1. Unit 4: Formula Stoichiometry

2. What is stoichiometry?
Deals with the quantitative information in chemical formula or chemical reaction.
Using our reference tables we will learn how to calculate the following:
Mass of an atom
Mass of a compound
Moles of an atom or compound (molar mass)
Percent Composition in a compound

3. Key Terms Review
Subscripts - tell you the amount of atoms in a formula
Coefficients - tell you the amount of molecules of that substance

4. How many oxygen atoms in each?
NH4NO3
C8H8O4 O3
C3H5(NO3)3
(3)
(4)
(3)
(9)

5. Counting Atoms Practice
Example #1:
(NH4)2CO3
* First list the types of atoms and then count each. N: H: C: O: 2 8 1 3

6. Counting Atoms Practice
Example #2:
CaCrO4
* First list the types of atoms and then count each.
Ca:
Cr: O: 1 4

7. Counting Atoms Practice
Example #3:
Ca3(PO4)2
* First list the types of atoms and then count each.
Ca: P: O: 3 2 8

8. Mass Ex) Ne Ex) CHCl3 Ex) CaCl2
- All measured in amu (atomic mass units)
Atomic Mass
Molecular Mass
Formula Mass
Mass of one atom of one element from periodic table.
Mass of all of the atoms in a “Covalent” or “Molecular” compound (non-metal atoms only)
Mass of all of the atoms in an ionic compound (metal + non-metal ions)
Ex) Ne Ex) CHCl3 Ex) CaCl2

9. Gram Formula Mass & Gram Molecular mass
The same as before, but instead of “amu” Mass expressed in grams (g)
H2O O +
H H = + +
Molecular Mass = amu amu amu
Gram Molecular Mass = g

10. How to calculate molecular or formula mass…
First: Identify and count atoms in the compound.
Second: Locate atomic mass of each element.
Third: Multiply mass by number of atoms, then get the total of all elements.

11. Gram Molecular Mass Example (Covalent Compound)
Example: H2O
H- 2 x 1
O- 1 x 16
Molecular Mass = 18 g

12. Gram Formula Mass Example (ionic compound)
Example: NaCl
Na- 1 x 23
Cl- 1 x 35
Formula Mass = 58 g

13. Percent Composition by Mass
Used to calculate how much of a substance’s mass is made up of a particular element…
Formula is given on Table T in reference table
Mass of Part
Mass of Whole
= % Composition “by Mass”
X 100

14. Example 1:
In a sample of calcium carbonate, what is the percent composition by mass of calcium?
Formula = CaCO3
Ca: 1
C: 1
O: 3
x 40 = 40 g
x 12 = 12 g
x 16 = 48 g
} total = 100 g
40 g
100 g
% Composition Calcium =
X 100
= 40 % Ca

15. Example 2:
A compound containing carbon and hydrogen has a mass of 16 grams. When decomposed, 12.0 grams are found to be carbon. What is the percent by mass of carbon in the compound?
Formula = C?H?
} total = 16 g
C = 12 g
O = __ g
12 g
16 g
% Composition Carbon =
X 100
= 75 % Carbon

16. Hydrated Crystals:
Some ionic compounds have surrounding water molecules.
These are called Hydrated Crystals.
Percent water hydration can be found using the formula for % composition…
Example: CuSO4 . 5H2O
5 H2O molecules are attached to 1 CuSO4

17. Percent Water Hydration (cont.)
% Hydration = Total mass of Water x 100
Formula Mass
Formula = CuSO4 . 5H2O
Cu = 1 x 64 = 64 g
S = 1 x 32 = 32 g
O = 4 x 16 = 64 g
Separately H2O:
O = 1 x 16 = 16 g
H = 2 x 1 = 2 g
_________________
H2O = 18 x 5 = 90 g g ✔

18. The Mole Pt. 1
A mole is essentially one complete unit of any particular substance
Think of it like the term “dozen”
Meaning 12 items of any substance make up a dozen of that substance
1 mole is always equal to 6.02 x 1023 particles of a substance

19. The Mole Pt. 2 Avagadro’s Number is 6.02 x 1023
Avagadro’s Number is the number of:
- Atoms in 1 mole
of any element
- Molecules in 1 mole
of any compound
- Particles in 1 mole
of any substance

20. The Mole Pt. 3
The mass of 1 mole will be different for all substances.
Just like a dozen eggs being different from a dozen tires (size and weight)
Gram Atomic / Formula / Molecular Mass =
Molar Mass - because it is the mass of one Units (g/mol) mole of any substance.
Example: 1 mole of H2O = 18g
Molar Mass of Water = 18 g/mol

21. Mole Relationships or Equalities
1 mole = formula mass in grams
1 mole = 6.02 x 1023 particles
1 mole = 22.4L (for gases)
So…
Formula mass = 6.02x1023 particles
22.4L of a gas = 6.02x1023 particles
Formula mass = 22.4L of a gas

22. Conversion Problems Some problems involve mass-mole conversions
you can use the “Mole Calculation” formula on Table T.
You will usually have to calculate the gram formula mass of the compound…
# of Moles = given mass (g)
gram formula mass

23. Example 1
- What is the mass of 1.75 moles of oxygen gas (Hint: oxygen is diatomic)?
1st find the molecular mass of oxygen remember oxygen is a diatomic element (O2)
O: 2 x 16 g= 32 g
Then use formula on table T:
Moles = Given Mass (g)
Gram Formula Mass
1.75 = ?
32 g ✔
? = 56 g

24. Example 2 - How many moles are in 42 g of water? ✔
1st find the molecular mass of H2O:
H: 2 x 1 g = 2 g
O: 1 x 16 g = 16 g
= 18 g
Convert from grams to moles…
? Moles = 42 g
18 g ✔
? =2.3 mol

25. Example 3
- How many moles of potassium chromate are in a 500 g sample?
K2CrO4
1) Find the formula of Potassium Chromate:
2) Calculate formula Mass:
K:2 x 39 g = 78 g
Cr: 1 x 52 g = 52 g
O: 4 x 16 g = 64 g
= 194 g
3) Calculate the Moles using the formula on table T:
? Moles = 500 g
194 g
? = 2.6 mol

26. Empirical Formula Review
Empirical Formula is when the subscripts in a chemical formula are simplified to their lowest common multiples.
Ex) C4H8 vs. CH2
N2O4 vs. NO2
NaCl vs. ?
Molecular
Empirical
Molecular and Empirical

28. Empirical Mass
- Is the mass of the empirical formula of a compound
In a question you are given the molecular mass and empirical formula…
1st you must find the empirical mass
Then divide molecular mass by empirical mass
Multiply all subscripts in the empirical formula by your answer to get molecular formula

29. Empirical Mass Example
Molecular Mass = 180
Empirical Formula = CH2O
Step 1: CH2O Empirical Mass
C – 1 x 12 = 12
H – 2 x 1 = 2
O – 1 x 16 = 16
30 amu
Step 2: Divide (Molecular mass/Empirical mass)
180/30 = 6
Step 3: Multiply Empirical Formula by Step 2
6(CH2O) = C6H12O6

30. Moles in Chemical Equations
In problems involving chemical reactions we balance chemical reactions using coefficients
These coefficients also represent the mole ratios in a reaction
Ex)
2C2H6 + 7O2  4CO2 + 6H2O
Moles C2H6
Moles O2
Moles CO2
Moles H2O 2 7 4 6 1
3.5 3

31. Moles in Chemical Equations Pt. 2
After balancing we can determine the amount of a reactant or product involved in the reaction
As long as your given the amount of one ingredient…
Ex)
4Al + 3O2  2Al2O3
What is the Minimum # of moles of Oxygen gas needed to make 1 mole of aluminum oxide?

32. Moles in Chemical Equations Pt. 3
The mole ratio in a balanced reaction can also be used to calculate the mass of reaction ingredients
Ex) Based on the reaction below a student calculates 20g of H2 will react with 20g Cl2 to form 40g HCl, is this true?
H2 + Cl2  2HCl

33. Moles in Chemical Equations Pt. 4
We can use these mole ratios to calculate the volume of reaction ingredients also (only in gas reactions)
Ex)
Mole ratio = 2 : 7 : 4 : 6
2C2H6 (g)+ 7O2 (g) 4CO2 (g)+ 6H2O (g)
How many Liters of CO2 (g) will be produced from the combustion of 30 Liters of C2H6 (g) ?
30 L 2
= 15 L
Then,
15 L x 4 = 60 L

34. A pound of meth is worth 40,000 $
1 lb = ~454 g
Balance the reaction and figure out the amount of reactants needed to synthesize 454 g of meth…
_C9H10O + _CNH5  _C10H13N + _H2O
? g + ? g  g