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Stoichiometry and the Mole

Stoichiometry and the Mole. SOL Review. Stoichiometry and the Mole. The Mole. The Mole. Not everything that counts can be counted, and not everything that can be counted counts . Albert Einstein. What is a mole?. The mole is a counting unit. Like . . . 1 dozen = 12 each

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Stoichiometry and the Mole

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  1. Stoichiometryand the Mole SOL Review

  2. Stoichiometryand the Mole The Mole

  3. The Mole Not everything that counts can be counted, and not everything that can be counted counts. Albert Einstein

  4. What is a mole? • The mole is a counting unit. • Like . . . • 1 dozen = 12 each • 1 yard = 3 feet • 1 cup = 8 ounce • So then ... • 1 mol = 6.022 x 1023 particles That’s Avogadro’s Number!

  5. Where did it come from? • Mole (n) is the SI unit for the number of particles • Amedo Avogadro determined the number of particles in a mole • The mole is the measure of the amount of a substance whose number of particles is the same as 12 grams of Carbon - 12

  6. Calculations • Using dimensional analysis you can determine the number of particles in a mole • 1 mole = 6.022 x 1023 particles, molecules, etc. • 6.022 x 1023 particles = 1 mole

  7. So, let’s count . . . • 1 mol of Ag = • 6.022 x 1023 atoms Ag • 1 mol of CO2 = • 6.022 x 1023 molecules of CO2 • 1 mol of pizza = • 6.022 x 1023 pizzas!

  8. The Mole Road Map

  9. Moles to Mass Conversions • 1 Moles = Molar Mass (g) = Molar mass/Mole or (g/mol) Now let’s apply that knowledge!

  10. The Mole: Molar Mass Calculations Reminder – Finding Molar Mass: The molar mass = the sum of all the atomic masses. You try one: What is the gram formula mass (molar mass) of Mg3(PO4)2? Mg = 3(24.305) P = 2(30.97376) O = 8(15.9994) 262.86 grams Example: Ca(NO3)2 Ca = 40.08 N = 2(14.01) O = 6(16.00) 164.10 grams

  11. Molar or Formula Mass • Chemical compounds are written as an empirical formula. • Ex. H2SO4 is Sulfuric Acid • Calculating atomic mass, add each atom. • H = 1.008 x 2 = 2.016 • S = 32.07 x 1 = 32.07 • O = 15.999 x 4 = 63.996 Total Atomic Mass = 2.016+32.07+63.996 = 98.08 amu

  12. The Mole and Mole Calculations • One mole = 6.02 x 1023 representative particles • One mole = 22.4 Liters of gas at 0°C and one atmosphere of pressure • One mole = the atomic mass listed on the periodic table. • For example: one mole of Helium contains 6.02 x 1023 atoms of Helium and it has a mass of 4.00260 grams. At 0°C and one atmosphere of pressure, it would occupy 22.4 Liters. • Sample problem: • How many liters would 2.0 moles of Neon occupy? • 2.0 moles Ne x 22.4 Liters Ne = 44.8 Liters Ne • 1.0 moles Ne

  13. Calculating MolesOne Step How many moles are in 3.011 x 1023 atoms of Oxygen? 3.011 x 1023 atoms O21 mol Cu 6.02 X 1023 atoms = • 0.5 moles of Oxygen

  14. More One-Step Conversions Ex 1) Convert 4.3 grams of NaCl to moles. Mass  mol 4.3 g NaCl x 1 mol NaCl = 7.4 x 10-2 mol NaCl 58.45 g NaCl Ex 2) Convert 0.00563 mol NH3 to grams. Mol -> mass 0.00563 mol NH3 x 17 g NH3=9.57 x 10 –2 g NH31 mol NH3

  15. The Mole and Mole Calculations • Sample problem: • How many moles are in 15.2 grams of Lithium? • Answer: • 15.2 g Li x 1 mole Li = 2.19 mole Li • 6.941 g Li • REMINDER: • One mole = 6.02 x 1023 representative particles • One mole = 22.4 Liters of gas at 0°C and one atmosphere of pressure • One mole = the atomic mass listed on the periodic table. • Sample problem: • How many liters would 14 grams of Helium occupy? • Answer: • 14 g He x 1 mole He x 22.4 L He = 78 Liters He • 4.0026 g He 1 mole He

  16. Mole Calculations How many atoms of Cu are present in 35.4 g of Cu? Molar mass of Cu is 63.55 g/mol 35.4 g Cu 1 mol Cu 6.02 X 1023 atoms Cu 63.5 g Cu 1 mol Cu = 3.4 X 1023 atoms Cu

  17. And another . . . • What mass would 4.52 x 1024 molecules of water have? = 135 g H2O 4.52 x 1024 mlcs of H20 1 mol 18.02 g 6.02 X 1023 mlcs1 mol

  18. Stoichiometryand the Mole Stoichiometry

  19. Stoichiometry • Stoichiometry means that if you know one piece of information about ONE compound in an equation, you can determine EVERYTHING else! • If you have 3L of Nitrogen, how many liters of ammonia will you produce? N2 + 3H2 2NH3

  20. Stoichiometry • Let’s look at that last reaction again. N2(g) + 3H2(g) 2NH3(g) • If you start out with 1 mole of Nitrogen gas and 3 moles of Hydrogen gas, you will make 2 moles of Ammonia gas. • It is important in industry to know the exact proportions of your ingredients so that you will not have excess waste in your product.

  21. How many grams of silver chloride can be produced from the reaction of 17.0 g silver nitrate with excess sodium chloride solution? Stoichiometry 1. Write the balanced equation 17.0g ?g AgNO3 + NaCl  AgCl + NaNO3 2. Given and asked for 3. Moles of given 17.0g AgNO3 x 1 mol = 170 g 0.100 mol AgNO3

  22. StoichiometryMass-Mass Problem

  23. StoichiometryMass-Mass Problem

  24. StoichiometryMass-Mass Problem AgNO3 + NaCl  AgCl + NaNO3 4. Moles asked for 0.100 mols AgNO3 x 1 mol AgCl = 1 mol AgNO3 0.100 mol AgCl 5. Convert your answer 0.100 mol AgCl x 144 g AgCl = 1 mol AgCl 14.4 g AgCl

  25. Standard Temperature & Pressure 0°C and 1 atm C. Molar Volume at STP 1 mol of a gas=22.4 L at STP

  26. Limiting Reactant Problems Stoichiometryand the Mole

  27. Limiting Reactant Problems: Given the following reaction: 2Cu + S  Cu2S • What is the limiting reactant when 82.0 g of Cu reacts with 25.0 g S? • What is the maximum amount of Cu2S that can be formed? • How much of the other reactant is wasted?

  28. Limiting Reactant Problems: • Our 1st goal is to calculate how much S would react if all of the Cu was reacted. • From that we can determine the limiting reactant (LR). • Then we can use the Limiting Reactant to calculate the amount of product formed and the amount of excess reactant left over. • 82g Cu • mol Cu • mol S • g S

  29. 2Cu + S  Cu2S • So if all of our 82.0g of Copper were reacted completely it would require only 20.7 grams of Sulfur. • Since we initially had 25g of S, we are going to run out of the Cu, the limiting reactant) & end up with 4.3 grams of S 1molCu 1mol S 32.1g S 82.0gCu 63.5gCu 2molCu 1mol S =20.7 g S

  30. ________ Limiting Reactant Problems: • Copper being our Limiting Reactant is then used to determine how much product is produced. • The amount of Copper we initially start with limits the amount of product we can make. 1molCu 159gCu2S 2molCu2S 82.0gCu 1molCu2S 63.5gCu 1molCu2S = 103 g Cu2S

  31. Limiting Reactant Problems: • So the reaction between 82.0g of Cu and 25.0g of S can only produce 103g of Cu2S. • The Cu runs out before the S and we will end up wasting 4.7 g of the S.

  32. Stoichiometryand the Mole Percent Yield

  33. Calculating Percent Yield • In theory, when a teacher gives an exam to the class, every student should get a grade of 100%. Sadly, this is not always true. The calculation for percent yield is similar. • We already know that we do not get a 100% yield of products in an reaction.

  34. Calculating Percent Yield • Consider the Following Reactions: Mg + 2HCl  MgCl2 + H2 5.0 g Mg is reacted with an excess of HCL. How much MgCl2 will be produced. = 21.6 g of MgCl2 5.0g Mg 1molMg 1mol MgCl2 105.2 g MgCl2 24.3 g Mg 1mol MgCl2 1mol Mg

  35. Calculating Percent Yield • You might assume that using stoichiometry to calculate that our reaction will produce 21.6 g of MgCl2, but we will actually only recover 15.2 g of MgCl2 in the lab. • 21.6 g of MgCl2 is the value representing the theoretical yield(theoretical yield is the maximum amount of product that could be formed). • The 15.2 g of MgCl2 is called the actual yield (the actual yield is less than the theoretical yield).

  36. Calculating Percent Yield • The percent yield is the ratio of the actual yield to the theoretical yield as a percent • It measures the measures the efficiency of the reaction. measured in lab actual yield Percent yield= x 100 theoretical yield calculated on paper

  37. Calculating Percent Yield • Why do reactions not go to completion. • Impure reactants and competing side rxns may cause unwanted products to form. • Actual yield can also be lower than the theoretical yield due to a loss of product during filtration or transferring between containers. • If a wet precipitate is recovered it might weigh heavy due to incomplete drying, etc.

  38. Calculating Percent Yield Calcium carbonate is synthesized by heating, as shown in the following equation: CaO + CO2 CaCO3 • What is the theoretical yield of CaCO3 if 24.8 g of CaO is heated with 43.0 g of CO2? • What is the percent yield if 33.1 g of CaCO3 is produced? • Determine which reactant is the limiting and then decide what the theoretical yield is.

  39. 24.8gCaO • molCaO • mol CaCO3 • gCaCO3 • 24.8gCaO • molCaO • mol CO2 • gCO2 24.8 g CaO 1molCaO 1mol CO2 44 g CO2 56g CaO 1mol CaO 1molCO2 LR = 19.5gCO2 1mol CaO 100g CaCO3 24.8 g CaO 1molCaCO3 56g CaO 1mol CaO 1molCaCO3 = 44.3 g CaCO3

  40. _____________ Calculating Percent Yield • CaO is our LR, so the reaction should theoretically produce 44.3 g of CaCO3 (How efficient were we?) • Our percent yield is: 33.1 g CaCO3 Percent yield= x 100 44.3 g CaCO3 Percent yield = 74.7%

  41. GOOD LUCK!! Mrs. Armani Mr. Buchanan Ms. Nichols Mr. Smith

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