1. Stoichiometry: Mass-Volume and Mass-Energy Relationships

2. Objectives 2. Solve problems based on mass-volume of solution relationships in chemical reactions. 2. Solve problems based on mass-volume of solution relationships in chemical reactions. 3. Solve problems based on mass-energy relationships in chemical reactions 3. Solve problems based on mass-energy relationships in chemical reactions

3. Molarity Laypeople discuss the concentration of a solution in percentages: Laypeople discuss the concentration of a solution in percentages: E.g. 5% acetic acid (grocery store vinegar) E.g. 5% acetic acid (grocery store vinegar) E.g. 50% bleach E.g. 50% bleach E.g. 1% hydrogen peroxide E.g. 1% hydrogen peroxide Chemists discuss the concentration of a solution in moles. Chemists discuss the concentration of a solution in moles. How many moles of solute are in 1 Liter of a solution? How many moles of solute are in 1 Liter of a solution? If there are 0.5 moles NaCl in 1L of H 2 O, then it is a 0.5 Molar NaCl solution. If there are 0.5 moles NaCl in 1L of H 2 O, then it is a 0.5 Molar NaCl solution. Molar is abbreviated M, so we write 0.5M NaCl Molar is abbreviated M, so we write 0.5M NaCl N.B. 1L = 1000 cm 3 N.B. 1L = 1000 cm 3

4. EXAMPLE 1: Molarity What is the molarity of a solution if there are 0.504 g NaOH in 250. mL? What is the molarity of a solution if there are 0.504 g NaOH in 250. mL? ? Mol NaOH = 0.504 g NaOH * 1 mol NaOH * 1000 mL solution ? Mol NaOH = 0.504 g NaOH * 1 mol NaOH * 1000 mL solution L solution 250 mL 40.007 g NaOH 1 L solution L solution 250 mL 40.007 g NaOH 1 L solution 0.0504 M NaOH 0.0504 M NaOH This is a bit cumbersome, so it might be easier to know how many millimoles/Liter (mM) the solution is: This is a bit cumbersome, so it might be easier to know how many millimoles/Liter (mM) the solution is: ? mmol = 0.0504 mol * 1000 mmol ? mmol = 0.0504 mol * 1000 mmol L L 1 mol L L 1 mol 0.0504 M NaOH = 50.4 mM NaOH 0.0504 M NaOH = 50.4 mM NaOH

5. Mole-Volume (Solution) Conversion For reactions that occur with aqueous reactants or products, the mass of the solution doesn’t tell the mass of the chemical in question For reactions that occur with aqueous reactants or products, the mass of the solution doesn’t tell the mass of the chemical in question This is because you’re interested in the mass of the solute, but the solvent likely comprises the majority of the solution’s total mass! This is because you’re interested in the mass of the solute, but the solvent likely comprises the majority of the solution’s total mass! It also doesn’t tell you the number of moles of the chemical in question It also doesn’t tell you the number of moles of the chemical in question Since balanced chemical equations deal in molar ratios, you need to know both the volume and molarity of your solution to determine the number of moles of solute you have. Since balanced chemical equations deal in molar ratios, you need to know both the volume and molarity of your solution to determine the number of moles of solute you have. E.g. How many moles HCl are in 500. mL of a 0.375 M HCl solution? E.g. How many moles HCl are in 500. mL of a 0.375 M HCl solution? ? Mol = 500. mL HCl * 1 L HCl * 0.375 mol HCl ? Mol = 500. mL HCl * 1 L HCl * 0.375 mol HCl 1000 mL HCl 1 L HCl 1000 mL HCl 1 L HCl 0.188 mol HCl 0.188 mol HCl

6. Mass-Volume (Solution) Conversion If you can convert solution volume into moles, you can convert between mass and solution volume in order to predict theoretical yields! If you can convert solution volume into moles, you can convert between mass and solution volume in order to predict theoretical yields! Zn (s) + 2HCl (aq)  ZnCl 2 (s) + H 2(g) Zn (s) + 2HCl (aq)  ZnCl 2 (s) + H 2(g) How many grams of zinc chloride can be produced by the reaction of 45.0 mL of 0.25 M hydrochloric acid solution? How many grams of zinc chloride can be produced by the reaction of 45.0 mL of 0.25 M hydrochloric acid solution? ? g ZnCl 2 = 45.0 mL HCl * 1 L HCl * 0.25 mol HCl * 1 mol ZnCl 2 * 136.29 g ZnCl 2 ? g ZnCl 2 = 45.0 mL HCl * 1 L HCl * 0.25 mol HCl * 1 mol ZnCl 2 * 136.29 g ZnCl 2 1000 mL HCl 1 L HCl 2 mol HCl 1 mol ZnCl 2 1000 mL HCl 1 L HCl 2 mol HCl 1 mol ZnCl 2 0.77 g ZnCl 2 0.77 g ZnCl 2

7. Mass-Volume (Solution) Conversion Zn (s) + 2HCl (aq)  ZnCl 2 (s) + H 2(g) What volume of 0.300 M hydrochloric acid solution would be needed to react completely with 4.00 g of zinc? What volume of 0.300 M hydrochloric acid solution would be needed to react completely with 4.00 g of zinc? ? mL 0.300 M HCl = 4.00 g Zn * 1 mol Zn* 2 mol HCl * 1 L HCl * 1000 mL HCl ? mL 0.300 M HCl = 4.00 g Zn * 1 mol Zn* 2 mol HCl * 1 L HCl * 1000 mL HCl 136.29 g 1 mol Zn 0.300 mol HCl 1 L HCl 136.29 g 1 mol Zn 0.300 mol HCl 1 L HCl 196 mL 0.300 M HCl 196 mL 0.300 M HCl

8. Exothermic and Endothermic Reactions Chemical reactions generally either produce or use heat during the course of the reaction. This energy change doesn’t change either the composition of the reactants and products, or their molar ratios. But it is something we track. Chemical reactions generally either produce or use heat during the course of the reaction. This energy change doesn’t change either the composition of the reactants and products, or their molar ratios. But it is something we track. Exothermic Reactions produce heat (kJ) as a product Exothermic Reactions produce heat (kJ) as a product i.e. they get hot! i.e. they get hot! kJ will be listed on the product side of the reaction kJ will be listed on the product side of the reaction Endothermic Reactions use heat (kJ) as a reactant Endothermic Reactions use heat (kJ) as a reactant i.e. they get cold! i.e. they get cold! kJ will be listed on the reactant side of the reaction kJ will be listed on the reactant side of the reaction

9. Mass-Energy Conversions We can calculate the amount of energy consumed/released in the reaction of a particular quantity of a reactant We can calculate the amount of energy consumed/released in the reaction of a particular quantity of a reactant Or Or We can calculate the quantity of a reactant needed to consume/release a particular amount of energy We can calculate the quantity of a reactant needed to consume/release a particular amount of energy

10. Mass-Energy Conversion Calculate the amount of heat released, in kJ, by a living system when 1.000 g glucose, C 6 H 12 O 6, decomposes as follows. Calculate the amount of heat released, in kJ, by a living system when 1.000 g glucose, C 6 H 12 O 6, decomposes as follows. C 6 H 12 O 6 (s) + 6O 2 (g)  6CO 2 (g) + 6H 2 O (g) + 2816 kJ C 6 H 12 O 6 (s) + 6O 2 (g)  6CO 2 (g) + 6H 2 O (g) + 2816 kJ ? kJ = 1.000 g C 6 H 12 O 6 * 1 mol C 6 H 12 O 6 * 2816 kJ ? kJ = 1.000 g C 6 H 12 O 6 * 1 mol C 6 H 12 O 6 * 2816 kJ 180.2 g C 6 H 12 O 6 1 mol C 6 H 12 O 6 180.2 g C 6 H 12 O 6 1 mol C 6 H 12 O 6 Solution 15.63 kJ Solution 15.63 kJ

11. Dr. Wolfe’s Moles, Chemical Reactions, & Stoichiometry Practice Problems http://www.mindspring.com/~drwolfe/Mol es.htm http://www.mindspring.com/~drwolfe/Mol es.htm http://www.mindspring.com/~drwolfe/Mol es.htm http://www.mindspring.com/~drwolfe/Mol es.htm