1. Experiment #5 Avagadro’s Number 6.02 x 10 23 And The Standard Deviation x  S

2. 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.

3. Key Ideas ► ► Amu Vs. Gram – ► ► Atom Vs. Ion – ► ► Atom Vs. Mole – ► ► Formula Weight – ► ► GFW,GAW,GMW – ► ► Dimensional Analysis – ► ► Standard Deviation –

4. 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.

5. 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

6. 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

7. 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

8. 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

9. 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.

10. Dimensional Analysis Convert 10 m to mm Remember 1 m = 1000mm They both mean the same thing it all depends on what you need.

11. 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.

12. Keep in mind X going and / going

13. Ex#1 pg 51 Find the # of moles in the sample of Tin. Given the sample has a mass of 29.04g

14. 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

15. 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

16. 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

17. 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.

18. 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.