2. Friday, April 15th: “A” Day Monday, April 18th: “B” Day Agenda
Homework questions/problems?
Quiz over section 7.2
Begin 7.3: “Formulas & Percentage Composition”
In-Class Assignments:
Practice pg. 243: #1-4
Practice pg. 245: #1-3

3. Quiz 7.2: “Relative Atomic Mass and Chemical Formulas”
You can use your book and your guided notes for this walk-talk quiz…
Remember: answer only the question that corresponds to the month you were born in.
If you were born in November, answer question #2. If you were born in December, answer question # 7.
Once everyone has answered their question, get up, walk around, and talk/compare answers with others.

4. 7.3: “Formulas and Percentage Composition”
The percentage composition is the percentage by mass of each element in a compound.
Percentage composition helps verify
a substance’s identity.
Percentage composition can also be used to compare the ratio of masses contributed by the elements in two different substances.

5. Percent Composition of Iron Oxides

6. Empirical Formula
An actual formula shows the actual ratio of elements or ions in a single unit of a compound.
Empirical formula: a chemical formula that shows the simplest ratio for the relative numbers and kinds of atoms in a compound.
For example, the empirical formula for hydrogen peroxide is HO, while the actual formula is H2O2

7. Determining Empirical Formulas
You can use the percentage composition for a compound to determine its empirical formula.
Convert the percentage of each element in the compound to grams.
Convert from grams to moles using the molar mass of each element as a conversion factor.
Compare these amounts in moles to find the simplest whole-number ratio among the elements.

8. Determining Empirical Formulas
To find the simplest whole-number ratio, divide each amount in moles by the smallest of all the amounts of moles.
This will give a subscript of 1 for the atoms present in the smallest amount.
Finally, you may need to multiply all of the amounts of moles by a number to convert all subscripts to small, whole numbers.
The final numbers you get are the subscripts in the empirical formula.

9. Determining an Empirical Formula form Percentage Composition (Sample Problem G, pg. 242)
Chemical analysis of a liquid shows that it is 60.0% C, 13.4% H, and 26.6% O by mass. Calculate the empirical formula of this substance.
1. Assume that you have a 100 g sample so that each percentage is the same as the amount in grams:
for C: % = 60.0 g C
for H: 13.4% = 13.4 g H
for O: 26.6% = 26.6 g O

10. Sample Problem G, continued…
2. Use the molar mass to convert each amount in grams to amount in moles:

11. Sample Problem G, continued…
Divide each number of moles found by the smallest number of moles found. (1.66 moles O)
Carbon: mol = mol C
1.66 mol
Hydrogen: mol= mol H
Oxygen: mol = 1 mol O
These numbers are within experimental error to be considered whole numbers so the empirical formula is: C3H8O

12. Additional Practice
Find the empirical formula given the following composition:
26.58% K, % Cr, and % O
Assume 100 g sample:
26.58 g K
35.35 g Cr
38.07 g O

13. Additional Practice Use molar mass to convert from grams to moles.
26.58 g K X 1 mole K = mol K
39.10 g K
35.35 g Cr X 1 mole Cr = mol Cr
52.00 g Cr
38.07 g O X 1 mole O = mol O
16.00 g O

14. Additional Practice
Divide each number of moles found by the smallest number of moles found (.6798 mol).
.6798 mol K = 1 mol K
.6798 mol
.6798 mol Cr = 1 mol Cr
2.379 mol O = 3.5 mol O

15. Additional Practice K2Cr2O7
Since these are not whole numbers, multiply each one by 2 to get whole numbers.
1 mol K (2) = mol K
1 mol Cr (2) = mol Cr
3.5 mol O (2) = 7 mol O
These ARE whole numbers, so the empirical formula is:
K2Cr2O7

16. Molecular Formulas are Multiples of Empirical Formulas
The formula for an ionic compound shows the simplest whole-number ratio of the large numbers of ions in a crystal of the compound.
A molecular formula is a whole-number multiple of the empirical formula.
The molar mass of any compound is equal to the molar mass of the empirical formula times a whole number, n.
n (empirical formula) = molecular formula

17. Comparing Empirical and Molecular Formulas
Compound
Empirical Formula
Molecular Formula
Formaldehyde
CH2O
Acetic Acid
C2H4O2
2X the empirical formula
n = 2
Glucose
C6H12O6
6X the empirical formula
n = 6

18. Determining a Molecular Formula from an Empirical Formula (Sample Problem H, pg. 245)
The empirical formula for a compound is P2O5. Its experimental molar mass is 284 g/mol. Determine the molecular formula of the compound.
1. Use the periodic table to find the molar mass of the empirical formula:
For P: 2(30.97) = g/mol
For O: 5(16.00) = g/mol
Molar mass of P2O5 = g/mol

19. Sample Problem H, continued…
Find the multiplier, n:
n = experimental molar mass of compound
molar mass of empirical formula
n = 284 g/mol = Hint: the bigger #
g/mol always goes on top!
3. To find the molecular formula, simply multiply the empirical formula by 2:
2 (P2O5) = P4O10

20. Additional Practice Determine the molecular formula for the following:
Molar mass: g/mol
Empirical formula: OCNCl
Find molar mass of empirical formula:
O = g/mol
C = g/mol
N = g/mol
Cl = 35.5 g/mol = g/mol

21. Additional Practice 2. Find the multiplier, n:
n = experimental molar mass of compound
molar mass of empirical formula
n = g/mol = 3
77.52 g/mol
3. To find the molecular formula, simply multiply the empirical formula by 3:
3 (OCNCl) = O3C3N3Cl3

22. In-Class Assignments You Must SHOW WORK!
Practice pg. 243: #1-4
Practice pg. 245: #1-3
We will finish this section next time…