: substance remaining when the water is removed

from a hydrate, usually through heating

2. Anhydrous forms are often used as drying agents because they can

absorb water easily

3. Some are used to store solar energy

B. Naming Hydrates

1. Formulas of hydrates contain the formula of the crystal followed

by a dot and the number of water molecules contained for each

formula unit

a. Examples:

Na2CO3 ᐧ 10H2O contains 10 water molecules per formula unit

NiSO3 ᐧ 6H2O contains 6

2. To name hydrates:

a. Name compound

b. Choose appropriate prefix

for the number of water molecules

c. Combine prefix with the word “-hydrate”

d. Ex: CaSO3 ᐧ 2H2O

calcium sulfite dihydrate

Prefix # Prefix #

mono 1 hexa 6

di 2 hepta 7

tri 3 octa 8

tetra 4 nona 9

penta 5 deca 10

C. Finding Formulas of Hydrates

1. In a lab environment, a hydrated compound will be massed,

heated to drive off the water, then massed again to determine the

mass of the water.

a. Usually both of these masses will be given

b. If you are given percent composition instead of the masses, assume

the sample is 100 g

2. Convert grams of each component (compound & water) to moles

3. Find the ratio of compound to water by dividing both amounts of

moles by the number of moles of the compound (1:n where n is

the # of water molecules per formula unit)

4. Write the formula of the hydrate with n as the coefficient on H2O!

5. Examples:

a. A 10.4 g sample of hydrated barium iodide (BaI2) is heated strongly to

drive off the water. The dry sample has a mass of 9.52 g. Find the

formula of the hydrate.

• Find mass of water: 10.4 g – 9.52 g = 0.88 g H2O

• Convert masses to moles:

= 0.02434 mol BaI2

= 0.04883 mol H2O

• Divide both #s by the # of moles of compound – 0.02434 in this case

• Ratio is 1:2, so 2 becomes the coefficient on H2O

• Formula: BaI2 ᐧ 2H2O

• Name: barium iodide dihydrate

9.52 g BaI2 1 mol

391.1 g

0.88 g H2O 1 mol

18.02 g

/ 0.02434 = 1

/ 0.02434 = 2

b. Find the formula of a hydrate with the following percentages by mass: 23%

zinc; 11% sulfur; 22% oxygen; 44% water.

• Same setup as finding empirical formula; do NOT break up water into H & O

23.0 g Zn 1 mol

65.39 g Zn

11.0 g S 1 mol

32.06 g S

= 0.3517 / 0.3431 = 1

/ 0.3431 = 1

22.0 g O 1 mol

16.00 g O / 0.3431 = 4

= 0.3431

= 1.375

44.0 g H2O 1 mol

18.02 g H2O = 2.442 / 0.3431 = 7

formula: ZnSO4 ᐧ 7H2O name: zinc sulfate heptahydrate

Homework

• p. 340 #63-64

• p. 341 #65-69

• Composition of Hydrates Worksheet (back of the one

about molecular formulas)

Review

• p. 345: Vocabulary (all) -- you already did this!

• p. 346: Concept Mapping #70 -- DRAW IT!

• p. 346: Mastering Concepts #71-88

• p. 346: Mastering Problems #89-156

• HONORS ONLY p. 349: Thinking Critically #170-173

• p. 351: Standardized Test Practice #1-10