1. Chapter 3 Stoichiometry

2. Atomic Mass Carbon-12 is assigned a mass of exactly 12.00 atomic mass units (amu) Masses of other elements are compared to Carbon-12 Comparisons are made using a mass spectrometer –Compares ratios of isotopes present

3. Mass Spectrometer

4. Average Atomic Mass An average of the atomic masses of all naturally occurring isotopes of an element Average atomic mass is the mass value on the periodic table No atom has the exact mass listed on the table Carbon = 12.01 amu –This means the majority of C is Carbon-12 Used to weigh large numbers of atoms

5. Example Lead occurs in 4 main isotopes 1.40% has a mass of 203.973amu. 24.10% has a mass of 205.975amu. 22.10% has a mass of 206.976, and 52.40% has a mass of 207.977amu. What is average atomic mass of Lead?

6. The Mole 1 mole = 6.022x10 23 particles Avogadro’s Number –He did not determine the number! The average atomic mass of an element expressed in grams contains 1 mole of atoms of that element. 22.99 g of Sodium is one mole of sodium 40.08 g of Calcium is one mole of calcium

7. Example How many moles of sodium atoms are in 2.83 grams of sodium?

8. Example How many atoms of sodium are there in 2.83 grams?

9. Molar Mass The mass in grams of one mole of an element or compound –Units of grams/mole (g/mol) Molar mass of sodium = 22.99 g/mol Molar mass of compound is the sum of all atoms present Molar Mass of Water (H 2 O) 2(1.01) + 16.00 = 18.02 g/mol

10. Example Determine the molar mass of Mg 3 N 2 C 2 H 5 O 2 N Ca(NO 3 ) 2

11. Example 18.1g Calcium Nitrate = ? moles

12. Example How many nitrate ions are present in 18.1 grams of Calcium Nitrate?

13. Percent Composition The percent by mass each element contributes to the compound (# of atoms of element)(atomic mass) X 100 (molar mass of compound)

14. Example What is the percent composition of Calcium Nitrate?

15. Homework P. 123 #’s 21,34,36,40,44,49,52,53

16. Objectives Compare Empirical and Molecular Formulas Calculate Empirical and Molecular Formulas Balance and Interpret chemical equations Perform stoichiometry calculations

17. Empirical and Molecular Formulas Empirical Formula –A chemical formula with the simplest whole number ratio CH 2 O (30.03 g/mol) Molecular Formula –The exact chemical formula for a compound C 3 H 6 O 3 (90.09 g/mol)

18. Calculating Empirical Formulas Assume 100 g of compound –(Percents become grams) Or use given masses Covert grams to moles Divide each by the smallest # of moles If you don’t have whole #’s multiply by an integer to get whole numbers –3.33 * 3 4.5 * 2 8.25 *4

19. Example A compound contains 49.31%C, 43.79%O and the rest H. The molar mass is 146 g/mol. What is the empirical and molecular formula?

20. Combustion Analysis Compounds containing C,H, and O are often burned to determine the chemical makeup. –The products of combustion are CO 2 and H 2 O These are collected and massed –Carbon and Hydrogen come from compound –Oxygen comes from compound and air

21. Combustion Analysis

22. Example The combustion of 17.27 grams of a compound which contains only C,H and O yields 44.06 grams of CO 2 and 11.27 grams of H 2 O. What is the empirical formula of the compound?

24. Balanced Chemical Equations Gives the recipe for a reaction Reactants on left / Products on the right Coefficients tell us the atomic/mole ratios State symbols tell what state of matter (s) solid, (l) liquid, (g) gas, (aq) solution 2H 2 (g) + O 2 (g)  2H 2 O(l) Means – 2 mol of gaseous Hydrogen reacts with 1 mol of gaseous Oxygen to form 2 mol of liquid Water

25. Balancing Equations Use coefficients to get the same number of atoms on both sides of the equation –Due to Law of Conservation of Mass –Coefficient – number in front of a formula Balance the equation Fe 2 S 3 + HCl  FeCl 3 + H 2 S

26. Cont. Combustion Equations –When there are an even number of Carbons in the reactant you may have to double the reaction. C 2 H 6 + O 2  CO 2 + H 2 O

27. Stoichiometry A process for comparing amounts in reactions Compares reactant to reactant or reactant to product Start with a balanced equation Process Grams A  Moles A  Moles B  Grams B Molar Mass AMolar RatioMolar Mass B

28. Cont. Could Include –Density –Percent Could start or end with moles Read the problem and determine the best way to proceed

29. Example 6.00 grams of hydrochloric acid is reacted with barium hydroxide to form barium chloride and water. How many grams of barium chloride are formed?

30. Example 4.82 grams of propane (C 3 H 8 ) is burned with excess oxygen to form carbon dioxide and water. What volume of water vapor is formed if the density is.8044 g/L?

31. Homework P. 125 #’s 66,74,79,86

32. Objectives Describe limiting and excess reactants Be able to calculate the limiting reactant Compare theoretical and actual yield Describe and calculate percent yield

33. Limiting Reactant The reactant that runs out in a chemical reaction. When one reactant is used up the reaction stops Based on mole ratio NOT on mass

34. Example What is the limiting reactant when 16.00g of Hydrogen gas reacts with 64.00g of oxygen gas to produce water

35. Example What mass of carbon dioxide gas is produced when 4.00g of methane (CH 4 ) reacts with 32.00g of oxygen gas

36. Theoretical and Actual Yield Theoretical yield is the amount of product you should get based on calculations Actual yield is the amount of product you get when the reaction is run –Actual yield cannot be more than the theoretical yield. (Conservation of mass) –Actual yield is usually less than theoretical b/c of product loss

37. Percent Yield Measures how much product a reaction produces High percent yields are good Percent Yield = Actual X 100 Theoretical

38. Example 1000. grams of calcium phosphate reacts with 1000. grams of 98% sulfuric acid to produce calcium sulfate and phosphoric acid. When the reaction is run 1100. g of calcium sulfate was produced. What is the percent yield of the reaction?

39. Homework Stoichiometry Worksheet