1. Formula & Equation Stoichiometry

2. What is stoichiometry?
Deals with the specifics of QUANTITY in chemical formula or chemical reaction.

3. Review: Counting Atoms
Remember subscripts tell you how many atoms…
… Coefficients tell you how many TOTAL molecules compounds

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

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

7. Counting Atoms Practice Ditto cont.
Example #3:
Ca3(PO4)2
* First list the types of atoms and then count each.
Ca: P: O: 3 2 8
**Do the rest on your own!**

8. Atomic Mass
Mass of an atom of one element based on percent abundances and masses of all isotopes (on PT)
Units: amu

9. Molecular Mass
Mass of a molecular compound (ex: H2O, CO2)
Units: amu

10. Formula Mass
Mass of an ionic compound (ex: NaCl, MgI2)
Units: amu

11. Gram Formula Mass
Mass expressed in grams

12. Gram Molecular Mass = H2O O + H + H + +
Molecular Mass = amu amu amu
Gram Molecular Mass = g

13. How do I calculate molecular or formula mass?
First: Identify and count atoms in the compound.
Second: Locate atomic mass of each element.
Third: Multiply mass x number of atoms and total all elements.
Voila!

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

15. Formula Mass Example (ionic compound)
Example: NaCl
Na- 1 x 23
Cl- 1 x 35.5
Formula Mass = 58.5 g

16. Formula Mass Practice

17. Percent Composition by Mass
Experimentally-
Use masses given in problem
Part x100
Whole
Mass of Element x 100
Mass of Compound

18. Example:
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?

19. Percent Composition by Atomic Mass
Theoretically-
Use Atomic Masses
Formula
Part x 100
Whole
(#atoms of element) x (atomic mass) x100
Formula mass of compound

20. Hydrated Crystals
Some ionic compounds are found to have surrounding water molecules.
These compounds are called Hydrated Crystals.
The percent water hydration can be found.
Example
CuSO4 . 5H2O
The dot shows that 5 H2O molecules are attached to 1 CuSO4 molecule.

21. Percent Water Hydration
% Hydration = Total mass of Water x 100
Formula Mass
CuSO4 . 5H2O
Cu – 1 x 63.5 = 63.5
S – 1 x 32 = 32
O – 4 x 16 = 64
H2O – 5 x 18 = 90
Formula Mass = 249.5
% Hydration = (90/249.5) x 100
= 36% H2O

22. % Hydration Example Problems
MgCl2 . 6H2O
CaCl2 . 2H2O
Ba(OH)2 . 8H2O
CoSO4 . 7H2O
NH4Al(SO4)2 . 12H2O

23. The Mole
Avagadro’s Number is the number of atoms (for monatomic elements) or molecules (for diatomic elements & compounds) in 1 mole of any substance.
Avagadro’s Number is 6.02 x 1023
That means that in 1 mole of any sample of matter there are 6.02x1023 particles. (atoms or molecules depending on the substance)

24. The Mole
There are 6.02 x 1023 particles in one mole of any substance (element or compound).
One particle can be one atom or one molecule.
That’s a BIG number!!!!!
Even though for any substance the number of particles in 1 mole is always the same different substances have different masses so the mass of 1 mole will be different for all substances.
Atomic / Formula / Molecular Mass = Molar Mass because it is the mass of 1 mole of any substance.
Example: 1 mole of H2O = 18g

25. Mole Relationships or Equalities
1 mole = gram formula mass (molecule)
1 mole = atomic mass (element)
1 mole = 6.02 x 1023 particles
1 mole = 22.4L (for gases)
Formula mass = 6.02x1023 particles
22.4L of a gas = 6.02x1023 particles
Formula mass = 22.4L of a gas

26. Table T : Mole Calculations
For questions that involve mass (gram) to mole conversions use the “Mole Calculation” formula on Table T.
In order to use this formula you will need to plug in what is given and solve for the unknown value.
If solving for moles then plug in given mass and gram formula mass (GFM)
If solving for a mass then plug in moles and GFM and solve for given mass
You will usually have to calculate the gram formula mass of the substance unless it is given.
See Table T

27. Examples
Example #1: What is the mass of 1.75 moles of oxygen gas (Hint: oxygen is diatomic)?
Convert from moles to grams using the mole equation on Table T by solving for given mass
# of moles = given mass
gram-formula mass x 32
1st find the gram formula mass (GFM)of oxygen remember oxygen is a diatomic element (O2).
O: 2 x 16 g= 32 g
1.75 =
X= 56 g

28. Examples Example #2: How many moles are in 42 g of water?
1st find the GFM of H2O.
H: 2 x 1 = 2
O: 1 x 16 = 16
= _18 g_
1 mol
Use mole calculation equation solve for # of moles
# of moles = given mass
gram-formula mass 42 18
X =
2.3 mol

29. Examples
Example #3: How many moles of potassium chromate are in a 500 g sample?
1st find the molecular mass of potassium dichromate (K2CrO4).
K:2 x 39 = 78
Cr: 1 x 52 = 52
O: 4 x 16 = 64
= _194 g_
1 mol
Use mole calculation equation solve for moles
500
194
X =
2.6 mol

30. Example 4
How much does 2.0 moles of carbon dioxide weigh?

31. Example 5
What is the mass of 0.50 moles of sulfur dioxide at STP?

32. Example 6 How many moles of helium gas are in 20.0 grams of helium?
Try the Practice problems

33. Empirical & Molecular Formulas
Empirical Formula: formula that contains the lowest whole number ratio of atoms.
Molecular Formula: the actual formula for a compound (does not have to be empirical but it can)
The subscripts in any formula represents the number of moles of atoms in a 1 mole sample.
The subscripts in an empirical formula can tell you the mole ratio of atoms in a compound.
Example: In 1 mole of H2O there are 2 moles of Hydrogen atoms and 1 mole of oxygen atoms.

34. Calculating the Molecular Formula Given Formula Mass and Empirical Formula
Since the empirical formula is the lowest mole ratio of atoms in a compound the molecular formula mass must be a multiple of the empirical formula mass.

35. Step 1: Find the empirical formula mass.
Example 1: What is the molecular formula for a compound that has an empirical formula of CH2O and the molecular mass is 180g.
Step 1: Find the empirical formula mass.
Step 2: Divide molecular mass given by the empirical mass you calculated in step 1.
Step 3: Multiply all of the subscripts in the empirical formula by the multiple you calculated in step 2

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

37. Example 2: A mass of a given compound is 9.2g. It is determined that 2.8g is nitrogen and 6.4g is oxygen. What is the empirical formula of this compound?
Step 1: Convert the given mass of each element to moles by dividing by the atomic mass.
Step 2: Plug moles values from step 1 into the formula and divide both by the lowest to get a whole number ratio. If ratio is still not whole number then multiply the whole formula by 2.

38. N: 2.8/14 = .2mol. O: 6.4/16 = .4mol. N.2/.2 O.4/.2 Answer = NO2
Example 2: A mass of a given compound is 9.2g. It is determined that 2.8g is nitrogen and 6.4g is oxygen. What is the empirical formula of this compound?
N: 2.8/14 = .2mol.
O: 6.4/16 = .4mol.
N.2/.2 O.4/.2
Answer = NO2

39. Example 3: A compound is found to be 25. 9% nitrogen and 74. 1% oxygen
Example 3: A compound is found to be 25.9% nitrogen and 74.1% oxygen. What is the emeprical formula of this compound?
Step 1: Treat the percentage values as mass values (assume sample size is 100g) and convert to moles.
Step 2: Plug mole values into the equation and divide both by the lowest to get whole number ratios. If ratio is still not whole number then multiply the entire formula by 2.

40. Example 3: A compound is found to be 25. 9% nitrogen and 74. 1% oxygen
Example 3: A compound is found to be 25.9% nitrogen and 74.1% oxygen. What is the emeprical formula of this compound?
N: 25.9/14 = 1.85mol.
O: 74.1/16 = 4.63mol.
N1.85/1.85O4.63/1.85 = 2(NO2.5)
Answer = N2O5

43. Mole Relationships in Balanced Equations
Balanced chemical equations tell us the number and type of substances involved in a chemical reaction.
The coefficients indicate the amounts of reactants that go into a reaction and the amount of product formed.
In chemistry we also use the coefficients in a balanced equation to identify the mole ratios of reactants and products.

44. Mole Ratios in a Balanced Equation
2H2 + O2 → 2H2O
In the balanced equation for the synthesis of water it shows that for every 2 hydrogen molecules you need 1 oxygen molecule and 2 water molecules are produced.
It can also be said that for every 2 moles of hydrogen and 1 mole of oxygen 2 moles of water are produced.
The coefficient can be read as molecules or moles, in lab we use moles because it is a measureable amount.

45. Solving Equation Stoichiometry Problems
Remember that balanced equations are mole ratios of reactants and products.
Ex 1: Synthesis of water
2H2 + O2 → 2H2O
Means 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water.
The Ratio of reactants and products are 2:1:2.
We can use the molar ratios from a balanced equation to determine the amounts of reactants used or products produced with varying amounts.

46. Examples Problem 2H2 + O2 → 2H2O Ex 2: How many moles of water are
Examples Problem H2 + O2 → 2H2O Ex 2: How many moles of water are produced if 2 moles of oxygen reacted? Set up a proportion Solve for unknown

47. How many moles of carbon dioxide will be produced by the complete combustion of 2.0 moles of glucose (C6H12O6)
First: write the equation and balance it. C6H12O6 + 6O2 ⟶ 6CO2 + 6H2O Next: set up a proportion Finally: solve for the number of moles

48. How many moles of ammonia can be produced from 9
How many moles of ammonia can be produced from 9.0 moles of hydrogen reacting with nitrogen?
Try the practice!!!