2.  The study of the quantitative relationships between reactants and products in a reaction  It is used to answer questions like; If I have this much reactant, how much product can I make?  If I want this much product, how much reactant do I need?  These questions have real life application, particularly in manufacturing.  It allows us to convert the mass of a substance to the number of particles (atoms, ions or molecules) it contains.  These numbers can be really large, so they are counted in groups  Much like when we count a lot of pennies we stack them in 10’s and count by 10

3.  Atoms are very tiny, so small that the grouping we use to count them must be very large  MOLE; the group (unit of measure) used to count atoms, molecules, formula units or ions of a substance  1 mole of a substance has a particular number of particles in it!  Much like 1 dozen always means 12; whether it is 12 eggs 12 oranges or 12 gold bars

4. The number of particles in a mole = 6.02 x 10 23 or 602,000,000,000,000,000,000,000 ! This is known as Avogadro’s Number Using this, We can easily count the number of particles in all kinds of things !

5. There are 6.02 x 10 23 Carbon atoms in a mole of carbon There are 6.02 x 10 23 CO 2 molecules in a mole of CO 2 There are 6.02 x 10 23 sodium ions in a mole of sodium There are 6.02 x 10 23 marbles in a mole of marbles That’s a lot of marbles! The Size of a mole of a substance changes, the bigger the substance the more space a mole of the substance takes up, but the number of particles in a mole is always the same!

7.  Chemicals do not come bundled in moles, like a dozen eggs comes in a 1 dozen or 1 ½ dozen package so we use the mole as a grouping unit. The mass of 1 mole of a pure substance called it’s molar mass  If I want to produce 500g of methanol using the following equation, CO2 +3H2  CH3OH + H20 how many grams of CO2 and H2 do I need?  These are the questions stoichiometry answers  These are the questions stoichiometry answers !

8. If I want to produce 500g of methanol using the following equation; CO 2 +3H 2  CH 3 OH + H 2 0 How many grams of CO 2 and H 2 do I need? This equation relates the molecules of reactants and products, NOT THEIR MASSES!  1 molecule of CO 2 and 3 molecules of H 2 will make 1 molecule of CH 3 OH We need to relate the masses to the number of molecules.

9. Remember; The average atomic masses of the elements are found on the Periodic Table! Remember; The average atomic masses of the elements are found on the Periodic Table!  We can use the atomic masses on the PT to relate the mass of the compound to the mass of a mole!

10. Molar mass: The mass (in grams)of one mole of a molecule or a formula unit Molecular mass: mass in atomic mass units of just one molecule Formula Mass: mass in atomic mass units of one formula unit of an ionic compound

11. Steps 1. Find the average Atomic Mass of the element on the PT. (state it in grams instead of atomic units) a) Example: molar mass of Fe = 55.847 g b) Example: molar mass of Pt = 195.08 g 2. If the element is a molecule, count the number of atoms in the molecule then multiply the atomic mass by the number of atoms. a) Example: O 2, the mass of O =16.0g There are 2 atoms of O in the O 2 molecule, 2 atoms X 16.0g = 32.00g is the molar mass of the molecule.

12. Calculate the molar mass of each of the following: 1. N 2 2. Cl 2 3. Br 2 4. I 2 5. H 2 6. F 2

13. Calculate the molar mass of each of the following: 1. N 2 = 14.007g X 2 =28.014 g/mol 2. Cl 2 = 35.453g X 2 =70.906 g/mol 3. Br 2 = 79.904g X 2 =159.808 g/mol 4. I 2 = 126.904g X 2 =253.808 g/mol 5. H 2 = 1.008g X 2 =2.016 g/mol 6. F 2 = 18.998g X 2 =37.996 g/mol

14. Steps 1. Count the number and type of atoms 2. Find the Atomic Mass of each atom type, on the periodic table. Write it in grams. 3. Multiply the mass times the # of Atoms. Then add the totals

15. 1. Count the number and type of atoms Ethanol (C2H5OH) 2. Find the Atomic Mass of each atom type, on the periodic table. Write it in grams. 3. Multiply The mass X the # of Atoms. Then add the totals. Atom typeAmount of each atom C2 H6 O1 Atom typeAmount of atomAve. Atomic Mass in g C212.0 H61.00 O116.0 Atom typeAmount of atomAve. Atomic Mass in gTotal C212.0=24.0 H61.00=6.0 O116.0=16.0 Molar Mass Of Ethanol (C 2 H 5 OH)= 46.0g/mole

16. Atom Types Amount of Atoms Ave. Atomic Mass in g Total Ca140.1 Cl235.571.0 Mass of 1 mol of CaCl 2 (molar mass)111.1 g/mole Example: Calcium Chloride (CaCl 2 )

17. What is the molar mass of each of the following? 1. Fe 2 O 3 2. H 2 O 3. CO 2 4. NaCl 5. NH 3 6. BaI 2

18. Fe 2 O 3 = 55.85g X 2= 111.7 g 16.0g X 3 = 48.0g = 159.7 g/mol _______________________________________________ H 2 O = 1.01g X 2 = 2.02 16.0g X 1 = 16.0 = 18.02 g/mol _______________________________________________ CO 2 = 12.01g X 1 = 12.01 16.0g X 2 = 32.0 = 44.01 g/mol ________________________________________________ NaCl = 22.99 gX1 = 22.99 35.45g X1 = 35.45 = 58.44 g/mol ________________________________________________ NH 3 =14.01g X 1 = 14.01 1.01g X 3 = 3.03 = 17.04 g/mol ________________________________________________ BaI 2 = 137.33g X 1 = 137.33 126.90g X 2 = 253.80 = 391.13 g/mol

19. If I want to produce 500g of methanol using the following equation; 6CO 2 +17H 2  3C 2 H 5 OH + 9H 2 0 How many grams of CO 2 and H 2 do I need? The Molar Mass Of Ethanol (C 2 H 5 OH) = 46.0g/mole Now we need to find the number of atoms in the sample. How many molecules of methanol are in 500g?

20. Steps to finding the number of atoms in a given mass of a sample 1. Use PT to find the molar mass of the substance 2. Convert the mass of the substance to number of moles in the sample (convert using mass of one mole as conversion factor) 3. Use the number of atoms in a mole to find the number of atoms in the sample 4. Solve and check answer by canceling out units

21. The mass of an iron bar is 16.8g. How many iron(Fe) atoms are in the sample? Step 1: Use PT to find the molar mass of the substance : The molar mass of Fe =55.8g/mole Step 2: Convert the given mass of the substance to number of moles in the sample: Fe =55.8g/mole (16.8g Fe) (1 mol Fe) (6.022 X 10 23 Fe atoms) = 1.81 X 10 23 Fe atoms (55.8g Fe) (1 mol Fe) Step 3: Use the number of atoms in a mole to find the number of atoms in the sample = 1.18 X 10 23

22. 1. 25.0 g silicon, Si 2. 1.29 g chromium, Cr

23. ( 25.0 g Si ) ( 1 mol Si ) ( 6.02 X 10 23 Si atoms ) 1 28.1g Si 1 mol Si = 5.36 X10 23 atoms Si ( 1.29 g Cr ) ( 1 mol Cr ) ( 6.02 X 10 23 Cr atoms ) 1 52.0g Cr 1 mol Cr = 1.49 X10 22 atoms Cr

24. 1. 98.3g mercury, Hg 2. 45.6g gold, Au 3. 10.7g lithium, Li 4. 144.6g tungsten, W

25. 1. ( 98.3 g Hg ) ( 1 mol Hg )( 6.02 X 10 23 Hg atoms ) 1 200.6g Hg 1 mol Hg = 2.95 X10 23 atoms Hg 2. ( 45.6 g Au ) ( 1 mol Au )( 6.02 X 10 23 Au atoms ) 1 197.0g Au 1 mol Au = 1.39 X10 23 atoms Au 3. ( 10.7 g Li ) ( 1 mol Li )( 6.02 X 10 23 Li atoms ) 1 6.94g Li 1 mol Li = 9.28 X10 23 atoms Li 4. ( 144.6 g W ) ( 1 mol W )( 6.02 X 10 23 W atoms ) 1 183.8g W 1 mol W = 4.738 X10 23 atoms W

26. Steps 1. Use the PT to calculate the molar mass of one formula unit 2. Convert the given mass of the compound to the number of molecules in the sample (use the molar mass as the conversion factor) 3. Multiply the moles of the compound by the number of the formula units in a mole (Avagadro’s number) and solve 4. Check by evaluating the units

27. 1. Calculate the molar mass (Fe 2 O 3 ) 2 Fe atoms 2X 55.8 = 111.6 3 O atoms 3 X 16.0 = +48.0 molar mass 159.6 g/mol 2. (change given mass  mole per mass  atoms per mole) ( 16.8 g Fe 2 O 3 ) ( 1 mol Fe 2 O 3 )( 6.02 X 10 23 Fe 2 O 3 Formula units ) 1 159.6g Fe 2 O 3 1 mol Fe 2 O 3 = 6.34 X10 22 Fe 2 O 3 Formula units

28. 1. 89.0g sodium oxide (Na 2 O) 2. 10.8g boron triflouride ( BF 3 )

29. 1. 89.0g sodium oxide (Na 2 O) Calculate the molar mass (Na 2 O) 2 Na atoms 2X 23.0 = 46.0 1 O atoms 1 X 16.0 = +16.0 molar mass 62.0 g/mol (change given mass  mole per mass  molecules per mole) ( 89.0 g Na 2 O ) ( 1 mol Na 2 O )( 6.02 X 10 23 Na 2 O Molecules ) 1 62.0g Na 2 O 1 mol Na 2 O = 8.64 X10 23 Na 2 O molecules

30. 2. 10.8g boron triflouride ( BF 3 ) Calculate the molar mass (Na 2 O) 1 B atom 1X 10.8 = 10.8 3 F atoms 3 X 19.0 = +57.0 molar mass 67.8 g/mol (change given mass  mole per mass  molecules per mole) ( 10.8 g BF 3 ) ( 1 mol BF 3 )( 6.02 X 10 23 BF 3 Molecules ) 1 67.8g BF 3 1 mol BF 3 = 9.59 X10 22 BF 3 molecules

31. Steps 1. Determine the molar mass 2. Change given mass to moles by using molar mass as the conversion factor.

32. Calculate the number of moles in 6.84g sucrose (C 12 H 22 O 11 ) 12 C atoms 12 X 12.0 = 144.0 22 H atoms 22 X 1.0 = 22.0 11 O atoms 11 X 16.0 = +176.0 molar mass 342.0 g/mol (change molar mass  mole per mass) ( 6.84 g sucrose ) ( 1 mol sucrose ) 1 342.0g sucrose = 2.0 X10 -02 moles of sucrose

33. 1. 16.0g sulfur dioxide, SO 2 2. 68.0g ammonia, NH 3 3. 17.5g copper(II) oxide, CuO

34. 1. 0.250 mol SO 2 2. 4.00 mol NH 3 3. 0.22 mol CuO