1. The Mole
the mole is the SI unit representing atomic amounts.
ie. 1 dozen donuts = 12 donuts
1 pair = 2
1 year = 365 days
1 ream = 500 sheets
1 mole =
6.022 x 1023
particles

2. 6.022 x 1023 is referred to as Avogadro’s Number.
Molar Mass
The mass of one mole of any substance
is called the molar mass.
(same value as the atomic mass)
ie: Molar mass of He =
Molar mass of Hg =
Molar mass of Li =
4 g/mol
g/mol
6.94 g/mol

3. Mole Conversions x molar mass = = (1/molar mass) x x Avogadro’s # =
Amount in moles
Mass in grams
= (1/molar mass) x
x Avogadro’s # =
Amount in moles
Number of atoms
= (1/Avogadro’s #) x

4. If you are given moles, you……………MOLE-TIPLY!

5. Examples:
What is the mass in grams of 3.50 mol of the element copper?
3.50 mol Cu x (63.5 g Cu / mol Cu) = 222 g Cu
A chemist produced 11.9 g of aluminum. How many moles of aluminum have been produced?
11.9 g Al x (1 mol Al/ 27.0 g Al) = mol Al
How many moles of silver are there in 3.01 x 1023 atoms?
3.01 x 1023 atoms Ag x (1 mole Ag/ 6.02 x 1023 atoms) =
0.500 mol Ag

6. What do I want to know?
Answer: cost per party ($/party)
$ pizza slices persons X X X
pizza slices person party
= $73.90 per party

7. Answer: kilograms of Ca(NO3)2?
What do I want to know?
Answer: kilograms of Ca(NO3)2?
164 g Ca(NO3) mol Ca(NO3) molecule Ca(NO3)2 X X
1 mol Ca(NO3) x 1023 molecules Ca(NO3) atoms Oxygen
Did you
Get it?
5.00 x 1025 atoms Oxygen kg X
sample g
= kg Ca(NO3)2
sample

8. How about How about a quiz? a quiz?
Remember The Fun We Had With Mole Conversions?!?!?
MOLES
x molar mass
x Avogadro's #
/ molar mass
/ Avogadro's #
MOLECULES
MASS
How about
a quiz?
How about
a quiz?

9. Percent Composition
to find the percentage by mass of each element in a compound:
Mass of element in sample of compound
Total mass of compound
x 100
Ex: Find the percentage composition of copper (I) sulfide.
copper (I) sulfide  Cu2S
% Cu = 2(63.55 amu) / amu = %
% S = amu / amu = %
Ex: Calculate the percent composition of sodium nitrate.
sodium nitrate  NaNO3
% Na = 23 amu / 85 amu = 27.1 %
% N = 14 amu / 85 amu = 16.5 %
% O = 3(16 amu) / 85 amu = 56.4%

10. Calculate the percent composition of silver sulfate.
silver sulfate  Ag2SO4
% Ag = 69.2 % % S = 10.3 % % O = 20.5 %

11. Percent Error
a calculation used to see how accurate your measurements are.
Try this: A student measures the mass of a sample
as 9.67 g. Calculate the percent error, given that the
correct mass is 9.82 g.
Answer: 1.5%

12. Percent Error
Try this: A handbook gives the density of calcium as
1.54 g/cm3. What is the percent error of a density
calculation of 1.25 g/cm3 based on lab measurements?
Answer: 19%
What is the pecent error of a length measurement of
0.229 cm if the correct value is cm?
Answer: -2%

13. x 100 Total mass of compound H20: 5 x 18 = 90 Cu: 1 x 63.5 S: 1 x 32.0
Mass of molecule in sample of compound
Total mass of compound
x 100
H20: 5 x 18 = 90
Cu: 1 x 63.5
S: 1 x 32.0
O: 4 x 16.0
= 63.5
= 32.0
= 64.0
159.5
Mass of molecule in sample of compound = 90
Total mass of compound =
= 249.5
90/249.5 = x 100 = 36.07%

14. DETERMINATION OF EMPIRICAL & CHEMICAL FORMULAS
empirical formula: a formula that shows the smallest whole-number ratio of the different atoms in the compound.
ex: C6H12O6  CH2O
To find the empirical formula use % compositions.
Ex: Find the empirical formula of a substance containing 78.1 % B and 21.9 % H.
Step1: Convert percentages to grams.
Assume 100g  thus, 78.1 g B and 21.9 g H

15. Easy right????? Step 2: Find the moles of each.
B: 78.1 g / g/mol = 7.22 mol B
H: 21.9 g / 1.01 g/mol = 21.7 mol H
Step 3: Divide by the lowest mole value.
B: 7.22 mol / 7.22 mol = 1
H: 21.7 mol / 7.22 mol = 3.01
Step 4: Show the ratio in a formula (empirical formula)
empirical formula  BH3
Easy right?????

16. Percent to mass Mass to mole Divide by small Multiply 'til whole!

17. Ex: Find the empirical formula of a compound that is 32. 38g Na, 22
Ex: Find the empirical formula of a compound that is 32.38g Na, 22.65g S, and 44.99g O.
Step 1: Already have masses!
Step 2: Convert to moles.
Na: 32.38g / 23g/mol = 1.41 moles
S: 22.65g / 32.1g/mol = 0.71 moles
O: 44.99g / 16.0g/mol = 2.81 moles
Step 3: Divide by the lowest mole value.
Na: 1.41 moles / 0.71 moles = 1.99  2
S: 0.71 moles / 0.71 moles = 1
O: 2.81 moles / 0.71 moles = 3.96  4
Step 4: Use step 3 to make the empirical formula.
Na2SO4

18. A compound with a mass of 10. 15g contains P and O. There is 4
A compound with a mass of 10.15g contains P and O. There is 4.43g of P, what is the empirical formula of this compound?
Step 1: Find the grams of each.
P: 4.43g O: 10.15g – 4.43 = 5.72g
Step 2: Find the moles of each.
P: 4.43g / g/mol = 0.14 mol
O: 5.72g / 16 g/mol = 0.36 mol
Step 3: Find the mole ratio.
P: 0.14 mol / 0.14 mol = 1
O: 0.36 mol / 0.14 mol = 2.5  Be careful here!!
Need to double anything that ends in a .5
P: 2 O: 5
Step 4: Find the empirical formula.
P2O5

19. Calculating Molecular Formulas
an empirical formula may or may not be the correct molecular formula.
to determine the molecular formula of a compound, you must know the compound’s formula mass.
Ex: The empirical formula of a compound was found to be P2O5. Experimentation shows that the molar mass of this compound is g/mol. What is the compound’s molecular formula?
Need to know the empirical formula’s molar mass:
P2O5: 2(30.97g/mol) + 5(16g/mol) = g/mol
2. Compare the molecule’s molar mass to the empirical formula’s molar mass:
( g/mol) / ( g/mol) = 2
Find the molecular formula:
(P2O5)2 P4O10

20. Ex: A sample of a compound with a formula mass of 34
Ex: A sample of a compound with a formula mass of amu is found to consist 0.44g H and 6.92g O. Find its molecular formula.
Find the empirical formula:
H: 0.44g H /1.01 g/mol = mol
O: 6.92g O / 16 g/mol = mol
Divide by smallest mole amount:
H: mol / mol = 1.007
O: mol / mol = 1
Empirical Formula  HO (molar mass = g/mol)
Compare e.f. to m.f. using molar masses:
34.00 g/mol / g/mol = 2
Write molecular formula:
(HO)2  H2O2

21. Ex: If 4.04 g N combine with g O to produce a compound with a formula mass of amu, what is the molecular formula of this substance?
Find the empirical formula.
N: 4.04g / g/mol = mol
O: 11.46g / g/mol = mol
Ratio of N:O
N: mol / mol = 1
O: mol / mol = 2.49  2.5!!
Empirical Formula
N2O5 (formula mass = amu)
Compare the empirical formula mass to the molecular formula mass.
108.0 amu / amu = 1.00
Molecular Formula  N2O5

22. An unknown compound is found to be 91. 8% silicon and 8. 2% hydrogen
An unknown compound is found to be 91.8% silicon and 8.2% hydrogen. The molar mass of the compound is 122 g/mol. Find the molecular formula of the compound.
Find the empirical formula.
Si: 91.8 g / g/mol = 3.27 mol
H: 8.2 g / 1.01 g/mol = 8.1 mol
Ratio of moles.
Si: 3.27 mol/ 3.27 mol = 1
H: 8.1 mol/ 3.27 mol = 2.48  2.5
Si2H5 (molar mass = g/mol)
Find the molecular formula.
122 g/mol / g/mol = 2
Si4H10