2. The Mole 6.02 X 10 23

3. The Mole A counting unit Similar to a dozen, except instead of 12, it’s 602 billion trillion 602,000,000,000,000,000,000,000 6.02 X 10 23 (in scientific notation) Amedeo Avogadro (1776 – 1856)This number is named in honor of Amedeo Avogadro (1776 – 1856), who studied quantities of gases and discovered that no matter what the gas was, there were the same number of molecules present

4. Just How Big is a Mole? Enough soft drink cans to cover the surface of the earth to over 200 miles deep Avogadro's number of unpopped popcorn kernels spread across the U.S. would be popcorn to a depth of over 9 miles Counting atoms at the rate of 10 million per second would take about 2 billion years to count the atoms in one mole

5. Everybody Has Avogadro’s Number! But Where Did it Come From? It was NOT just picked! It was MEASURED. One of the better methods of measuring this number was the Millikan Oil Drop Experiment Since then we have found even better ways of measuring using x- ray technology

6. The Mole 1 dozen cookies = 12 cookies 1 mole of cookies = 6.02 X 10 23 cookies 1 dozen cars = 12 cars 1 mole of cars = 6.02 X 10 23 cars 1 dozen Al atoms = 12 Al atoms 1 mole of Al atoms = 6.02 X 10 23 atoms Note that the NUMBER is always the same, but the MASS is very different! Mole is abbreviated mol

7. = 6.02 x 10 23 C atoms = 6.02 x 10 23 H 2 O molecules = 6.02 x 10 23 NaCl formula units 6.02 x 10 23 Na + ions and 6.02 x 10 23 Cl – ions A Mole of Particles A Mole of Particles Contains 6.02 x 10 23 particles 1 mole C 1 mole H 2 O 1 mole NaCl

8. The Mass of 1 mole (in grams) Equal to the numerical value of the average atomic mass (get from periodic table) 1 mole of C atoms= 12.01 g 1 mole of Mg atoms =24.31 g 1 mole of Cu atoms =63.55 g Molar Mass

9. Other Names Related to Molar Mass AMU’S: masses listed on the periodic table are for one single atom Molar Mass terms have same numerical value as amu: only the units are different. The units are in grams. Gram atomic Mass / atomic mass / atomic weight: refers to a mole of a single kind of atom Gram molecular mass / molecular mass / molecular weight: refers to 1 mole of molecules. Gram formula mass / formula mass / formula weight: applies to ionic compounds or ions. THE POINT: You may hear all of these terms which mean the SAME NUMBER… just different units

10. Find the molar mass A.1 mole of Br atoms B.1 mole of Sn atoms =79.90 g/mole = 118.71 g/mole

11. Mass in grams of 1 mole equal numerically to the sum of the atomic masses A) 1 mole of CaCl 2 = 110.98g/mol 1 mole Ca x 40.08 g/mol + 2 moles Cl x 35.45 g/mol = 110.98 g/mol CaCl 2 B)1 mole of N 2 O 4 = 92.02 g/mol Molar Mass of Molecules and Compounds

12. A.Molar Mass of K 2 O = ? Grams/mole B. Molar Mass of antacid Al(OH) 3 = ? Grams/mole

13. A.Molar Mass of K 2 O = ? Grams/mole = 94.2 g/mol B. Molar Mass of antacid Al(OH) 3 = ? Grams/mole = 78.01 g/mol

14. The artificial sweetener aspartame (Nutra-Sweet) formula C 14 H 18 N 2 O 5 is used to sweeten diet foods, coffee and soft drinks. How many moles of aspartame are present in 225 g of aspartame?

16. The artificial sweetener aspartame (Nutra-Sweet) C 14 H 18 N 2 O 5 = ? g/mol carbon = 12.01 g/mol x 14 = 168.14 g hydrogen = 1.01 g/mol x 18 = 18.18 g nitrogen = 14.01 g/mol x 2 = 28.02 g oxygen = 16.00 g/mol x 5 = 80.00 g 294.34 g/mol First find molar mass

18. Given 225 g of aspartame find the number of moles: 225g x 1 mole = 225 mole = 0.76mole 1 294.34g 294.34

20. 0.76mole × 6.02 X 10 23 molecules = 1 1 mole = ( 0.76 × 6.02) X 10 23 molecules = 4.58 X 10 23 molecules

21. The Mole-Volume Relationship  Many of the chemicals we deal with are in the physical state as: gases. - They are difficult to weigh (or mass).  But, we may still need to know how many moles of gas we have.  Two things effect the volume of a gas: a) Temperature and b) Pressure  We need to compare all gases at the same temperature and pressure.

22. Standard Temperature and Pressure  0ºC and 1 atm pressure - is abbreviated “STP”  At STP, 1 mole of any gas occupies a volume of 22.4 L - Called the molar volume  This is our fourth equality: 1 mole of any gas at STP = 22.4 L

23.  What is the volume of 4.59 mole of CO 2 gas at STP?  How many moles is 5.67 L of O 2 at STP?  What is the volume of 8.8 g of CH 4 gas at STP? = 103 L = 0.253 mol = 12.3 L

24. Density of a gas  D = m / V (density = mass/volume) - for a gas the units will be: g / L  We can determine the density of any gas at STP if we know its formula.  To find the density we need: 1) mass and 2) volume.  If you assume you have 1 mole, then the mass is the molar mass (from periodic table)  And, at STP the volume is 22.4 L.

25.  Find the density of CO 2 at STP. D = 44g/22.4L = 1.96 g/L  Find the density of CH 4 at STP. D = 16g/22.4L = 0.714 g/L

26. Another way:  If given the density, we can find the molar mass of the gas.  Consider that you have 1 mole at STP, so V = 22.4 L. modify: D = m/V to show:  “m” will be the mass of 1 mole, since you have 22.4 L.  What is the molar mass of a gas with a density of 1.964 g/L?  Molar mass with density of 2.86 g/L? = 44.0 g/mol = 64.0 g/mol m = D x V

27. Summary These four items are all equal: a) 1 mole b) molar mass (in grams/mol) c) 6.02 x 10 23 representative particles (atoms, molecules, or formula units) d) 22.4 L of gas at STP We can make conversion factors from these 4 values!

28. Percent Composition and Chemical Formulas –Describe how to calculate the percent by mass of an element in a compound. –Interpret an empirical formula. –Distinguish between empirical and molecular formulas.

29. Calculating Percent Composition of a Compound  Like all percent problems: part whole 1)Find the mass of each of the components (the elements), 2)divide by the total mass of the compound; then x 100 x 100 % = percent

30.  Calculate the percent composition of a compound that is made of 29.0 grams of Ag with 4.30 grams of S 29.0 g Ag 33.3 g total X 100 = 87.1 % Ag 4.30 g S 33.3 g total X 100 = 12.9 % S Total = 100 %

31.  If we know the formula, assume you have 1 mole,  then you know the mass of the elements and the whole compound (these values come from the periodic table!).

32.  Calculate the percent composition of C 2 H 4 ?  How about Aluminum carbonate?  We can also use the percent as a conversion factor 85.6% C, 14.4 % H 23.1% Al, 15.4% C, and 61.5 % O

33. Formulas Example: molecular formula for benzene is C 6 H 6 (note that everything is divisible by 6) Therefore, the empirical formula = CH (the lowest whole number ratio) Empirical formula: the lowest whole number ratio of atoms in a compound. Molecular formula: the true number of atoms of each element in the formula of a compound.

34. Formulas for ionic compounds are ALWAYS empirical (the lowest whole number ratio = cannot be reduced). Examples: NaClMgCl 2 Al 2 (SO 4 ) 3 K 2 CO 3

35. Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio). Molecular: H2OH2O C 6 H 12 O 6 C 12 H 22 O 11 Empirical: H2OH2O CH 2 OC 12 H 22 O 11 (Correct formula) (Lowest whole number ratio)

36. Calculating Empirical Formula Just find the lowest whole number ratio C 6 H 12 O 6 CH 4 N It is not just the ratio of atoms, it is also the ratio of moles of atoms In 1 mole of CO 2 there is 1 mole of carbon and 2 moles of oxygen In one molecule of CO 2 there is 1 atom of C and 2 atoms of O

37. Calculating Empirical Formula Pretend that you have a 100 gram sample of the compound. That is, change the % to grams. Convert the grams to moles for each element. Write the number of moles as a subscript in a chemical formula. Divide each number by the least number. Multiply the result to get rid of any fractions.

38. Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N. Assume 100 g so 38.67 g C x 1mol C = 3.219 mole C 12.01 g C 16.22 g H x 1mol H = 16.09 mole H 1.01 g H 45.11 g N x 1mol N = 3.219 mole N 14.01 g N

39. 3.219 mole C 16.09 mole H 3.219 mole N C 3.219 H 16.09 N 3.219 If we divide all of these by the smallest one It will give us the empirical formula

40. The ratio is 3.220 mol C = 1 mol C 3.219 mol N 1 mol N The ratio is 16.09 mol H = 5 mol H 3.219 mol N 1 mol N C 1 H 5 N 1 is the empirical formula

41. 43.64 g P x 1mol P = 1.41 mole P 30.97 g P 56.36 g O x 1mol O = 3.52 mole O 16 g O P 1.41 O 3.52 A compound is 43.64 % P and 56.36 % O What is the empirical formula?

42. Divide both by the lowest one The ratio is 3.5 mol O = 2.5 mol O 1.4 mol P 1 mol P P 1.4 O 3.5 P 1 O 2.5

43. Multiply the result to get rid of any fractions. (P 1 O 2.5 ) 2 X = P 2 O 5

44. Calculating Empirical  Just find the lowest whole number ratio C 6 H 12 O 6 CH 4 N  A formula is not just the ratio of atoms, it is also the ratio of moles.  In 1 mole of CO 2 there is 1 mole of carbon and 2 moles of oxygen.  In one molecule of CO 2 there is 1 atom of C and 2 atoms of O. = CH 2 O = this is already the lowest ratio.

45. Calculating Empirical  We can get a ratio from the percent composition. 1)Assume you have a 100 g sample - the percentage become grams (75.1% = 75.1 grams) 2)Convert grams to moles. 3)Find lowest whole number ratio by dividing each number of moles by the smallest value.

46.  Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N.  Assume 100 g sample, so  38.67 g C x 1mol C = 3.22mole C 12.01 g C  16.22 g H x 1mol H = 16.22 mole H 1.01 g H  45.11 g N x 1mol N = 3.22 mole N 14.01 g N Now divide each value by the smallest value

47.  The ratio is 3.22 mol C = 1 mol C 3.22 mol N 1 mol N  The ratio is 16.22 mol H = 5 mol H 3.22 mol N 1 mol N = C 1 H 5 N 1 which is = CH 5 N  A compound is 43.64 % P and 56.36 % O. What is the empirical formula?  Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula? = P 2 O 5 = C 4 H 5 N 2 O

48. Empirical to molecular  Since the empirical formula is the lowest ratio, the actual molecule would weigh more.  By a whole number multiple.  Divide the actual molar mass by the empirical formula mass – you get a whole number to increase each coefficient in the empirical formula  Caffeine has a molar mass of 194 g. what is its molecular formula? = C 8 H 10 N 4 O 2