1. Chapter 9
Empirical Formulas

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

3. Empirical Formulas
Write your own one-sentence definition for each of the following:
Empirical formula
Molecular formula

4. 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.

5. Learning Check EF-1 1) CH2O 2) C2H4O2 3) C3H6O3
A. What is the empirical formula for C4H8?
1) C2H ) CH ) CH
B. What is the empirical formula for C8H14?
1) C4H7 2) C6H ) C8H14
C. What is a molecular formula for CH2O?
1) CH2O 2) C2H4O ) C3H6O3

6. Solution EF-1 1) CH2O 2) C2H4O2 3) C3H6O3
A. What is the empirical formula for C4H8?
2) CH2
B. What is the empirical formula for C8H14?
1) C4H7
C. What is a molecular formula for CH2O?
1) CH2O 2) C2H4O ) C3H6O3

7. Learning Check EF-2
If the molecular formula has 4 atoms of N, what is the molecular formula if SN is the empirical formula? Explain.
1) SN
2) SN4
3) S4N4

8. Solution EF-2
If the molecular formula has 4 atoms of N, what is the molecular formula if SN is the empirical formula? Explain.
3) S4N4
If the actual formula has 4 atoms of N, and S is related 1:1, then there must also be 4 atoms of S.

9. Empirical and Molecular Formulas
molar mass = a whole number = n
simplest mass
n = 1 molar mass = empirical mass
molecular formula = empirical formula
n = 2 molar mass = 2 x empirical mass
molecular formula =
2 x empirical formula
molecular formula = or > empirical formula

10. Empirical Formula
Molecular Formula
Empirical Mass
Molecular
Mass

11. Learning Check EF-3 2) C6H8O6 3) C9H12O9
A compound has a formula mass of and an empirical formula of C3H4O3. What is the molecular formula?
1) C3H4O3
2) C6H8O6
3) C9H12O9

12. Solution EF-3
A compound has a formula mass of and an empirical formula of C3H4O3. What is the molecular formula?
2) C6H8O6
C3H4O3 = g/EF
176.0 g =
88.0

13. Learning Check EF-4 1) C7H6O4 2) C14H12O8 3) C21H18O12
If there are g of O in the molecular formula, what is the true formula if the EF is C7H6O4?
1) C7H6O4
2) C14H12O8
3) C21H18O12

14. Solution EF-4 3) C21H18O12 192 g O = 3 x O4 or 3 x C7H6O4
If there are g of O in the molecular formula, what is the true formula if the EF is C7H6O4?
3) C21H18O12
192 g O = 3 x O4 or 3 x C7H6O4
64.0 g O in EF

15. Finding the Molecular 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. State mass percents as grams in a g sample of the compound.
Cl g C g H g

16. 2. Calculate the number of moles of each element.
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

17. Why moles?
Why do you need the number of moles of each element in the compound?

18. 3. Find the smallest whole number ratio by dividing each mole value by the smallest mole values:
Cl: = 1 Cl
2.02
C: = 1 C
H: = H
4. Write the simplest or empirical formula
CH2Cl

19. 5. EM (empirical mass)
= 1(C) + 2(H) + 1(Cl) =
6. n = molar mass/empirical mass
Molar mass = g/mol = n = 2
E M g/EM
7.Molecular formula
(CH2Cl)2 = C2H4Cl2

20. Learning Check EF-5
Aspirin is 60.0% C, 4.5 % H and 35.5 O. Calculate its simplest formula. In 100 g of aspirin, there are g C, 4.5 g H, and g O.

21. Solution EF-5 60.0 g C x ___________= ______ mol C
4.5 g H x ___________ = _______mol H
35.5 g O x ___________ = _______mol O

22. Solution EF-5 60.0 g C x 1 mol C = 5.00 mol C 12.0 g C
4.5 g H x mol H = mol H
1.01 g H
35.5 g O x 1mol O = mol O
16.0 g O

23. 5.00 mol C = ________________ ______ mol O
Divide by the smallest # of moles.
5.00 mol C = ________________
______ mol O
4.5 mol H = ________________
2.22 mol O = ________________
Are are the results whole numbers?_____

24. Are are the results whole numbers?_____
Divide by the smallest # of moles.
5.00 mol C = ___2.25__
2.22 mol O
4.5 mol H = ___2.00__
2.22 mol O = ___1.00__
Are are the results whole numbers?_____

25. Finding Subscripts
A fraction between 0.1 and 0.9 must not be rounded. Multiply all results by an integer to give whole numbers for subscripts.
(1/2) x 2 = 1
(1/3) x 3 = 1
(1/4) x 4 = 1
(3/4) x 4 = 3

26. C9H8O4 Multiply everything x 4 C: 2.25 mol C x 4 = 9 mol C
H: 2.0 mol H x 4 = 8 mol H
O: mol O x 4 = 4 mol O
Use the whole numbers of mols as the subscripts in the simplest formula
C9H8O4

27. Learning Check EF-6
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?

28. Solution EF 6 0.853 mol S /0.853 = 1 S 0.857 mol N /0.853 = 1 N
1.71 mol Cl / = 2 Cl
Empirical formula = SNCl2 = g/EF
Mol. Mass/ Empirical mass /117.1 = 3
Molecular formula = S3N3Cl6

29. IMA = dE dR
Effort Distance
Resistance
AMA = FR FE
IMA = dE dR