1. MOLE AND STOIKIOMETRI We measure mass in grams. g
We measure volume in liters. L
We count pieces in MOLES. mol

2. MOLE CONCEPT A mole of Carbon atoms has a mass of 12 grams
A Carbon-12 atom
has a mass of 12amu C
= 12amu
A mole of Carbon atoms
has a mass of 12 grams C

3. Gram Atomic Mass Equals the mass of 1 mole of an element
Written on the Periodic Table
12.01 grams of C has the same number of pieces as grams of H and grams of iron.
We can write this as g C = 1 mole C
We can count things by weighing them.

4. Atomic Mass Mole of atoms in grams 27.0g A mole of aluminum =
The Mole Concept
Atomic Mass
in grams
Mole of atoms
27.0g
A mole of aluminum =
A mole of gold =
A mole of silver =
A mole of boron =
197g
108g
10.8g

5. If a Helium atom is 4 times heavier than a Hydrogen atom,
then a dozen Helium atoms is 4 times
heavier than a dozen Hydrogen atoms. He H

6. And a mole of Helium atoms is 4 times
heavier than a mole of Hydrogen atoms.

7. The number of atoms in a mole is called,
So, the number of atoms in a mole
is always the same
The number of atoms in a mole is called,
Avogadro's Number
Avogadro's Number = x 1023

8. Mole – Particle Conversions
The mole concept allows us to count atoms, molecules, formula units and ions.
1 mole of Al = x 1023 atoms of Al
So how many atoms in 3 mol of Al?

9. Examples How much would 2.34 moles of carbon weigh?
How many moles of magnesium is g of Mg?
How many atoms of lithium is 1.00 g of Li?

10. What about compounds?
In 1 mole of H2O molecules there are 2 moles of H atoms and 1 mole of O atoms
To find the mass of one mole of a compound
determine the moles of the elements they have
Find out how much they would weigh
add them up

11. Types of Molar Mass
Gram Molecular Mass (GMM)- the total number of atoms in a molecular compound
Cl2 = 2 x 35.4 g = 70.8 g
CH4=
Gram Formula Mass (GFM)- the total number of atoms in an ionic compound
NaCl = (1 x 23.0g) + (1 x 35.4g) = 58.4g
Ca(OH)2=
(1 x 12.0g) + (4 x 1.0g) = 16g
(1 x 40.0g) + (2 x (16.0g+1.0g))= 74g

12. Molar Mass
Molar mass is the generic term for the mass of one mole of any substance (in grams)
The same as:
1) gram molecular mass
2) gram formula mass
3) gram atomic mass

13. = 78.1 g = 92.0 g = 164.1 g = 180.0 g = 149.0 g Examples Na2S N2O4
Calculate the molar mass of the following and tell what type it is:
Na2S
N2O4
Ca(NO3)2
C6H12O6
(NH4)3PO4
= 78.1 g
= 92.0 g
= g
= g
= g

14. Gases Many of the chemicals we deal with are gases.
They are difficult to weigh.
Need to know how many moles of gas we have.
Two things effect the volume of a gas
Temperature and pressure
We need to compare them at the same temperature and pressure.

15. Standard Temperature and Pressure
0ºC and 1 atm pressure
0oC = 32oF= 273 K
1 atm= kPa = 760 mmHg (torr)
At STP 1 mole of gas occupies 22.4 L
Called the molar volume
1 mole = 22.4 L of any gas at STP

16. Examples What is the volume of 4.59 mole of CO2 gas at STP?
How many moles is L of O2 at STP?
4.59 mol CO2
22.4 L CO2
=103 L CO2
1 mol CO2
5.67 L O2
1 mol O2
=0.253 mol O2
22.4 L O2

17. Stoichiometry “Stoichiometry” is Greek for “Measuring Elements”
It starts with a balanced equation
2H2 + O2 ® 2H2O
2 moles of hydrogen reacts with
1 mole of oxygen
Forming 2 moles of water.

18. Stoichiometry 3 g of reactants can’t make only 2 g of products
The coefficients tell us how many moles of each substance. Not Grams!
For example:
2H2 + O2 ® 2H2O
2 g of H2 + 1 g of O2 = 2 g of H2O
3 g of reactants can’t make only 2 g of products

19. Mass of a Product 2 moles H2 2.02 g H2 = 4.04 g H2 1 mole H2 1 mole O2
The Law of Conservation of Mass applies
Convert the moles to grams and the equation does work.
2H2 + O2 ® 2H2O
2 moles H2
2.02 g H2 =
4.04 g H2
1 mole H2
1 mole O2
32.00 g O2 =
32.00 g O2
1 mole O2
36.04 g H2+O2

20. Mass of a Product 2H2 + O2 ® 2H2O 2H2 + O2 ® 2H2O 18.02 g H2O
2 moles H2O
18.02 g H2O
36.04 g H2O =
1 mole H2O
2H2 + O2 ® 2H2O
36.04 g H2 + O2 =
36.04 g H2O

21. 2Na + Cl2  2NaCl 2Na (s) + Cl2(g) 2NaCl (s)
How many grams of sodium are needed to react with .071g of chlorine gas?
2Na (s) + Cl2(g) 2NaCl (s)
Mole Ratio
Mass Ratio
23 x 2= 46
35.5 x 2 x 1= 71
58.5 x 2=
117
1 mol Cl2
2 mol Na
0.071 g Cl2
46 g Na
71 g Cl2
1 mol Cl2
2 mol Na
= g Na

22. Chemical Yield Actual Yield x 100 Theoretical Yield
3. Percent yield – a percentage/ratio between the actual yield and the theoretical yield.
% Yield =
% yield tells us how “efficient” a reaction is.
% yield can not be bigger than 100 %.
Actual Yield
x 100
Theoretical Yield

23. Example 11.8 g NaCl x 100 13.6 g NaCl Answer = 86.7 %
According to your calculations the theoretical yield for the production of NaCl is 13.6 grams. In the laboratory your actual yield is 11.8 grams of NaCl. What is the percent yield?
11.8 g NaCl
x 100
13.6 g NaCl
Answer = 86.7 %

24. Calculating Percent Composition of a Compound
Like all percent problems:
Part whole
Find the mass of each component,
then divide by the total mass.
x 100 %

25. Calculating Percent Composition of a Compound
Find the mass percent for the elements in Sodium hydrogen carbonate:
NaHCO3
-molar mass is 84 g/mol
-mass of: Na= 23g H= 1g
C= 12g O= 3 x 16= 48g
%C = (12g/84g) x 100= 14.3%
%Na = (23g/84g) x 100= 27.4%
%H = (1g/84g) x 100= 1.2%
%O = (48g/84g) x 100= 57.1%

26. The Empirical Formula
The lowest whole number ratio of elements in a compound.
The molecular formula = the actual ratio of elements in a compound.
The two can be the same.
CH2 is an empirical formula
C2H4 is a molecular formula
C3H6 is a molecular formula
H2O is both empirical & molecular

27. Calculating Empirical
Just find the lowest whole number ratio
C6H12O6
CH4N
It is not just the ratio of atoms, it is also the ratio of moles of atoms.
In 1 mole of CO2 there is 1 mole of carbon and 2 moles of oxygen.
In one molecule of CO2 there is 1 atom of C and 2 atoms of O.

28. Calculating Empirical
We can get a ratio from the percent composition.
Assume you have a 100 g.
The percentages become grams.
Convert grams to moles.
Find lowest whole number ratio by dividing by the smallest mole value

29. Example
Calculate the empirical formula of a compound composed of % C, % H, and %N.
Assume 100 g
38.67 g C x 1mol C = mole C g C
16.22 g H x 1mol H = mole H g H
45.11 g N x 1mol N = mole N g N

30. Example The ratio is 3.220 mol C = 1 mol C 3.219 mol N 1 mol N
The ratio is mol H = 5 mol H mol N mol N
= C1H5N1 = CH5N
Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula?