1. Stoichiometry

2. Atomic Mass
It is impossible to weigh a single atom, but one can easily establish the weight of an atom relative to another
We must establish a standard:
An atom of carbon-12 (six protons, six neutrons, six electrons) has a mass of exactly 12 atomic mass units (u)
Therefore, 1 atomic mass unit (u) is equivalent to exactly a twelfth of the mass of an atom of carbon 12
If it is found that a hydrogen atom has only 8.400% of the mass of a carbon-12 atom, the atomic mass of hydrogen must be: x 12 u = u

3. Average Atomic Mass Most natural elements have more than one isotope
e.g.; there are two isotopes of carbon; carbon-12 (98.89%) and carbon-13 (1.11%, atomic mass = u), and the average atomic mass is:
(0.9889)( u) + (0.0111)( ) = u
Since carbon-12 is the dominant isotope, the average atomic mass is approximately 12
N.B u is an average value, and no single carbon atom has this mass, i.e., it’s either u or u

4. Average Atomic Mass
The atomic masses of the two stable isotopes of boron, boron-10 (19.78 %) and boron-11 (80.22 %), are u and u, respectively. Calculate the average atomic mass of boron.
Solution: atomic mass of B = (0.1978)( u) + (0.8022)( u)
= u

5. Avogadro’s Number
One mole describes a particular number of objects, just like a pair (2), a dozen (12), or a gross (144)
A mole is the number of carbon-12 atoms in exactly 12 g of carbon-12
The current accepted value is mole = x 1023 particles

6. Molar Mass
The molar mass of an atom or molecule is the mass of one mole of these atoms or molecules
The molar mass of carbon-12 is thus exactly 12 g, by definition
The numerical value of the molar mass (in grams) for an atom is the same as the numerical value of the average atomic mass (in u)
e.g.; For Na, the average atomic mass = u and the molar mass = g

7. The Mass of an Atom
We know that, by definition, 12 g de carbon-12 contains x 1023 atoms of carbon-12
The mass of a single atom of carbon-12 is therefore
A single atomic mass unit is 12 times smaller, and therefore
1 u = x g, and 1 g = x 1023 u

8. The Molar Mass of an Element and Avogadro’s Number
Example: What is the mass (in grams) of an atom of iodine?
Solution:
The molar mass of iodine = g/mol, therefore

9. Molecular Mass
The molecular mass is the sum of the atomic masses (in atomic mass units) of the atoms that make up a molecule
ex.; the molecular mass of water is
(2)(1.008 u) + (1)(16.00 u) = u
The molar mass of a molecule (in grams) has a numerical value equal to its average molecular mass (in atomic mass units)
e.g.; the molar mass of water is g, and thus g of water contains one mole (or x 1023 molecules) of water

10. Molecular Mass
Example: What is the molecular mass of methanol (CH3OH)?
Solution: molecular mass of CH3OH = (4)(1.008 u) (1)(12.01 u) (1)(16.00 u) = u

11. Molar Mass
Example: Calculate the number of moles in 198 g of chloroform (CHCl3).
Solution:

12. Mass Spectrometry
An electron beam ionizes the molecules, and a magnet affects the trajectory of the ions produced according to their charge : mass ratio
The signal produced by a particular ion is proportional to its abundance

13. Mass Spectrometry
In this example, we can determine the relative abundance of each Ne isotope
Mass spectrometry can also be used to identify the molecular mass of a compound

14. Percent Composition
The percent composition is the mass percentage of each element contained within a compound
The percent composition allows us to establish the empirical formula of a compound
With the molecular mass, obtained from mass spectrometry, or the ideal gas law in the case of a gas, we can then determine the molecular formula

15. Percent Composition
Example: The percent composition of hydrogen peroxide is 5.94% H and 94.06% O. What is its empirical formula? If the molecular mass is u, what is the molecular formula?
Solution:
If we have 100 g of hydrogen peroxide, we’d have g of O and 5.94 g of H, and therefore (94.06 g)/(16.00 g/mol) = 5.88 mol of O and (5.94 g)/(1.008 g/mol) = 5.89 mol of H
This gives a H:O ratio of about 1:1, and therefore, the empirical formula is HO
If the molecular formula was HO, the molecular mass would be u (twice as small)
Therefore, the molecular formula is H2O2

16. Percent Composition
Example: Determine the empirical formula of a compound with the following percent composition: % K, 34.77% Mn, and 40.51% O.
Solution:
If we have 100 g of this compound, we’d have g of K, g of Mn, and g of O, or (24.75 g)/(39.10 g/mol) = mol of K, (34.77 g)/(54.94 g/mol) = mol of Mn, and (40.51 g)/(16.00 g/mol) = mol of O
This gives a K:Mn:O ratio of about 1:1:4
Therefore, the empirical formula is KMnO4

17. The percent composition of a compound is 38. 08% C, 31. 71% O, 25
The percent composition of a compound is 38.08% C, 31.71% O, 25.42% S, and 4.79% H. The molecular mass of the compound is approximately g/mol. What is the empirical formula of this compound? What is the molecular formula? What is the mass (in g) of a molecule of this compound?

18. Balancing Chemical Equations
Once all of the reactants and products are known (this is not always trivial), we can balance the chemical equation so as to respect the law of conservation of mass
We can balance the equation by playing with the coefficients in front of each reactant or product (we can not play with the molecular formula!!!)
It is often easier to start balancing an element that appears in only one place on each side of the reaction

19. Balancing Chemical Equations
e.g.; KClO KCl + O2
K and Cl are balanced (for now)
We balance O KClO KCl + 3 O2
O is now balanced
K et Cl are not balanced anymore KClO KCl + 3 O2
The equation is now balanced

20. Balancing Chemical Equations
N.B. the chemical equations KClO KCl + (3/2) O KClO KCl + 6 O are both acceptable and balanced
However, it is common to balance equations using the smallest possible integers as coefficients

21. Balancing Chemical Equations
e.g.; C2H6 + O CO2 + H2O
None of the elements are balanced
We balance the C C2H6 + O CO2 + H2O
We balance the H C2H6 + O CO2 + 3 H2O
We balance the O C2H6 + (7/2) O CO2 + 3 H2O
The equation is now balanced
We double the coefficients in order to obtain integers C2H6 + 7 O CO2 + 6 H2O

22. Balancing Chemical Equations
e.g.; Fe2O3 + CO Fe + CO2

23. Calculating Masses of Reactants and Products
Stoichiometry is the study of the relationships between the masses of the reactants and products in a chemical reaction
If we know the amount of reactants (products), we can calculate the amount of products (reactants) that are formed (required)
To simplify the process, we use the moles method, i.e., we work directly with the moles (rather than grams or liters)

24. Calculating Masses of Reactants and Products
For the reaction 2 CO(g) + O2(g) CO2(g) at least two moles of CO (g) are needed to produce two moles of CO2 (g) and at least one mole of O2 (g ) is also necessary
In this example, we say that 2 mol of CO are stoichiometrically equivalent to 1 mol of O2(g) as well as 2 mol of CO2(g)
In the same way, we say that 1 mol of O2(g) is stoichiometrically equivalent to 2 mol of CO2(g)

25. The Mole Method Step (1): Identify all of the reactants and products
Step (2): Balance the chemical equation
Step (3): Convert all the known quantities into moles
Step (4): Use the coefficients of the balanced chemical equation to determine the number of moles of the quantities sought
Step (5): If necessary, convert the number of the quantities sought into grams, litres, etc.

26. The Mole Method

27. The Mole Method
Example: The reaction between NO and O2 is a key step in the formation of photochemical smog: NO(g) + O2(g) NO2(g) (a) How many moles of NO2(g) are formed by the complete reaction of mol of O2(g)? (b) What mass, in grams, of NO2(g) is obtained by the complete reaction of 1.44 g of NO(g)?

28. The Mole Method
Solution: (a) The equation is already balanced and we know that et mol of O2(g) reacts. Thus, we produce twice as much NO2. We produce mol.
(b) 1.44 g of NO is (1.44 g)/(14.01 g/mol g/mol) = mol of NO This will produce mol of NO This has a mass of ( mol) x [14.01 g/mol + (2)(16.00 g/mol)] = 2.21 g

29. Limiting Reagents and the Yield of Reactions
The reagent which is first exhausted is called the limiting reagent
The maximum amount of product depends on the initial amount of the limiting reagent
The other reactants are called the excess reagents

30. Limiting Reagents and the Yield of Reactions
e.g.; For the reaction S(l) + 3 F2(g) SF6(g)
We have 4 moles of S and 9 moles of F2
4 moles of S will react with 12 moles of F2
But we only have 9 moles of F2
The F2 will be used up before all of the S can react
The F2 is the limiting reagent
N.B. We arrive at the same conclusion when saying that the 9 moles of F2 will react with only 3 moles of S, thus leaving 1 mole of S in excess
3 moles of SF6 is produced

31. What mass of POCl3 is produced when 790. 0 g of PCl3(l), 345
What mass of POCl3 is produced when g of PCl3(l), g of Cl2(g), and g of P4O10(s) reacts together in the following fashion? PCl3(l) + 6 Cl2(g) + P4O10(s) →10 POCl3(l)