1. 5-4 Coefficients and Subscripts
And you

2. A coefficient is the number in front of chemical formula and represents the number of non-bonded, separate species. In 3H2O, the coefficient 3 indicates there are 3 water molecules.

3. A subscript is the lowered number within a chemical formula that represents the number of bonded species of item that precedes the number. In H2O there are 2 H atoms bonded to the O atom.

4. Building the Concept of Particle Ratios:
1 H2O molecule = 2 H atoms O atom
2 H2O molecules = 4 H atoms O atoms
10 H2O molecules = H atoms O atoms
1000 H2O molec. = 2000 H atoms O atoms
Do you see the pattern here??????

5. Great JOB!!!!!!!! 106 H2O molec. = 2 x 106 H atoms , 1 x 106 O atoms
6.022 x (6.022 x 1023) 1(6.022 x 1023)
H2O molecules = H atoms O atoms
1 mole H2O molec = 2 moles H atoms 1 mole O atoms
5 moles H2O molec = 10 moles H atoms 5 moles O atoms
Great JOB!!!!!!!!

6. As you can see, the coeffients and subscripts can represent atoms, molecules, or moles. In other words, coefficients and subscripts are merely ratios, representing small things like atoms and molecules, or a large number of these small things, like a mole. Note: coefficients and subscript DO NOT give mass ratios, so we’ll have to convert mass to moles for formulas.
Be careful with diatomics, which can be confusing:

7. 6.022 x 1023 atoms of hydrogen =
1 mole H =
1.01 g
6.022 x 1023 molecules of hydrogen =
1 mole H2 =
2.02 g =
2 moles of H atoms

8. Mole Bridge Extension
We can now extend the mole bridge to link molecules to atoms:
 Example:
How many H atoms are present in g H2O?
 Grams →
Moles →
Molecules →
Atoms

9. Grams → Moles → Molecules → Atoms
0.200 g H2O x 1 mole H2O x x 1023 molecules H2O x 2 atoms H
18.02 g mole H2O H2O molecule
See I told you this would be fun!!!

10. 5-5 Percent Composition (Section 8.5)
An elemental analysis is the first step in determining a chemical’s formula and gives results called percent composition. Percent composition is typically reported by mass and is determined by dividing the mass of an element in a formula by the total mass of the compound.

11. Note how the % values should add to 100%
Example 1: A compound contains 7.00 g of nitrogen out of a total sample mass of 23.0 g. The rest of the mass is oxygen. Calculate the % composition by mass of nitrogen and oxygen. 
Since the total mass is 23.0 g and 7.00 g of this are nitrogen, 16.0 g are oxygen.
% N = 7.00/23.0 x 100 = 30.4 %
% O = 16.0/23.0 x 100 = 69.6 %
Note how the % values should add to 100%

12. Example 2: Determine the % by mass of each element in Ba(OH)2•8H2O
Ba(OH)2•8H2O Molar mass = g/mol
Ba: / x = %
O: 2(16.00)/ x = %
H: 1(1.01)/ x = %
H2O 8(18.02)/ x = %
Note how the % values should add to 100%.