The Mole

Diatomic elements – elements that do not exist on their own, they must be in a covalent bond with themselves to be more stable.

N2 O2 F2 Cl2 Br2 I2 H2

Dimensional Analysis Review How many seconds are in 5.0 hours? 5.0 hr 5.0 hr x 60 min x 60 sec = 18000 sec 1 hr 1 min

Dimensional Analysis Review Calculate the number of inches in 26 yards 26 yards 26 yards x 3 ft x 12 inches = 940 inches 1 yd 1 ft

The Mole Chemists need a convenient method for counting accurately the number of atoms, molecules, or formula units in a sample of a substance. The mole, commonly abbreviated mol, is the SI base unit used to measure the amount of a substance.

A mole is used to represent the amount of a substance or compound. H2 means 1 mole of H2 NaCl means 1 mole of NaCl. Hydrates – compounds that contain water. Written as NaCl • 2H2O

The Mole A mole of anything contains 6.02 x 1023 representative particles. A representative particle is any kind of particle such as atoms, molecules, formula units, electrons, or ions. 6.02 x 1023 is called Avogadro’s number

Molecular or Formula Mass Calculations You have to add up the masses of every element found in the compound or molecule. Ex. (NH4)3PO4 Ammonium Phosphate

Make a list of how many of each element you have. P = 1 O = 4

Multiply how many of each element you have by their mass Multiply how many of each element you have by their mass. (Look it up on the periodic table.) Then add up the total. The units always grams.

N = 3 x 14 = 42 H = 12 x 1 = 12 P = 1 x 31 = 31 O = 4 x 16 = 64 149 g

example NaCl Na = 23g Cl = 35 g Total 58 g

Hydrates- are just a molecule or compound that has a certain number of water molecules attached to it. They are written like NaCl • 5H2O.

When you are asked to calculate the mass you have to understand that there are 5 H2O molecules attached to 1 NaCl. That means: 10 H & 5 O in addition to the Na & Cl.

H = 10 x 1 = 10 O = 5 x 16 = 80 Na = 1 x 23 = 23 Cl = 1 x 35 = 35 Total 148 amu or grams

Conversion Factor # 1 1 mole Molecular weight (g) The molecular mass comes from the periodic table!

Mole – Mass Calculations What is the mass of 4.21 moles of iron (III) oxide? Start with your given: 4.21 moles Fe2O3 Draw your line 4.21 moles Fe2O3 x _____________

Mole – Mass Calculations Place conversion factors 4.21 moles Fe2O3 x 159.70 g Fe2O3 1 moles Fe2O3 Cross out units & work the problem 672 g Fe2O3

Another Example Calculate the mass of 1.630 moles of Na 1.630 moles x 22.9g 1 mole 37.47 g Na

Mole – Mass Calculations How many moles of Ca(OH)2 are in 325 grams? 325 g x 1 mole 74 g 4.39 moles Ca(OH)2

Another Example How many moles are in 62.17 g of sodium chloride? 62.17g x 1 mole 58 g 1.064 mol NaCl

Conversion Factor #2

Representative Particles Anything - Representative particles Elements – Atoms Covalent Compounds – Molecules Ionic Compounds – Formula Units Ions - Ions

Mole – Representative Particle Calculations Calculate the number of atoms in 3.50 moles of copper Start with your given 3.50 mol Cu Draw your line 3.50 mol Cu x _________

Mole – Representative Particle Calculations Place the conversion factors 3.50 mol Cu x 6.02 x 1023 atoms 1 mol Cu Work the problem 2.11 x 10 24 atoms Cu

Another Example Calculate the number of molecules in 2.6 moles of H2O 2.6 mol H2O x 6.02 x 1023 mlcs = 1.6 x 1024 1 mol H2O mlcs H2O

Another Example Calculate the number of formula units in 5.23 moles of NaCl 5.23 mole x 6.02 x 1023 f. units 1 mole 3.15 x 10 24 formula units NaCl

Mole – Representative Particle Calculations How many moles of MgO are in 9.72 x 10 23 molecules of MgO? 9.72 x 10 23 mlcs x 1 mole 6.02 x 10 23 mlcs 1.61 moles MgO

Another Example How many moles are in 4.50 x 1024 atoms of Zinc? 4.5 x 1024 atoms x 1 mole 6.02 x 10 23 atoms 7.48 moles Zn

Mass – Particle Conversions How many atoms of gold are in 25.0 g of gold? 25.0 g Au x 1 mole Au x 6.02 x 1023 atoms Au 196.79 g Au 1 mole Au 7.65 x 1022 atoms of Au

Mass – Particle Conversions How many grams of He are in 5.50 x 1022 atoms of He? 5.50 x 10 22 atoms x 1 mole x 4 g 6.02 x 10 23 atoms 1 mole 0.366 g He

example How many formula units are in 35.6 g of NaCl? 3.70 x 10 23 formula units

moles Al2S3 means you have 1 mole of the compound Al2S3. It is composed of 2 moles of Al atoms and 3 moles of S atoms

Ions in a compound How many Cl- ions are there in 35.6 g of AlCl3? 35.6 g AlCl3 x 1 mole AlCl3 x 3 mole Cl3 x 6.02 x 10 23 ions 132 g 1 mole AlCl3 1 mole Cl3 Remember ions is just a word, atoms particles, representative units. Anything works

% Composition, Empirical Formulas, & Molecular Formulas

% composition Mass of the element x 100 Mass of the compound

% composition- gives the relative mass of each atom in a compound. Steps 1. Calculate formula mass of compound 2. Determine % of each atom Ex: Calculate the % comp of N in NH4NO3 N x 2 = 28.0 H x 4 = 4.0 O x 3 = 48.0 80.0 g % N = 28.0 g/ 80.0 g x 100 = 35%

Calculate the % composition of each atom in a. Fe2O3 b. Ag2O c. HgO

% Composition Calculate the % Composition of iron (III) oxide % Fe = 69.94% % O = 30.06%

Molecular formulas- Used to tell exactly how many of each atom is present in the compound. Empirical Formulas- The simplest formulas possible indicating the ratio of atoms found in the molecule. Think of it as reducing a fraction to the least common denominator.

Empirical Formula What is the empirical formula for H2O2? HO What is the empirical formula for C6H12O6? CH2O

C6H6 C2H2 C6H12O6 C4H10 P4O10 SO3 N2O4 NO2 Ag2C4H4O6

Empirical Formulas-calculations The smallest whole number ratio of elements in a compound. It is determined by calculating the: Calculate the moles present of each element in the compound. Set up ratios for each element by dividing each mole amount by the smallest mole amount. Use the ratio from step # 2 as the subscripts to write out the empirical formula.

Ex: Determine the empirical formula for a compound with 36. 5 % Na, 25 Ex: Determine the empirical formula for a compound with 36.5 % Na, 25.4% S, 38.1% O Since each elements % adds up to 100% treat each % as a gram amount

Na = 36.5 g Na x 1 mole Na = 1.59 mole Na S = 25.4 g S x 1 mole S = .791 mole S 32.1 g S O = 38.1 g O x 1 mole O = 2.23 mole O 16.0 g O

Na S O 1.59 = 2.01 .791 = 1 2.23 = 3.01 .791 .791 .791 Ratio = 2:1:3 Na2SO3

Empirical Formula Problem Calculate the empirical formula of a compound containing 40.05 % S and 59.95 % O. 40.05 g S x 59.95 g O x 1 mol S = 32.07 g S 1 mol O = 16.00 g O 1.249 mol 3.747 mol / 1.249 mol = 1 1.249 mol = 3 SO3

Ratio problems Occasionally one part of your ratio will not be a whole number ( 1.5 : 1) It has to be!!!!! Since H1.5O can’t exist you have got to fix it. When this occurs you need to multiply by factor that will make it a whole # (like 2) * what you do to one you must do to all, so every mole amount gets multiplied by 2!!

Empirical Formula Problem Calculate the empirical formula for a compound containing 48.64 g C, 8.16 g H, and 43.20 g O. 48.64g x 1 mole C = 4.04 mole/ 2.7 =1.5 x 2 = 3 12 g 8.16 g x 1 mole H = 8.16 mole/2.7 = 3.02 x 2 = 6 1 g 43.20 x 1 mole O = 2.7 mole/2.7 = 1x 2 = 2 16 g C3H6O2

Steps for Calculating Molecular Formula Calculate the empirical formula Get the molecular mass of the empirical formula that you just determined Divide the experimentally determined molecular mass (given) by the molecular mass of the empirical formula You will get a whole number Multiply everything in the empirical formula by this number

Molecular Formula Problem Calculate the molecular formula of a compound containing 40.68%C, 5.08%H, and 54.25%O with an experimentally determined molecular weight of 118.1 g/mol

Molecular Formula Problem 40.68g C x 1 mol C = 3.387 / 3.387 = (1)2 = 2 12.01 g C 5.08g H x 1 mol H = 5.04 / 3.387 = (1.5)2= 3 1.01 g H 54.25 g Ox 1 mol O = 3.390 / 3.387 = (1)2 = 2 16.00 g O C2H3O2

Molecular Formula Problem C2H3O2 Molecular Mass = 59.04 EDMM / EFMM = 118.1 / 59.04 = 2 Molecular Formula C4H6O4

Molecular Formula Problem Calculate the molecular formula of a compound containing 57.84 g C, 3.64 g H, and 38.52 g O with an experimentally determined molecular mass of 249.21 g/mol C12H9O6

Stoichiometry Stoichiometry is just a long word for changing units in chemistry Just remember to ALWAYS start with your given! If you can do Dimensional Analysis, you can do stoichiometry