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Chapter 3 Chapter 2 : Composition stoichiometry – the relative ratios of different elements within one particular compound or molecule Chapter 3 : Reaction stoichiometry – the relative ratios between different substances as they react with each other Reactants. products. | | change | | NaOH + HCl à NaCl + H 2 O | not? | chemical equation Law of conservation of matter – a balanced chemical equation must always include the same # of each kind of atom on both sides of the equation.
Chapter 3 Ex. 1 ) CH 4 + 2 O 2 à CO 2 + 2 H 2 O Ex. 2 ) The combustion of propane (‘burning’) C 3 H 8 + O 2 à CO 2 + H 2 O C 3 H 8 + 5 O 2 à 3 CO 2 + 4 H 2 O What if we write 2 C 3 H 8 + 10 O 2 à 6 CO 2 + 8 H 2 O
- does this equation obey the law of conservation of matter? yes
- is this a ‘correct’chemical equation for propane combustion? no
Chapter 3 Chemical equations are a powerful ‘planning’tool: If I want to make NaCl, I now have a recipe! NaOH + HCl à NaCl + H 2 O If I want to make 27 formula units of NaCl, how many units of the following ingredients do I need? NaOH 27 HCl 27 but chemists work with large amounts – use moles.
Chapter 3 Calculations involving balanced equations. Consider the following cooking analogy: chicken breasts + veggies + oil + rice à stir fry we can use the recipe to write a balanced equation: 4 chicken 8 cups 3 tbsp 4 cups 8 servings breasts + veggies + oil + rice à stir fry But what if we only have 4 dinner guests (+ yourself) and we want to cut back on the amount of food we make? How much chicken do we need? How many cups of vegetables? We set up a ratio that relates the ingredients to one another. 5 servings x 4 chicken breasts = 2. 5 chicken 8 servings breasts 5 servings x 8 cups veggies = 5 cups 8 servings veggies
Chapter 3 CaCl 2 + 2 AgNO 3 à Ca(NO 3 ) 2 + 2 AgCl
- 0 g CaCl 2 yields how many grams of AgCl? Strategy: see next slide Molar mass CaCl 2 = 40. 08 + 2 ( 35. 45 ) = 110. 98 g/mol Molar mass AgCl = 107. 9 + 35. 45 = 143. 4 g/mol
- 0 g CaCl 2 x 1 mol CaCl 2 x 2 mol AgCl x 143. 4 g AgCl = 25. 8 g AgCl
- 0 g CaCl 2 1 mol CaCl 2 1 mol AgCl
Chapter 3 Problem solving strategy:
- what do we want to know? # g AgCl
- what do we know? #g CaCl 2
- how can we get from # 1 to # 2? g CaCl 2 g AgCl ? How can you get AgCl from CaCl 2? Use the balanced equation to create a mole ratio mol à mol CaCl 2 AgCl ¦ how can I do this kind of problem when the quantities that are given and/or requested are in grams? Convert to moles! Calculate the molar masses of AgCl and CaCl 2 Why? to relate grams to moles
- 00 g à mol à mol à? g CaCl 2 molar mass CaCl 2 mole ratio AgCl molar mass AgCl
Chapter 3 What is the maximum mass of sulfur dioxide which can be obtained from the reaction of 95. 6 g of carbon disulfide with
- g of oxygen? Strategy: see next slide Sulfur dioxide SO 2 Carbon disulfide CS 2 Oxygen O 2 Write equation: CS 2 + O 2 à SO 2 need one more product – CO 2 CS 2 + O 2 à SO 2 + CO 2 next balance CS 2 + 3 O 2 à 2 SO 2 + CO 2 Carbon disulfide CS 2 molar mass = 76. 14 g/mol Oxygen O 2 molar mass = 32. 00 g/mol
- g O 2 x 1 mol O 2 = 3. 4375 mol O 2
- 00 g O 2
- g CS 2 x 1 mol CS 2 = 1. 2556 mol CS 2
- 14 CS 2 3 O 2 = 3. 0 3. 4375 mol O 2 = 2. 74 limiting reactant is O 2 CS 2 1. 2556 mol CS 2 Sulfur dioxide SO 2 molar mass = 64. 07 g/mol
- g O 2 x 1 mol O 2 x 2 mol SO 2 x 64. 07 g SO 2 = 147 g SO 2
- 00 g O 2 3 mol O 2 1 mol SO 2 147 g is the maximum mass (there isn’t enough O 2 to make 161 g SO 2 ).
Chapter 3 Problem solving strategy:
- don’t panic.
- what do we want to know? ‘maximum mass’sulfur dioxide
- what do we know? (^) #g carbon disulfide, #g oxygen
- how can we get from # 1 to # 2?
- write the formulas for the names
- figure out if the compounds are reactants or products
- write an equation & balance it
- find the molar mass of CS 2 and O 2
- calculate the # moles of each
- write a reactant ratio of CS 2 and O 2 from the balanced equation
- calculate a ratio for the actual # moles of CS 2 and O 2
- compare the two ratios – if the ‘actual’ratio is bigger than the one from the equation, the limiting reactant is on the bottom (if smaller, the limiting reactant is the one on top)
- limiting reactant is O 2 -
- calculate the molar mass of SO 2
- set up a calculation to go from
- g à mol à mol à? g oxygen oxygen sulf. diox. sulfur dioxide
Chapter 3 We have done a lot of calculations between moles and grams. However, a lot of reactions happen in water. Solution – a homogeneous mixture of two or more substances (ex. NaCl in H 2 O) Solute – a substance that is dissolved in another substance (NaCl) Solvent – the substance in which the other substance is dissolved (H 2 O, liquid) Aqueous solution – when solvent is H 2 O For a given quantity of a solution, how can we know:
- amount of solute?
- amount of solvent? Concentration – the amount of solute per solution
Chapter 3 Percent by mass: Percent solute = mass of solute x 100 mass of solution ex. if we have 20. 0 g NaOH dissolved in H 2 O to make 250. g of solution, what is the percent by mass of NaOH?
- 0 g NaOH = 8. 00 % NaOH ( 230 g H 2 O) 250 g solution ex. if we have 32. 0 g of an 8. 00 % aqueous solution of NaOH, how much NaOH do we have?
- 0 g solution x 0. 0800 = 2. 56 g NaOH However, we often measure volume when we are working with liquids, not weight.
Chapter 3 Molarity (another concentration) Figure 3. 2 Molarity = # moles solute.
Liters solution
ex. calculate the molarity of 40. 0 g NaOH in 0. 50 L of solution. Molar mass NaOH = 40. 00 g/mol
- 0 g NaOH x 1 mol NaOH x 1 = 2. 0 M solution
- 00 g NaOH 0. 50 L Note – not 0. 50 L of H 2 O.
- 50 L of H 2 O & NaOH combined What if we already have a solution, and we use it to make a less concentrated solution? (initial molarity/volume) M 1 V 1 = M 2 V 2 (final molarity/volume) Ex. What volume of 18. 0 M sulfuric acid is required to prepare
- 50 L of 2. 40 M H 2 SO 4 solution? We are solving for V 1. ( 18. 0 mol/L)x = ( 2. 40 mol/L)( 2. 50 L) x = ( 2. 40 mol/L)( 2. 50 L)/( 18. 0 mol/L)
