Presentation is loading. Please wait.

Presentation is loading. Please wait.

Chapter 9 Chemical Quantities. 2 cup brownie mix + ½ c H 2 O + ¼c oil + 2 eggs  24 brownies What other items require a recipe? What do the numbers in.

Similar presentations


Presentation on theme: "Chapter 9 Chemical Quantities. 2 cup brownie mix + ½ c H 2 O + ¼c oil + 2 eggs  24 brownies What other items require a recipe? What do the numbers in."— Presentation transcript:

1 Chapter 9 Chemical Quantities

2 2 cup brownie mix + ½ c H 2 O + ¼c oil + 2 eggs  24 brownies What other items require a recipe? What do the numbers in front of each ingredient signify? What units are being used in this recipe? 2H 2 + O 2  2H 2 O

3 Stoichiometry The study of the quantitative or measureable relationships that exist in chemical formulas and chemical reactions The method by which scientists measure the amounts of substances that react and are formed in a chemical reaction

4 Information Given by the Chemical Equation Balanced equations show the relationship between the relative numbers of reacting molecules and product molecules. C + O 2  CO 2 1 C atom reacts with 1 O 2 molecule to produce 1 CO 2 molecule

5 C + O 2  CO 2

6 N 2(g) +3H 2  2NH 3(g) 1 molecule N 2 3 molecules H 2 2 molecules NH 3 1 dozen 3 dozen 2 dozen 6.02 X 10 23 1.81 X 10 24 1.20 X 10 24 molecules N 2 molecules H 2 molecules NH 3 1 mole N 2 3 moles H 2 2 moles NH 3 22.4L 3(22.4)= 67.2L 2(22.4L)= 44.8L 28.02gN 2 3(2.02g)= 6.06gH 2 2(17.04g)= 34.08g NH 3 mass reactants= 34g mass products= 34g

7 Coefficients in a balanced equation represent the relative number of particles (molecules, atoms and formula units) or relative number of moles ∴ will need to compare one substance with another in terms of moles ∴ need to understand mole ratios between components in a balanced equation Information Given by the Chemical Equation

8

9 N 2(g) +3H 2  2NH 3(g) What is the mole ratio between : N 2 and NH 3 1:2 1 mol N 2 forms 2 moles NH 3 N 2 and H 2 1:3 1 mol N 2 needs 3 moles H 2 H 2 and NH 3 3:2 3 mol H 2 forms 2 moles NH 3

10 mole-mole calculations How many moles of nitrogen are needed to make 7.24 moles of ammonia? Need a balanced equation N 2(g) +3H 2  2NH 3(g) Identify the given and the unknown N 2(g) +3H 2  2NH 3(g) ? (moles) 7.24 moles

11 mole-mole calculations How many moles of nitrogen are needed to make 7.24 moles of ammonia? N 2(g) +3H 2  2NH 3(g) ? (moles) 7.24 moles 7.24 mole NH 3 1 mole N 2 ————× ————— = 3.62 mole N 2 1 2 mole NH 3 Mole ratio from balanced equation

12 mole-mole calculations How many moles of ammonia are produced when 0.600moles of nitrogen reacts? N 2(g) +3H 2  2NH 3(g) 0.600moles ? (moles) 0.600mole N 2 2 mole NH 3 ————× ————— = 1.20 mole NH 3 1 1 mole N 2

13 POD How many moles of oxygen are needed to produce 19.8moles of SO 3 gas at STP? 2SO 2 (g) + O 2(g)  2SO 3(g) ? (moles) 19.8moles moles O 2 moles SO 3

14 Mass-mass calculations What mass of ammonia is produced when 5.40g of hydrogen reacts with excess nitrogen? N 2(g) +3H 2  2NH 3(g) 5.40g ? (g) Strategy: moles H 2 moles NH 3 Need to know the following: 1 mole H 2 = 2.02g 1mole NH 3 = 17.04g 3mole H 2 = 2 mole NH 3

15 Mass-mass calculations

16 How many grams of nitrogen are needed to produce 30.6g of ammonia? N 2(g) +3H 2  2NH 3(g) ? (g) 30.6 (g) moles N 2 moles NH 3 Need to know the following: 1 mole N 2 = 28.02g 1mole NH 3 = 17.04g 1mole N 2 = 2 mole NH 3

17 POD Determine the number of mass of carbon monoxide required to react with 48.0g of oxygen, and determine the grams of carbon dioxide produced, in the following reaction: 2 CO + O 2  2 CO 2 Use the coefficients to find the mole relationship: 2 moles CO = 1 mol O 2 = 2 moles CO 2 Since mass is involved, determine the molar mass of substances involved in the question: 1 mol CO = 28.01 g 1 mol O 2 = 32.00 g 1 mol CO 2 = 44.01 g

18 POD 2 CO + O 2  2 CO 2 ?mass CO 48.0g ? mass CO 2 moles CO moles O 2 moles CO 2

19 Mass-volume calculations How many milliliters of NO 2 are produced when 4.85mg of oxygen reacts? 2NO (g) + O 2(g)  2NO 2(g) 4.85mg) ? (mL) moles O 2 moles NO 2 Need to know the following: 1 mole O 2 = 32.00g 1mole NO 2 = 22.4L 1mole O 2 = 2 mole NO 2

20 Mass-volume calculations How many liters of oxygen are needed to produce 19.8L of SO3 gas at STP? 2SO 2 (g) + O 2(g)  2SO 3(g) ? (L) 19.8L moles O 2 moles SO 3 Need to know the following: 1 mole O 2 = 22.4L 1mole SO 3 = 22.4L 1mole O 2 = 2 mole SO 3

21 Moles A Volume of Gas A at STP Mass A in Grams # of Particles A Atoms, Molecules, Formula Units Moles B Volume of Gas B at STP # of Particles B Atoms, Molecules, Formula Units Mass B in Grams A  B Mol A to Mol B Bridge

22 2 cup brownie mix + ½ c H 2 O + ¼c oil + 2 eggs  24 brownies 1.What is the maximum # brownies I can make if I have ½ cup brownie mix and plenty of the other ingredients? 2. What is the maximum # brownies I can make if I have 1cup oil and plenty of the other ingredients? 3. What is the maximum # brownies I can make if I have 1egg and plenty of the other ingredients?

23 Limiting and Excess Reactants Limiting reactant: a reactant that is completely consumed when a reaction is run to completion Excess reactant: a reactant that is not completely consumed in a reaction Theoretical yield: the maximum amount of a product that can be made by the time the limiting reactant is completely consumed

24 Example: Limiting Reagent Determine the number of moles of carbon dioxide produced when 3.2 moles oxygen reacts with 4.0 moles of carbon monoxide 2 CO + O 2  2 CO 2 4.0 moles CO 3.2moles ? Moles CO 2.

25 Example: Limiting Reagent 2 CO + O 2  2 CO 2 4.0 moles CO 3.2moles ? Moles CO 2 Limiting Reactant Excess Reactant **Limiting Reagent (Reactant) determines Max amount of product that can be formed

26 2 CO + O 2  2 CO 2 Determine the mass of carbon dioxide produced when 48.0 g of oxygen reacts with 56.0 g of carbon monoxide. 2 CO + O 2  2 CO 2 56.0gCO 48.0g ? g CO 2

27 2 CO + O 2  2 CO 2 2 CO + O 2  2 CO 2 56.0g 48.0g ? g CO 2 Limiting Reactant Excess Reactant

28 2 CO + O 2  2 CO 2 Determine the mass of excess reagent left over after the reaction is complete 48.0 g of oxygen reacts with 56.0 g of carbon monoxide. a) Determine which substance is the excess reagent b) Calculate how much of the excess reagent was consumed (used up) in the reaction c) Calculate the amount of excess reagent left over

29 2 CO + O 2  2 CO 2 56.0g 48.0g Excess Reactant How much O 2 consumed (used up)? How much O 2 left over? Initial mount of O 2 - amount of O 2 used = amount left over O 2 48.00g - 31.99g = 16.01g

30 POD Silver tarnishes in the presence of hydrogen sulfide because of the reaction: 4Ag + 2H 2 S + O 2  2Ag 2 S + 2H 2 O How many grams of silver sulfide could be obtained from a mixture of 0.950g Ag, 1.40g H 2 S, and 0.0800 of oxygen gas? How many grams of the other substances are left unreacted?

31 Percent Yield Most reactions do not go to completion. Actual yield: the amount of product made in an experiment Theoretical yield: The amount of product calculated by stoichiometric calculations Percent yield: the percentage of the theoretical yield that is actually made Percent Yield = Actual Yield Theoretical Yield x 100%

32 2 CO + O 2  2 CO 2 Determine the Percent Yield of the reaction if 72.0 g of carbon dioxide is actually produced when 56.0 g of carbon monoxide reacts with excess oxygen. actual yield of CO 2 = 72.0g theoretical yield of CO 2 = 88.0g

33 Solution Stoichiometry Many reactions occur in solution. Therefore you need to be able to predict amounts of reactants and products in terms of concentrations and volumes as well as masses. Basic strategy is the same –Use molarity and volume to calculate moles molarity = moles of solute liters of solution

34 Solution Concentration Molarity Moles of solute per 1 liter of solution –Used because it describes how many moles of solute in each liter of solution –If a sugar solution concentration is 2.0 M, 1 liter of solution contains 2.0 moles of sugar. 2M 2moles sugar = 1 L molarity = moles of solute liters of solution

35 Moles A Volume of Gas A at STP Mass A in Grams # of Particles A Atoms, Molecules, Formula Units Moles B Volume of Gas B at STP # of Particles B Atoms, Molecules, Formula Units Mass B in Grams A  B Mol A to Mol B Bridge Molarity, Volume of A Volume of B

36 Example Calculate the mass of solid NaCl required to precipitate all the Ag +1 ions from 1.50 L of a 0.100 M AgNO 3 solution. AgNO 3 (aq) + NaCl(s)  AgCl(s) + NaNO 3 (aq) 1.50L,0.100M ?

37 When aqueous solutions of Na 2 SO 4 and Pb(NO 3 ) 2 are mixed, PbSO 4 precipitates. Calculate the mass of PbSO 4 formed when 1.25 L of 0.0500 M Pb(NO 3 ) 2 and 2.00 L of 0.0250 M Na 2 SO 4 are mixed. Example #2

38 Quantitative analysis of a mixture A substance of unknown composition can be converted to one or more substances of known composition. The amounts of the known can be determined and related to the original unknown substance

39 3TiO 2 (s) + 4BrF 3 (l)  3TiF 4 (s) + 2Br 2 (l) + 3O 2 (g) One method for determining the purity of a sample of titanium (IV) oxide TiO 2, an important industrial chemical, is to combine the sample with bromine trifluoride. Oxygen gas is evolved quantitatively. The gas can be captured readily, and its mass can be determined. Suppose 2.367g of TiO 2 - containing sample evolves 0.143g of O 2. What is the mass percent of TiO 2 in the sample?

40

41 Combustion analysis A sample of the hydrocarbon is burned The amount of each of the products, CO 2 and H 2 O, that are produced is measured. An empirical formula for the hydrocarbon can be determined.

42 Combustion analysis A 0.523g sample of an unknown compound CxHy is burned in air to give 1.612g of CO 2 and 0.7425g of H 2 O. A separate experiment gave a molar mass of CxHy of 114g/mol. Determine the empirical and molecular formulas for the hydrocarbon. CxHy + O 2  CO 2 + H 2 O ? Emp formula 1.612g 0.7425g

43 CxHy + O 2  CO 2 + H 2 O ? Emp formula 1.612g 0.7425g ∴ C 4 H 9


Download ppt "Chapter 9 Chemical Quantities. 2 cup brownie mix + ½ c H 2 O + ¼c oil + 2 eggs  24 brownies What other items require a recipe? What do the numbers in."

Similar presentations


Ads by Google