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CHEMISTRY

CHEMISTRY. The Mole. Atomic or molecular Mass The molar mass of an element is the same as the atomic mass with the units called atomic units. Molecules. An Atom. A Formula Unit. Does it add up?. What is the atomic mass of Sulfur? Look on the periodic table.

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CHEMISTRY

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  1. CHEMISTRY The Mole

  2. Atomic or molecular Mass The molar mass of an element is the same as the atomic mass with the units called atomic units. Molecules An Atom A Formula Unit

  3. Does it add up? • What is the atomic mass of Sulfur? • Look on the periodic table. • S atomic mass = 32.06 au (atomic units) • Three Significant Figures = 32.1 au Silver Indium Chromium Ag In Cr 108 115 52.0

  4. What about molecules? • Molecular mass is the sum of the atomic masses in the formula of a compound. • H2O • K2SO4 Hydrogen 1.01 x 2 = 2.02 au Oxygen 16.0 au 2.02 + 16.0 = 18.0 au K = 2( 39.1) = 78.2 au S = 32.1 au O = 4( 16.0) = 64.0 au 78.2 + 32.1 + 64.0 = 174 au 3 sig figures

  5. Percent mass • The percentage of an element in a compound is determined by the mass of the element and the mass of the compound. • % Ele = atomic mass Ele x number atoms x 100% • molecular mass

  6. Calculations • Find the percent of each element in magnesium nitrate • Mg(NO3)2 has a molecular mass of 148 au • % Mg = At. Mass Mg x # atoms (subscript) • molecular mass Mg(NO3)2 • % Mg = 24.3 x 1 X 100 % = 16.4 % • 148 • % N = 14.0 x 2 X 100 % = 18.9 % • 148 • % O = 3 x 16.0 x 2 X100 % = 64.9 % • 148

  7. What about the MOLE I almost wasn’t born One day may dad ( Hidden) was showing pictures of his family and he said “ This is Enee, Meene,” and Minee and John. The other man said “ Where is Mole.” My Dad said “ There ain’t going to be no mole.” Hi, I’m Mole I have three brothers Enee, Meene, and Minee.

  8. Mole Action Ok gang lets Rock and Mole. Lets make mountains out of Mole hills, put on a little mole of Olay and learn some Chemistry. Boy, I had a great game of Golf today. I got a Mole in one.

  9. Deductive reasoning • Do equal volumes of two different gases at the same temperature and pressure have the same mass or a different mass? Volumes are equal GAS A Pressures are equal GAS B Temperatures are equal Mass Are the masses equal? Mass

  10. Let’s do an Experiment to answer the question • Apparatus and Material • 1. gal size plastic bag • 2. One connector • 3. Rubber hose

  11. Assemble • Determine the mass of the bag empty on a scale. • Fill with a gas and determine the 2nd mass. Gas connector connector Rubber tubing Plastic bag mass Tank of gas He Scale

  12. Procedure • 1. Determine the mass of the empty bag with connector device. • 2. Fill the bag with methane ( natural gas ) until the pressure is approximately the same as the room, then place on the cap. • 3. Determine the mass of the bag + gas. • 4. Expel all the methane and fill the bag with helium ( He ). • 5. Determine the mass with He

  13. What about Buoyancy? • This is Archimedes • High School This is Archimedes’s Principal Let’s talk about another Principle

  14. Archimedes Principle • When an object is immersed in a fluid it is buoyed up by a force equal to the amount of fluid displaced. • What the HECK does this mean? • Well, lets illustrate.

  15. Water is a fluid • Lets experiment. <--- Spring Scale Lower the Weight into the water and it will weighs less. Fill with water until it reaches the spout -Metal Weight Water will overflow into the beaker. The mass of the water is equal to the apparent loss of mass of the weight. Can with a spout  <-Small beaker previously weighed.

  16. Some Calculations • Mass of weight in Air 50.0 g • Mass of weight in water 45.0 g • Mass of empty beaker 26.3 g • Mass of beaker with water 31.3 g • 31.3 g • -26.3 g • Mass of water displaced 5.0 g • 50.0 g • -45.0 g • Apparent loss of mass 5.0 g Subtract beaker masses with water and empty Subtract the water mass from the air mass of the weight

  17. So what about the Experiment? • When we weigh a bag of one gas (methane) in another gas ( air ) we must take into account the buoyant force of air. • In order to find the mass of air displaced, we need the volume of air displaced so we can use the density of air to calculate it mass. • Density = mass ; mass = d x V • volume

  18. Get the Volume • We will use water displacement to determine volume of the bag which is equal to the volume of air displaced. o Gas 0 <-2 liter bottle 0 0 hose water Plastic bag

  19. Lets Do This • When we invert the 2 liter bottle (not allowing any remaining water to escape) we can determine the volume of the gas trapped. Gas from bag -> o <-Measure water volume from bottle Water left over -> 2 liter bottle Graduated cylinder

  20. Volume of Bag • We can measure the remaining water in the bottle and calculate the volume of the gas. • Volume of bottle when full 2.12 liters • Volume of water left in bottle 0.87 liters • Volume of gas in bottle 1.25 liters The density of air at room temperature is 1.18 g/liter Mair disp = dair x Vol = (1.18g/l)(1.25 l) = 1.48 g

  21. The mass of the gas • The apparent mass of the gas is what was determined from the scale. • Mass of CH4 = app mass + mass air disp. = -0.33 g + 1.48 g = 1.15 g CH4 • Mass of He = app mass + mass air disp. = -1.20 g + 1.48 g = 0.28 g CO2 Mass Ratio = CH41.15 = 4.11 He 0.28 Actual Ratio = 4.00 % error = H – L x 100% H = 4.11-4.00 x 100% 4.11 = 2.68 %

  22. What is the Scenario • Deductive Reasoning MASS Gas A Gas B ? Same Mass Different Mass

  23. Same Mass • Different number of molecules Same number of molecules Gas A Gas B Gas A Gas B If gas A had the same mass as gas B and gas A had more molecules that would mean that each molecule of gas B is heavier than each molecule of gas A by a definite ratio. If the two gasses have the same mass and number of molecules, then each molecule of gas A has the same mass as each molecule of gas B.

  24. Different Mass • Different number of molecules Same number of molecules Gas A Gas B Gas A Gas B If they had the same number of molecules but different masses, then each molecule of the heavier gas would be greater the each molecule of the lighter gas by a definite ratio. If they had a different mass and a different number of molecules then each molecule of gas A has the same mass as each molecule as gas B. Gas B would have the heaviest mass because it has more molecules.

  25. Different or Same • Different number of molecules Same number of molecules Gas A Gas B Gas A Gas B In order to determine weather or not they had different numbers of molecules or the same number we need to do an experiment.

  26. We Need A Reaction • NH3 + HCl - NH4Cl • This is a one on one reaction NH3 HCl NH4Cl Why is this important? Let’s see.

  27. An Experiment • Lets make due. NH4Cl solid Allow the HCl to react HCl NH3 3.0 ml 3.0 ml Mercury 10.0 ml 10.0 ml Force the gas into the chamber Let in the gas to 10.0 ml

  28. Avogadro's Hypothesis • Equal Volumes of two gases at the same temperature and pressure have the same number of molecules, but different masses. Same Volume Same temperature CO2 O2 Same pressure Same Number of molecules mass Different masses mass

  29. What are those Masses? • First we need to define the conditions of same temperature and pressure. • The standard for pressure is the average barometric pressure at sea level. • 760 mm Hg = 1.00 atm = 101.3 kpa • The standard temperature is 00C the freezing point of water at sea level. • STP conditions means Standard Temperature and Pressure.

  30. What about Standard Volume? • Oxygen was used as a standard. O2 at STP The volume of 32.0 g of O2 at STP = 22.4 liters What about the number of molecules? 32.0 g

  31. Lets Count Escaping Gas • Apparatus Pin sized hole with known diameter Thin piece of Aluminum foil  Stop Watch Tank of Gas Pressurized bottle  By timing escape rate and volume its possible to calculate numbers

  32. So what are the Results? • 32.0 g of oxygen at STP has 6.02 x 1023 molecules This is called Avogadro's number Named in honor of the man who discovered it

  33. What is a mole? • 1 mole = molar mass = 22.4 liters @ STP = 6.02 X 1023 particles I have a mass My volume is 22. 4 liters And I have 6.02 x 1023 molecules The mass of a mole of a substance is in grams

  34. 1 Mole of Oxygen ( O2 )1 Mole of Carbon dioxide ( CO2 ) 1.00 mol Oxygen 1.00 mol Carbon dioxide Volume of 22.4 liters # molecules = 6.02 x 1023 Mass = 44.0 g Volume of 22.4 liters # molecules = 6.02 x 1023 Mass = 32.0 g

  35. Mole Calculations • How many molecules are in .250 moles of Carbon dioxide? • CO2 is the formula • 0.250 moles CO2x 6.02 x 1023 molecules = 1.51 x 1023 molecules • 1.00 moles CO2

  36. More Moles • How many moles are in 5.46 X 1024 molecules of SO3 ? • Solution: • 5.46 X 1024 molecules of SO3 X 1.00 Moles = 9.10 mol • 6.02 X 1023 molecules

  37. Mole Tree ? • Vol @ STP • | • Grams1----moles1 • | • #’s particles1

  38. Problem • Vol @ STP • What is the volume @ STP of 55.6 grams of C02? | • Molar mass = 12.0 + 2( 16.0 ) = 44.0 g/mole g1-----mole1 • | • 55.6 g CO2 x #’s1 • 55.6 g CO2 x moles CO2 • 55.6 g CO2 x moles CO2 Xliters @ STP • 55.6 g CO2 x moles CO2 Xliters @ STP • g CO2 moles CO2 • 55.6 g CO2 x moles CO2 x 22.4 liters @ STP = 28.3 liters @ STP • 44.0 g CO2 1.00 moles CO2

  39. Mole, Mole, Mole • How many grams are contained in 1.34 moles of Strontium Phosphate? • 1. Use the correct formula and find the molar mass • 3( 87.6 ) + 2( 31.0 +3( 16.0) ) = 421 g +2 ( ) Sr PO4 -3 Sr3(PO4)2

  40. Solve • How many grams are contained in 1.34 moles of Strontium Phosphate? • 421 g • 1.34 moles Sr3(PO4)2 Sr3(PO4)2 g Sr3(PO4)2 421 = 564 g ----------------------------- 1.00 Moles Sr3(PO4)2

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