CHAPTER 10

THE MOLE

DIMENSIONAL ANALYSIS

Also known as factor label method

Problem solving method that focuses on the UNITS that are used to describe matter

Unit labels are treated as FACTORS that can be divided out

Uses conversion factors– Example: 12 inches = 1 foot

Problem:– 5 ft - _____________ inches

5 ft 12 in

1 ft

5 X 12

160 in

10 mm = 1 cm

100 cm = 1 m

14 m = ___________ mm

14 m

1 m

100 cm

1 cm

10 mmmm

14 x 100 x 10 1

14,000

1000 L= 1 kL

1000 mL = 1 L

4 kL = ____________ mL

4kL 1000 L 1000 mL1 kL 1 L

4,000,000 mL

6.02 x 10 23 bananas = 1 mole

4 moles bananas = ___________ bananas

4 mole bananas

1 mole banana

6.02 x 1023 bananas

4 x (6.02 x 1023)1

2 x 1024 bananas

AVOGADRO’S CONSTANT

The number of particles in a moleEqual to 6.02 x 1023

Mole = a unit for the chemical quantity of a substance– A term that represents a certain # of particles– Need a large number because atoms and

molecules are so small– 1 mole = 6.02 x 1023 of anything (particles,

atoms, molecules etc.)

MASS AND THE MOLE

Atomic mass = mass of one atom of a substance

Measure in atomic mass unit (amu)

1 hydrogen atom = 1.0079 amu

MOLAR MASS

Mass in grams of a mole of any pure substance

Use the atomic mass (off the periodic table); change unit to grams/mole

1 atom hydrogen = 1.0079 amu

1 mole hydrogen = 1.0079 grams

M0LAR MASS OF COMPOUNDS AND MOLECULES

Sum of the masses of all of the atoms of a substance

Example: H2O

H 2 x 1.0 = 2.0

O 1 x 16.0 = 16.0

18.0 g/molemasses

KMnO4

– K 1 x 39.1 = 39.1– Mn 1 x 54.9 = 54.9– O 4 x 16.0 = 64.0

158.0 g/mole

DON’T FORGET SIG DIGS!!

Molar Mass of Hydrates (added H2O)

– CoCl2·6H2O• Co 1 x 58.9 = 58.9

• Cl 2 x 35.5 = 71.0

• H2O 6 x 18.0 = 108.0

237.9 g/mole

PERCENT COMPOSITION

A way to determine the formula of newly invented compounds (ones that you don’t know the oxidation numbers of)%composition is a calculation of the % by mass of each element in a compound (part from whole)Use the formula of the compound and the molar mass of each part

K2CrO4

– K 2 x 39.1 = 78.2– Cr 1 x 52.0 = 52.0– O 4 x 16.0 = 64.0

molar mass = 194.2 g /mole% K = 78.2 / 194.2 x 100 = 40.3 %%Cr = 52.0 / 194.2 x 100 = 26.8%%O = 64.0 / 194.2 x 100= 32.9%

CONVERSION OF THE MOLE

Remember:– 1 mole = 6.02 x 1023 particles, atoms etc– Molar mass = _________ g/mole

• ( g= 1 mole)

Know:

Grams moles particles,etcMolar mass

g/mole

6.02 x 1023

Particles etc./moleTo go from: grams to moles use molar massTo go from: moles to molecules use Avogadro’s #To go from: grams to molecules use both

GRAMS TO MOLES

Calculate the moles of 1.5 g of sodium– Use molar mass sodium = 23.0 g/mole

1.5 g Na

23.0 g Na

1 mole Na.065 moles Na

MOLES TO GRAMS

Calculate the mass in grams of 50.0 moles of KCl– Use molar mass KCl 74.6 g/mole

(K= 39.1; Cl = 35.5)

50.0 mole KCl

1 mole KCl

74.6 g KCl 3730 g KCl

MOLES TO PARTICLES(ATOMS, MOLECULES, FORMULA UNITS)

How many formula units are in 4.9 moles of MgSO4?

– Use: 6.02 x 1023 formula units/mole

4.9 moles MgSO4

1 mole MgSO4

6.02 x 1023 formula units

2.9 x 1024 formula units ofMgSO4

PARTICLES TO MOLES

How many moles are in 2.3 x 1025 molecules of CO2?

– Use: 6.02 x 1023 molecules/mole

2.3 x 1025 molecules CO2

6.02 x 1023 molecules CO2

1 mole CO2

= 38 moles CO2

GRAMS TO PARTICLES

How many atoms are in 14.0 g of barium?– Use: molar mass Ba = 137.3 g/mole– Use: 6.02 x 1023 atoms/mole

14.0 g Ba 1 mole Ba

137.3 g Ba

6.02 x 1023 atoms Ba

1 mole Ba

= 6.14 x 1022 atoms Ba

PARTICLES TO GRAMS

How many grams are in 4.5 x 1024

molecules of H2O?

– Use: 6.02 x 1023 molecules/mole

– Use: molar mass H2O 18.0 g/mole

4.5 x 1024 molc. H2O

6.02 x 1023 molecules H2O

1 mole H2O 18.0 g H2O

1 mole H2O

= 134.55 or 130 g H2O

How many grams are in 3.29 x 1024 molecules of chlorine gas?– Use: 6.02 x 1023 molecules/mole– Use: molar mass of chlorine 71.0 g/mole

3.29 x 1024 molecule Cl2 1 mole Cl2

6.02 x 1023 moleculeCl2

71.0 g Cl2

1 mole Cl2

= 388 g Cl2

EMPIRICAL FORMULAS

Definition: the simplest whole # ratio for a formula

You can calculate this if you know the % composition of each element of a newly made compound

STEPS FOR WRITING EMPIRICAL FORMULAS

1. Given the grams or % composition (cross off the % and change unit to grams)

2. Convert to moles for each element (use molar mass)

3.Find ratio of moles compared to other elements (divide by smallest # of moles)

4.From ratios, write the empirical formula

Example:

What is the empirical formula for a compound that contains .900 g of Ca and 1.60 g of Cl?– 1. Determine moles from molar mass for each

element.900g Ca 1 mole Ca

40.1 g Ca = .0224 mole Ca

1.60 g Cl 1 mole Cl 35.5 g Cl = .0451 mole Cl

– 2. Find simplest ratio (take smallest # of moles found in step 1 and divide each by that)

• Ca = .0224 .0224 = 1• Cl = .0451 .0224 = 2

– 3. Write empirical formula using ratios as subscripts

CaCl2

FRACTIONS: if the ratios divide out to be fractions, do the following:– 1.5:1 ratio difference is .5 multiply both by 2– If the difference is .33 or .67 multiply both by 3– If the difference is .25 or .75 multiply both by 4

What is the empirical formula of a compound that is 66.0 % Ca and 34.0 % P?

40.1 g Ca

66.0g Ca 1 mole Ca 1.65 mole

34.0 g P

31.0 g P

1 mole P 1.10 mole

Ca 1.65/1.10 = 1.5

P 1.10/1.10 = 1

Since there is a fraction, multiply both by 2

Ca 1.5 x 2 = 3

P 1 x 2 = 2

The formula is then Ca3P2 a 3:2 ratio

MOLECULAR FORMULA

Shows the actual # of atoms of each element– Example: Empirical = HO

Molecular = H2O2

To solve, you need one more piece of information the molar mass of the molecular formulaUse that to find the whole # multiplier

The empirical formula of a compound is found to be CH2O. The molar mass of the molecular formula is 120.1 g/mole. What is the molecular formula of the compound.

Example:

Calculate the empirical formula and molecular formula for a compound that is 56.4% P and 43.7% O. The molar mass of the molecular formula is 220 g/mole.

56.4 g P 1 mole

31.0 g P= 1.82 moles

43.7 g O 1 mole

16.0 g 0= 2.73 moles

Since 1.82 is the smallest amount of moles, we divide both by that– P 1.82 / 1.82 = 1– O 2.73 / 1.82 = 1.5

Since we have a fraction, multiply both by 2– P 1 x 2 = 2– O 1.5 x 2 = 3– The empirical formula is then P2O3

To find the molecular formula:– Get molar mass of Empirical:

• P2O3 62 + 48.0 = 110g/mole

– Divide that into molar mass of molecular which is given in the problem

• 220/110 = 2 which equal the whole # multiplier

Empirical = P2O3

Molecular = P4O6

HYDRATES

Crystals that contain water

Forms when crystals form in water solution (water molecules stick to the crystal in a specific ratio)

Ex: CuSO4 · 5H2OCuSO4

H2OH2O

H2OH2O

H2O

Copper (II) sulfatepentahydrate

Hydrate problems

Write the formula for the hydrate with 89.2% BaBr2 and 10.8% H2O.89.2 g BaBr2 1 mole BaBr2

297.0 g BaBr2

.300 .300

.3001

10.8 g H2O 1 mole H2O

18.0 g H2O.600

.600

.3002

BaBr2 · 2 H20 Barium bromide dihydrate

ANHYDROUS

Without water

Determine the hydrate formula for:

.391 g of Li2SiF6 and .0903 g H20

Determine the hydrate formula for:

76.9% CaSO3 and 23.1% H20

Ch. 10 Test15 multiple choice: definitions, conversions

9 calculations– Molar mass

– Percent composition

– Grams<-> moles, moles<-> particles, grams<-> particles

– Empirical and molecular formulas

– Hydrate formulas

Periodic table and conversion “cheat” formula will be provided

I am a Mole

- Videos, more practice and tutorials on webpage – Chapter Practice section

Popcorn lab (video slow mo) (how its made)

Lab prep

Data (2 brands- get another groups info)

Calculations (show work for both groups)

Questions 1-6 (use articles

Conclusion (full- what learned, error, application)

Calculations (show work-both brands)

1. Mass of unpopped kernels= Data #2 - Data #1

2. Mass of popped kernels = Data #4 - Data #1

3. Difference in volume = Data #5 - Data#3

4. Mass of water in popcorn= Calc. #1 – Calc #2

5. % composition of water= Calc #4 / Calc#1 x 100