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# 14 the mole!!!

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• 1. The MOLE!!!
• 2. The MOLE!!!People use words to represent specific quantities all the timeDozen eggs Pair of gloves Six-Pack
• 3. The MOLE!!!1 mole = 6.02214199 x 1023 particles 1 mole = 6.02 x 1023 (the short form for mole is “mol”)
• 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. The MOLE!!!1 mole = 602214199000000000000000 molecules A very big number!
• 6. The MOLE!!!Converting moles to number of particles: Number of moles N = n X NANumber of Avogadro’s particles number
• 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. 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. 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. 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. 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. 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. 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. MOLAR MASS Number of moles m=nxMmass Molar mass m n M
• 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. MOLAR MASSRearranging the formula… n=m M
• 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. PERCENTAGE COMPOSITION
• 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. ANALYTICAL MACHINES FOR CHEMISTRY
• 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. CARBON-HYDROGEN COMBUSTION ANALYZER1)Weigh H2O absorber and CO2 absorber before experiment2)Sample is burned3)Absorbers are weighed again
• 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. 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. 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. 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. 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. 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