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14 the mole!!!

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14 the mole!!!

1. 1. The MOLE!!!
2. 2. The MOLE!!!People use words to represent specific quantities all the timeDozen eggs Pair of gloves Six-Pack
3. 3. The MOLE!!!1 mole = 6.02214199 x 1023 particles 1 mole = 6.02 x 1023 (the short form for mole is “mol”)
4. 4. The MOLE!!! 1 mol = 6.02 x 1023Called the Avogadro constant or Avogadro’s number (devised through experiments thatdetermined how many carbon atoms were present in exactly 12 grams of carbon)
5. 5. The MOLE!!!1 mole = 602214199000000000000000 molecules A very big number!
6. 6. The MOLE!!!Converting moles to number of particles: Number of moles N = n X NANumber of Avogadro’s particles number
7. 7. The MOLE!!! N = n X NAA sample contains 1.25 mol of NO2.a) How many molecules are in thesample?b) How many atoms are in thesample?
8. 8. The MOLE!!! N = n X NA a) A sample contains 1.25 mol of NO2. How many molecules are in the sample? N = n X NA N = (1.25mol) X (6.02 x 1023 molecules/mol) N = 7.52 x 1023 molecules.: there are 7.52 x 1023 molecules in 1.25 mol of NO2
9. 9. The MOLE!!! N = n X NA A sample contains 1.25 mol of NO2. b) How many atoms are in the sample? (7.52 x 1023 molecules) x (3 atoms/molecule) = 2.26 x 1024 atoms.: there are 2.26 x 1024 atoms in 1.25 mol of NO2
10. 10. The MOLE!!!Rearranging the formula… n=N NA How many moles are present in a sample of CO2 made up of 5.83 x 1024 molecules?
11. 11. The MOLE!!!How many moles are present in a sample of CO2 made up of 5.83 x 1024 molecules? n= N NA = (5.83 x 10 24 molecules CO2) (6.02 x 1023 molecules/mol) = 9.68 mol CO 2.: there are 9.68 mol of CO2 in the sample
12. 12. MOLAR MASS M = molar mass Molar mass of H =1.0079 grams per mole Molar mass of Li =6.941 grams per mole MNa = 22.990g/mol
13. 13. MOLAR MASS Molar mass of compoundsMBeO = 9.01g/mol + 16.00g/mol = 25.01g/mol MCO = 2 12.01g/mol + 2x16.00g/mol = 44.01g/mol
14. 14. MOLAR MASS Number of moles m=nxMmass Molar mass m n M
15. 15. MOLAR MASS A flask contains 0.750 mol of CO2. What mass of CO2 is in this sample?GIVEN: n = 0.750mol M = 12.01g/mol + 2 x 16.00g/mol = 44.01g/mol m=? m = nxM = (0.750mol) x (44.01g/mol) = 33.0g .: the mass of CO2 is 33.0g
16. 16. MOLAR MASSRearranging the formula… n=m M
17. 17. MOLAR MASSHow many moles of CH3COOH are in a 23.6g sample?GIVEN: m = 23.6g M = (2 x 12.01g/mol C) + (4 x 1.008g/mol H) + (2 x 16.00g/mol O) = 60.06g n=? n= m M = (23.6g)/(60.06g/mol CH3COOH) = 0.393mol CH3COOH.: there are 0.393mol of CH3COOH in 23.6g of CH3COOH
18. 18. PERCENTAGE COMPOSITION
19. 19. PERCENTAGE COMPOSITIONLaw of Definite Proportions:The elements in a compound are always presentin the same proportions by massExample:Water = 11.2% hydrogen, 88.8% oxygenMH O = 18.016g/mol MH = 1.008g/mol 2% of H in H2O = mass of H/mass of water = (1.008g/mol x 2)/(18.016g/mol) = 0.112  11.2%
20. 20. PERCENTAGE COMPOSITION A compound with a mass of 48.72g contains32.69g of Zn & 16.03g of S. What is the percent composition of the compound?%Zn = 32.69g/48.72g = 0.6710  67.10% .: the percentage composition for Zn is 67.10% and the%S = 16.03g/48.72g percentage composition = 0.3290  32.90% for S is 32.90%
21. 21. EMPIRICAL FORMULAA compound is 81.9% carbon, 6.12% hydrogen, and 12.1% oxygen by mass. What is the empirical formula of the compound? Molecular formula = Actual formula of compound Empirical formula = simplest formula (shows the lowest number ratio of the elements) Example: Benzene Molecular formula = C6H6 Empirical formula = CH
22. 22. EMPIRICAL FORMULAA compound is 81.9% carbon, 6.12% hydrogen, and 12.1% oxygen by mass. What is the empirical formula of the compound?STEP1: Assume the sample is 100gSTEP2: Find the number of moles of each elementSTEP3: Divide all answers from STEP2 by the LOWESTanswer from STEP2
23. 23. EMPIRICAL FORMULA A compound is 81.9% carbon, 6.12% hydrogen, and 12.1% oxygen by mass. What is the empirical formula of the compound?STEP1: Assume the sample is 100gSo... C = 81.9g H = 6.12g O = 12.1g
24. 24. EMPIRICAL FORMULAA compound is 81.9% carbon, 6.12% hydrogen, and 12.1% oxygen by mass. What is the empirical formula of the compound?STEP2: Find the number of moles of each elementC = 81.9g/12.01g/mol H = 6.12g/1.008g/mol = 6.819mol = 6.0714molO = 12.1g/16.00g/mol = 0.75625mol SMALLEST ANSWER
25. 25. EMPIRICAL FORMULAA compound is 81.9% carbon, 6.12% hydrogen, and 12.1% oxygen by mass. What is the empirical formula of the compound?STEP3: Divide all answers by the smallest answerC = 6.819/0.75625 H = 6.0714/0.75625 = 9.0168 = 8.028O = 0.75625/0.75625 =1
26. 26. EMPIRICAL FORMULAA compound is 81.9% carbon, 6.12% hydrogen, and 12.1% oxygen by mass. What is the empirical formula of the compound? C = 6.819/0.75625 H = 6.0714/0.75625 = 9.0168 = 8.028 O = 0.75625/0.75625 =1 = C9H8O
27. 27. MOLECULAR FORMULAThe empirical formula of ribose is CH2O. The molar mass of this compound was determined to be150g/mol. What is the molecular formula of ribose?GIVEN: Empirical formula (1 C, 2 H, 1 O) M = 150g/mol STEP1: Determine the molar mass of the empirical formula STEP2: Divide the given molar mass by your answer from STEP1 STEP3: Multiply your empirical formula by your answer from STEP2
28. 28. MOLECULAR FORMULAThe empirical formula of ribose is CH2O. The molar mass of this compound was determined to be150g/mol. What is the molecular formula of ribose?STEP1: Determine the molar mass of the empirical formula12g/mol + 2 x 1.008g/mol + 16g/mol = 30g/molSTEP2: Divide the given molar mass by your answer from STEP1150g/mol / 30g/mol = 5
29. 29. MOLECULAR FORMULAThe empirical formula of ribose is CH2O. The molar mass of this compound was determined to be150g/mol. What is the molecular formula of ribose?STEP3: Multiply your empirical formula by your answerfrom STEP2 C1x5H2x5O1x5 = C5H10O5
30. 30. ANALYTICAL MACHINES FOR CHEMISTRY
31. 31. MASS SPECTROMETER1) Upload sample2) Same is vapourized3) Sample is ionized4) Ions accelerated by electric field5) Detection to mass-to-charge ratio based on details of motion6) Ions assorted according to mass-to-charge ratio
32. 32. CARBON-HYDROGEN COMBUSTION ANALYZER1)Weigh H2O absorber and CO2 absorber before experiment2)Sample is burned3)Absorbers are weighed again
33. 33. CARBON-HYDROGEN COMBUSTION ANALYZER A 1.000g sample of pure compound, containing only carbon and hydrogen, was combusted in a carbon-hydrogen combustion analyzer. The combustion produced 0.6919g of water and 3.338g of carbon dioxide.a)Calculate the masses of the hydrogen and the carbonb)Find the empirical formula of the compound.
34. 34. CARBON-HYDROGEN COMBUSTION ANALYZER A 1.000g sample…The combustion produced 0.6919g of water and 3.338g of carbon dioxide.a) Calculate the masses of the hydrogen and the carbonMass of H = 2.02g/mol H2 x 0.6919g H2O always 18.02g/mol H2O 0.112 This gives you the percent = 0.07756g H2 composition of hydrogen in waterMass of C = 12.01g/mol C x 3.338g CO2 always 44.01g/mol CO2 0.27289 = 0.9109g C Therefore there was 0.0775g of H and 0.911g of C
35. 35. CARBON-HYDROGEN COMBUSTION ANALYZER A 1.000g sample…The combustion produced 0.6919g of water and 3.338g of carbon dioxide.b) Find the empirical formula of the compoundMoles of H = 0.07756g Moles of C = 0.9109g 1.008g/mol 12.01g/mol = 0.07694mol = 0.07584molEmpirical formula = C0.07584/0.07584H0.07694/0.07584 SMALLEST ANSWER = C1.0H1.0 = CH Therefore the empirical formula is CH
36. 36. HYDRATED SALTSA 50.0g sample of Ba(OH)2·XH2O contains 27.2g of Ba(OH)2.a)Calculate the percent, by mass, of water in the sampleb)Find the value of X
37. 37. HYDRATED SALTSA 50.0g sample of Ba(OH)2·XH2O contains 27.2g of Ba(OH)2.a) Calculate the percent, by mass, of water in the sample= (total mass of sample) – (mass of Ba(OH)2 in sample) x 100% (total mass of sample)= 50.0g – 27.2g x 100% 50.0g= 45.6%Therefore the percent by mass of water is 45.6%
38. 38. HYDRATED SALTSA 50.0g sample of Ba(OH)2·XH2O contains 27.2g of Ba(OH)2.b) Find the value of XnBa(OH)2 = m/M = (27.2g) (171.3g/mol) = 0.159 mol Ba(OH)2 SMALLEST ANSWERnH2O = m/M = (50.0g – 27.2g) (18.02g/mol)  this is the molar mass of H2O = 1.27mol H2O 0.159/0.159 mol Ba(OH)2·1.27/0.159 mol H2OBa(OH)2·8H2O Therefore X is 8