Chapter 2: Composition stoichiometry the relative ratios of different elements within one particular compound or molecule : Reaction stoichiometry the relative ratios between different substances as they react with each other Reactants. products. change NaOH + HCl 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. 1
Ex.1) CH 4 + 2O 2 CO 2 + 2H 2 O Ex. 2) The combustion of propane ( burning ) C 3 H 8 + O 2 CO 2 + H 2 O C 3 H 8 + 5O 2 3CO 2 + 4H 2 O What if we write 2C 3 H 8 + 10O 2 6CO 2 + 8H 2 O -does this equation obey the law of conservation of matter? yes -is this a correct chemical equation for propane combustion? no 2
Ex.3) The combustion of methanol CH 3 OH + O 2 CO 2 + H 2 O CH 3 OH + O 2 CO 2 + 2H 2 O 2CH 3 OH + 3O 2 2CO 2 + 4H 2 O Balancing equations is usually a trial and error process, so the best way to learn is from lots of practice. Some helpful hints: 1) look for elements that are only in one component on each side of the iron (Fe in previous example) 2) if free elements occur on either side (O 2 in combustion reactions), balance last 3) check to see if coefficients are all divisible by 2 or 3 to give whole numbers 3
Chemical equations are a powerful planning tool: If I want to make NaCl, I now have a recipe! NaOH + HCl 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. 4
Calculations involving balanced equations. following cooking analogy: Consider the chicken breasts + veggies + oil + rice 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 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 5
We use balanced equations to set up mole ratios to do the same kind of calculations. Ex. if we want to make 0.750 moles of AgCl, how much CaCl 2 do we need? How much AgNO 3? CaCl 2 + 2AgNO 3 Ca(NO 3 ) 2 + 2AgCl 0.750 mol AgCl x 1 mol CaCl 2 = 0.375 mol CaCl 2 2 mol AgCl 0.750 mol AgCl x 1 mol AgNO 3 = 0.375 mol AgNO 3 2 mol AgCl How much Ca(NO 3 ) 2 will we also make? 0.750 mol AgCl x 1 mol Ca(NO 3 ) 2 = 0.375 mol AgNO 3 2 mol AgCl This example involves moles. But what about grams? 6
CaCl 2 + 2AgNO 3 Ca(NO 3 ) 2 + 2AgCl 10.0 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 10.0g CaCl 2 x 1 mol CaCl 2 x 2 mol AgCl x 143.4g AgCl = 25.8 g AgCl 110.0g CaCl 2 1 mol CaCl 2 1 mol AgCl 7
Problem solving strategy: 1. what do we want to know? # g AgCl 2. what do we know? #g CaCl 2 3. 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 CaCl 2 mol 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 1.00 g mol mol? g CaCl 2 molar mass CaCl 2 mole ratio AgCl molar mass AgCl 8
In these calculations, we have been assuming that we have plenty of the other reactants - plenty of AgNO 3 for CaCl 2 but what if I have limited quantities of both reactants? Which quantity determines the amount of product? Limiting reactant a substance that limits the amount of product(s) that can be formed. 2 oz. Cherry syrup + 1 can coke = 1 cherry coke 8 oz. Cherry syrup 6-pack coke how many cherry cokes? 8 oz. Syrup x 1 cherry coke = 4 cherry cokes 2 oz. cherry syrup 6 cokes x 1 cherry coke = 6 cherry cokes 1 coke only have enough to make 4 cherry cokes syrup is limiting reactant. 9
What is the maximum mass of sulfur dioxide which can be obtained from the reaction of 95.6 g of carbon disulfide with 110. g of oxygen? Strategy: see next slide Sulfur dioxide SO 2 Carbon disulfide CS 2 Oxygen O 2 Write equation: CS 2 + O 2 SO 2 need one more product CO 2 CS 2 + O 2 SO 2 + CO 2 next balance CS 2 + 3O 2 2SO 2 + CO 2 Carbon disulfide CS 2 molar mass = 76.14 g/mol Oxygen O 2 molar mass = 32.00 g/mol 110. g O 2 x 1 mol O 2 = 3.4375 mol O 2 32.00g O 2 95.6. g CS 2 x 1 mol CS 2 = 1.2556 mol CS 2 76.14 CS 2 3O 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 110. g O 2 x 1 mol O 2 x 2 mol SO 2 x 64.07g SO 2 = 147 g SO 2 32.00g O 2 3 mol O 2 1 mol SO 2 147g is the maximum mass (there isn t enough O 2 to make 161 g SO 2 ). 10
Problem solving strategy: 1. don t panic. 2. what do we want to know? maximum mass sulfur dioxide 3. what do we know? #g carbon disulfide, #g oxygen 4. how can we get from #1 to #2? 1. write the formulas for the names 2. figure out if the compounds are reactants or products 3. write an equation & balance it 4. find the molar mass of CS 2 and O 2 5. calculate the # moles of each 6. write a reactant ratio of CS 2 and O 2 from the balanced equation 7. calculate a ratio for the actual # moles of CS 2 and O 2 8. 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 2-9. calculate the molar mass of SO 2 10. set up a calculation to go from 110. g mol mol? g oxygen oxygen sulf. diox. sulfur dioxide 11
But what if we had only gotten 100. g of SO 2 with those amounts of CS 2 and O 2? - reaction did not go to completion - several competing reactions involving CS 2 and/or O 2 were going on at the same time. Theoretical yield the amount of product we should get when a reactant is completely consumed according to our chemical equation Actual yield the amount of product actually measured The actual yield is always equal to or lower than the theoretical yield Percent yield actual yield x 100 theoretical yield the percent yield of SO 2 is 100. g SO 2 x 100 = 68.1 % 146.8 g SO 2 12
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 13
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? 20.0 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? 32.0 g solution x 0.0800 = 2.56 g NaOH However, we often measure volume when we are working with liquids, not weight. 14
Density mass/volume g/ml or g/cm 3 density of a pure substance can change depending on - phase changes (solid, liquid, gas) - temperature density of a solution depends on - the identities and relative amounts of solute/solvent ex. calculate the mass of NaOH in 300.0 ml of a 8.00 % solution if the density = 1.09 g/ml. 300.0 ml x 1.09 g x 8.00 g NaOH = 26.2 g NaOH 1 ml 100.g solution ex. what volume of 12.0 % KOH solution contains 40.0 g KOH? the density = 1.11 g/ml. 40.0 g KOH x 100g solution x 1 ml = 300. ml KOH solution 12.0 g KOH 1.11 g But what if we want to do reactions with solutions? To do calculations involving reaction stoichiometry, we need to use moles. We can add another step to go from grams to moles in these calculations, but there is another way of expressing concentration that is more convenient for this purpose. 15
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 40.0 g NaOH x 1 mol NaOH x 1 = 2.0 M solution 40.00g NaOH 0.50L Note not 0.50 L of H 2 O. 0.50 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 2.50 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) 16