2. MEASURING MASS  A moleis a quantity of things, just as… 1 dozen= 12 things 1 gross = 144 things 1 mole= 6.02 x 10 23 things  “Things” usually measured in moles are atoms, molecules, ions, and formula units 2

3.  You can measure mass, or volume, or you can count pieces  We measure mass in grams  We measure volume in liters  We count pieces in MOLES 3

4. A MOLE…  is an amount, defined as the number of carbon atoms in exactly 12 grams of carbon- 12  1 mole = 6.02 x 10 23 of the representative particles  Treat it like a very large dozen 6.02 x 10 23 is called: Avogadro’s number 4

5.  Similar Words for an amount:  Pair: 1 pair of shoelaces = 2 shoelaces  Dozen: 1 dozen oranges = 12 oranges  Gross: 1 gross of pencils= 144 pencils  Ream: 1 ream of paper= 500 sheets of paper 5

6. What are Representative Particles (“RP”)?  The smallest pieces of a substance: 1. For a molecular compound: it is the molecule. 2. For an ionic compound: it is the formula unit (made of ions) 3. For an element: it is the atom  Remember the 7 diatomic elements? (made of molecules) 6

7. Practice Counting Particles  How many oxygen atoms in the following? 1. CaCO3 3 atoms of oxygen 2. Al2(SO4)3 12 (4 x 3) atoms of oxygen  How many ions in the following? 1. CaCl2  3 total ions (1 Ca2+ ion and 2 Cl1- ions) 2. NaOH  2 total ions (1 Na1+ ion and 1 OH1- ion) 3. Al2(SO4)3  5 total ions (2 Al3+ + 3 SO4 ions) 7

8. CONVERSION FACTOR  RPs x ____1 mole___ = MOLES 6.02 x 10 23 RPs 8

9. EXAMPLES: ATOMS  MOLES  How many moles of B are in 3.15 x 10 23 atoms of B?  Conversion: 1 mole B = 6.02 x 10 23 atoms B (b/c the atom is the RP of boron) 1 mole B 3.15 x 10 23 atoms of B 6.02 x 10 23 atoms B = 0.523 mole 9

10. EXAMPLES: MOLES  ATOMS  How many atoms of Al are in 1.5 mol of Al?  Conversion: 1 mole = 6.02 x 10 23 atoms 1.5 mol of Al6.02 x 10 23 atoms Al 1 mole Al = 9.03 x 10 23 atoms of Al 10

11. CAUTION: Identify RPs Carefully!  See next slide! 11

12. EXAMPLES: MOLECULES  MOLES How many atoms of H are there in 3 moles of H 2 O? (HINT: Are atoms the RP for water?) Conversions: 1 mole = 6.02 x 10 23 molecules (b/c molecules are the RP for H 2 O) 3 moles of H 2 O 6.02 x 10 23 molec H 2 O 1 mole H 2 O 2 atoms H 1 H 2 O molecule = 3.612 x 10 24 atoms H H 2 O molecule = 2 atoms of Hydrogen 12

13. MOLAR MASS Def: The mass of a mole of representative particles of a substance. Each element & compound has a molar mass. 13

14. MOLAR MASS OF AN ELEMENT Determined simply by looking at the periodic table Molar mass (g) = Atomic Mass (amu) Ca 20 40.08 * Thus, 1 mol Ca = 40 g 1 atom of Ca weighs 40.08 amu 1 mole of Ca atoms weighs 40.08 grams 14

15. MOLAR MASS FOR COMPOUNDS  To calculate the molar mass of a compound, find the number of grams of each element in one mole of the compound  Then add the masses within the compound Example: H 2 O H= 1.01 2 (1.01) + 1 (15.999)= 18.02 g/mol O= 15.999 15

16. SOME PRACTICE PROBLEMS  How many atoms of O are in 3.7 mol of O?  2.2 X 10 24 atoms of oxygen  How many atoms of P are in 2.3 mol of P?  1.4 x 10 24 atoms of phosphorus  How many atoms of Ca are there in 2.5 moles of CaCl 2 ?  1.5 x 10 24 atoms Ca  How many atoms of O are there in 1.7 moles of SO 4 ?  4.1 x 10 24 atoms of oxygen 16

17. Remember!!!!  The molar mass of any substance (in grams) equals 1 mole  This applies to ALL substance: elements, molecular compounds, ionic compounds  Use molar mass to convert between mass and moles  Ex: Mass, in grams, of 6 mol of MgCl 2 ? mass of MgCl 2 = 6 mol MgCl 2 92.21 g MgCl 2 1 mol MgCl 2 = 571.26 g MgCl 2 17

18. VOLUME AND THE MOLE  Volume varies with changes in temperature & pressure  Gases are predictable, under the same physical conditions  Avogadro’s hypothesis helps explain: equal volume of gases, at the same temp and pressure contains equal number of particles  Ex: helium balloon 18

19.  MOLAR VOLUME  Gases vary at different temperatures, makes it hard to measure  To make it easier to measure gases, we use something called “STP”  “Standard Temperature and Pressure”  Temperature = 0° C  Pressure = 1 atm (atmosphere) 19

20. Molar Volume  At STP:1 mole, 6.02 x 10 23 atoms, of any gas has a volume of 22.4 L 1 mole gas = 22.4 L gas  Use it to convert between # of moles and vol of a gas @ STP  Ex: what is the volume of 1.25 mol of sulfur gas? 1.25 mol S 22.4 L S = 28.0 L 1 mol S 20

21. MOLAR MASS FROM DENSITY  Different gases have different densities  Density of a gas measured in g/L @ a specific temperature  Can use the following formula to solve : grams = grams X 22.4 L mole L 1 mole  Ex: Density of gaseous compound containing oxygen and carbon is 1.964 g/ L, what is the molar mass?  grams = 1.964 g X 22.4 L then you solve mole 1 L 1 mole= 44.o g/mol 21

22. Atoms, molecules, etc. 22

23. Molarity  Def: the concentration of a solution. How many moles/liter  Can be used to calculate # of moles of a solute  Ex: Household laundry bleach is a dilute aqueous solution of sodium hypochlorite (NaClO). How many moles of solute are present in 1.5 L of 0.70 M NaClO? 23

24. Calculating Percent Composition of a Compound  Like all percent problems: a part ÷ the whole 1. Find the mass of each of the components (the elements) 2. Next, divide by the total mass of the compound 3. Then X 100 % = percent Formula: % Composition = Mass of element X 100% Mass of compound 24

25. Method #1: % Comp When Actual Masses are Given A compound is formed when 9.03 g of Mg combines completely with 3.48 g of N. What is the percent composition of the compound? 1. First add the 2 mass of the 2 compounds to reach the total mass 9.03 g Mg + 3.48 g N = 12.51 g Mg 3 N 2 2. Find the % of each compound % Mg= 9.03 g Mg X 100 = 72.2 % 12.51 g Mg 3 N 2 % N= 3.48 g N X 100= 27.8 % 12.51 g Mg 3 N 2 25

26. Method 2: % Comp When Only The Formula is Known  Can find the percent composition of a compound using just the molar mass of the compound and the element  % mass=mass of the element 1 mol cmpd X100% molar mass of the compound  Example: Find the percent of C in CO 2 12.01 g C X 100 = 27.30% C 44.01 g CO 2 26

27. Using % Composition  Can use % composition as a conversion factor just like the mole  After finding the % comp. of each element in a cmpd. can assume the total compound = 100g  Example: C= 27.3% 27.3 g C O= 72.7 % 72.7 g O  In 100 g sample of compound there is 27.3 g of C & 72.7 g of O How much C would be contained in 73 g of CO 2 ? 73 g CO 2 27.3 g C= 19.93 g C 100 g CO 2 27

28. EMPIRICAL FORMULAS  Empirical formulas are the lowest WHOLE number ratios of elements contained in a compound 28

29. REMEMBER…  Molecular formulas tells the actual number of of each kind of atom present in a molecule of the compound  Ex: H 2 O 2 HO MolecularEmpiricalFormula CO 2 MolecularEmpirical For CO 2 they are the sameFormula 29

30.  Formulas for ionic compounds are ALWAYS empirical (the lowest whole number ratio = can not be reduced)  Examples: NaCl MgCl2 Al2(SO4)3 K2CO3 Simplest whole number ratio for NaCl 30

31.  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  Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio)  Molecular: H 2 O C 6 H 12 O 6 C 12 H 22 O 11 (Correct formula)  Empirical: ( Lowest whole H 2 O CH 2 O C 12 H 22 O 11 number ratio) 31

32.  STOP HERE!!! 32

33. 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 33

34. Example calculations  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 1 mol C = 12.0 g C 16.22 g H x 1 mol H = 1.0 g H 45.11 g N x 1 mol N = 14.0 g N *Now divide each value by the smallest value 3.22 mole C 3.22 mole N 16.22 mole H 34

35. …Example 1  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 35

36. MORE PRACTICE  A compound is 43.64 % P and 56.36 % O What is the empirical formula? PO3  Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O What is its empirical formula? C 4 H 5 N 2 O 36

37. EMPIRICAL TO MOLECULAR  Since the empirical formula is the lowest ratio, the actual molecule would weigh more  Divide the actual molar mass by the empirical formula mass – you get a whole number to increase each coefficient in the empirical formula 37

38. EXAMPLE  Caffeine has a molar mass of 194 g, what is its molecular formula? 1. Find the mass of the empirical formula, C 4 H 5 N 2 O 2. Divide the molar mass by the empirical mass: 194.0 g/mol = 97.1 g/mol 3. Now multiply the entire empirical formula by 2 2(C 4 H 5 N 2 O) = final molecular formula 2 C 8 H 10 N 4 O 2 38