HW #6: 6.30, 6.32, 6.38, 6.40, 6.42, 6.48, 6.50, 6.54, 6.62, 6.72, 6.76

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1 Chemistry 121 Lecture 13: Avogadro s Number and the Mole; Gram-Mole Conversions; Stoichiometry - Mass/Mole Relationships in Chemical Reactions; Limiting Reagent and Percent Yield Chapter 6 in McMurry, Ballantine, et. al. 7 th edition HW #6: 6.30, 6.32, 6.38, 6.40, 6.42, 6.48, 6.50, 6.54, 6.62, 6.72, 6.76 Objectives: 1. Define the mole as a scale-up number used to make convenient measurements of groups of elements, ionic compounds or molecules 2. Determine the formula weight or molecular weight (if appropriate) of a compound 3. Use formula weight or molecular weight to determine the mass of a compound needed to obtain a given number of moles 4. Revisit the balanced chemical equation and reinforce why the numbers of every type of atom must be the same on the reactant and product sides of the equation based on conservation of mass 5. Given a balanced chemical equation, determine the mass of reactant needed to completely convert another reactant 6. Given a balanced chemical equation, determine the mass of reactants needed to form a given amount of product 7. Given a balanced chemical equation, determine the mass of product formed from a given amount of reactant Define and employ the concept of limiting reagent 8. Given a balanced chemical equation, determine the % yield for a reaction based on mass of product actually obtained 1

2 Chapter 6: Chemical Reactions Mole & Mass Relationships 6.1: Avogadro s Number and the Mole A mole is just a number it just happens to be a very large number 6.02 x and is employed so that we have a convenient scale to manipulate compounds Avogadro s Avocados Revisited Avogadro s number = 6.02 x A medium sized California avocado weighs 173 g Avogadro s avocados: 173 g/avocado (6.02 x avocados/mole) = 1.04 x g 1.04 x g(1 kg/1000g)(metric ton/1000 kg) = 1.04 x metric tons A heavy duty rail car is 60 and can carry approximately 100 metric tons 1.04 x metric tons(rail car/100 metric tons) = 1.04 x rail cars If we had an enormously powerful locomotive traveling at 60 /s (40.9 mi/hr), a 60 rail car would go by every second and it would take 1.04 x s(1 min/60 s)(1 hr/60 min)(1 day/24 hr)(1 yr/365 day) = 3.30 x years for one mole of avocados to pass. In order for Avogadro s avocado train to finish passing today, it would have had to have left the station before there was a station best estimates of Earth s age is 4.6 billion years of course this is ridiculous since flowering plants of which the avocado is included - did not appear until about the time of the late Triassic, a mere 200 million years ago If you stood in front of the train every day for the whole of your life and you lived 100 years, you would have witnessed % of 1 mole pass 6.02 x is a VERY large number 2

3 So then, how do we get a mole of something? We have developed a relative scale such that 12g of 12 C is exactly 1 mole; we compare everything else to this the beauty being if you want a mole of something all you have to do is weigh out the formula weight of it in grams 6.2: Gram to Mole Conversions Formula Weight and Molecular Weight Formula weight is general for ionic and molecular compounds; the term molecular weight should properly be reserved for covalent species. The formula weight of a given compound is found by adding up the atomic weight of all the atoms in the compound: Na NaOH HCl BaSO4 H2SO4 Fe2(SO4)3 C3H8 C6H12O6 NaC2H3O2 Another term for formula weight is molar mass, the mass [in grams] needed to obtain 1 mole of a compound The formula weight (or molar mass) gives you the conversion factor you need for mass to mole conversion and visa versa Question: How many mol NaOH are in 40 g of pure NaOH? Question: How many mol HCl are in 36.4 g HCl gas? 3

4 6.3, 6.4: Stoichiometry Stoichiometry the quantitative mass relationships in a chemical reaction All you need to remember is the total number of each type of atom must be the same on either side of a chemical equation; since you cannot change the number of atoms within a given molecule, you must change the number of molecules you do this with the stoichiometric [lead] coefficient. Again, mass is conserved in a [non-nuclear] chemical reaction and the atoms contain the mass, so there must be the same number of each type of atom on both the reactant and product side Question: How many mol NaOH would be required to completely react with 36.4 g HCl? For this, we need the balanced chemical equation: NaOH + HCl H2O + NaCl Answer: Since we have the number of moles (1.00) in 36.4 g HCl from above, all we need to determine is the molar ratio in which NaOH and HCl react Question: What mass of NaOH would be required to completely react with 36.4 g HCl? Answer: Convert the number of moles above to the corresponding mass: Question: What mass of NaOH would be required to completely react with 98 g H2SO4? 2NaOH + H2SO4 2H2O + Na2SO4 4

5 More on balancing simple chemical equations Both of the above (acid-base) reactions may be efficiently viewed as a swap between H + and the sodium cation Na + It is handy to be able to identify the polyatomic ions and see them as stable units this greatly simplifies your approach to balancing the equations o Sulfuric acid is sulfate with 2 available H + to donate which would allow 2 OH - to reform water, so 2 NaOH are necessary Again, in cases where an elemental species is involved, balance the elemental species last In order to get a better feel for the need to balance chemical reactions, let s consider the combustion of methane (CH4) vs. the combustion of propane (C3H8). Combustion of methane liberates 212 kcal/mol, and propane liberates 530 kcal/mol. CH4 + O2 C3H8 + O2 Question: Which gives the most energy on a gram basis? Question: Which gives more H2O on a gram basis? Question: Which gives more CO2 on a gram basis? 5

6 From the balanced chemical equation we can determine 1. How much reactant is needed to completely convert another reactant, assuming complete reaction What mass of oxygen is needed to completely combust 116 g of butane (C4H10)? 2. The number of moles of product (and thus mass of product) formed from a given mass of reactants What mass of CO2 is generated in the combustion of 116 g (C4H10)? 3. How much reactant is needed to produce a given number of moles of product (and thus mass of product), assuming complete reaction How much butane would have to be burned to give 44 g CO2? 6

7 6.5: Limiting Reagent and Percent Yield 4. Determine the % yield for a reaction, by comparing actual mass of product obtained to maximum mass possible for that product If combustion of 116 g of butane with 52 g O2 gave 44 g CO2, what would the % yield be? Answering this question requires 2 important concepts: theoretical yield and limiting reagent The theoretical yield is the maximum mass possible based on the masses reacted and the balanced chemical equation. In a perfect world all reactions would be quantitative; sadly, this is not the case and we communicate how well a given reaction converts reactants to products by o % yield = (actual yield/theoretical yield) x 100 If you have an excess of 1 reagent the other will limit the amount of product possible and thus the theoretical yield. The reagent you run out of first, naturally, is the limiting reagent o Since you can do no better than what the limiting agent allows where yield is concerned, base your theoretical yield on the limiting reagent 7

Chemical Quantities: Stoichiometry and the Mole

Chemical Quantities: Stoichiometry and the Mole This is trying to summarize what we have learned up to this point: formulas, names, conversions, moles, quantities, reaction types, balancing equations,