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Unit 12 Chemistry Langley

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1 Unit 12 Chemistry Langley
STOICHIOMETRY BASICS Unit 12 Chemistry Langley **Corresponds to Chapter 12 in the Prentice Hall Chemistry Textbook

2 INTRODUCTION Stoichiometry
Dervived from two Greek words meaning measured and element 4 types Mole-Mole Mass-Mole Mass-Mass Volume-Volume Mathematically based Balanced equations very important; first place to start

3 Interpreting Chemical Reaction
From a balanced chemical equation, you can describe a chemical reaction in terms of quantities of products and reactants. These quantities include the number of molecules, atoms, moles, mass, and volume of the compounds used as well as each element used to make the respective compounds

4 Interpreting a Chemical Reaction
N2 + 3H2  2NH3 (Figure 12.3, pg. 357) Number of atoms: 2 atoms of nitrogen, 6 atoms of hydrogen, 8 atoms of ammonia Number of molecules: 1 molecule of nitrogen, 3 molecules, 2 molecules of ammonia (1:3:2 ratio) Number of relative moles: 1 mole of nitrogen, 3 relative moles of hydrogen, 2 relative moles of ammonia Molecules and moles are not the same thing Mass: g of nitrogen, 6.0 g of hydrogen, 34 g of ammonium Volume: 22.4L of nitrogen, 67.2L of hydrogen, 44.8L of ammonia Because 1 mol of any gas at STP occupies a volume of 22.4L

5 Molar Mass of Compounds
The molar mass (MM) of a compound is determined the same way, except now you add up all the atomic masses for the molecule (or compound) Ex. Molar mass of CaCl2 Avg. Atomic mass of Calcium = 40.08g Avg. Atomic mass of Chlorine = 35.45g Molar Mass of calcium chloride = g/mol Ca + (2 X 35.45) g/mol Cl  g/mol CaCl2 20 Ca  17 Cl 35.45

6 Molar Mass Calculation
Calculate the Molar Mass of calcium phosphate Formula = (NH4)3(PO4) Masses elements: Molar Mass =

7 Atoms or Molecules Moles Mass (grams) Divide by 6.02 X 1023
Multiply by 6.02 X 1023 Moles Multiply by atomic/molar mass from periodic table Divide by atomic/molar mass from periodic table Mass (grams)

8 Everything must go through Moles!!!
Calculations molar mass Avogadro’s number Grams Moles particles Everything must go through Moles!!!

9 Chocolate Chip Cookies!!
1 cup butter 1/2 cup white sugar 1 cup packed brown sugar 1 teaspoon vanilla extract 2 eggs 2 1/2 cups all-purpose flour 1 teaspoon baking soda 1 teaspoon salt 2 cups semisweet chocolate chips Makes 3 dozen How many eggs are needed to make 3 dozen cookies? How much butter is needed for the amount of chocolate chips used? How many eggs would we need to make 9 dozen cookies? How much brown sugar would I need if I had 1 ½ cups white sugar?

10 Recipes and Equations Just like chocolate chip cookies have recipes, chemists have recipes as well Instead of calling them recipes, we call them reaction equations Furthermore, instead of using cups and teaspoons, we use moles Lastly, instead of eggs, butter, sugar, etc. we use chemical compounds as ingredients

11 Chemistry Recipes Looking at a reaction tells us how much of something you need to react with something else to get a product (like the cookie recipe) Be sure you have a balanced reaction before you start! Example: 2 Na + Cl2  2 NaCl This reaction tells us that by mixing 2 moles of sodium with 1 mole of chlorine we will get 2 moles of sodium chloride What if we wanted 4 moles of NaCl? 10 moles? 50 moles?

12 Starting Point-Balanced Equation
Write the balanced reaction for hydrogen gas reacting with oxygen gas. 2 H2 + O2  2 H2O How many moles of reactants are needed? What if we wanted 4 moles of water? What if we had 3 moles of oxygen, how much hydrogen would we need to react and how much water would we get? What if we had 50 moles of hydrogen, how much oxygen would we need and how much water produced?

13 Mole-Mole Calculations
These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical Example: How many moles of chlorine is needed to react with 5 moles of sodium (without any sodium left over)? 2 Na + Cl2  2 NaCl 5 moles Na 1 mol Cl2 2 mol Na = 2.5 moles Cl2

14 Mole-Mole Calculations
How many moles of ammonia are produced when 0.60 moles of nitrogen reacts with hydrogen? N2(g) + 3H2 (g)  2NH3(g) 2.0 mol NH3 0.6 mol N2 = 1.2 mol NH3 1.0 mol N2

15 Mass-Mole Calculations
Sometimes you are going to start with mass and will have to convert to moles of product or another reactant We use molar mass and the mole ratio to get to moles of the compound of interest Calculate the number of moles of ethane (C2H6) needed to produce 10.0 g of water 2 C2H6 + 7 O2  4 CO2 + 6 H20 10.0 g H2O 1 mol H2O 2 mol C2H6 18.0 g H2O 6 mol H20 = mol C2H6

16 Mass-Mole Calculations
Calculate how many moles of oxygen are required to make 10.0 g of aluminum oxide

17 Mole-Mass Calculations
Most of the time in chemistry, the amounts are given in grams instead of moles We still go through moles and use the mole ratio, but now we also use molar mass to get to grams Example: How many grams of chlorine are required to react completely with 5.00 moles of sodium to produce sodium chloride? 2 Na + Cl2  2 NaCl 5.00 moles Na 1 mol Cl g Cl2 2 mol Na 1 mol Cl2 = 177g Cl2

18 Mole-Mass Calculations
Calculate the mass in grams of Iodine required to react completely with 0.50 moles of aluminum.

19 Mass-Mass Calculations
Most often we are given a starting mass and want to find out the mass of a product we will get (called theoretical yield) or how much of another reactant we need to completely react with it (no leftover ingredients!) Now we must go from grams to moles, mole ratio, and back to grams of compound we are interested in

20 Mass-Mass Calculations
Ex. Calculate how many grams of ammonia are produced when you react 2.00g of nitrogen with excess hydrogen. N2 + 3 H2  2 NH3 2.00g N mol N mol NH g NH3 28.02g N2 1 mol N mol NH3 = 2.4 g NH3

21 Mass-Mass Calculations
How many grams of calcium nitride are produced when 2.00 g of calcium reacts with an excess of nitrogen?

22 Volume-Volume Calculations
1 mole of any substance = 22.4L Nitrogen monoxide and oxygen gas combine to form the brown gas nitrogen dioxide, which contributes to photochemical smog. How many liters of nitrogen dioxide are produced when 34 L of oxygen reacts with an excess of nitrogen monoxide? Assume conditions of STP.

23 Volume-Volume Calculations
2NO + O2(g)  2NO2(g) Know: volume of oxygen = 34L O2 2 mol NO2/1 mol O2 1 mol O2 = 22.4 L O2 (at STP) 1 mole NO2 = 22.4 L NO (at STP) 34 L O2 1 mol O2 2 mol NO2 22.4 L NO2 = 64 L O2 22.4 L O2 1 mol O2 1 mol NO2

24 Limiting Reactants 1 cup butter 1/2 cup white sugar
1 cup packed brown sugar 1 teaspoon vanilla extract 2 eggs 2 1/2 cups all-purpose flour 1 teaspoon baking soda 1 teaspoon salt 2 cups semisweet chocolate chips Makes 3 dozen If we had the specified amount of all ingredients listed, could we make 4 dozen cookies? What if we had 6 eggs and twice as much of everything else, could we make 9 dozen cookies? What if we only had one egg, could we make 3 dozen cookies?

25 Limiting Reactant Most of the time in chemistry we have more of one reactant than we need to completely use up other reactant. That reactant is said to be in excess (there is too much). The other reactant limits how much product we get. Once it runs out, the reaction This is called the limiting reactant.

26 Limiting Reactant To find the correct answer, we have to try all of the reactants. We have to calculate how much of a product we can get from each of the reactants to determine which reactant is the limiting one. The lower amount of a product is the correct answer. The reactant that makes the least amount of product is the limiting reactant. Once you determine the limiting reactant, you should ALWAYS start with it! Be sure to pick a product! You can’t compare to see which is greater and which is lower unless the product is the same!

27 Limiting Reactant: Limiting Reactant
10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced? 2 Al + 3 Cl2  2 AlCl3 Start with Al: Now Cl2: 10.0 g Al 1 mol Al mol AlCl g AlCl3 27.0 g Al mol Al mol AlCl3 = 49.4g AlCl3 35.0g Cl mol Cl mol AlCl g AlCl3 71.0 g Cl mol Cl mol AlCl3 = 43.9g AlCl3

28 Limiting Reactant We get 49.4g of aluminum chloride from the given amount of aluminum, but only 43.9g of aluminum chloride from the given amount of chlorine. Therefore, chlorine is the limiting reactant. Once the 35.0g of chlorine is used up, the reaction comes to a complete

29 Limiting Reactant 15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made.

30 Finding the Amount of Excess
By calculating the amount of the excess reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess. Can we find the amount of excess potassium in the previous problem?

31 Finding Excess 15.0 g of potassium reacts with 15.0 g of iodine. 2 K + I2  2 KI We found that Iodine is the limiting reactant, and 19.6 g of potassium iodide are produced. 15.0 g I mol I mol K g K 254 g I mol I mol K = 4.62 g K USED! 15.0 g K – 4.62 g K = g K EXCESS Given amount of excess reactant Amount of excess reactant actually used Note that we started with the limiting reactant! Once you determine the LR, you should only start with it!

32 Limiting Reactant: Recap
Convert ALL of the reactants to the SAME product (pick any product you choose.) The lowest answer is the correct answer. The reactant that gave you the lowest answer is the LIMITING REACTANT. The other reactant(s) are in EXCESS. To find the amount of excess, subtract the amount used from the given amount. You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT. If you have to find more than one product, be sure to start with the limiting reactant. You don’t have to determine which is the LR over and over again!

33 Empirical vs. Molecular Formulas
Empirical Formula of a compound shows the smalles whole number ration of the atoms in the compound A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the emprical formula of the compound? 25.9 g N * (1 mol/14.0g) = 1.85 mol N 74.1 g O * (1 mol O/16.0g) = 4.63 mol O Empirical Formula = N2O5

34 Empirical vs. Molecular Formulas
Molecular Formula tells you the actual number of each kind of atom present in a molecule of the compound Calculate the molecular formula of a compound whose molar mass is 60.0 g and empirical formula is CH4N. Find empirical formula mass: =30g Molar mass/efm = 60/30 = 2 Molecular Formula = C2H8N2

35 Percent, Theoretical, and Actual Yield
Actual Yield – the amount produced in a reaction (amount of products) Theoretical Yield – the amount that should have been produced in a reaction Maximum amount of product that could be formed from given amount of reactants Percent Yield – (actual/theoretical) X 100% Often less than the theoretical

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