2. The Mole

5. It's not a spy, a machine for digging tunnels, a burrowing animal, or a spot of skin pigmentation - it's - a unit of measurement containing about 6.02 x 10 23 particles. Chemists created the mole as their own counting unit.

6. 1mole of atoms is 602,214,199,000,000,000,000,000 atoms!

7. Avogadro’s Number 6.02 x 10 23 particles / mole

8. Just How Big is a Mole? Enough soft drink cans to cover the surface of the earth to a depth of over 200 miles. If you had Avogadro's number of unpopped popcorn kernels, and spread them across the United States of America, the country would be covered in popcorn to a depth of over 9 miles. If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years to count the atoms in one mole.

9. Moles of Elements 1 mole element = (atomic mass) in grams of that element 1 mole C = 12.011 g C 1 mole B = 10.81 g B 1 mole Cu = 63.55 g Cu

10. 6.02 x 10 23 particles 1 mole or 1 mole 6.02 x 10 23 particles Note that a particle could be an atom OR a molecule! Avogadro’s Number as Conversion Factor

11. ParticlesMoleculesCompound Formula unit AtomElement

12. Equation to find number of particles! number of moles x = number of particles 6.02 x 10 23 molecules 1 mole Example: How many particles of sucrose are in 3.50 moles? 3.50 mol sucrose x = 2.11 x 10 24 molecules of sucrose 6.02 x 10 23 particles 1 mole

13. Equation to find number of moles! number of particles x = number of moles Example: How many moles are contained in 4.50 x 10 24 atoms of zinc? 4.50 x 10 24 atoms of zinc x = 7.48 mol Zn 1 mole 6.02 x 10 23 particles 1 mole Zn 6.02 x 10 23 particles

14. 1. Number of atoms in 0.500 mole of Al a) 500 Al atoms b) 6.02 x 10 23 Al atoms c) 3.01 x 10 23 Al atoms 2.Number of moles of S in 1.8 x 10 24 S atoms a) 1.0 mole S atoms b) 3.0 mole S atoms c) 1.1 x 10 48 mole S atoms Learning Check

15. Molar Mass 11.2

16. The Mass of 1 mole (in grams) Molar mass is equal to the atomic mass (look on the periodic table) 1 mole of C atoms= 12.01 g 1 mole of Au atoms = 196.97g 1 mole of Cu atoms = 63.55 g Always round to the hundredths place. Molar Mass

17. Mass in grams of 1 mole equal numerically to the sum of the atomic masses 1 mole of CaCl 2 Ca = 40.08 Cl = 35.453 x 2 70.906 40.08 + 70.906 = Molar Mass of Molecules and Compounds 110.99 g/mol

18. Molar Mass of Molecules and Compounds 1 mole of N 2 O 4 N = 14.007 O = 15.999 x 2 x 4 28.01463.996 28.014 + 63.996 = 92.01 g/mol

19. Mole to Mass Conversion Number of moles x number of grams = mass 1 mole Example: Calculate the mass in grams of 0.0450 moles of chromium. 0.0450 moles Cr x 52.0 g Cr = 2.34 g Cr 1 mol Cr Where will you find the number of grams per mole? The Periodic Table Look at your Periodic Table

20. Mass to Mole Conversion Mass x 1 mole = number of moles number of grams Example: How many moles of calcium are there in 525 grams of calcium? 525 g Ca x 1 mol Ca = 13.1 mol Ca 40.1 g Ca Where will you find the number of grams per mole? The Periodic Table Look at your Periodic Table

21. Mass to Atoms Conversion Two steps: 1.Use the mass to moles conversion 2.Now calculate the number of atoms using the moles to particles equation.

22. Example: How many atoms of gold are in a pure gold nugget having a mass of 25.0 g? Step 1 Mass x 1 mole = number of moles number of grams 25.0 g Au x 1 mol Au = 0.13 mol Au 197.0 g Au Round to the hundredths place!!!!!!

23. Step 2 number of moles x = number of particles 6.02 x 10 23 atoms 1 mole 0.13 mol Au x = 7.83 x 10 22 atoms Au 6.02 x 10 23 particles 1 mole Round to the hundredths place!!!!!!

24. Atoms/Molecules and Grams How many atoms of Cu are present in 35.4 g of Cu? 35.4 g Cu 1 mol Cu 6.02 X 10 23 atoms Cu 63.5 g Cu 1 mol Cu = 3.4 X 10 23 atoms Cu

25. Learning Check! How many atoms of K are present in 78.4 g of K? 78.4 g K 1 mol K 6.02 X 10 23 atoms K 39.1 g K 1 mol K = 1.20 X 10 24 atoms K

26. Empirical and Molecular Formulas 11.4

27. Percent Composition Mass of element x 100 = percent by mass Mass of compound Example: Determine the percent composition of NaHCO 3. Determine the mass of each element first! %mass element = mass of element in 1 mol compound x 100 molar mass of compound Percent Na = 22.99 g Na x 100 = 27.37% Na 84.01 g NaHCO 3 You have to complete this step for each element in the compound!!!!

28. lDivide the mass of each element by the molar mass of the compound and multiply by 100% Determine the Percent Composition from the Formula C 2 H 5 OH

29. Types of Formulas Empirical FormulaEmpirical Formula The formula of a compound that expresses the smallest whole number ratio of the atoms present. Molecular FormulaMolecular Formula The formula that states the actual number of each kind of atom found in one molecule of the compound.

30. ¬Convert the percentages to grams by assuming you have 100 g of the compound –Step can be skipped if given masses Determine the Empirical Formula of Acetic Anhydride if its Percent Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen

31. ­Convert the grams to moles Determine the Empirical Formula of Acetic Anhydride if its Percent Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen

32. ®Divide each by the smallest number of moles Determine the Empirical Formula of Acetic Anhydride if its Percent Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen

33. ¯If any of the ratios is not a whole number, multiply all the ratios by a factor to make it a whole number –If ratio is ?.5 then multiply by 2; if ?.33 or ?.67 then multiply by 3; if ?.25 or ?.75 then multiply by 4 Multiply all the Ratios by 3 Because C is 1.3 Determine the Empirical Formula of Acetic Anhydride if its Percent Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen

34. °Use the ratios as the subscripts in the empirical formula C4H6O3C4H6O3 Determine the Empirical Formula of Acetic Anhydride if its Percent Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen

35. Molecular Formulas The molecular formula is a multiple of the empirical formula To determine the molecular formula you need to know the empirical formula and the molar mass of the compound

36. uDivide the given molar mass of the compound by the molar mass of the empirical formula –Round to the nearest whole number Determine the Molecular Formula of Benzopyrene if it has a molar mass of 252 g and an empirical formula of C 5 H 3

37. uMultiply the empirical formula by the calculated factor to give the molecular formula (C 5 H 3 ) 4 = C 20 H 12 Determine the Molecular Formula of Benzopyrene if it has a molar mass of 252 g and an empirical formula of C 5 H 3