1. Molar Mass

2. Do Now You have a bag of blue and red marbles. You’re trying to figure out the mass of the bag, but you don’t have a balance/ scale. What might you need to know to solve this problem? How might you be able to find the mass?

3. Do Now- con’t. Your bag contains 10 blue marbles and 15 red marbles. What else may we need to know?

4. Do Now- con’t. Your bag contains 10 blue marbles and 15 red marbles. Each blue marble weighs 2 grams, and each red marbles weighs 3 grams. What is the mass of the bag?

5. Mass on the Molecular Level Just like we weren’t able to directly weigh the bag of marbles, it’s impossible to “weigh” one molecule. – But, we can find the mass of a molecule by adding up the mass of its individual atoms Molar Mass: the mass (in grams) of one mole of molecules – This is based on the atomic mass in the periodic table

6. Determining Molar Mass 1.From the formula, write down each of the different elements that are present. MgCl 2 :Mg Cl 2.Identify how many atoms of each element are present using the subscripts (no subscript = just one atom) Mg: 1 Cl: 2 3.Multiply the number of atoms of each element by that element’s atomic mass, then find the total mass by adding all the elements’ masses together Mg: 1 x 24.3050 g = 24.3050g Cl: 2 x 35.4527 g = + 70.9054g 95.2104g

7. Practice! Determine the molar mass of H 2 SO 4 (sulfuric acid).

8. Practice! Determine the molar mass of H 2 SO 4 (sulfuric acid). H 2 SO 4 H: 2 x 1.0079g = 2.0158g S: 1 x 32.066g = 32.066g O: 4 x 15.999 = 63.996g 98.0778g

9. Practice! What is the molar mass of (H 3 O) 3 PO 4 ?

10. Practice! What is the molar mass of (H 3 O) 3 PO 4 ? Note: when you have parentheses in the formula, the subscript outside them is distributed (multiplied) by all of the subscripts inside!

11. Practice! What is the molar mass of (H 3 O) 3 PO 4 ? H: (3 x 3) = 9 O: (1 x 3) + 4 = 7 P: 1 H: 9 x 1.0079 = 9.0711 O: 7 x 15.999 = 111.993 P: 1 x 30.974 = 30.974 152.038g

12. What exactly is a mole? Avogadro’s Number: 6.022 x 10 23 The number of carbon atoms in exactly 12g of pure 12C 1 mole of ANYTHING = 6.022 x 10 23 units of that substance How many teachers are in 1 mole of teachers? How many pencils are in 1 mole of pencils? How many candy bars are in 1 mole of candy bars?

13. What exactly is a mole? Avogadro’s Number: 6.022 x 10 23 The number of carbon atoms in exactly 12g of pure 12C 1 mole of ANYTHING = 6.022 x 10 23 units of that substance How many teachers are in 1 mole of teachers? How many pencils are in 1 mole of pencils? How many candy bars are in 1 mole of candy bars? 6.022 x 10 23 !!!

14. If we know moles… The molar mass is the grams of that substance that it takes to make up 1 mole So, if we know how many moles we started with, we can determine the mass of our substance OR, We also know the number of particles in one mole (Avogadro’s number) So, we can determine the number of molecules of our substance

15. Our map… Grams Moles# of Particles Molar Mass Avogadro’s Number

16. Going between Grams and Moles We have 24g of carbon. We know that the molar mass is 12.011g/mol (there are 12.011 grams of carbon in one mole) SET UP THE CONVERSION! Start with what you know. The molar mass is our conversion factor.

17. Going between Grams and Moles We have 24g of carbon. We know that the molar mass is 12.011g/mol (there are 12.011 grams of carbon in one mole) SET UP THE CONVERSION! Start with what you know. The molar mass is our conversion factor. 24g C x 1 mole = 24g C mol = 1.998 mol C 1 12.011g C 12.011g C

18. Going between Grams and Moles We have 3 moles of sulfur. How many grams do we have?

19. Going between Grams and Moles We have 3 moles of sulfur. How many grams do we have? 3 mol S x 32.066g S = 96.198 g S 11 mol S

20. Higher Order Questions What is another real-life application of stoichiometry? What is happening on a molecular level when two substances react to form new products? THINK-WRITE-PAIR-SHARE! Pick one topic to contemplate with your partner.

21. CO 2(g) + 2LiOH (s) -> Li 2 CO 3(s) + H 2 O (l) In a spacecraft, the carbon dioxide exhaled by astronauts can be removed by its reaction with lithium hydroxide, LiOH, according to the above chemical equation. How many grams of carbon dioxide are produced if the average person breathes out 20 mol of CO 2 per day?

22. You Practice! Convert from grams to moles: 125g phosphorus (P) 75g NaCl Convert from moles to grams: 5 moles of carbon (C) 7 moles of HBr

23. You Practice! Convert from grams to moles: 125g phosphorus (P) = 4.04 mol P 75g NaCl = 1.28 mol NaCl Convert from moles to grams: 5 moles of carbon (C) = 60.055g C 7 moles of HBr = 566.4g HBr

24. Going Between Moles and Particles We have 15 x 10 23 carbon atoms. How many moles is this? SET UP THE CONVERSION! Start with what you know. Avogadro’s number is our conversion factor.

25. Going Between Moles and Particles We have 15 x 10 23 carbon atoms. How many moles is this? SET UP THE CONVERSION! Start with what you know. Avogadro’s number is our conversion factor. 15 x 10 23 C atoms x 1 mol 6.022 x 10 23 atoms = 2.49 mol C

26. Going Between Moles and Particles We have 5.5 mol of NaCl. How many molecules is this?

27. Going Between Moles and Particles We have 5.5 mol of NaCl. How many molecules is this? 5.5 mol NaCl x 6.022 x 10 23 molecules 1 mol = 33.121 x 10 23 molecules NaCl = 3.3121 x 10 24 molecules NaCl (correct scientific notation)

28. Going Between Grams and Particles This is just another conversion! We don’t know the relationship between grams and particles, so what could we go to first?

29. Going Between Grams and Particles This is just another conversion! We don’t know the relationship between grams and particles, so what could we go to first? MOLES!!! Moles are like our base in metric conversions- when in doubt, convert to moles!

30. Going Between Grams and Particles We have 600 x 10 23 carbon atoms. How many grams is this?

31. Going Between Grams and Particles We have 600 x 10 23 carbon atoms. How many grams is this? 600 x 10 23 atoms C x 1 mol C x 6.022 x 10 23 atoms C 12.011g C= 119.7g C 1 mol C

32. Going Between Grams and Particles We have 30g of pure copper. How many atoms of copper is this?

33. Going Between Grams and Particles We have 30g of pure copper. How many atoms of copper is this? 30g Cu x 1 mol Cu x 6.022 x 10 23 atoms Cu = 63.546g Cu1 mol Cu = 2.8 x 10 23 atoms Cu

34. Density Density: a physical property of a substance, representing the mass per unit volume d = m/V m = mass (in grams) V = volume (in mL) Note: 1mL = 1cm 3 We can use the density to determine the mass if it isn’t given!

35. Using Density to Determine Mass Determine the mass of mercury (Hg) in 30mL if the density of mercury is 13.55g/mL. Set up the equation! Fill in what you know, then solve.

36. Using Density to Determine Mass Determine the mass of mercury (Hg) in 30mL if the density of mercury is 13.55g/mL. d = m/V 13.55g = m mL 30mL 13.55g x 30 = m 406.5g = m

37. Using Density to Determine Mass Turpentine has a density of 0.85g/mL. How many grams are in 75mL?

38. Using Density to Determine Mass Turpentine has a density of 0.85g/mL. How many grams are in 75mL? d = m/V 0.85g = m mL 75mL 0.85g x 75 = m 63.75g = m

39. Exit Ticket #1

40. Solve the following problems, showing all work for full credit. 1.What is the molar mass of acetic acid (HC 2 H 3 O 2 )? (1 point) 2.How many pennies are in one mole of pennies? (1 point) 3.How many moles are in 50g of magnesium? Show your work! (2 points) 4.How many grams are in 3.5mol of CaO? Show your work! (3 points) 5.How many molecules are in 15g of HF? Show your work! (3 points)

41. Percent Composition You always bring a lunch to school. You wanted to know how heavy your lunch was, so you decided to weigh each of the foods you packed. Your sandwhich was 550g. Your apple was 200g. Your cookie was 50g. Your yogurt was 200g. What percent of the mass of your lunch was from the sandwich?

42. Percent Composition Percent Composition: a given element’s portion of the molecule’s mass = molar mass of the element x 100% molar mass of the whole molecule 1. Find the mass of that element in the molecule 2. Find the total mass of the molecule 3. Enter values into the above equation **If you add up the percentages for all elements in the compound, you should get 100%

43. Percent Composition: Example What percent of the molecule’s mass is hydrogen in H 2 SO 4 ?

44. Percent Composition: Example What percent of the molecule’s mass is hydrogen in H 2 SO 4 ? H: 2 x 1.0079g = 2.0158g S: 1 x 32.066g = 32.066g O: 4 x 15.999g = 63.996g = 98. 078g g H x 100% = 2.0158g x 100% = 0.0206 x 100% g H 2 SO 4 98.078g = 2.06%

45. Percent Composition: Example 2 Still using H 2 SO 4, determine the percentage of the mass for sulfur and oxygen.

46. Percent Composition: Example 2 Still using H 2 SO 4, determine the percentage of the mass for sulfur and oxygen. S: 32.066g/ 98. 078g x 100% = 32.69% S O: 63.996g/ 98. 078g x 100% = 65.25% O

47. Empirical Formulas Empirical Formula: the simplest whole-number ratio of atoms in a compound 1.Assume that you have 100g of substance 2.Determine the number of grams of each element from the percent composition (ex. 50% carbon = 50g carbon) 3.Convert each element from grams to moles 4.Divide the moles of each element by the smallest value to get the simplest whole-number ratio If you have a decimal of 0.5, multiply EVERYTHING by 2 If you have a decimal of 0.333, multiply EVERYTHING by 3 If you have a decimal of 0.25, multiply EVERYTHING by 4 If the number is VERY close to a whole number (ex. 0.999), it is acceptable to round

48. Empirical Formula Example Determine the empirical formula for a compound consisting of the following: 71.65% Cl24.27% C4.07% H

49. Empirical Formula Example Determine the empirical formula for a compound consisting of the following: 71.65% Cl24.27% C4.07% H Cl: 71.65g x 1mol/35.45g = 2.021mol/ 2.021 = 1 C: 24.27g x 1mol/ 12.01g = 2.021mol/ 2.021 = 1 H: 4.07g x 1mol/1.008g = 4.04mol/ 2.021 = 2 ClCH 2

50. Molecular Formula Molecular Formula: the exact formula of a molecule – May or may not be the same as the empirical formula – You need to know the molar mass of the compound! 1.Determine the empirical formula 2.Find the molar mass of the empirical formula 3.Divide the molar mass of the compound by the molar mass from the empirical formula 4.Multiply the subscripts in the empirical formula by the above ratio

51. Molecular Formula Example Earlier, we determined the empirical formula was ClCH 2. Given that the molar mass of the compound is 98.96g/mol, determine the molecular formula.

52. Molecular Formula Example Earlier, we determined the empirical formula was ClCH 2. Given that the molar mass of the compound is 98.96g/mol, determine the molecular formula. ClCH 2 : 35.45g + 12.01g + 2.02g = 49.48g/mol 98.96g/mol = 2 49.48g/mol (ClCH 2 ) 2 = Cl 2 C 2 H 4

53. Practice Determine the empirical and molecular formulas for a compound containing 40.0% carbon, 6.7% hydrogen, and 53.5% oxygen. The compound has a molar mass of 180.156g/mol.

54. Practice Determine the empirical and molecular formulas for a compound containing 40.0% carbon, 6.7% hydrogen, and 53.5% oxygen. The compound has a molar mass of 180.156g/mol. C: 40.0g x 1mol/12.0107g = 3.33mol/ 3.33 = 1 H: 6.7g x 1mol/ 1.00794g = 6.65mol/ 3.33 = 2 O: 53.3g x 1mol/15.9994g = 3.33mol/ 3.33 = 1 Empirical: CH 2 O = 30.026g/mol 180.156 = 6 : CH 2 O  Molecular: C 6 H 12 O 6 30.026

55. Exit Ticket #2

56. Exit Ticket #2- Show ALL Work! 1. What is the percent composition of oxygen in H 2 O? (2 points) 2. Which of the following is an empirical formula? (1 point) a. C 6 H 12 O 6 b. C 2 H 4 NO c. C 2 H 6 d. All of the above. 3. A compound’s empirical formula is C 3 H 7 and its molecular weight is 86 g/mole. Find its molecular formula. (2 points) 4. A compound consists of 72.2% magnesium and 27.8% nitrogen by mass. What is its empirical formula? (2 points) 5. A compound is 58.8 % C, 9.9 % H, and 31.3 % O. If its molecular mass is 306 g/mole, what is the molecular formula? (3 points)

57. Exit Ticket #3

58. Balance the following reactions. Write a coefficient 1 wherever needed so I can tell you didn’t just leave it blank. (2 points each) 1.S 8 + NO 2  SO 2 + N 2 2.C 3 H 8 + O 2  CO 2 + H 2 O 3.C 6 H 6 + HNO 3  C 6 H 5 NO 2 + H 2 O 4.C 5 H 10 + O 2  CH 2 O 5.N 2 + C 2 H 6  N 2 H 4 + C 2 H 2

59. Stoichiometry Stoichiometry: the area of chemistry that considers the quantities of materials consumed and produced in chemical reactions The coefficients in a balanced reaction tell you the RATIO of MOLES for each compound that are needed for the reaction

60. Reading the Reaction Equation Cl 2 + 2 KBr  2 KCl + Br 2 – How many moles of Cl 2 will react with 2 moles of KBr? P 4 + 5 O 2 + 6 H 2 O  4 H 3 PO 4 – How many moles of O 2 are needed to produce 4 moles of H 3 PO 4 ? 2 Fe + 3 Cl 2  2 FeCl 3 – How many moles of Fe will produce 2 moles of FeCl 3 ?

61. Reading the Reaction Equation Cl 2 + 2 KBr  2 KCl + Br 2 – How many moles of Cl 2 will react with 2 moles of KBr? 1 mole P 4 + 5 O 2 + 6 H 2 O  4 H 3 PO 4 – How many moles of O 2 are needed to produce 4 moles of H 3 PO 4 ? 5 moles 2 Fe + 3 Cl 2  2 FeCl 3 – How many moles of Fe will produce 2 moles of FeCl 3 ? 2 moles

62. Stoichiometry P 4 + 5 O 2 + 6 H 2 O  4 H 3 PO 4 What is the mole ratio between O 2 and H 3 PO 4 ? (How many moles of O 2 per how many moles of H 3 PO 4 ?)

63. Stoichiometry P 4 + 5 O 2 + 6 H 2 O  4 H 3 PO 4 What is the mole ratio between O 2 and H 3 PO 4 ? (How many moles of O 2 per how many moles of H 3 PO 4 ?) There are 5 mol of O 2 per 4 mol of H 3 PO 4 5 mol O 2 or 4 mol H 3 PO 4 4 mol H 3 PO 4 5 mol O 2

64. Stoichiometry P 4 + 5 O 2 + 6 H 2 O  4 H 3 PO 4 What is the mole ratio between P 4 and O 2 ? What is the mole ratio between P 4 and H 2 O? What is the mole ratio between P 4 and H 3 PO 4 ?

65. Stoichiometry P 4 + 5 O 2 + 6 H 2 O  4 H 3 PO 4 What is the mole ratio between P 4 and O 2 ? 1mol P 4 / 5 mol O 2 What is the mole ratio between P 4 and H 2 O? 1mol P 4 / 6mol H 2 O What is the mole ratio between P 4 and H 3 PO 4 ? 1mol P 4 /4mol H 3 PO 4

66. Stoichiometry P 4 + 5 O 2 + 6 H 2 O  4 H 3 PO 4 What is the mole ratio between O 2 and H 2 O? What is the mole ratio between H 2 O and H 3 PO 4 ?

67. Stoichiometry P 4 + 5 O 2 + 6 H 2 O  4 H 3 PO 4 What is the mole ratio between O 2 and H 2 O? 5mol O 2 / 6mol H 2 O What is the mole ratio between H 2 O and H 3 PO 4 ? 6mol H 2 O/ 4mol H 3 PO 4

70. Moles to Moles We use the mole ratio (from the balanced reaction) as a conversion factor P 4 + 5 O 2 + 6 H 2 O  4 H 3 PO 4 If we started with 6mol of O 2, how many moles of H 3 PO 4 would we produce? (Set up the conversion!)

71. Moles to Moles P 4 + 5 O 2 + 6 H 2 O  4 H 3 PO 4 If we started with 6mol of O 2, how many moles of H 3 PO 4 would we produce? (Set up the conversion!) 6mol O 2 x 4mol H 3 PO 4 = 24mol H 3 PO 4 = 4.8mol 15mol O 2 5mol  4.8 mol of H 3 PO 4

72. Moles to Moles to Grams P 4 + 5 O 2 + 6 H 2 O  4 H 3 PO 4 We determined that if we start with 6mol O 2, we produce 4.8mol H 3 PO 4. How many grams is this? (Continue with the same conversion set- up! Round molar mass to the nearest whole number.)

73. Moles to Moles to Grams P 4 + 5 O 2 + 6 H 2 O  4 H 3 PO 4 We determined that if we start with 6mol O 2, we produce 4.8mol H 3 PO 4. How many grams is this? 6mol O 2 x 4mol H 3 PO 4 x 98g H 3 PO 4 = 470.4g 15mol O 2 1mol H 3 PO 4

74. Grams to Moles to Moles to Grams P 4 + 5 O 2 + 6 H 2 O  4 H 3 PO 4 If we start with 25g of O 2, how many grams of H 3 PO 4 will we produce? (Round molar masses to the nearest whole number)

75. Grams to Moles to Moles to Grams P 4 + 5 O 2 + 6 H 2 O  4 H 3 PO 4 If we start with 25g of O 2, how many grams of H 3 PO 4 will we produce? 25g O 2 x 1mol O 2 x 4mol H 3 PO 4 x 98g H 3 PO 4 = 132g5mol O 2 1mol H 3 PO 4 =61.25g H 3 PO 4

76. Exit Ticket #4

77. Limiting Reactant Limiting Reactant: the reactant that limits the amount of product that can be formed in a reaction Hint: when the word “excess” is used in a problem, it means there is MORE than enough of that reactant, so that reactant will not be limiting

78. Finding the Limiting Reactant For each of your reactants, multiply the number of MOLES you start with by the coefficient in front of that compound in the balanced reaction Whichever compound has the smaller number is the limiting reactant

79. Finding the Limiting Reactant 1 C 3 H 8 + 5 O 2  3 CO 2 + 4 H 2 O You have 3mol C 3 H 8 of and 1.8mol of O 2. What is the limiting reactant?

80. Finding the Limiting Reactant 1 C 3 H 8 + 5 O 2  3 CO 2 + 4 H 2 O You have 3mol C 3 H 8 of and 1.8mol of O 2. What is the limiting reactant? C 3 H 8 : 3mol x 1 = 3mol  Limiting O 2 : 1.8mol x 5 = 9mol

81. Finding the Limiting Reactant 1 C 3 H 8 + 5 O 2  3 CO 2 + 4 H 2 O You have 7.6mol C 3 H 8 of and 2.3mol of O 2. What is the limiting reactant?

82. Finding the Limiting Reactant 1 C 3 H 8 + 5 O 2  3 CO 2 + 4 H 2 O You have 7.6mol C 3 H 8 of and 2.3mol of O 2. What is the limiting reactant? C 3 H 8 : 7.6mol x 1 = 7.6mol  Limiting O 2 : 2.3mol x 5 = 11.5mol

83. Finding the Limiting Reactant 1 C 3 H 8 + 5 O 2  3 CO 2 + 4 H 2 O You have 15.9mol C 3 H 8 of and 3.3mol of O 2. What is the limiting reactant? C 3 H 8 : 15.9mol x 1 = 15.9mol O 2 : 3.1mol x 5 = 15.5mol  Limiting

84. Finding the Limiting Reactant 1 C 3 H 8 + 5 O 2  3 CO 2 + 4 H 2 O You have 7.5mol C 3 H 8 of and 1.5mol of O 2. What is the limiting reactant? C 3 H 8 : 7.5mol x 1 = 7.5mol O 2 : 1.5mol x 5 = 7.5mol  No limiting reactant

85. Finding the Limiting Reactant 1 C 3 H 8 + 5 O 2  3 CO 2 + 4 H 2 O You have 45g C 3 H 8 of and 40g of O 2. What is the limiting reactant?

86. Finding the Limiting Reactant 1 C 3 H 8 + 5 O 2  3 CO 2 + 4 H 2 O You have 45g C 3 H 8 of and 40g of O 2. What is the limiting reactant? * Since we’re given grams, we must convert to moles!

87. Finding the Limiting Reactant 1 C 3 H 8 + 5 O 2  3 CO 2 + 4 H 2 O You have 45g C 3 H 8 of and 40g of O 2. What is the limiting reactant? C 3 H 8 : 45g x 1mol = 1.02mol x 1 = 1.02mol (limiting) 44.097g O 2 : 40g x 1mol = 1.25mol x 5 = 6.25mol 31.998g

88. Determining Yield Theoretical Yield: the amount of product that a reaction produces if it reaches completion under perfect conditions – Once you know the limiting reactant, you can calculate how much of your product the reaction should produce – To set up the conversion, use whatever amount of the limiting reactant was given to you, then set up the conversion from there

89. Determining Yield 1 C 3 H 8 + 5 O 2  3 CO 2 + 4 H 2 O You have 3mol C 3 H 8 of and 1.8mol of O 2. How many grams of H 2 O will be produced? C 3 H 8 : 3mol x 1 = 3mol  Limiting O 2 : 1.8mol x 5 = 9mol

90. Determining Yield 1 C 3 H 8 + 5 O 2  3 CO 2 + 4 H 2 O You have 3mol C 3 H 8 of and 1.8mol of O 2. How many grams of H 2 O will be produced? C 3 H 8 : 3mol x 1 = 3mol  Limiting O 2 : 1.8mol x 5 = 9mol 3mol C 3 H 8 x 4mol H 2 O x 18g H 2 O = 216g H 2 O 1 1mol C 3 H 8 1mol H 2 O

91. Determining Yield 1 C 3 H 8 + 5 O 2  3 CO 2 + 4 H 2 O You have 7.6mol C 3 H 8 of and 2.3mol of O 2. How many grams of CO 2 are produced? C 3 H 8 : 7.6mol x 1 = 7.6mol  Limiting O 2 : 2.3mol x 5 =11.5mol

92. Determining Yield 1 C 3 H 8 + 5 O 2  3 CO 2 + 4 H 2 O You have 7.6mol C 3 H 8 of and 2.3mol of O 2. How many grams of CO 2 are produced? C 3 H 8 : 7.6mol x 1 = 7.6mol  Limiting O 2 : 2.3mol x 5 =11.5mol 7.6mol C 3 H 8 x 3mol CO 2 x 44g CO 2 = 1,003g H 2 O 1 1mol C 3 H 8 1mol CO 2

93. Determining Yield 1 C 3 H 8 + 5 O 2  3 CO 2 + 4 H 2 O You have 200g C 3 H 8 of and 200g of O 2. How many grams of H 2 O are produced?

94. Determining Yield 1 C 3 H 8 + 5 O 2  3 CO 2 + 4 H 2 O You have 200g C 3 H 8 of and 200g of O 2. How many grams of H 2 O are produced? C 3 H 8 : 200g x 1mol = 4.55mol x 1 = 4.55mol: Limiting 44g O 2 : 200g x 1mol = 6.25mol x 5 = 31.25mol 32g 200g C 3 H8 x 1mol C 3 H 8 x 4mol H 2 O x 18g H 2 O = 327.3g 1 44g C 3 H 8 1mol C 3 H 8 1mol H 2 O

95. Percent Yield (Recovery) Percent Yield (Percent Recovery): the percent of the theoretical yield that was obtained during an experiment – Will always be less than 100% Percent Yield = Actual Yield x 100% Theoretical Yield

96. Percent Yield (Recovery) Previously, we calculated that 216g H 2 O would be produced if we started with 3mol C 3 H 8 and 1.8mol of O 2. If we recovered 200g, what was our percent yield (percent recovery)?

97. Percent Yield (Recovery) Previously, we calculated that 216g H 2 O would be produced if we started with 3mol C 3 H 8 and 1.8mol of O 2. If we recovered 200g, what was our percent yield (percent recovery)? % Yield = 200g x 100 = 92.6% 216g

98. Percent Yield (Recovery) Previously, we calculated that 1,003g H 2 O would be produced if we started with 7.6mol C 3 H 8 and 2.3mol of O 2. If we recovered 578g, what was our percent yield (percent recovery)? % Yield = 578g x 100 = 57.6% 1,003g

99. Summary of Stoichiometry 1.Balance the reaction. 2.Find the limiting reactant by multiplying starting moles by the coefficients in the reaction. 3.Determine the theoretical yield (conversions). 4.If given actual yield, determine percent yield.

100. Making Solutions: Molarity Molarity: the concentration of a solution, measured in moles of solute per liter of solvent M = mol solute / L solvent Solute: what you’re dissolving Solvent: what does the dissolving (usually a liquid)

101. Making Solutions If you have 10mol of HCl dissolved in 0.75L of water, what is the molarity of the solution?

102. Making Solutions If you have 10mol of HCl dissolved in 0.75L of water, what is the molarity of the solution? M = mol/ L M = 10mol/ 0.75L = 13.3M

103. Making Solutions If you have a 6M solution of HCl that is 300mL, how many moles of HCl are dissolved?

104. Making Solutions If you have a 6M solution of HCl that is 300mL, how many moles of HCl are dissolved? M = mol/ L 300mL x 1L/ 1,000mL = 0.3L 6M = mol/ 0.3L *0.3L = *0.3L  1.8mol

105. Making Solutions You’re trying to make a 1.25M solution of HCl. If you have 500g of HCl, what volume must the solution be in L? In mL?

106. Making Solutions You’re trying to make a 1.25M solution of HCl. If you have 500g of HCl, what volume must the solution be in L? In mL? M = mol/L 500g x 1mol/ 36g = 13.9mol 1.25M = 13.9mol/ L  11.1L  11.1L x 1,000mL/ 1L = 11, 100L

107. Dilutions Dilution: adding solvent to a solution to decrease the molarity (concentration) The product of the initial volume and molarity = the product of the final volume and molarity M 1 V 1 = M 2 V 2 M = concentration (molarity) V = volume

108. Dilutions Example You have a 250mL stock solution of 5M HCl. You need to make a dilution to 1.5M HCl. What must the final volume be? How much water will you have to add?

109. Dilutions Example You have a 250mL stock solution of 5M HCl. You need to make a dilution to 1.5M HCl. What must the final volume be? How much water will you have to add? M 1 V 1 = M 2 V 2 5M*250mL = 1.5MV 2 1,250M*mL = 1.5MV 21.5M 833.3mL = V 2

110. Molarity of Sugar in Sodas Activity Your task is to determine the molarity (concentration) of sugar in various beverages. To do this, you must determine: – The amount of sugar in one serving – The volume of one serving We will make the following assumptions: – All sugar in the beverage is fructose: C 6 H 12 O 6 – There are 30mL per 1 fl. oz. When you finish, read the article on sugar and tooth decay, and relate the molarity of sugar that you found to the article.

111. Exit Ticket #6

112. Show all work, including the original equation. The molar mass of HCl is 36.5g/mol. Round your answers to the nearest tenth. 1.How many mol of HCl are in a 2L solution with a concentration of 6.5M? (1 point) 2.How many grams of HCl are required to make a 2M, 350mL solution? (2 points) 3.What volume of HCl is needed to create a 3.5M solution from a starting mass of 150g? (3 points) 4.You dilute a 5.5M solution of HCl from 100mL to 250mL. What is the final concentration? (1 point) 5.You leave a 200mL solution of 3M HCl out overnight, and some of the water evaporates. If the final volume is 125mL, what is the final concentration? (2 points) 6.You have a 3M solution of HCl. You add water to the solution until you reach a final concentration of 1.2M. If you started with 175mL, how much water did you add? (3 points)