mass reactants= mass products Stoichiometry This word sounds scary, but it’s really just a big word for the relationships between reactants and products in a chemical reaction. -based on law of conservation of mass mass reactants= mass products

Proportional Relationships 2 1/4 c. flour 1 tsp. baking soda 1 tsp. salt 1 c. butter 3/4 c. sugar 3/4 c. brown sugar 1 tsp vanilla extract 2 eggs 2 c. chocolate chips Makes 5 dozen cookies. I have 5 eggs. Assuming I have plenty of the other ingredients, how many cookies can I make? Ratio of eggs to cookies 5 dozen cookies 2 eggs 5 eggs = 12.5 dozen cookies

Proportional Relationships Stoichiometry mass relationships between substances in a chemical reaction based on the mole ratio Mole Ratio indicated by coefficients in a balanced equation 2 Mg + O2  2 MgO

Mole Ratios 2 Mg + O2  2 MgO Relationships between coeffecients are used as conversion factors in stoichiometry! A mole ratio is a ratio between the numbers of moles of any 2 substances.

2 Mg + O2  2 MgO Mole Ratios Total of 6 mole ratios! What are the mole ratios in this equation? 2 molMg 2 molMg 2 mol MgO 1 mol O2 1 mol O2 1 mol O2 2 mol Mg 2 mol MgO 2 mol MgO 2 mol MgO 2 mol Mg 1 mol O2 Total of 6 mole ratios!

2KClO3→2KCl + 3O2 Mole Ratios Total of 6 mole ratios! What are the mole ratios in this equation? 2 molKClO3 2 molKClO3 3 mol O2 2 mol KCl 2 mol KCl 2 mol KCl 2 mol KClO3 3 mol O2 3 mol O2 3 mol O2 2 mol KClO3 2 mol KCl Total of 6 mole ratios!

Stoichiometry Steps Core step in all stoichiometry problems!! 1. Write a balanced equation. 2. Identify known & unknown. 3. Line up conversion factors. Mole ratio - moles  moles Molar mass - moles  grams Molarity - moles  liters soln Molar volume - moles  liters gas Mole ratio - moles  moles Core step in all stoichiometry problems!! 4. Check answer.

Mole ratios CO2(g) + 2LiOH(s) → Li2CO3(s) + H2O What is the ratio of CO2 molecules to LiOH molecules? CO2 molecules to Li2CO3 molecules Li2CO3 molecules to LiOH molecules

Molar Conversions MASS IN GRAMS MOLES NUMBER OF PARTICLES Molar Mass (g/mol) 6.02  1023 particles/mol

Chemical Equations and the Mole P4O10 + 6 H2O  4 H3PO4 1 molecule of P4O10 reacts with 6 molecules of H2O to give 4 molecules of H3PO4 1 mole of P4O10 reacts with 6 moles of H2O to give 4 moles of H3PO4

Mole to Mole conversions -use mole ratios from last notes Formula used: Moles of known Moles of unknown Moles of known = moles of unknown

Mole to Mole conversions Steps in Solving Determine moles given in problem Find mole ratio showing unknown on top and known on bottom Fill into formula above and calculate moles of unknown

Stoichiometry Problems How many moles of KClO3 must decompose in order to produce 9 moles of oxygen gas? 2KClO3  2KCl + 3O2 ? mol 9 mol 9 mol O2 2 mol KClO3 3 mol O2 = 6 mol KClO3

Stoichiometry and Chemical Equations The coefficients in a chemical equation not only describe the number of atoms/molecules being reacted/produced, they also describe the number of moles of atoms/molecules being reacted/produced. N2(g) + 3H2(g) 2NH3(g) Suppose 5.46 mol N2 completely react in this fashion: 5.46 mol N2 3 mol H2 1 mol N2 = 16.4 mol H2 are needed to react 5.46 mol N2 2 mol NH3 1 mol N2 = 10.9 mol NH3 must be produced

Stoichiometry Practice Bismuth(III) chloride will react with hydrogen sulfide to form bismuth(III) sulfide and hydrochloric acid. Write the balanced equation for this reaction, then calculate how many moles of acid would be formed if 15.0 mol of hydrogen sulfide react. 2 BiCl3 + 3 H2S Bi2S3 + 6 HCl 15.0 mol H2S 6 mol HCl 3 mol H2S = 30.0 mol HCl

Mole to Mass conversions -use if you know # moles of one thing and you want to know mass of another thing Steps in solving Convert moles of known to moles of unknown. Convert moles of unknown to grams of unknown

Stoichiometry Practice Sodium nitride can be formed by reacting sodium metal with nitrogen gas. Write the balanced equation for this reaction, then calculate how many moles of sodium nitride can be produced from 25.0 g of sodium. 6 Na + N2 2 Na3N 25.0 g Na 1 mol Na 2 mol Na3N 22.99 g Na 6 mol Na = 0.362 mol Na3N

Stoichiometry Practice When magnesium burns in air, it combines with oxygen to form magnesium oxide according to the following equation: What mass (in grams) of magnesium oxide is produced from 2.00 mol of magnesium? 2Mg(s) + O2(g) 2MgO(s) 2.0 mol Mg 2 mol MgO 40.3g MgO 2 mol Mg 1 mol MgO = 80.6 g MgO

Mass to Mass conversions Use if you know the mass of 1 thing and want to find mass of another Steps in solving Convert mass of known to moles of known Do mole-mole conversion Convert moles of unknown to mass of unknown

2KClO3 → 2KCl + 3O2 Calculate the mass of O2 produced if 2 2KClO3 → 2KCl + 3O2 Calculate the mass of O2 produced if 2.5 g of KClO3 is completely decomposed when burned 2.5 g KClO3 1 mol KClO3 3 mol O2 32 g O2 122.6g KClO3 2 mol KClO3 1 mol O2 = .98g O2 Mole ratio

NH4NO3 → N2O + 2H2O Determine the mass of water produced from the decomposition of 25.0g of solid ammonium nitrate 25.0 g NH4NO3 1 mol NH4NO3 2 mol H2O 18.02 g H2O 80.04 g NH4NO3 1 mol NH4NO3 1 mol H2O = 11.26g H2O Mole ratio

Summary of Conversions Mole to Mole: 1 step Mass to Mole: 2 steps Mass to Mass: 3 steps What setup do all conversions require? Mole:mole ratio

Limiting Reactants

How many bicycles can be assembled from the parts shown? From eight wheels four bikes can be constructed. From three pedal assemblies three bikes can be constructed. From four frames four bikes can be constructed. The limiting part is the number of pedal assemblies.

Likewise, in a chemical reaction, it is possible for one reactant to be used up before the other(s). A limiting reactant is a substance that is completely used up in a chemical reaction and causes the reaction to stop. Therefore, it is the limiting reactant that also determines how much product can actually be formed.

The bicycle wheels and frames were in excess in our example An excess reactant are the left-over reactants that are in abundance in a reaction. The bicycle wheels and frames were in excess in our example

To determine which reactant is limiting in a given reaction: 1. Convert grams of each reactant to moles Divide both #’s by the smallest # to determine actual mole ratio Determine balanced mole ratio using moles in balanced equation Which reactant isn’t as large as it’s supposed to be? THAT is your limiting reactant.

Limiting Reactant Example S8 + 4Cl2 → 4S2Cl2 If we are given 100 grams of Cl2 and 200 grams of S8, what is the limiting reactant? Step 1: Convert grams into moles 100 g Cl2 1 mol Cl2 = 1.4 mol Cl2 70 g Cl2 200 g S8 1 mol S8 = 0.8 mol S8 256 g S8

For Cl2: 1.4 mol Cl2 For S8: 0.8 mol S8 = 1.75 =1 0.8 mol S8 Step 2: Divide both #’s by the smallest # to determine actual mole ratio For Cl2: 1.4 mol Cl2 For S8: 0.8 mol S8 = 1.75 =1 0.8 mol S8 0.8 mol S8 Divide by smallest #

Step 3: Determine balanced mole ratio using moles in balanced equation S8 + 4Cl2 → 4S2Cl2 Balanced mole ratio Actual mole ratio 1 mol S8 1 mol S8 4 mol Cl2 1.75 mol Cl2

Step 4: Determine limiting reactant based on which reactant isn’t as large as it’s supposed to be Balanced mole ratio Actual mole ratio 1 mol S8 1 mol S8 4 mol Cl2 1.75 mol Cl2 Cl2 is limiting b/c it’s actual mole ratio is smaller that the balanced mole ratio!!!

Calculating the amount of product from a limiting reactant: 1. Determine the limiting reactant Determine the amount of limiting reactant (given in problem) Multiply amount of limiting reactant by the mole ratio of product/limiting rgnt. Convert moles of product into grams (multiply by molar mass)

S8 + 4Cl2 → 4S2Cl2 Step 1: Determine the limiting reactant We know it’s Cl2 already! Step 2: Determine the amount of limiting reactant- given in problem 1.4 mol Cl2

S8 + 4Cl2 → 4S2Cl2 Step 3: Multiply amount of limiting reactant by the mole ratio of product/limiting reactant from balanced equation Step 4: Multiply by molar mass 1.4 mol Cl2 4 mol S2Cl2 135 g S2Cl2 =190.4g S2Cl2 4mol Cl2 1 mol S2Cl2

Calculating amount of excess reactant: 1. Determine the limiting reactant Determine the amount of limiting reactant (given in problem) Multiply amount of limiting reactant by the mole ratio of excess reactant/limiting rgnt. Convert moles of excess reactant into grams (multiply by molar mass) Subtract mass in #4 from original mass

1.4 mol Cl2 S8 + 4Cl2 → 4S2Cl2 Step 1: Determine limiting reactant Step 2: Determine amount (in moles of limiting reactant) 1.4 mol Cl2

S8 + 4Cl2 → 4S2Cl2 Step 3: Multiply amount of limiting reactant by the mole ratio of Excess reactant Limiting reactant Step 4: Multiply by molar mass of excess reactant 1.4 mol Cl2 1 mol S8 256.5g S8 = 90.42 g S8 4mol Cl2 1 mol S8

Step 5: Subtract mass in #4 from original mass given at the beginning of the problem. That is the amount left over 200g S8 - 90.42g S8 = 109.6g S8 in excess

Limiting Reactant Practice Sodium metal will react with gaseous ammonia to produce solid sodium amide, NaNH2. The unbalanced equation for this reaction is… Na + NH3 NaNH2 + H2 If 60.0 g of sodium are mixed with 48.0 g of ammonia, how many grams of sodium amide can be produced? Which substance is the limiting reactant?

Limiting Reactant Practice 2 Na + NH3 2 2 NaNH2 + H2 smallest 60.0 g Na 1 mol Na 2 mol NaNH2 = 2.61 mol NaNH2 22.99 g Na 2 mol Na Na is limiting 48.0 g NH3 1 mol NH3 2 mol NaNH2 = 2.82 mol NaNH2 17.04 g NH3 3 mol NH3

Theoretical, Actual & Percent Yields Theoretical yield: the amount of product you would get if the reaction occurs with complete efficiency. Actual yield: what you “actually” get when you perform the reaction. Percent yield: the actual yield divided by the theoretical yield multiplied by 100. % Yield = (Actual/Theoretical) x 100

Percent Yield measured in lab calculated on paper

When 45. 8 g of K2CO3 react with excess HCl, 46. 3 g of KCl are formed When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g ? g actual: 46.3 g

K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g ? g actual: 46.3 g Theoretical Yield: 45.8 g K2CO3 2 mol KCl 1 mol K2CO3 74.55 g KCl 1 mol KCl 1 mol K2CO3 138.21 g = 49.4 g KCl

K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g 49.4 g actual: 46.3 g Theoretical Yield = 49.4 g KCl 46.3 g 49.4 g  100 = 93.7% % Yield =