 # Unit 08 – Moles and Stoichiometry I. Molar Conversions.

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Unit 08 – Moles and Stoichiometry I. Molar Conversions

A. What is the Mole? b A counting number (like a dozen) b Avogadro’s number (6.02  10 23 particles) (SI unit) b 1 mol = molar mass A large amount!!!!

b 1 mole of hockey pucks would equal the mass of the moon! A. What is the Mole? b 1 mole of pennies would cover the Earth 1/4 mile deep! b 1 mole of basketballs would fill a bag the size of the earth!

B. Molar Mass o Molar Mass- the mass of a mole of any element or compound (in grams) o Round to 2 decimal places o Also called: o Gram Formula mass – sum of the atomic masses of all the atoms in a formula of a compound o Formula weight

B. Molar Mass Examples b water b sodium chloride H 2 O (1.01g x 2) + 16.00g = 18.02 g NaCl 22.99g + 35.45g = 58.44 g

B. Molar Mass Examples b sodium hydrogen carbonate b sucrose NaHCO 3 22.99g + 1.01g + 12.01g + (16.00g x 3) = 84.01 g C 12 H 22 O 11 (12.01g x12) + (1.01g x 22) + (16.00g x11)= 342.34 g

C. Number of Particles in a Mole What is a representative particle? How the substance normally exists: 1 mole = 6.02 × 10 23 representative particles (also called Avogadro’s Number) 1.Atom- rep. particle for most elements 2.Ions – if atom is charged 3.Molecule- rep. particle for covalent compounds and diatomic molecules “BrINCl HOF” 4.Formula unit- rep. particle for ionic compounds

D. Volume of a Mole of Gas b The Volume of a gas varies with a change in temperature or pressure. b Measured at standard temperature and pressure (STP)  0°C at 1 atmosphere (atm)  1 mole of any gas occupies a volume of 22.4L

The Mole Road Map Atoms (ions) Molecule Formula unit

E. Molar Conversion Examples b How many moles of carbon are in 26 g of carbon? 26 g C 1 mol C 12.01 g C = 2.2 mol C

E. Molar Conversion Examples b How many molecules are in 2.50 moles of C 12 H 22 O 11 ? 2.50 mol 6.02  10 23 molecules 1 mol = 1.51  10 24 molecules C 12 H 22 O 11

E. Molar Conversion Examples b Find the number of molecules of 12.00 L of O 2 gas at STP. 12.00 L 1 mol 22.4 L = 3.225 x 10 23 molecules 6.02 x 10 23 molecules 1 mol

II. Stoichiometric Calculations

A. Proportional Relationships b I have 5 eggs. How many cookies can I make? 3/4 c. brown sugar 1 tsp vanilla extract 2 eggs 2 c. chocolate chips Makes 5 dozen cookies. 2 1/4 c. flour 1 tsp. baking soda 1 tsp. salt 1 c. butter 3/4 c. sugar 5 eggs5 doz. 2 eggs = 12.5 dozen cookies Ratio of eggs to cookies

A. Proportional Relationships b Stoichiometry mass relationships between substances in a chemical reaction based on the mole ratio b Mole Ratio indicated by coefficients in a balanced equation 2 Mg + O 2  2 MgO

B. Stoichiometry Steps 1. Write a balanced equation. 2. Identify known & unknown. 3. Convert known to moles (IF NECESSARY) Line up conversion factors. 4. Use Mole ratio – from equation 5. Convert moles to unknown unit (IF NECESSARY) 6. Calculate and write units.

C. Stoichiometry Problems b How many moles of KClO 3 must decompose in order to produce 9 moles of oxygen gas? 9 mol O 2 2 mol KClO 3 3 mol O 2 = 6 mol KClO 3 2KClO 3  2KCl + 3O 2 ? mol9 mol

b How many grams of KClO 3 are required to produce 9.00 L of O 2 at STP? 9.00 L O 2 1 mol O 2 22.4 L O 2 = 32.8 g KClO 3 2 mol KClO 3 3 mol O 2 122.55 g KClO 3 1 mol KClO 3 ? g9.00 L C. Stoichiometry Problems 2KClO 3  2KCl + 3O 2

C. Stoichiometry Problems b How many grams of silver will be formed from 12.0 g copper? 12.0 g Cu 1 mol Cu 63.55 g Cu = 40.7 g Ag Cu + 2AgNO 3  2Ag + Cu(NO 3 ) 2 2 mol Ag 1 mol Cu 107.87 g Ag 1 mol Ag 12.0 g? g

b Calculate the number of grams of NH 3 produced by the reaction of 5.40 g of hydrogen with an excess of nitrogen. N 2 + 3H 2 → 2NH 3 C. Stoichiometry Problems

b Acetylen gas (C 2 H 2 ) is produced by adding water to calcium carbide (CaC 2 ). CaC 2 + 2H 2 O  C 2 H 2 + Ca(OH) 2 How many grams of acetylene are produced by adding water to 5.00 g CaC 2 ? Using the same equation, determine how many moles of CaC 2 are needed to react completely with 49.0 g H 2 O.

III. Stoichiometry in the Real World

A. Limiting Reactants b Available Ingredients 4 slices of bread 1 jar of peanut butter 1/2 jar of jelly b Limiting Reactant bread b Excess Reactants peanut butter and jelly

A. Limiting Reactants b Limiting Reactant used up in a reaction determines the amount of product b Excess Reactant added to ensure that the other reactant is completely used up cheaper & easier to recycle

A. Limiting Reactants 1. Write a balanced equation. 2. For each reactant, calculate the amount of product formed. 3. Smaller answer indicates: limiting reactant amount of product

A. Limiting Reactants b How many moles of ammonia (NH 3 ) can be produced from the reaction of 28.2 L of nitrogen and 25.3 L of hydrogen? 28.2 L? mol 25.3 L N 2 + 3H 2  2NH 3 Using the following equation identify the limiting reagent.

A. Limiting Reactants 28.2 L N 2 1 mol N 2 22.4 L N 2 = 2.5 mol NH 3 2 mol NH 3 1 mol N 2 28.2 L? mol 25.3 L N 2 + 3H 2  2NH 3

A. Limiting Reactants 25.3 L H 2 1 mol H 2 22.4 L H 2 = 0.753 mol NH 3 2 mol NH 3 3 mol H 2 28.2 L? mol 25.3 L N 2 + 3H 2  2NH 3

A. Limiting Reactants N 2 : 2.5 mol NH 3 H 2 : 0.753 mol NH 3 Limiting reactant: H 2 Excess reactant: N 2 Product Formed: 0.753 mol NH 3

Limiting Reactants How many grams of magnesium chloride are produced from the reaction of 2.08 mol of Mg and 2.08 mol of HCl? Mg + 2HCl → MgCl 2 + H 2

B. Percent Yield calculated on paper measured in lab

B. Percent Yield b When 45.8 g of K 2 CO 3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2 45.8 g? g actual: 46.3 g

B. Percent Yield 45.8 g K 2 CO 3 1 mol K 2 CO 3 138.21 g K 2 CO 3 = 49.4 g KCl 2 mol KCl 1 mol K 2 CO 3 74.55 g KCl 1 mol KCl K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2 45.8 g? g actual: 46.3 g Theoretical Yield:

B. Percent Yield Theoretical Yield = 49.4 g KCl % Yield = 46.3 g 49.4 g  100 = 93.7% K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2 45.8 g49.4 g actual: 46.3 g