1. Empirical and Molecular Formulas
6.2
Empirical and Molecular Formulas

2. Molecular/True Formula
Types of Formulas
The formulas for compounds can be expressed as an empirical formula and as a molecular (true) formula.
Empirical Formula
Molecular/True Formula
Name CH
C2H2
acetylene
C6H6
benzene
CO2
Carbon dioxide
CH2O
C5H10O5
ribose

3. An empirical formula represents the simplest whole number ratio of the atoms in a compound.
The molecular formula is the true or actual ratio of the atoms in a compound.

4. Learning Check 1) CH2O 2) C2H4O2 3) C3H6O3
A. What is the empirical formula for C4H8?
1) C2H4 2) CH ) CH
B. What is the empirical formula for C8H14?
1) C4H7 2) C6H12 3) C8H14
C. What is a molecular formula for CH2O?
1) CH2O 2) C2H4O ) C3H6O3

5. Calculating Empirical Formulas
Basic Chemistry Copyright © 2011 Pearson Education, Inc.

6. Learning Check
A compound contains 7.31 g Ni and 20.0 g Br. Calculate its empirical (simplest) formula.
Basic Chemistry Copyright © 2011 Pearson Education, Inc.

7. Solution STEP 1 Determine the moles of each element.
7.31 g Ni x 1 mol Ni = mol of Ni
58.69 g Ni
20.0 g Br x 1 mol Br = mol of Br
79.90 g Br
STEP 2 Divide by the smallest number of moles.
0.125 mol Ni = 1 mol of Ni
0.125
0.250 mol Br = 2 mol of Br
Basic Chemistry Copyright © 2011 Pearson Education, Inc.

8. Solution (continued)
STEP 3 Use the lowest whole-number ratio of moles as subscripts.
NiBr2
Basic Chemistry Copyright © 2011 Pearson Education, Inc.

9. Converting Decimals to Whole Numbers
When the number of moles for an element is a
decimal
greater than 0.1, but less than 0.9
multiply the moles by a small integer to obtain whole numbers
Basic Chemistry Copyright © 2011 Pearson Education, Inc.

10. Percent Composition Using 100 g
Basic Chemistry Copyright © 2011 Pearson Education, Inc.

11. Finding the Empirical Formula
“Percent to mass
Mass to mole
Divide by small
Multiply ‘til whole”

12. Finding the Empirical Formula
A compound is Cl 71.65%, C %, and H 4.07%. What are the empirical and molecular formulas? The molar mass is known to be
99.0 g/mol.
1. “Percent to Mass” - state mass percentages as grams in a g sample of the compound.
Cl g C g H g

13. 2. “Mass to Moles”
71.65 g Cl x 1 mol Cl = mol Cl
35.5 g Cl
24.27 g C x 1 mol C = mol C
12.0 g C
4.07 g H x mol H = mol H
1.01 g H

14. 4. Write the simplest or empirical formula CH2Cl
“Divide by Small”
Find the smallest whole number ratio by dividing each mole value by the smallest mole values:
Cl: mol = 1 Cl
2.02 mol
C: mol = 1 C
H: mol = H
4. Write the simplest or empirical formula
CH2Cl
(“Multiply ‘til whole”)

15. Learning Check
Aspirin is 60.0% C, 4.5 % H, and 35.5 % O. Calculate its empirical (simplest) formula.
“Percent to mass
Mass to mole
Divide by small
Multiply ‘til whole”
Basic Chemistry Copyright © 2011 Pearson Education, Inc.

16. “Percent to Mass & Mass to Mole”
STEP 1 Calculate the moles of each element.
100.0 g aspirin contains 60.0% C or 60.0 g of C, 4.5% H or 4.5 g of H, and 35.5% O or 35.5 g of O.
60.0 g C x 1 mol C = mol of C
12.01 g C
4.5 g H x 1 mol H = mol of H
1.008 g H
35.5 g O x 1mol O = mol of O
16.00 g O
Basic Chemistry Copyright © 2011 Pearson Education, Inc.

17. “Divide by Small” STEP 2 Divide by the smallest number of moles.
5.00 mol C = mol of C
2.22
4.5 mol H = 2.0 mol of H
2.22 mol O = mol of O
Basic Chemistry Copyright © 2011 Pearson Education, Inc.

18. “Multiply ‘til Whole”
STEP 3 Use the lowest whole-number ratio of moles as subscripts.
To obtain whole numbers of moles, multiply by a factor, in this case x 4.
C: mol of C x 4 = 9 mol of C
H: 2.0 mol of H x 4 = 8 mol of H
O: mol of O x 4 = 4 mol of O
Using these whole numbers as subscripts, the simplest formula is
C9H8O4
Basic Chemistry Copyright © 2011 Pearson Education, Inc.

19. Finding the Molecular Formula
Multiplier: molar mass = a whole number
empirical mass
To get Molecular Formula, first calculate Empirical Formula, then multiply all subscripts by the multiplier
Note: If your multiplier = 1, the molecular formula = empirical formula
eg. if multiplier = 2, the MF = 2 x EF

20. Back to the problem….
We calculated Empirical Formula = CH2Cl & were given the molar mass = 99.0 g/mol
5. Calculate EM (empirical mass)
= 1(C) + 2(H) + 1(Cl) = g/mol
6. Calculate Multiplier:
Molar mass = g/mol = multiplier = 2
Empirical mass g/mol
7. Molecular formula = Empirical Formula x multiplier
(CH2Cl) x 2 = C2H4Cl2

21. Learning Check
A compound is 27.4% S, 12.0% N and 60.6 % Cl. If the compound has a molar mass of 351 g/mol, what is the molecular formula?

22. Solution S = 0.853 mol & divide by 0.853 = 1 S
N = mol & divide by = 1 N
Cl = 1.71 mol & divide by = 2 Cl
Empirical Formula (EF) = SNCl2
Empirical Mass (EM) = g/mol
Given Molar Mass = 351 g/mol
Multiplier = g/mol = 3,  so MF = S3N3Cl g/mol