CHAPTER 10 THE MOLE
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