1. The Mole
Chapter 6

2. The Mole Remember it is a measurement for anything
1 mole means x 1023 of anything
This is called Avogadro’s Number
A sample of an element with a mass equal to that element’s average atomic mass expressed in grams contains 1 mol of atoms
26.98 g of Aluminum has x 1023 atoms = 1mol

3. How can this help us?
If you know how many grams make 1 mole, you can figure out how many moles you have in a given mass
If you have g of Silicon how many atoms do you have?
Know that 1 mol Si is g and 1 mole is x 1023 atoms, these are your conversion factors
So g x 1 mol/28.09 g = mol Si
mol Si x x 1023 atoms/ 1 mol = 1.22 x 1020 Si atoms

4. Molar Mass
What is the mass of 1 mole of a compound? Simply add the mass of the elements moles to get the compounds molar mass
CH4 is made up of 1 mol of C and 4 mol of H
1 mol C = 1 x g = g
4 mol H = 4 x g = g
So 1 mol of CH4 = g g = g
Try to find molar mass of CaCO3
Answer =
Now figure out how many grams of CaCO3 are in 4.86 mol

5. Practice Problem
How many moles and how many molecules of C7H14O2 is in a bee sting if 1 gram is released?
First find Molar Mass
Answer
Second find how many moles in 1 gram
Finally change moles to molecules

6. Enough for 1 day, Watch the Video “The Mole”

7. Percent composition of Compounds
Take mass of 1 element and divide it by the mass of entire compound and multiply by 100
Example: C2H5OH (ethanol) What is the mass percent of Carbon in this compound?
What is mass of 1 mole of ethanol?
Answer
What is the mass percent of carbon in ethanol?
Now do Hydrogen and Oxygen
Answers

8. 6.6 Formulas of Compounds
Empirical Formula – simplest formula and expresses the smallest whole-number ratio of the atoms present
C6H12O6 and C4H8O4 both have the same empirical formula CH2O
Molecular Formula – the actual formula of a compound that tells you the composition of the molecules that are present
C6H12O6 is the molecular formula for glucose

9. 6.7 Calculation of Empirical Formulas
Important to learn the chemical formula of a new compound
First figure out relative masses
Second convert masses to number of moles
Divide by the smallest number of moles found
Multiply the numbers from 3rd step by smallest number that will make them all whole numbers, this is the empirical formula

10. Example Problem
An oxide of aluminum is formed by the reaction of g of aluminum with g of oxygen. What is the empirical formula of the compound formed?
Relative Masses must be converted to moles
4.151 g Al x 1 mol Al/ g Al = mol of Al atoms
3.692 g O x 1 mol O/ 16.oo g O= mol of O atoms
Divide by smallest number of moles
mol Al / = mol Al
mol O / = mol O
Multiply by smallest number to make them all whole
1.500 O x 2 = 3 O atoms
1.000 Al x 2 = 2 Al atoms
So formula is Al2O3

11. Practice Problem
In a lab experiment it was observed that g of lead combines with g of chlorine to form a binary compound. What is the empirical formula of this compound?
Step 1 Relative Masses
Step 2 Convert to moles
Step 3 Divide by smallest number
Step 4 Multiply by smallest number that makes them all whole numbers
Rules apply no matter how many elements are in the compound

12. Calculation of Molecular Formulas
Must know percent composition and the molar mass
It’s always a multiple of the empirical formula (Empirical Formula)n = molecular formula or Empirical Formula x n = molecular formula
Compare the empirical formula to molar mass

13. Example
If Empirical formula is P2O5 and molar mass is g, what is compound’s molecular formula?
Know we have 2 moles P and 5 moles O
2 mol P = 2 x g = g
5 mol O = 5 x g = g
1 mol P2O5 = g
Know empirical formula x n = molecular formula and that the molecular formula = molar mass
This tells us n = molar mass/ empirical formula
So g / = n = 2
So (P2O5)2 means P4O10

14. Practice Problem
What are the empirical and molecular formulas of a compound with these percent compositions 71.65% Cl, 24.27% C, 4.07% H and a molar mass of g?