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Chemistry Unit 3! Class 2

Chemistry Unit 3! Class 2 . “Without passion, you don’t have energy; without energy, you have nothing. Nothing great in the world has been accomplished without passion” – Donald Trump - . More Calculations:. The true meaning of the ‘mole’…. Ie / 1 mole of peas = 6.022 x 10 23 peas.

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Chemistry Unit 3! Class 2

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  1. Chemistry Unit 3!Class 2 “Without passion, you don’t have energy; without energy, you have nothing. Nothing great in the world has been accomplished without passion” – Donald Trump -

  2. More Calculations: • The true meaning of the ‘mole’… Ie/ 1 mole of peas = 6.022 x 1023 peas. Anything with 6.022 x 1023 units = 1 mol Avogadro’s number KEYWORDS: ‘How many…atoms/ molecules are there’

  3. Also … • We also know that m = n x Mr • This lets us find out how many molecules/atoms there are from the MASS! Mosquitoes have 47 teeth • Eg/ Sam weighed out 5.6g of potassium dichromate (K2Cr2O7 ). a) How many molecules are there? b) How many dichromate molecules? c) How many potassium ions are there? d) How many oxygen atoms are there? • e) How many atoms are there?

  4. Stoichiometry…

  5. Questions: • Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid?

  6. Another famous mole formula Note: We will do this in more detail as we start volumetric analysis. n = CV n= mol.C = concentration (Molarity – mol/L Why?)V = volume (L) Eg/ Tracey had 22.5 ml of hydrochloric acid it was at a concentration of 5M. How many mole of HCl is there?

  7. Limiting / Excess Reactions: • Sometimes the reactants are not mixed in EXACT ratio. This is when we have to do all the calculations with the LEAST amount If I have 2 mole of X and 3 mole of Y I cannot use 3 mole in my calculations Since X runs out it is ‘impossible’ to create Z and Y.

  8. How to picture this… 1) Each molecule will collide… once it does it creates the ‘pink pill’ + = But there are only 3 ‘boxes’ and 6 ‘circles’

  9. 5 Steps to find Limiting Reactant

  10. A problem:

  11. How to find out how much reactant is left Over

  12. Example • What is the mass of excess left? 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)The mass of superoxide is 45 g and water is 32 g.

  13. Theoretic yield vs. Actual Yield - Theoretic yield is the amount of product formed in theory. • - Actual yield is the amount of product ACTUALLY made. - It will be different due to loss in precipitate, or other unavoidable error.

  14. A Problem… • Suppose the theoretical yield for an experiment was calculated to be 19.5 grams, and the experiment was performed, but only 12.3 grams of product were recovered. Determine the % yield.

  15. Percentage by Mass

  16. Unit Conversion µg Gg ng

  17. Worked Example: • 0.0008 mg into grams. • 7458 μg into grams • 0.04 mg into μg • 560493 ng into mg • 456 k-ants to M-antsHint:In chem. you generally work will SI units (grams for mass, L for volume) unless otherwise stated in the exam

  18. Students that get 99 + Students with 85 - 95 Students below 70 Students 70-85

  19. Gravimetric Analysis

  20. Precipitate = cloudy

  21. The Steps of Gravimetric analysis

  22. Why must there be excess ? • We need excess AgNO3 to make sure that all the salt within the kebab undergoes the reaction • The more AgNO3 we add the greater the probability that there will be a collision with the salt. • This will increase the chance that 100% of the salt turns to AgCl (s) • This way error is decreased.

  23. The Kebab story….Part 2: • The NaCl from the dissolved kebab will react with the excess AgNO3 • This will create a precipitate – this precipitate must be STABLE as if it reacts it will look like there is less of NaCl • This will create a precipitate – this precipitate must be STABLE as if it reacts it will look like there is less of NaCl • We then filter the solution…

  24. Kebab’s Adventure Part 3… • All chemists care about is the solid precipitate because that’s the only PRACTICLE thing they can measure – i.e. weighing • But before weighing – • 1) The precipitate must be washed: To remove contaminants such as free ions that can effect weight • 2) The precipitate must be dried: Water on the AgCl will increase mass – we only care about AgCl … all the water must be removed.

  25. A Few Notes on the precipitate: • The precipitate must be inert (i.e. non reactive!!!) otherwise you cannot accurately measure it. • The precipitate must have a large molar mass – The larger it is the less error there will be in the calculations. • The precipitate must be FULLY INSOLUBLE – some are partial DO NOT use them as you will not get an accurate answer. • The precipitate must be stable when heated and must have a KNOWN molar mass.

  26. How do we get rid of the water? • 1) Weigh the precipitate + water (initial) • 2) The precipitate must be heated in an oven • This way the water will evaporate • Nothing should happen to the precipitate (no shrinking, charring etc) • 3) Take it out – weigh + record • 4) Put it back into oven – weigh + record • 5) Repeat cycle 3) & 4) until constant mass.

  27. A Problem… • Jack wanted to calculate the salt content of a packet of fries from Mc Donald’s:a) Write a flow chart of what he should do.b) Jack decides to heat the solution when the precipitate is forming. Why?

  28. What do we used the weighed precipitate for?

  29. A Problem: • Jenny wants to calculate the amount of salt in pizza. The pizza weighs 0.78 kg. Her doctor advised her that she shouldn’t eat any foods that have 3.5% of salt by mass (her blood pressure is too high). If the mass of the precipitate weighed 1097 mg … can Jenny eat the pizza?

  30. Other types of questions: • Calculate the grams analyte of the precipitate for the following: i) P (at wt =30.97) in Ag3PO4 (FW = 711.22)ii) Bi2S3 (FW 514.15) in BaSO4 (FW = 233.40) • Trick: Look for the common element. Make sure that the elements are equivalent in number!!!

  31. Another Example: • Phosphate in a 0.2711 g sample was precipitated giving 1.1682 g of (NH4)2PO4.12 MoO3 (FW = 1876.5). i) Find percentage P (at wt = 30.97) ii) Find the percentage P2O5 (FW = 141.95) in the sample.

  32. Yet another: • Manganese in a 1.52 g sample was precipitated as Mn3O4 (FW = 228.8) weighing 0.126 g.i) Find percentage Mn2O3 (FW = 157.9) ii) Find percentage of Mn (at wt = 54.94) in the sample

  33. And another…… • A mixture containing only FeCl3 (FW = 162.2) and AlCl3 (FW = 133.34) weighs 5.95 g. The chlorides are converted to hydroxides and ignited to Fe2O3 (FW = 159.7) and Al2O3 (FW = 101.96). The oxide mixture weighs 2.26 g. Calculate:i) the percentage Fe (at wt = 55.85) ii) the % of Al (at wt = 26.98) in the sample.

  34. A 0.4960 g sample of a CaCO is dissolved in an acidic solution. The calcium is precipitated as CaC2O4 .H2O and the dry precipitate is found to weigh 0.6186 g. What is the percentage of CaO in the sample?

  35. What I want you to do: • The questions Set 2 • Go over the lectures (both) • Attempt the topic test for lecture 1 • Study for at least 30 mins. a day for chemistry. • If you get any SAC’s send a PDF file to my email and I will go over the important things you must include in email.

  36. Next Week We go on with: • Gravimetric analysis – we finish it • Empirical formulae • Gas equations • Begin volumetric analysis

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