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Experiment #5 Avagadro’s Number 6.02 x 10 23 And The Standard Deviation x S
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Introduction ► ► Purpose: Relate a gram formula weight to Avogadro’s number by using the dimensional analysis approach and determine the correct number of significant figures in your applications. Compute the standard deviation for the volume of water delivered by a 10 mL Graduate Cylinder.
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Key Ideas ► ► Amu Vs. Gram – ► ► Atom Vs. Ion – ► ► Atom Vs. Mole – ► ► Formula Weight – ► ► GFW,GAW,GMW – ► ► Dimensional Analysis – ► ► Standard Deviation –
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Amu Vs. Gram Amu – We use if we are talking about the mass of an atom. Ex. the mass of an iron atom is 55.8 amus. Gram – We use if we are talking about the mass of a mole or fraction of a mole. Ex. The mass of one mole of iron is 55.8 grams.
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Atom Vs. Ion Ion Have a charge; can be positive or negative. Not on the Periodic Table React Can be more then one atom Unequal number of Protons and Electrons Atoms Do not have a charge On Periodic Table Do not React Only one atom Equal number of Protons & Electrons
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Atom Vs. Mole Mole 6.02X10 23 Atoms Larger mass measured in grams Larger in size Can be seen Atom 1 Atom Very small mass measured in Amus Very Small in size Can’t be seen
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Formula weight – the mass of the collection of atoms represented by a chemical formula. For Ex. water contains two hydrogen atoms and one oxygen atom. 1 x 16.00 (mass of O) = 16.00 2 x 1.01 (mass of H) = + 2.02 Formula Weight 18.02 The formula weight tells us the mass of one mole of our substance. Formula Weight
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GFW – Gram Formula Weight This refers to the mass of an ion or ionic compound Ex. Mass of a Cl - GAW – Gram Atomic Weight This refers to the mass of a atom Ex. Mass of a Cl atom GMW Gram Molecular Weight This refers to the mass of a molecule or molecular compound Ex. Mass of one molecule of H 2 O
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How do Avagadro’s number, Mole and GAW Apply to each other? ► ► One mole of Tin has a mass of 118.69g. ► ► One atom of Tin has a mass of 118.69 Amu. ► ► One mole of Tin contains 6.02X10 23 atoms.
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Dimensional Analysis Convert 10 m to mm Remember 1 m = 1000mm They both mean the same thing it all depends on what you need.
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Dimensional Analysis # 1 # 3 # 2 Mult. left to right divide top to bottom What you are given always over 1 The 1 is a filler. The unit you want to eliminate is placed in the lower right so it will cancel out with your original unit.
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Keep in mind X going and / going
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Ex#1 pg 51 Find the # of moles in the sample of Tin. Given the sample has a mass of 29.04g
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Chart on pg 55 To determine the g formula wt. determine the mass of all atoms present. To determine #g / molecule think about the units Think how can we compare grams and molecules? We need to use moles. So # grams per mole / number of parts in 1 mole To determine the number of atoms per mole add up the number of atoms in the formula
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Standard Deviation ► ► This is how much your value is off from the actual answer. Can be used as a correction factor. ► ► Ex. A 1mL pipette delivers 1mL + or -.007mL So the pipette delivers.9993mL or 1.007 mL. X = Avg
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Standard Deviation Ex. 1 mL Pipette 1.02gM H 2 O1.01g 1.02g Density of H 2 0 = 0.9982 g/mL 1.02mL 1.01mL1.02mL1.01mLVol. of pipet Use V= M/D to determine volume delivered Mass of Beaker & water 109.50g111.53g109.50g 111.53g 110.51g Mass of Beaker 110.51g
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N= number of trials ∑ (Xi-X avg.) 2 Xi-Xavg. (a)1.01-1.02=.01 (b)1.02-1.02= 0 (c)1.01-1.02=.01 (d)1.02-1.02= 0 (e)1.02-1.02= 0 X avg.= 1.02 mL #1 #2 #3 ( Xi-Xavg. ) 2 (a).01 =1.0X10 -4 (b) 0 = 0 (c).01 =1.0X10-4 (d) 0 = 0 (e) 0 = 0 #4 #5∑(Xi-X avg.) 2 /N-1 #6 Take Square Root of Step #5 Final Answer: +/- 0.007 mL.
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Due Next Week ► ► Pages 51 & 53 we are determining the number of atoms / ions or moles that are asked for. ► ► YOU MUST SHOW ALL WORK FOR CREDIT!! ► ► Page 55 complete the table and show all work. ► ► Pages 61 determine the St. Dev. for 1 mL pipette ( I do as a example) ► ► Page 63 determine the St. Dev. for a 10mL. graduated cylinder.
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