1. Ch 11: The Mole

2. Section 11.1 Measuring Matter
Dozen= 12
Ream=
500
Pair= 2
Gross=
144

3. What is a mole?
It is the SI base unit used to measure the amount of a substance
Abbreviated mol
Represents particles
Is also called Avogadro’s number
6.02 x 1023 (3 significant figures)

4. Remember conversion factors?
12 roses = 1 dozen
So, 3.5 dozen= ? Roses
= 42
Well, 6.02 x 1023 atoms (or any representative paticles) = 1 mol

5. Review Scientific Notation
Review Rounding

6. Work practice problems 1-3
Example:
3.50 mol sucrose has how many molecules?
Work practice problems 1-3
How many moles are in 3.58 X 1020 atoms of Ca?
Work practice problems 4-7

7. closure How is a mole similar to a dozen?
What is the relationship between avagadro’s number & one mole?
Why do chemists use moles?
**worksheet: The mole & Avogadro’s number
STOP

8. Section Mass & the Mole
Different particles (atoms) have different masses.
Remember atomic mass.
Each element has its own specific mass.
Therefore each compound has its own specific mass.

9. Molar mass (g/mol)-mass in grams of any pure substance
The molar mass of any element is numerically equal to its atomic mass.
Thus…1 mol Mn = g/mol Mn = 6.02 x 1023 atoms Mn

10. Example:
1 mol Zn =
65.4 g/mol
1 mol O2 =
32.0 g/mol

11. Practice
3 mol Zinc =
196 g/mol Zn
1 mol H2O =
18.0g/mol H2O
1 mol sulfuric acid =
98.1 g/mol H2SO4

12. Mass  Mole conversions
Example:
Calculate the mass in grams of mol Cr.
Work Practice problems 1-4

13. Hydrate: CuSO4·5H2O
4) Calculate the mass in grams of 2.45 mol of CaCl2·2H2O

14. Mole Mass conversions
Example:
Determine the number of moles for 25.5 g Ag. (mass mol)
Work practice problems 5-7.
**worksheet: moles & mass

15. Mass  Atom Conversion Example:
How many atoms of gold (Au) are in a pure nugget having the mass of 25.0g?
Practice problems 1-5

16. Atom  Mass Conversion Example
A party balloon contains 5.50 x 1022 atoms of helium (He) gas. What is the mass in grams of the helium?
Practice problems 6-10.

17. Mole  Mole
Example
According to the following balanced equation, how many moles of O2 is produced from 3.00 moles of CuO?
2CuO  2Cu + O2
Practice problems 1 & 2.

18. Section 11.4 Empirical & Molecular Formula
Percent Composition is the percent by mass of each element in a compound.
Example:
If we had 100 g of a sample of some new compound contains 55g of element X & 45 g of element Y. What is the % of element X & Y?

19. Mass of element in 1 mol of compound X 100
If we already know the _chemical formula for a compound, you can calculate its percent composition.
% by mass=
Mass of element in 1 mol of compound X 100
Molar mass of compound

20. Ex. Determine the percent composition of H2O.
(If you had 350. g of water, then how much is oxygen?)
Practice problems 1-3.
311g

21. Empirical Formula is the smallest whole number ratio of the elements.
Calculating Empirical formula from percent composition:
Directions:
The percent should be assumed to be 100 g & converted to moles. Then we figure out the mole ratio by dividing each by the smallest mole.

22. POEM:
Percent to Mass
Mass to Mole
Divide by small
Multiply til whole.

23. Example 1
The percent composition of an oxide of sulfur is 40.05% S & 59.95% O.
Example 2
Determine the empirical formula for methyl acetate which has the following chemical analysis:48.64% C, 8.16% H, & 43.20% O.
Practice Problems

24. **worksheet: determining empirical formulas