1. Empirical and Molecular Formulas and Percent Composition
Unit 7: Stoichiometry – Part III
Mrs. Callender

2. Lesson Essential Questions…..
What is percent composition and
how do I calculate it?
What is the difference between an
Empirical and Molecular Formula?
How do I calculate Empirical
and Molecular Formulas?
Lesson Essential Questions…..

3. What does the term “Percent” mean?
Calculate the Percent Composition of AlAsO4?
1. Find the formula mass of the compound.
What does the term “Percent” mean?
26.98 g g + 4(16.00 g) = g
Percent means a “part of the whole”.
26.98
165.90
Al =
x 100 = 16.26%
74.92
165.90
As =
x 100 = 45.16%
64.00
165.90
x 100 = 38.58%
O = %

4. 1. Find the formula mass of the compound.
Calculate the Percent Composition of Glucose C12H22O11?
12(12.01) g + 22(1.01) g + 11(16.00 g) = g
1. Find the formula mass of the compound.
144.12
342.34
C =
x 100 = 42.10%
22.22
342.34
H =
x 100 = 6.49%
176.00
342.34
x 100 = 51.41%
O = %

5. F O R M U L A S There are two types of chemical formulas:
EMPIRICAL FORMULA
The lowest whole number ratio of elements in a compound.
MOLECULAR FORMULA
The true number of atoms of each element in a compound or formula.
Molecular formula = (Empirical Formula)n
Molecular Formula of Benzene - C6H6 or (CH)6
Empirical Formula of Benzene - CH

6. Practice…. 1. C2H4 CH2 4. C2H6O2 CH3O 2. C8H18 C4H9 5. X39Y13 X3Y
What are the empirical formulas for the following?
1. C2H4
CH2
4. C2H6O2
CH3O
2. C8H18
C4H9
5. X39Y13
X3Y
C14H18N2O5
3. WO2
WO2
6. C28H36N4O10
IONIC compounds should already be empirical formulas because of how the charge is balanced.
4. Na2SO4
Na2SO4

7. Law of Definite Proportions
law stating that every pure substance always contains the same elements combined in the same proportions by weight.
Ratios of subscripts will always be in small whole numbers.
Ex. CO2 always has a ratio of one carbon to two oxygen's.

8. Lets Start with an easy problem….
A compound with an empirical formula of C2OH4 and a molar mass of 88 grams per mole. What is the molecular formula of this compound?
Lets Start with an easy problem….
Step 2: Divide the molecular formula molar mass by
the empirical formula molar mass.
Step 1: Find the molar mass (formula mass) of the
empirical formula.
Step 3: Take that number and multiply it by the
empirical formula.
Step 4: Check molar mass of molecular formula.
2(12.01 g) g + 4(1.01 g) = g
4(12.01 g) + 2(16.00 g) + 8(1.01 g) = g
Is the empirical formula mass the same as the molecular formulas molar mass? 88
= 2
44.06
This mass matches closely to the molecular formula mass so you have successfully solved the problem. NO
2 X
C2OH4
C4O2H8

9. The local CSI found and unknown compound to contain 13. 5 g Ca, 10
The local CSI found and unknown compound to contain 13.5 g Ca, 10.8 g O and g of H. Calculate the empirical formula for this compound.
13.5 g Ca x 1
1 mole Ca
40.08 g Ca
= mole Ca
10.8 g O x 1
1 mole O
16.00 g O
= mole O
0.675 g H x 1
1 mole H
1.01 g H
= mole H
Step 3: Divide all moles by the smallest number of moles.
Step 4: How can we write this formula better?
Step 2: Write down the elements into the form of
a compound with the mole amount.
Step 1: Start with the grams of each element and convert to moles by using molar mass. Ca O H Ca (O H) 2 1 2 2
.337
.675
.668
.337
.337
.337

10. NutraSweet is 57. 14% C, 6. 16% H, 9. 52% N, and 27. 18% O
NutraSweet is 57.14% C, 6.16% H, 9.52% N, and 27.18% O. Calculate the empirical formula of NutraSweet and find the molecular formula. (The molar mass of NutraSweet is g/mol)
57.14 g C x 1
1 mole C
12.01 g C
= 4.76 mole C
14(12.01 g) + 18(1.01 g) +
2(14.01 g) + 5(16.00 g) = g
6.16 g H x 1
1 mole H
1.01 g H
= 6.10 mole H
Step 6: Since the molar mass of the molecular formula is the same as the molar mass of the empirical formula, they are one in the same and you have successfully answered the problem.
Step 5: Find the molar mass of your empirical formula. Compare this to the molar mass given in the problem.
9.52 g N x 1
1 mole N
14.01 g N
= mole N
27.18 g O x 1
1 mole O
16.00 g O
= 1.70 mole O
Step 4: The 2.5 can NOT be rounded so it must be multiplied by 2 to get rid of the decimal.
Step 1: Start by changing % to grams by thinking
you have 100 g of the sample. Change
the grams to moles.
Step 3: Divide all moles by the smallest number of moles.
Step 2: Write down the elements into the form of
a compound with the mole amount. C
C14H18N2O5 H N O
2 x 7 9 1
2.5 14
4.76 18
6.10
.680 2
1.70 5
.680
.680
.680
.680