2Big IdeasOBJECTIVES:Calculate the amount of reactants required, or product formed, in a chemical reaction.Interpret balanced chemical equations in terms of interacting moles, representative particles, masses, and gas volume at STP.
3Remember The Mole 1 mole = 6.02 X 1023 particles 1 mole = the atomic weight in grams of the element1 mole = the atomic weight in grams of the compound1 mole = 22.4 liters of a gas at STP
4What is a Particle? Element Made of atoms Molecules Molecular compound 2 non-metalsFormula UnitsIonic CompoundsMetal and non-metal
5Let’s Review How many formula units are in 2.75 moles of CaCl2? 1.66x1024 formula unitsHow many total atoms are in 2.75 moles of CaCl2?4.97x1024 atoms
6Cookies?When baking cookies, a recipe is usually used, telling the exact amount of each ingredientIf you need more, you can double or triple the amountThus, a recipe is much like a balanced equation
7Stoichiometry Greek for “measuring elements” The calculations of quantities in chemical reactions using a balanced equationWe can interpret balanced chemical equations several ways
8In terms of Particles 2H2 + O2 ® 2H2O Two molecules of hydrogen and one molecule of oxygen form two molecules of water.Al2S3 ® 2Al + 3S1 formula unit of Al2S3 decomposes to 2 atoms of Al + 3 atoms of S2Na + 2H2O ® 2NaOH + H2
9In terms of MolesThe coefficients tell us how many moles of each substance2 Al2O3 ® 4Al + 3O22 moles of Al2O3 decompose to 4 moles Al + 3 moles O22Na + 2H2O ® 2NaOH + H2
10In terms of MassThe Law of Conservation of Mass appliesWe can check using moles2H2 + O2 ® 2H2O2.02 g H22 moles H2=4.04 g H21 mole H232.00 g O21 mole O2=32.00 g O21 mole O236.04 g H2+O2
11In terms of Mass 2H2 + O2 ® 2H2O 2H2 + O2 ® 2H2O 18.02 g H2O 2 moles H2O=1 mole H2O2H2 + O2 ® 2H2O36.04 g H2 + O2=36.04 g H2O
12Practice:Show that the following equation follows the Law of Conservation of Mass:2 Al2O3 ® 4Al + 3O2
13Chemical Calculations OBJECTIVES:Construct mole ratios from balanced chemical equations, and apply these ratios in mole-mole stoichiometric calculations.
14Chemical Calculations OBJECTIVES:Calculate stoichiometric quantities from balanced chemical equations using units of moles, mass, representative particles, and volumes of gases at STP.
15Mole to Mole conversions 2 Al2O3 ® 4Al + 3O2Each time we use 2 moles of Al2O3 we make 3 moles of O2Possible conversion factors:2 moles Al2O33 moles O2or3 mole O22 moles Al2O3
16Mole to Mole conversions How many moles of O2 are produced when 3.34 moles of Al2O3 decompose?2 Al2O3 ® 4Al + 3O23 mol O23.34 mol Al2O3=5.01 mol O22 mol Al2O3
17Practice:2C2H2 + 5 O2 ® 4CO2 + 2 H2OIf 3.84 moles of C2H2 are burned, how many moles of O2 are needed?(9.60 mol)How many moles of C2H2 are needed to produce 8.95 mole of H2O?(8.95 mol)If 2.47 moles of C2H2 are burned, how many moles of CO2 are formed?(4.94 mol)
19Mass Stoichiometry Calculations We do not measure moles directly, so what can we do?We can convert between grams and molesUse Periodic Table for molar mass
20EXTREMELY IMPORTANT!!! Label what substance your factor is for 1 mol 1 mol NaCl
21Mass to Mole Calculations 2 step conversions:1st step: mass to molesDivide by molar mass of what’s given2nd step: moles of one to moles of anotherUse mole ratio
22Mass to Mole Calculations 2Na + 2H2O ® 2NaOH + H2If 23.5g of NaOH react, how many moles of H2 are produced?23.5g NaOH0.294 mol H21 mol H21 mol NaOH40.00g NaOH2 mol NaOH
23Mole to Mass Calculations 2 step conversions:1st step: moles of one to moles of anotherUse mole ratio2nd step: moles to massMultiply by molar mass of new substance
24Mole to Mass Calculations 2Na + 2H2O ® 2NaOH + H2If 3.47 moles of H2 are produced, what mass of NaOH reacted?3.47 mol H2278g NaOH40.00g NaOH2 mol NaOH1 mol H21 mol NaOH
25Mass to Mass Conversions 1st step: convert grams to molesDivide by molar mass of what’s given2nd step: moles of one to moles of the otherUse mole ratio3rd step: Convert the moles back to gramsMultiply by molar mass of new substance
26Mass to Mass Conversions 2Fe + 3CuSO4 ® Fe2(SO4)3 + 3CuIf 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how much solid copper would form?Answer = 17.2 g Cu
27Mass to Mass Conversions 2 Al2O3 ® 4Al + 3O2What mass of aluminum can be formed from 22.8g of aluminum oxide?Answer: 12.1g AlIf 3.47g of O2 are produced, what mass of Al2O3 reacted?Answer: 7.37g Al2O3
28Limiting Reagent & Percent Yield OBJECTIVES:Identify and use the limiting reagent in a reaction to calculate the maximum amount of product(s) produced, and the amount of excess reagent.
29Limiting Reagent & Percent Yield OBJECTIVES:Calculate theoretical yield, actual yield, or percent yield, given appropriate information.
30“Limiting” ReagentIf you are given one dozen loaves of bread, a gallon of mustard, and three pieces of salami, how many salami sandwiches can you make?The limiting reagent is the reactant you run out of first.The excess reagent is the one you have left over.The limiting reagent determines how much product you can make
31How do you find out? Do two stoichiometry problems. The one that makes the least product is the limiting reagent.For example:2Cu + S ® Cu2SIf 10.6 g of copper reacts with 3.83 g S how much copper (II) sulfide will be formed?
32If 10. 6 g of copper reacts with 3. 83 g S If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be formed?2Cu + S ® Cu2SCu is Limiting Reagent1 mol Cu1 mol Cu2Sg Cu2S10.6 g Cu63.55g Cu2 mol Cu1 mol Cu2S= 13.3 g Cu2S= 13.3 g Cu2S1 mol S1 mol Cu2Sg Cu2S3.83 g S32.06g S1 mol S1 mol Cu2S= 19.0 g Cu2S
33Another Example 2Al + 3CuSO4 ® Al2(SO4)3 + 3Cu If 10.3 g of aluminum are reacted with 51.7 g of CuSO4 how much copper will be produced?How much excess reagent will remain?
35Yield The amount of product made in a chemical reaction. There are three types:1. Actual yield- what you get in the lab when the chemicals are mixed2. Theoretical yield- what the balanced equation tells should be made3. Percent yield = Actual TheoreticalX 100
36Example6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate.2Al + 3 CuSO4 ® Al2(SO4)3 + 3CuWhat is the actual yield?What is the theoretical yield?What is the percent yield?
37Details Percent yield tells us how “efficient” a reaction is. Percent yield can not be bigger than 100 %.
39In terms of Volume 2H2 + O2 ® 2H2O At STP, 1 mol of any gas = 22.4 L (2 x 22.4 L H2) + (1 x 22.4 L O2) ® (2 x 22.4 L H2O)NOTE: mass and atoms are always conserved- however, molecules, formula units, moles, and volumes will not necessarily be conserved!
40Volume-Volume Calculations How many liters of CH4 at STP are required to completely react with 17.5 L of O2 ?CH4 + 2O2 ® CO2 + 2H2O22.4 L O21 mol O21 mol CH41 mol CH422.4 L CH41 mol O222.4 L CH417.5 L O222.4 L O22 mol O21 mol CH4= 8.75 L CH4
41Avogadro told us:Equal volumes of gas, at the same temperature and pressure contain the same number of particles.Moles are numbers of particlesYou can treat reactions as if they happen liters at a time, as long as you keep the temperature and pressure the same.
42Shortcut for Volume-Volume: How many liters of H2O at STP are produced by completely burning L of CH4 ?CH4 + 2O2 ® CO2 + 2H2O2 L H2O17.5 L CH4= 35.0 L H2O1 L CH4Note: This only works for Volume-Volume problems.
43More practice...How many liters of CO2 at STP will be produced from the complete combustion of 23.2 g C4H10 ?Answer = 35.8 L CO2What volume of Oxygen would be required?Answer = 58.2 L O2
44Another exampleIf 10.1 g of magnesium and 2.87 L of HCl gas are reacted, how many liters of gas will be produced?How many grams of solid?How much excess reagent remains?