1. Hi Guys & Gals! Warm-Up Time!
Draw the Lewis Structure and give the molecular shape and molecular polarity of the following molecules:
H2S
CHCl3
SO3 O2
Lower case L

2. Give the name or formula of the following provided the opposite.
Hi Guys & Gals! Quiz Time!
Give the name or formula of the following provided the opposite.
H3PO4
CF4
NaNO3
C2H4
SO2
Methane
Nitrogen trihydride
Ammonium hydroxide
Ethanol
Hydrochloric acid

3. Follow along in your text Chapter 7 Sections 1 & 2 Pages 224 - 240
The Mole
Follow along in your text Chapter 7 Sections 1 & 2 Pages

4. What is the Mole? A large amount!!!! A counting number (like a dozen)
Avogadro’s number (NA)
1 mol = 6.02  1023 particles
A large amount!!!!
VERY

5. HOW LARGE IS IT???
1 mole of hockey pucks would equal the mass of the moon!
1 mole of basketballs would fill a bag the size of the earth!
1 mole of pennies would cover the Earth 1/4 mile deep!

6. Molar Mass Mass of 1 mole of an element or compound.
Atomic mass tells the...
atomic mass units per atom (amu)
grams per mole (g/mol)
Usually rounded to 2 decimal places

7. Molar Mass Examples 12.01 g/mol 26.98 g/mol 65.39 g/mol
1 mol of carbon
1 mole of aluminum
1 mole of zinc

8. Molar Mass Examples H2O 2(1.01) + 16.00 = 18.02 g/mol NaCl
water
sodium chloride
H2O
2(1.01) = g/mol
NaCl
= g/mol

9. Molar Mass Examples NaHCO3
sodium bicarbonate
sucrose
NaHCO3
(16.00) = g/mol
C12H22O11
12(12.01) + 22(1.01) + 11(16.00) = g/mol

10. Relative Atomic Mass & Percentage Composition
Follow along in your text
Chapter 7 Sections 2 & 3
Pages

11. Relative Atomic Mass 100 = Average Mass of A
Is the same as Average Atomic Mass
The percent of each isotope of an element is used to find the average mass of that element.
(mass A1 x % of A1) + (mass A2 x % of A2)
100
= Average Mass of A
This was done earlier in the year.
Remember the Skittlesium Lab?

12. Relative Atomic Mass (84.91 x 72.17%) + (86.91 x 27.83%) = 85.47 amu
Calculate the average atomic mass of rubidium if 72.17% of its atoms have a mass of amu and 27.83% of its atoms have a mass of amu
(84.91 x 72.17%) + (86.91 x 27.83%) =
85.47 amu

13. Percentage Composition
the percentage by mass of each element in a compound

14. Percentage Composition
Find the % composition of Cu2S.
g Cu
g Cu2S
%Cu =
 100 =
79.852% Cu
32.07 g S
g Cu2S
%S =
 100 =
20.15% S

15. Percentage Composition
Find the percentage composition of a sample that is 28 g Fe and 8.0 g O.
28 g
36 g
%Fe =
 100 =
78% Fe
8.0 g
36 g
%O =
 100 =
22% O

16. Percentage Composition
How many grams of copper are in a 38.0-gram sample of Cu2S?
Cu2S is % Cu
(38.0 g Cu2S)( ) = 30.3 g Cu

17. Percentage Composition
Find the mass percentage of water in calcium chloride dihydrate, CaCl2•2H2O?
36.04 g g
%H2O =
 100 =
24.51%
H2O

18. Empirical Formula & Molecular Formula
Follow along in your text
Chapter 7 Sections 3
Pages

19. CH3 C2H6 Empirical Formula
Smallest whole number ratio of atoms in a compound
C2H6
reduce subscripts
CH3

20. Empirical Formula
1. Find mass of each element. If given percent, just say it’s a mass.
2. Find moles of each element.
3. Divide all moles by the smallest mole to find subscripts.
4. When necessary, multiply subscripts to get whole #’s.
If number ends in… multiply by… ≈ ≈ ≈

21. Empirical Formula
Find the empirical formula for a sample of 25.9% N and 74.1% O.
25.9 g
1 mol
14.01 g
= 1.85 mol N
= 1 N
1.85 mol
74.1 g
1 mol
16.00 g
= 4.63 mol O
= 2.5 O

22. N2O5 N1O2.5 Empirical Formula
Need to make the subscripts whole numbers
 multiply by 2
N2O5

23. C2H6 CH3 Molecular Formula
“True Formula” - the actual number of atoms in a compound
CH3
empirical
formula ?
C2H6
molecular
formula

24. Molecular Formula
1. Find the empirical formula (EF). 2. Find the empirical formula mass. 3. Divide the molecular formula (MF) mass by the empirical mass. 4. Multiply each subscript of the empirical formula by the answer from step 3.

25. (CH2)2  C2H4 Molecular Formula 28.1 g/mol 14.03 g/mol
The empirical formula for ethylene is CH2. Find the molecular formula if the molecular mass is g/mol?
empirical formula mass = g/mol
28.1 g/mol
14.03 g/mol
= 2.00
(CH2)2  C2H4