2. The Mole  Chapter 8.2

3. What is a mole? It is  "that equal number" of atoms arbitrarily chosen.  the number of atoms in the atomic weight in g of any element.  the number of atoms in 16 g of oxygen, in 4 grams of He, in 32 g of sulfur, etc.

4.  the number of molecules in the molecular weight, in grams, of any compound.  the number of molecules in 18 g of water, in 40 g of lithium carbide.  Much later, a mole, that equal number, was found to be 6.022 x 10 23.

5. What Can Be Expressed Using Moles?  A mole can be used to represent 6.02 x 10 23 : -atoms -molecules -ions(But not Grams) -compounds -particles -items (EX: students, homework assignments)

6. Test your understanding with the following questions: 1. How many bananas are in a mole of bananas? 2. How many kisses is a half mole of kisses? 6.02 x 10 23 ÷ 2 = 6.02 x 10 23 3.01 x 10 23

7. Calculating Molar Mass  Molar mass is the mass of 1 mole of a pure substance.  Molar mass may also be referred to as: -gram formula mass (ionic compounds) -formula mass (ionic compounds) -molecular mass (non metals)

8. Calculating Molar Mass  The mass of one mole (6.02 x 10 23 ) of atoms is based on the mass of one mole of Carbon- 12, which is exactly 12.0 grams.  The molar masses of elements can then be obtained from their respective atomic masses (rounded to 2 places after the decimal).  Use the unit g/mol to denote molar mass.

9. Calculating Molar Mass EX: Find the molar masses for the following:  Silver  Argon  Potassium  Oxygen  Hydrogen

10. Calculating Molar Mass  The molar mass of a molecule or formula can be determined by adding the molar masses of each of the elements present. EX: H 2 O Elements Moles of each Atomic mass H 2 x 1.01 = 2.02 O 1 x 16.00 =16.00 Molar mass of H 2 0  18.02 g/mol

11.  Calculate molar mass of the following: 1.Al 2 S 3 Al (2 x 26.98) 53.96 S (3 x 32.06) 96.18 150.14 g/mol Add hydrate example: CuCl 2 · 2H 2 O

12. Mole Conversions  The key to converting between any units is to label numeric values with appropriate units and work with these until the only one that will not cancel is the one you want.

13. Converting Moles  Molecules  Use Avogadro’s constant (6.02 x 10 23 ) when converting between moles and molecules. EX: How many atoms are in 3.4 moles of iron atoms? 6.02 x 10 23 small unit = 1 mole molec, atoms, ions, particles, formula units, ionic compounds

14. Calculation  1. Start with known value divided by 1.  2. Then line up units so that only the desired units are not cancelled.  3. Divide the products of the numerators by the products of the denominators. 3.4 moles 6.02 x 10 23 atoms = 2.1 x 10 24 11 moleatoms

15. Who Loves M&Ms?  Do you think you could eat ½ a mole of M&Ms in your lifetime?

16.  80 year lifespan  3.01x10 23 M&Ms

17.  Start eating:  1.193x10 14 per second  119,300,000,000,000 per second

18. Practice  2.54 x 10 21 molecules of CO 2 is equal to ___________ moles.  0.010 moles of HCl is equal to ____________ molecules of HCl.  2 moles of M & Ms is equal to __________ pieces. 4.22 x 10 -3 mol 6.0 x 10 21 molecules 1 x 10 24 pieces

19. Converting Moles  Mass Molar mass is the conversion bridge needed to convert between mass and mole of any substance. EX: 1.01 g/mol = 1.01 g = 1 mole 1 mole 1.01 g

20. Mole Conversions  What is the mass in grams of 2.5 mol of O 2 ? 2.5 mol (32.00 g) = 1 ( 1 mol ) 80. g of O 2 0.277 mol of H 2 O * Divide notepacket area into 2 sections for 2 nd problem.  Determine the number of moles in 5.00 g of H 2 O. 5.00 g ( 1 mol ) = 1 (18.02 g)

21. Convert Molec  Mass  1 mole of any substance is equivalent to 1 mole of another substance. 1 mole = 6.02 x 1023 particles.  Molar masses of substances are not equivalent. EX: NaCl = 58.44g/mol ; NaOH = 40.00 g/mol  There is no direct conversion between molecules and grams. So…Molec  Moles  Mass

22. Mole Conversions  Determine the number of molecules there are in a 5.45 g sample of CaCl 2. 5.45 g CaCl 2 x ( 1 mol ) ( 6.02 x 10 23 molecules) 1 (110.98 g ) ( 1 mol ) 2.96 x 10 22 molecules of CaCl 2

23. Percent Composition  Percent composition of a compound is a statement of the relative mass each element contributes to the mass of the compound as a whole.  Chemists often compare the percent compositions of unknown compounds to those of known compounds to identify the unknown.

24. Calculating % Composition  Find the % composition of the elements in NaCl. Step 1: calculate the molar mass of the substance. Step 2: divide the molar mass of each of the elements by the molar mass of the substance. Step 3: multiply by 100 to get into percent.

25.  Molar mass of NaCl is 58.44 g/mol  Na = 22.99g/mol x 100 = 58.44g/mol  Cl = 35.45g/mol x 100 = 58.44g/mol 39.34% 60.66%

26.  Na = 39.34 %  Cl = 60.66 % 100.00 %  Each formula unit of table salt is about 40 % sodium cation and 60 % chlorine anion by mass.  Refer to page 206 in text for additional info.

27. Empirical Formulas  Remember that the elements in a compound combine in whole number ratios such as 1:1, 1:2, 2:3, and so forth.  If elements combine in these whole number ratios, we can predict the same applies for moles of each atom.  We can use this principle to find the empirical (lowest whole number) formula for compounds based on the relative masses of each of the elements in the compound.

28. Calculating Empirical Formulas  What is the empirical formula for a compound if a 2.50 gram sample contains 0.900 g of calcium and 1.60 g of chlorine? Step 1: convert grams of each of the elements into moles. Step 2: obtain the simplest ratio by dividing the moles by the smallest number of moles. Step 3: round or multiply to express the ratio using whole numbers.

29. 0.900 g Ca 1 mol 1 40.08 g = 0.0224 mol 1.60 g Cl 1 mol 1 35.45 g = 0.0451 mol  The smallest mole is 0.0224 so we divide each by this number.

30.  Ca = 0.0224 moles 0.0224 moles = 1  Cl = 0.0451 moles 0.0224 moles = 2.01 (round to 2)

31.  Based on our calculations, there is 1 mole of calcium for every 2 moles of chlorine.  The formula is then: CaCl 2

32. Practice Problem  What is the empirical formula of a compound that is 66.0% Ca and 34.0% P? *When given only the percentages of each element, assume that there is 100 g of the compound.

33.  Sometimes dividing by the smallest number of moles will not lead to a whole number. EX: 66.0 g Ca1 mole 1 40.1 g = 1.65 mol 34.0 g P1 mole 131.0 g = 1.10 mol

34.  Ca = 1.65 mole 1.10 mole = 1.5  P = 1.10 mole 1.10 mole = 1  1.5 can not be rounded to the next highest number.  We must multiply each by 2 in order to turn 1.5 into a whole number.

35.  Ca = 1.5 x 2 = 3  P = 1 x 2 = 2  The empirical formula for this compound is: Ca 3 P 2

36.  Hints for rounding or multiplying until a whole number is achieved:  If.8 then round to whole number.  If 1.5, then multiply each substance in the problem by 2 to get 3.  If 1.25, then multiply each substance in the problem by 4 to get 5.  If 1.33, then multiply by 3 to get 4.

37. Empirical Formula Poetry % to Mass Mass to Mole Divide by Small Multiply til Whole

38. Molecular Formulas  Molecular formulas are true formulas for compounds. Steps: 1. Determine empirical formula. 2. Determine molar mass of empirical formula. 3. Divide molecular mass that was given by the empirical molecular mass. 4. Distribute whole number as subscript to empirical formula.

39. molecular mass = n empirical formula mass xnynxnyn  Find the molecular formula for CH 2 if the molecular mass is 42.09 g/mol. 42.09 g/mol = 3 =C 3 H 6 14.03 g/mol

40. Practice: 1. Vitamin C (ascorbic acid) contains 40.92 % C, 4.58 % H, and 54.50 % O, by mass. The experimentally determined molecular mass is 176 g/mol. What is the empirical and molecular formula for ascorbic acid? 2. NutraSweet is 57.14% C, 6.16% H, 9.52% N, and 27.18% O. Calculate the empirical formula of NutraSweet and find the molecular formula. (The molar mass of NutraSweet is 294.30 g/mol)