Unit 4 – Conservation of Mass and Stoichiometry Cartoon courtesy of NearingZero.net Unit 4 – Conservation of Mass and Stoichiometry

Stoichiometry “In solving a problem of this sort, the grand thing is to be able to reason backward. This is a very useful accomplishment, and a very easy one, but people do not practice it much.” Sherlock Holmes, in Sir Arthur Conan Doyle’s A Study in Scarlet Reaction Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions.

Review: Atomic Masses Carbon = 98.89% 12C 1.11% 13C <0.01% 14C Elements occur in nature as mixtures of isotopes Carbon = 98.89% 12C 1.11% 13C <0.01% 14C Carbon’s atomic mass = 12.01 amu

Review: The Mole The number equal to the number of carbon atoms in exactly 12 grams of pure 12C. 1 mole of anything = 6.022 ´ 1023 units of that thing

Review: Molar Mass A substance’s molar mass (molecular weight) is the mass in grams of one mole of the compound. CO2 = 44.01 grams per mole H2O = 18.02 grams per mole Ca(OH)2 = 74.10 grams per mole

Standard Molar Volume 1 mole of a gas occupies 22.4 liters of volume Equal volumes of all gases at the same temperature and pressure contain the same number of molecules. - Amedeo Avogadro At STP (Standard Temperature and Pressure): 1 mole of a gas occupies 22.4 liters of volume

The Mole

Review: Chemical Equations Chemical change involves a reorganization of the atoms in one or more substances. C2H5OH + 3O2 ® 2CO2 + 3H2O reactants products When the equation is balanced it has quantitative significance: 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water

Using Compound Masses Using Chemical Equations: Compound A to Compound B conversions Using Compound Masses Compound A Compound B

Mole Relations

Calculating Masses of Reactants and Products Balance the equation. Convert to moles. Set up mole ratios. Use mole ratios to calculate moles of desired substituent. Convert moles to desired unit.

Working a Stoichiometry Problem 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed. 1. Identify reactants and products and write the balanced equation. 4 Al + 3 O2 2 Al2O3 a. Every reaction needs a yield sign! b. What are the reactants? c. What are the products? d. What are the balanced coefficients?

Working a Stoichiometry Problem 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed? 4 Al + 3 O2  2Al2O3 6.50 g Al 1 mol Al 2 mol Al2O3 101.96 g Al2O3 = ? g Al2O3 26.98 g Al 4 mol Al 1 mol Al2O3 6.50 x 2 x 101.96 ÷ 26.98 ÷ 4 = 12.3 g Al2O3

Gas Stoichiometry #1 How many liters of ammonia can be produced when 12 liters of hydrogen react with an excess of nitrogen? 3 H2(g) + N2(g)  2NH3(g) 1 mole H2 22.4 L NH3 12 L H2 2 mol NH3 = L NH3 8.0 3 mole H2 1 mol NH3 22.4 L H2

Gas Stoichiometry #2 How many liters of oxygen gas, at STP, can be collected from the complete decomposition of 50.0 grams of potassium chlorate? 2 KClO3(s)  2 KCl(s) + 3 O2(g) 50.0 g KClO3 1 mol KClO3 3 mol O2 22.4 L O2 122.55 g KClO3 2 mol KClO3 1 mol O2 = L O2 13.7

Gas Stoichiometry #3 How many liters of oxygen gas, at 37.0C and 0.930 atmospheres, can be collected from the complete decomposition of 50.0 grams of potassium chlorate? 2 KClO3(s)  2 KCl(s) + 3 O2(g) 50.0 g KClO3 1 mol KClO3 3 mol O2 = “n” mol O2 0.612 mol O2 122.55 g KClO3 2 mol KClO3 = 16.7 L

Limiting Reactant The limiting reactant is the reactant that is consumed first, limiting the amounts of products formed.

Limiting Reagents - Combustion

Limiting Reagent Limiting reactant Limits or determines the amount of product that can be formed in a reaction. is the reagent (reactant) that you run out of first. When you run out then the reaction stops.

Excess Reagent More than enough to react with limiting reagent. You have extra Left overs

Cake recipe 3 cups flour 2 cups sugar 1 T baking soda 1.5 cups milk .5 cup butter 1.5 tsp vanilla 3 eggs

Recipe  I have  can make 6 cups flour 6 cups sugar 5 T baking soda 8 cups milk 4 cups butter 6 tsp vanilla 3 eggs 2 3 5 5+ 8 4 1 3 cups flour 2 cups sugar 1 T baking soda 1.5 cups milk .5 cup butter 1.5 tsp vanilla 3 eggs

Which do I run out of first? Eggs are my limiting reagent because I only have enough eggs to make one cake. I run out of eggs first. What is the maximum amount I can produce? ONE

Limiting Reagent Calculation s Convert each reagent to the SAME product. (It does not matter which product you use. However, if the question asks about a specific product, convert to that product.) The reagent/reactant that produces the LEAST amount of the product is the limiting reagent.

Limiting Reagent Example N2 + 3 H2  2 NH3 How many grams of NH3 can be produced from the reaction of 28g of N2 and 25g of H2?

Limiting Reagent K: 28 g N2 ; 25 g H2 P: gN2 mol N2  mol NH3  g NH3 N2 + 3 H2  2 NH3 U: limiting reagent  will either be N2 or H2 K: 28 g N2 ; 25 g H2 P: gN2 mol N2  mol NH3  g NH3 gH2 mol H2  mol NH3  g NH3 reagent that produced the least NH3 is limiting reagent LEAST amount of NH3 is the amount produced

Limiting Reagent = 34 g NH3 = 140 g NH3 28 g N2 2 g H2 N2 + 3 H2  2 NH3 S: 28g N2 1 mole N2 2 mole NH3 17 g NH3 = 34 g NH3 1 mole N2 28 g N2 1 mole NH3 25 g H2 1 mole H2 2 mole NH3 17 g NH3 = 140 g NH3 3 mole H2 2 g H2 1 mole NH3

Nitrogen is the limiting reagent because it N2 + 3 H2  2 NH3 S: 28g N2 1 mole N2 2 mole NH3 17 g NH3 = 34 g NH3 1 mole N2 28 g N2 1 mole NH3 25 g H2 1 mole H2 2 mole NH3 17 g NH3 = 140 g NH3 3 mole H2 2 g H2 1 mole NH3 Nitrogen is the limiting reagent because it produces the LEAST NH3

Excess Reagent K: 28 g N2 limiting; 25 g H2 N2 + 3 H2  2 NH3 U: excess reagent  how much H2 is left over K: 28 g N2 limiting; 25 g H2 P: gN2mol N2  mol H2  g H2 subtract the amount used based on limiting reagent from the amount of H2 that you have.

Excess Reagent - = = 6.0 g H2 used 28 g N2 N2 + 3 H2  2 NH3 S: 28g N2 1 mole N2 3 mole H2 2 g H2 = 6.0 g H2 used 1 mole N2 28 g N2 1 mole H2 19 g H2 left over 25 g H2 Have - 6.0 g H2 used =

Limiting Reagent Stop and work practice problems.

Solving a Limiting Reagent Stoichiometry Problem Balance the equation. Convert each reagent to the SAME product. (It does not matter which product you use. However, if the question asks about a specific product, convert to that product.) The reagent/reactant that produces the LEAST amount of the product is the limiting reagent

% yield = actual yield * 100% Percent yield Amount actually recovered/produced compared to what should have been recovered/produced. Tells what percent of the predicted amount you are able to produce. % yield = actual yield * 100% Theoretical yield

Percent yield U: % yield K: theoretical yield = 21.8g actual yield = 13.9 g P: % yield = actual yield * 100% Theoretical yield S: % yield = 13.9 g * 100% 21.8g % yield = 63.8%

Percent Yield Stop and work practice problems.