Empirical and Molecular Formulas

Empirical vs Molecular Formula

• The Molecular Formula (MF) gives the actual number of each type of atom present.

• The Empirical Formula (EF) gives the lowest whole-number ratio of the atoms present.

• Example: C2H6 and C3H9

– they have the same EF CH3 yet have very different MF

Timberlake LecturePLUS 3

Types of Formulas

The formulas for compounds can be expressed as an empirical formula and as a molecular(true) formula.

Empirical Molecular (true) Name CH C2H2 acetyleneCH C6H6 benzeneCO2 CO2 carbon dioxideCH2O C5H10O5 ribose

Timberlake LecturePLUS 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.

EF vs. MF

When you have an ionic compound: the EF = MF

For some molecular compounds:the EF = MF, but not always

Determining the EF

Remember: The EF is the lowest whole-number ratio of the moles of each atom present.

Example: CH4 has 1 mol C atoms

4 mol H atoms

Determining the EF

If a compound consists of:62.1% C 13.8% H 24.1% N

The percentages are based on MASS not MOLES

We can compare moles, not masses

Determining the EF

In order to determine the ratio of C:H:N, we need to know the mole ratio

Step 1: Convert % of each into gramsMake it easy on yourself, assume a sample size of 100.00g

62.1g C 13.8g H 24.1g N

Determining the EFNow that you know how many grams of each atom

you have:Step 2: Convert grams to moles using the molar

mass of each62.1g C

13.8g H

24.1 g N

Determining the EF

Now that you have the mol ratio, you need to make them Whole-Numbers.

Step 3: Divide each mol by the smallest mol value from step #2

5.17 mol C13.7 mol H1.72 mol N

Determining the EF

This results in 3 mol C, 8 mol H and 1 mol N

therefore the EF = C3H8N

Timberlake LecturePLUS 12

Learning Check EF-1

A. What is the empirical formula for C4H8?

1) C2H4 2) CH2 3) CHB. What is the empirical formula for C8H14?

1) C4H7 2) C6H12 3) C8H14

C. What is a molecular formula for CH2O?

1) CH2O 2) C2H4O2 3) C3H6O3

Timberlake LecturePLUS 13

Solution EF-1

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) C2H4O2 3) C3H6O3

Timberlake LecturePLUS 14

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) SN2) SN4 3) S4N4

Timberlake LecturePLUS 15

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.

Timberlake LecturePLUS 16

Empirical and Molecular Formulas

molar mass = a whole number = nsimplest 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

EF to MFTo go from the EF to the MF, you need two additional

pieces of information:1 – calculate the mass from your EF2 – You must be given the mass of the MF

(X) EFmass = MFmass

X = MFmass / EFmass

EF to MF

Example: You found the EF to be HO, and the MFmass = 34.02g

1. Calculate the EFmass = 17.01g

2. Calculate X = 34.02g / 17.01g3. X = 24. EF = HO MF = H2O2

Timberlake LecturePLUS 19

Learning Check EF-3

A compound has a formula mass of 176.0 and an empirical formula of C3H4O3. What is the molecular formula?

1) C3H4O3

2) C6H8O6

3) C9H12O9

Timberlake LecturePLUS 20

Solution EF-3

A compound has a formula mass of 176.0 and an empirical formula of C3H4O3. What is the molecular formula?2) C6H8O6

C3H4O3 = 88.0 g/EF

176.0 g = 2.00 88.0

Timberlake LecturePLUS 21

Learning Check EF-4

If there are 192.0 g of O in the molecular formula, what is the true formula if the EF is C7H6O4?

1) C7H6O4

2) C14H12O8

3) C21H18O12

Timberlake LecturePLUS 22

Solution EF-4

If there are 192.0 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

Timberlake LecturePLUS 23

Finding the Molecular Formula

A compound is Cl 71.65%, C 24.27%, 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 100.00-g sample of the compound.

Cl 71.65 g C 24.27 g H 4.07 g

Timberlake LecturePLUS 24

2. Calculate the number of moles of each element. 71.65 g Cl x 1 mol Cl = 2.02 mol Cl

35.5 g Cl

24.27 g C x 1 mol C = 2.02 mol C 12.0 g C

4.07 g H x 1 mol H = 4.04 mol H 1.01 g H

Timberlake LecturePLUS 25

Why moles?

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

Timberlake LecturePLUS 26

3. Find the smallest whole number ratio by dividing each mole value by the smallest mole values:Cl: 2.02 = 1 Cl

2.02

C: 2.02 = 1 C 2.02

H: 4.04 = 2 H 2.024. Write the simplest or empirical formula

CH2Cl

Timberlake LecturePLUS 27

5. EM (empirical mass)= 1(C) + 2(H) + 1(Cl) = 49.5

6. n = molar mass/empirical mass

Molar mass = 99.0 g/mol = n = 2 E M 49.5 g/EM

7.Molecular formula(CH2Cl)2 = C2H4Cl2

Timberlake LecturePLUS 28

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 60.0 g C, 4.5 g H, and 35.5 g O.

Timberlake LecturePLUS 29

Solution EF-5

60.0 g C x ___________= ______ mol C

4.5 g H x ___________ = _______mol H

35.5 g O x ___________ = _______mol O

Timberlake LecturePLUS 30

Solution EF-5

60.0 g C x 1 mol C = 5.00 mol C 12.0 g C

4.5 g H x 1 mol H = 4.5 mol H 1.01 g H

35.5 g O x 1mol O = 2.22 mol O 16.0 g O

Timberlake LecturePLUS 31

Divide by the smallest # of moles.5.00 mol C = ________________

______ mol O

4.5 mol H = ______________________ mol O

2.22 mol O = ______________________ mol OAre are the results whole numbers?_____

Timberlake LecturePLUS 32

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

2.22 mol O = ___1.00__2.22 mol OAre are the results whole numbers?_____

Timberlake LecturePLUS 33

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) 0.5 x 2 = 1(1/3) 0.333 x 3 = 1 (1/4) 0.25 x 4 = 1(3/4) 0.75 x 4 = 3

Timberlake LecturePLUS 34

Multiply everything x 4

C: 2.25 mol C x 4 = 9 mol CH: 2.0 mol H x 4 = 8 mol HO: 1.00 mol O x 4 = 4 mol O Use the whole numbers of mols as the subscripts in the simplest formula

C9H8O4

Timberlake LecturePLUS 35

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?

Timberlake LecturePLUS 36

Solution EF 6

0.853 mol S /0.853 = 1 S0.857 mol N /0.853 = 1 N1.71 mol Cl /0.853 = 2 ClEmpirical formula = SNCl2 = 117.1 g/EF

Mol. Mass/ Empirical mass 351/117.1 = 3

Molecular formula = S3N3Cl6