1. Chapter 3 Stoichiometry

2. Chemical Stoichiometry Stoichimetry from Greek “measuring elements”. That is “Calculation of quantities in chemical reactions” That is “Calculation of quantities in chemical reactions” Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions.

3. 3.2 Atomic Masses Atoms are so small, it is difficult to discuss how much they weigh in grams. Atomic mass units, amu, were used Mass of C-atom of C-12 was assigned a mass of exactly 12 amu Thus, an atomic mass unit (amu) is one twelth (1/12) the mass of a carbon-12 atom. The masses of all other atoms are given relative to this standard The decimal numbers on the table are atomic masses in amu.

4. Atomic Masses by mass spectrometer Mass Spectrometer –Atoms are passed into a beam of high speed electrons. –Electrons are knocked out and +Ve ions formed –Applied electric field accelerates +Ve ions into a magnetic field.

5. Atomic Masses by mass spectrometer Mass Spectrometer –Fast atomic ions (current) bend near magnet –Deflection varies inversely with masses Multiple isotopes differ in mass and so give multiple beam deflections. The ratio of the masses of C-13 and C-12 found to be 1.0836129 The mass of C-13 = mass of C-12 X 1.0836129 = 13.003355 amu = 13.003355 amu Masses of other atoms can be determined similarly Exact number by definition

6. Average atomic mass (6.941) Atomic masses are not whole numbers

7. The atomic masses are not whole numbers including Carbon, Why? Because they are based on averages of atoms and of isotopes. It could be possible to determine the average atomic mass from the mass of the isotopes and their relative abundance. add up the percent as decimals times the masses of the isotopes.

8. Natural lithium is: 7.42% 6 Li (6.015 amu) 92.58% 7 Li (7.016 amu) (7.42 x 6.015) + (92.58 x 7.016) 100 = ________ amu Average atomic mass of lithium: Average atomic mass = (%isotope X atomic mass)+(%isotope X atomic mass) 100

9. Atomic Weight of an Element from Isotopic Abundances Naturally occurring chlorine is 75.78% 35 Cl, which has an atomic mass of 34.969 amu, and 24.22% 37 Cl, which has an atomic mass of 36.966 amu. Calculate the average atomic mass (that is, the atomic weight) of chlorine. The average atomic mass is found by multiplying the abundance of each isotope by its atomic mass and summing these products.

10. The Mole The mole (mol) is a number equal to the number of carbon atoms in exactly 12.00 grams of 12 C Techniques such as mass spectrometry were used to count this number The number was found as 6.02214X10 23 This number was known as “Avogadro’s number” Thus, one mole of a substance contains of 6.022X10 23 units of that substance So, dozen of eggs is 12; a mole of eggs is Avogadro’s number of eggs.

11. How the mole is used in chemical calculations? 12 grams of 12 C contain Avogadro’s number of atoms = = 6.022X10 23 C atoms 12.01 g of natural C ( 12 C, 13 C, 14 C) contains 6.022X10 23 C atoms Atomic mass of C atom =12.01 amu

13. Express amu in grams

14. Practice Convert grams of atoms moles of atoms grams of atoms moles of atoms number of atoms number of atoms and visa versa

15. 3.4 Molar mass Molar mass is the mass of 1 mole of a substance. Often called molecular weight. To determine the molar mass of a compound: add up the molar masses of the elements (taking # moles of each element into consideration) that make it up.

16. Find the molar mass of CH 4 Mg 3 P 2 Ca(NO 3 ) 3 Al 2 (Cr 2 O 7 ) 3 CaSO 4 · 2H 2 O Convert mass molar mass and visa versa Conversion between mass, molar mass and number of molecules

17. 3.5 Percent Composition of Compounds It is the mass (weight) percent of each element a compound is composed of. Find the mass of each element, divide by the total mass, multiply by a 100. Use a mole of the compound for simplicity. n x molar mass of element molar mass of compound x 100%

18. Practice %C = 2 x (12.01 g) 46.07 g x 100% = 52.14%H = 6 x (1.008 g) 46.07 g x 100% = 13.13%O = 1 x (16.00 g) 46.07 g x 100% = 34.73% 52.14% + 13.13% + 34.73% = 100.0% Find the percent composition of each element in C 2 H 6 O

19. Determination of percent composition and simplest formula from experiment

20. Example

23. 3.6 Determining the formula of a compound molecular formula = (empirical formula) n molecular formula = (empirical formula) n [n = integer] [n = integer] molecular formula = C 6 H 6 = (CH) 6 molecular formula = C 6 H 6 = (CH) 6 empirical formula = CH empirical formula = CH Empirical formula: the lowest whole number ratio of atoms in a compound. Molecular formula: the true number of atoms of each element in the formula of a compound.

24. Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio). Examples: NaClMgCl 2 Al 2 (SO 4 ) 3 K 2 CO 3

25. Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio). Molecular: H2OH2O C 6 H 12 O 6 C 12 H 22 O 11 Empirical: H2OH2O CH 2 OC 12 H 22 O 11

26. Experimental Determination of the formula of a compound by elemental analysis A sample of a compound composed of carbon oxygen and hydrogen are combusted in a stream of O 2 to produce CO 2 and H 2 O. The H 2 O and CO 2 are trapped and the masses of each are measured. A compound of unknown composition is decomposed by heat. The elements are carefully trapped and the number of moles of each are analyzed.

27. Pure O 2 in CO 2 is absorbed H 2 O is absorbed Sample is burned completely to form CO 2 and H 2 O Sample O 2 and other gases Experimental Determination of the formula of a compound

28. The sample has a mass of 0.255g. When the reaction is complete, 0.561 g of CO 2 and 0.306g of H 2 O are produced. What is the empirical formula of the compound? 1.Determine the mass of C in the sample. For each mole of CO 2, there is one mole of C. Convert moles of C to grams of C. 0.561 g CO 2 x (1 mol CO 2 / 44.01 g CO 2 ) x (1 mol C / 1 mol CO 2 ) (12.01 g C/ mol C) = 0.153 g C Calculating empirical formula

29. 2. Determine the mass of H in the sample. There are 2 moles of hydrogen per mole of H 2 O. 0.306 g H 2 O x (1 mol H 2 O / 18.0 g H 2 O) x (2 mole H / mole H 2 O) x (1.01 g H / mol H) = 0.0343 g H Mass O = mass sample - mass H - mass C Mass sample = 0.255 g Mass O = 0.255 - 0.153 - 0.0343 = 0.068 g O

30. To get empirical formula, convert g back to moles 0.153 g C x ( 1 mol C / 12.01 g C ) = 0.0128 mol C 0.0343 g H x (1 mol H / 1.01 g H) = 0.0340 mol H 0.068 g O x (1 mol O / 16.0 g O) = 0.0043 mol O Divide each by 0.0043 to get ratio of each element to O C: 0.0128 mol C / 0.0043 mol O = 2.98 ~ 3 There are 3 moles of carbon for each mole of oxygen H: 0.0340 mol H / 0.0043 mol O = 7.91 ~ 8 There are 8 moles of hydrogen per mole of oxygen Empirical Formula C 3 H 8 O

31. g CO 2 mol CO 2 mol Cg C g H 2 O mol H 2 Omol Hg H g of O = g of sample – (g of C + g of H) Burn 11.5 g ethanol Collect 22.0 g CO 2 and 13.5 g H 2 O 6.0 g C = 0.5 mol C 1.5 g H = 1.5 mol H 4.0 g O = 0.25 mol O Empirical formula C 0.5 H 1.5 O 0.25 Divide by smallest subscript (0.25) Empirical formula C 2 H 6 O

32. Calculating Empirical formula from the percent composition We can get a ratio from the percent composition. Assume you have a 100 g. –The percentages become grams. Convert grams to moles. Find lowest whole number ratio by dividing by the smallest value.

33. Example Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N. Assume 100 g so 38.67 g C x 1mol C = 3.220 mole C 12.01 gC 16.22 g H x 1mol H = 16.09 mole H 1.01 gH 45.11 g N x 1mol N = 3.219 mole N 14.01 gN Now divide each value by the smallest value

34. The ratio is 3.220 mol C = 1 mol C 3.219 mol N 1 mol N The ratio is 16.09 mol H = 5 mol H 3.219 mol N 1 mol = C 1 H 5 N 1 = C 1 H 5 N 1

35. Determining molecular formula from empirical formula Since the empirical formula is the lowest ratio, the actual molecule would weigh more. –By a whole number multiple. Divide the actual molar mass by the empirical formula mass –you get a whole number to increase each coefficient in the empirical formula Caffeine has a molar mass of 194 g. what is its molecular formula?

36. molecular formula = (empirical formula) n n = integer] n = integer] molecular formula = C 6 H 6 = (CH)

37. 3.7 Chemical Equations Chemical change involves reorganization of the atoms in one or more substances. Chemical reactions occur when bonds between the outermost parts of atoms are formed or broken Chemical reactions involve changes in matter, the making of new materials with new properties, or energy changes. Atoms cannot be created or destroyed Chemical Reactions are described using a shorthand called a chemical equation

38. Chemical Equations A representation of a chemical reaction: C 2 H 5 OH + 3O2 2 CO2 + 3H 2 O reactants products

39. The chemical equation for the formation of water can be visualized as two hydrogen molecules reacting with one oxygen molecule to form two water molecules: 2H 2 + O 2  2H 2 O Products Reactants

40. The plus sign (+) means “react” and the arrow points towards the substance produce in the reaction. The chemical formulas on the right side of the equation are called reactants and after the arrow are called product. The numbers in front of the formulas are called stoichiometric coefficients. 2Na + 2H 2 O  2NaOH + H 2 Stoichiometric coefficients: numbers in front of the chemical formulas; give ratio of reactants and products. Reading Chemical Equations Coefficient ReactantsProducts

41. Understanding Chemical Equations Coefficients and subscripts included in the chemical formula have different effects on the composition.

42. Balanced chemical equation C 2 H 5 OH + 3O 2  2CO 2 + 3H 2 O The equation is balanced. 1 mole of ethanol reacts to produce with 3 moles of oxygen 2 moles of carbon dioxide and 3 moles of water 2 moles of carbon dioxide and 3 moles of water

43. Physical states in chemical equations State Symbol Solid(s) Liquid(l) Gas(g) Dissolved in water(aq) (in aqueous solution) HCl(aq) + NaHCO 3 (l) CO 2 (g) + H 2 O(l) + NaCl(aq) Catalysts, photons (h ), or heat (  ) may stand above the reaction arrow (  ).

44. 3.8 Balancing chemical equations  When balancing a chemical reaction coefficients are added in front of the compounds to balance the reaction, but subscripts should not be changed  Changing the subscripts changes the compound

45. Balancing Chemical Equations By inspection (Trial and error)  Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation. Ethane reacts with oxygen to form carbon dioxide and water C 2 H 6 + O 2 CO 2 + H 2 O  Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts. 2C 2 H 6 NOT C 4 H 12

46. Balancing Chemical Equations  Start by balancing those elements that appear in only one reactant and one product. C 2 H 6 + O 2 CO 2 + H 2 O start with C or H but not O 2 carbon on left 1 carbon on right multiply CO 2 by 2 C 2 H 6 + O 2 2CO 2 + H 2 O 6 hydrogen on left 2 hydrogen on right multiply H 2 O by 3 C 2 H 6 + O 2 2CO 2 + 3H 2 O

47. Balancing Chemical Equations  Balance those elements that appear in two or more reactants or products. 2 oxygen on left 4 oxygen (2x2) C 2 H 6 + O 2 2CO 2 + 3H 2 O + 3 oxygen (3x1) multiply O 2 by 7 2 = 7 oxygen on right C 2 H 6 + O 2 2CO 2 + 3H 2 O 7 2 remove fraction multiply both sides by 2 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O

48. Balancing Chemical Equations  Check to make sure that you have the same number of each type of atom on both sides of the equation. 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O 4 C (2 x 2)4 C 12 H (2 x 6)12 H (6 x 2) 14 O (7 x 2)14 O (4 x 2 + 6) 3.7

49. Balancing Chemical Equations 3.7 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O ReactantsProducts 4 C 12 H 14 O 4 C 12 H 14 O  Check to make sure that you have the same number of each type of atom on both sides of the equation.

50. Stoichiometry From Greek “measuring elements”. That is “Calculation of quantities in chemical reactions” Given an amount of either starting material or product, determining the other quantities. use conversion factors from –molar mass (g - mole) –balanced equation (mole - mole)

51. C 2 H 4 (g) + HCl(g)C 2 H 5 Cl(g) If we have 15.0 g of C 2 H 4 how many moles of HCl are needed to carry out the reaction to completion? The balanced equation tells us it takes 1 mole of HCl for each mole of C 2 H 4 reacted. We must begin by converting the number of grams of C 2 H 4 which has a molar mass of 28.08 g/mol, to moles 15.0 g C 2 H 4 x 1 mol C 2 H 4 = 0.534 mol C 2 H 4 28.08 g C 2 H 4 Example

52. C 2 H 4 (g) + HCl(g)C 2 H 5 Cl(g) 0.534 mol C 2 H 4 x 1mol HCl = 0.534 mol HCl 1mol C 2 H 4 What moles of HCl is needed to carry the reaction through to completion? 0.534 mol HCl x 36.5 g HCl = 19.6 g HCl 1mol HCl What mass of HCl is needed to carry the reaction through to completion?

53. C 2 H 4 (g) + HCl(g)C 2 H 5 Cl(g) How many moles of product are made when 15.0 g of C 2 H 4 is reacted with an excess of HCl? Example

54. C 2 H 4 (g) + HCl(g)C 2 H 5 Cl(g) 15.0 g? Molar mass of C 2 H 4 (ethylene) = 28.08 g/mol 15.0 g C 2 H 4 x 1mole C 2 H 4 = 0.534 mol C 2 H 4 28.06 g C 2 H 4 Stoichiometry of the balanced equation indicates a mole ratio of 1:1 reactants

55. C 2 H 4 (g) + HCl(g) C 2 H 5 Cl(g) 0.534 mol 0.534 mol (molar mass C 2 H 5 Cl = 64.51 g/mol) 0.543 mol C 2 H 5 Cl x 64.51 g C 2 H 5 Cl = 35.0 g C 2 H 5 Cl 1 mol C 2 H 5 Cl

56. How many grams of NH 3 will be produced from 4.8 mol H 2 ? N 2 (g) + H 2 (g) NH 3 (g) 32 Excess 4.8 mol ? 4.8 mol H 2 x 2 mol NH 3 = 3.2 mol NH 3 3 mol H 2 Begin with what you are given. You have 4.8 mol H 2 3.2 mol NH 3 x 17.0 g NH 3 = 54.4 g NH 3 1 mol NH 3 Example

57. If the reactants are not present in stoichiometric amounts, at end of reaction some reactants are still present (in excess). Limiting Reactant: one reactant that is consumed 3.10 Calculations involving a limiting reactants Limiting Reactant: Reactant that limits the amount of product formed in a chemical reaction

58. Limiting Reactant

59. Limiting Reactants/ Limiting Reactants/ Example Cu + 2AgNO 3 -> CO(NO 3 ) 2 + 2Ag when 3.5g of Cu is added to a solution containing 6.0g of AgNO 3 what is the limiting reactant? What is the mass Ag produced and what the mass of the excess reagent?

60. Cu + 2AgNO 3 -> CO(NO 3 ) 2 + 2Ag 1. 3.5g Cu x 1 mol Cu x 2 mol Ag = 0.11 mol of Ag 63.5g Cu 1 mol Cu 2. 6.0g AgNO 3 x 1mol AgNO 3 x 2mol Ag = 0.035mol Ag 170g AgNO3 2mol AgNO 3 The Limiting Reactant is AgNO 3. Limiting reactant is: The reactant that produces the least amount of product Calculations involving a limiting reactant

61. Determining the limiting reactant by comparing the mole ratio of Cu and AgNO 3 required by the balanced equation with the mole ratio actually present 3.5g Cu x 1 mol Cu x = 0.055 mol Cu 63.5g Cu 6.0g AgNO 3 x 1mol AgNO 3 = 0.0353mol AgNO 3 170g AgNO 3 Cu + 2AgNO 3 -> CO(NO 3 ) 2 + 2Ag Thus AgNO 3 is the limiting reactant

62. What is the mass of Ag produced? Take the limiting reactant: 6.0g AgNO 3 x 1mol AgNO 3 x 2mol Ag x 108g Ag 170g AgNO3 2mol AgNO 3 1molAg = 3.8g Ag - Limiting reactant is: The reactant that produces the least amount of product

63. Excess reagent Cu + 2AgNO 3 -> CO(NO 3 ) 2 + 2Ag Excess reagent is ???? 6.0g AgNO 3 x 1mol AgNO 3 x 1mol Cu x 63.5 Cu 170g AgNO3 2mol AgNO 3 1molCu = 1.12 g Cu Amount of Cu left (Excess reagent) = 3.5 – 1.12 = 2.38 g Cu Calculate amount of Cu needed to react with the limiting reactant

64. The Yield The reactant you don’t run out of is the excess reagent The amount of stuff you make is the yield. The theoretical yield is the amount you would make if everything went perfect. The actual yield is what you make in the lab.

65. The percent yield relates the actual yield (amount of material recovered in the laboratory) to the theoretical yield: The percent yield