1. 3 - 1 Stoichiometry Law of Conservation of Matter Balancing Chemical Equations Mass Relationships in Chemical Reactions Limiting Reactants Theoretical, Actual and Percent Yields Empirical Formulas Molecular and Structural Formulas Law of Conservation of Matter Balancing Chemical Equations Mass Relationships in Chemical Reactions Limiting Reactants Theoretical, Actual and Percent Yields Empirical Formulas Molecular and Structural Formulas

2. 3 - 2 Stoichiometry Stoichiometry The study of quantitative relationships between substances undergoing chemical changes. Law of Conservation of Matter In chemical reactions, the quantity of matter does not change. The total mass of the products must equal that of the reactants.

3. 3 - 3 Chemical equations Chemist’s shorthand to describe a reaction. It shows: All reactants and products The state of all substances Any conditions used in the reaction CaCO 3 (s) CaO (s) + CO 2 (g)  Reactant Products A balanced equation shows the relationship between the quantities of all reactants and products.

4. 3 - 4 Balancing chemical equations Each side of a chemical equation must have the same number of each type of atom. CaCO 3 (s) CaO (s) + CO 2 (g) Reactants Products 1 Ca 1 C 3 O

5. 3 - 5 Balancing chemical equations Step 1 Step 1Count the number of atoms of each element on each side of the equation. Step 2 Step 2 Determine which atom numbers are not balanced. Step 3 Step 3 Balance one atom at a time by using coefficients in front of one or more substances. Step 4 Step 4Repeat steps 1-3 until everything is balanced.

6. 3 - 6 Balance the following HCl + Ca _____ > CaCl 2 + H 2 - not balanced Step 1 & 2 1 H2 H - not balanced - not balanced 1 Cl2 Cl - not balanced 1 Ca1 Ca 2 Step 3 2HCl + Ca _____ > CaCl 2 + H 2 Step 4 2 H2 H 2 Cl2 Cl 1 Ca1 Ca The balanced equation 2 2HCl + Ca _____ > CaCl 2 + H 2

7. 3 - 7 Balance the following C 2 H 6 + O 2 _____ > CO 2 + H 2 O - not balanced Step 1 & 2 2 C1 C - not balanced - not balanced 6 H2 H - not balanced - not balanced 2 O3 O - not balanced Step 3 Balance biggest molecule first, C 2 H 6 23 C 2 H 6 + O 2 _____ > 2CO 2 + 3H 2 O Step 4 2 C2 C 6 H6 H - not balanced 2 O7 O - not balanced

8. 3 - 8 Balance the following 23 C 2 H 6 + O 2 _____ > 2CO 2 + 3H 2 O Step 3a Now let’s balance O 2 3.5 C 2 H 6 + 3.5O 2 _____ > 2 CO 2 + 3 H 2 O Step 4a 2 C2 C 6 H6 H 7 O7 O You can’t have 3.5 O 2 so double the equation! Balanced Equation 27 4 6 2 C 2 H 6 + 7 O 2 _____ > 4 CO 2 + 6 H 2 O

9. 3 - 9 Example. Decomposition of urea (NH 2 ) 2 CO + H 2 O ______ > NH 3 + CO 2 < not balanced 2 N1 N < not balanced < not balanced 6 H3 H < not balanced1 C2 O We need to double NH 3 on the right. 2 (NH 2 ) 2 CO + H 2 O ______ > 2NH 3 + CO 2

10. 3 - 10 Another example CH 3 OH + PCl 5 _____ > CH 3 Cl + POCl 3 + H 2 O 1 C1 C 4 H5 H 1 O2 O1 P 5 Cl4 Cl We need another Cl on the right. Increase CH 3 Cl then recheck.

11. 3 - 11 Another example. 2 CH 3 OH + PCl 5 _____ > 2CH 3 Cl + POCl 3 + H 2 O 1 C2 C 4 H8 H 1 O2 O1 P5 Cl Another C is needed on the left so double CH 3 OH.

12. 3 - 12 Another example. 2 2 2CH 3 OH + PCl 5 _____ > 2CH 3 Cl + POCl 3 + H 2 O 2 C2 C8 H2 O1 P5 Cl Now its balanced!

13. 3 - 13 2 H 2 + O 2 -----> 2 H 2 O You need a balanced equation and you WILL work with moles. You need a balanced equation and you WILL work with moles. Mass relationships in chemical reactions Stoichiometry Stoichiometry - The calculation of quantities of reactants and products in a chemical reaction.

14. 3 - 14 Stoichiometry, General steps. 1 1 Balance the chemical equation. 3 3 Convert masses to moles. 2 2 Calculate formula masses. 4 4 Use chemical equation to get the needed answer. Convert back to mass if needed. 5 5

15. 3 - 15 Mole calculations The balanced equation shows the reacting ratio between reactants and products. 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O For each chemical, you can determine the moles of each reactant consumed moles of each product made If you know the formula mass, mass quantities can be used.

16. 3 - 16 Mole-gram conversion How many moles are in 14 grams of N 2 ? Formula mass =2 N x 14.01 g/mol = 28.02 g /mol moles N 2 = 14 g x 1 mol /28.02 g = 0.50 moles

17. 3 - 17 Mass calculations We don’t directly weigh out molar quantities. We can use measured masses like kilograms, grams or milligrams. The formula masses and the chemical equations allow us to use either mass or molar quantities. We don’t directly weigh out molar quantities. We can use measured masses like kilograms, grams or milligrams. The formula masses and the chemical equations allow us to use either mass or molar quantities.

18. 3 - 18 Mass calculations How many grams of hydrogen will be produced if 10.0 grams of calcium is added to an excess of hydrochloric acid? 2HCl + Ca ______ > CaCl 2 + H 2 Note: We produce one H 2 for each calcium. There is an excess of HCl so we have all we need.

19. 3 - 19 Mass calculations 2HCl + Ca ____ > CaCl 2 + H 2 First - Determine the number of moles of calcium available for the reaction. Moles Ca= grams Ca / FM Ca = 10.0 g = 0.25 mol Ca 1 mol 40.08 g

20. 3 - 20 Mass calculations 2HCl + Ca _____ > CaCl 2 + H 2 10 g Ca = 0.25 mol Ca According to the chemical equation, we get one mole of H 2 for each mole of Ca. So we will make 0.25 moles of H 2. grams H 2 produced = moles x FW H 2 = 0.25 mol x 2.016 g/mol = 0.504 grams

21. 3 - 21 Mass calculations OK, so how many grams of CaCl 2 were made? 2HCl + Ca _____ > CaCl 2 + H 2 10 g Ca = 0.25 mol Ca We would also make 0.25 moles of CaCl 2. g CaCl 2 = 0.25 mol x FM CaCl 2 = 0.25 mol x 110.98 g / mol CaCl 2 = 27.75 g CaCl 2

22. 3 - 22 Limiting reactant In the last example, we had HCl in excess. Reaction stopped when we ran out of Ca. limiting reactant. Ca is considered the limiting reactant. Limiting reactant Limiting reactant - the material that is in the shortest supply based on a balanced chemical equation.

23. 3 - 23 Limiting reactant example

24. 3 - 24 Example For the following reaction, which is limiting if you have 5.0 g of hydrogen and 10 g oxygen? Balanced Chemical Reaction 2H 2 + O 2 ________ > 2H 2 O You need 2 moles of H 2 for each mole of O 2. Moles of H 2 5 g = 2.5 mol Moles of O 2 10g = 0.31 mol 1 mol 32.0 g 1 mol 2.0 g

25. 3 - 25 Example Balanced Chemical Reaction 2H 2 + O 2 2H 2 O You need 2 moles of H 2 for each mol of O 2 You have 2.5 moles of H 2 and 0.31 mol of O 2 Need a ratio of 2:1 but we have a ratio of 2.5 : 0.31 or 8.3 : 1. Hydrogen is in excess and oxygen is the limiting reactant.

26. 3 - 26 Theoretical, actual and percent yields Theoretical yield The amount of product that should be formed according to the chemical reaction. Actual yield The amount of product actually formed. Percent yield Ratio of actual to theoretical yield, as a %. Quantitative reaction When the percent yield equals 100%.

27. 3 - 27 Yield Less product is often produced than expected. Possible reasons A reactant may be impure. Some product is lost mechanically since the product must be handled to be measured. The reactants may undergo unexpected reactions - side reactions. No reaction truly has a 100% yield due to the limitations of equilibrium.

28. 3 - 28 Percent yield The amount of product actually formed divided by the amount of product calculated to be formed, times 100. % yield = x 100 In order to determine % yield, you must be able to recover and measure all of the product in a pure form. Actual yield Theoretical yield

29. 3 - 29 % Yield example Example. Example. The final step in the production of aspirin is the reaction of salicylic acid with acetic anhydride. 48.6 g of aspirin is produced when 50.0 g of salicylic acid and an excess of acetic anhydride are reacted. What is the % yield? HOC 6 H 4 COOH (s) + (CH 3 CO) 2 O (l) salicylic acid acetic anhydride CH 3 OC 6 H 4 COOH (s) + CH 3 COOH (l) aspirin acetic acid

30. 3 - 30 % Yield example Number of moles of salicylic acid used: One mole of aspirin should be produced for each mole of salicylic acid consumed. Number of grams of aspirin that should have been produced -- theoretical yield: (0.362 mol aspirin)( 180 g/mol) = 65.2 g aspirin = 0.362 mole of salicylic acid 1 mol 138 g 50.0 g

31. 3 - 31 % Yield example % Yield for this reaction Theoretical yield = 65.2 g Actual yield= 48.6 g % Yield = x 100 = 74.5% Yields less than 100% are very common in industrial processes. 48.6 g 65.2 g

32. 3 - 32 Empirical formula This type of formula shows the ratios of the number of atoms of each kind in a compound. For organic compounds, the empirical formula can be determined by combustion analysis. Elemental analyzer An instrument in which an organic compound is quantitatively converted to carbon dioxide and water -- both of which are then measured.

33. 3 - 33 Elemental analyzer furnace CO 2 trap H 2 O trap O2O2 sample A sample is ‘burned,’ completely converting it to CO 2 and H 2 O. Each is collected and measured as a weight gain. By adding other traps elements like oxygen, nitrogen, sulfur and halogens can also be determined.

34. 3 - 34 Elemental analysis Example A compound known to contain only carbon, hydrogen and nitrogen is assayed by elemental analysis. The following information is obtained. Original sample mass= 0.1156 g Mass of CO 2 collected= 0.1638 g Mass of H 2 O collected= 0.1676 g Determine the % of each element in the compound.

35. 3 - 35 Elemental analysis Mass of carbon Mass of hydrogen Mass of nitrogen 0.1638 g CO 2 12.01 g C 44.01 g CO 2 = 0.04470 g C0.1675 g H 2 O 2.016 g H 18.01 g H 2 O = 0.01875 g H 0.1156 g - 0.04470 g C - 0.01875 g H = 0.05215 g N

36. 3 - 36 Elemental analysis Since we know the total mass of the original sample, we can calculate the % of each element. % C = x 100% = 38.67 % % H = x 100% = 16.22 % % N = x 100% = 45.11 % 0.04470 g 0.1156 g 0.01875 g 0.1156 g 0.05215 g 0.1156 g

37. 3 - 37 Empirical formula The simplest formula that shows the ratios of the number of atoms of each element in a compound. Example Example - the empirical formula for hydrogen peroxide (H 2 O 2 ) is HO. We can use our percent composition information from the earlier example to determine an empirical formula.

38. 3 - 38 Empirical formula From our earlier example, we found that our compound had a composition of: If we assume that we have a 100.0 gram sample, then we can divide each percentage by the elements atomic mass and determine the number of moles of each. % C = 38.67 % % H = 16.22 % % N = 45.11 % % C = 38.67 % % H = 16.22 % % N = 45.11 %

39. 3 - 39 Empirical formula 16.22 g H 1 mol H 1.008 g H = 12.09 mol H 38.67 g C 1 mol C 12.01 g C = 3.220 mol C 45.11 g N 1 mol N 14.01 g N = 3.220 mol N ( )

40. 3 - 40 Empirical formula The empirical formula is then found by looking for the smallest whole number ratio. C3.220 / 3.220= 1.000 H16.09 / 3.220= 4.997 N3.220 / 3.220= 1.000 The empirical formula is CH 5 N

41. 3 - 41 Molecular formula Molecular formula Molecular formula - shows the actual number of each type of atom in a molecule. They are multiples of the empirical formula. If you know the molecular mass, then the molecular formula can be found. For our earlier example, what would be the molecular formula if you knew that the molecular mass was 62.12?

42. 3 - 42 Molecular formula Empirical formulaCH 5 N Empirical formula mass31.06 u Molecular mass62.12 Ratio:62.12 / 31.06= 2 The molecular formula is C 2 H 10 N 2 Note: This does not tell you have the atoms are arranged in the compound!

43. 3 - 43 Structural formula These are used to show how atoms are attached in a molecule.Example Both of the following structural formula would have a molecular formula of C 2 H 6 O ethyl alcoholdimethyl ether These chemicals have very different properties.

44. 3 - 44 Structural formula We use a variety of ways to represent structural formula. Condensed structural formulas are commonly used for organic molecules. They list a carbon and then what is attached to it. The next carbon in the chain is then listed. ethyl alcohol-CH 3 CH 2 OH dimethyl ether-CH 3 OCH 3

45. 3 - 45 Structural formula Models may also be used to help view a molecule. ethyl alcoholdimethyl ether Ball-and- stick Space- filling