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 1 60 in

10 mm = 1 cm 100 cm = 1 m 14 m = ___________ mm 14 m 100 cm 10 mm 14,000 mm 1 m 1 cm 14 x 100 x 10 1

1000 L= 1 kL 1000 mL = 1 L 4 kL = ____________ mL 4kL 1000 L 1000 mL

4 moles bananas = ___________ bananas 6.02 x 10 23 bananas = 1 mole 4 moles bananas = ___________ bananas 6.02 x 1023 bananas 4 mole bananas 1 mole banana 4 x (6.02 x 1023) 1 2 x 1024 bananas

AVOGADRO’S CONSTANT The number of particles in a mole Equal 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/mole masses

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: Know: Grams moles particles,etc 1 mole = 6.02 x 1023 particles, atoms etc Molar mass = _________ g/mole ( g= 1 mole) Know: Grams moles particles,etc Molar mass g/mole 6.02 x 1023 Particles etc./mole To go from: grams to moles use molar mass To 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 1 mole Na .065 moles Na 23.0 g 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 74.6 g KCl 3730 g KCl 1 mole 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 of MgSO4

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 1 mole CO2 6.02 x 1023 molecules 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 = 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 = 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 = 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

3. Write empirical formula using ratios as subscripts CaCl2 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 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 formula Use that to find the whole # multiplier

The empirical formula of a compound is found to be CH2O 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 · 5H2O CuSO4 H2O Copper (II) sulfate pentahydrate

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 1 .300 10.8 g H2O 1 mole H2O 18.0 g H2O .600 .300 2 .600 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 Test 15 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) Calculations (show work-both brands) Mass of unpopped kernels= Data #2 - Data #1 Mass of popped kernels = Data #4 - Data #1 Difference in volume = Data #5 - Data#3 Mass of water in popcorn= Calc. #1 – Calc #2 % composition of water= Calc #4 / Calc#1 x 100 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)