(4.6/4.7) Empirical and Molecular Formulas

SCH 3U

• 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.

Types of Formulas

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

eg. empirical formula = CH2O

Learning Check

A. What is the empirical formula for C4H8?

1) C2H4 2) CH2 3) 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) C2H4O2 3) C3H6O3

Finding the Empirical Formulaa) A compound is 71.65% Cl, 24.27% C, and 4.07% H. What is the empirical formula?

1. Assume a 100 g sample. Determine the mass, in grams, of each element present.

Cl 71.65 g C 24.27 g H 4.07 g

2. Calculate the number of moles of each element.

nCl = 71.65 g = 2.021 mol

35.45 g/mol

nC = 24.27 g = 2.021 mol

12.01 g/mol

nH = 4.07 g = 4.03 mol

1.01 g/mol

3. Divide each by the smallest number of moles to obtain the simplest whole number ratio.

**If whole numbers are not obtained, multiply subscripts by the smallest number that will give whole numbers**

Cl: 2.021 mol = 1.000 Cl (1) 2.021 mol

C: 2.021 mol = 1.000 C (1) 2.021 mol

H: 4.04 mol = 2.00 H (2) 2.021 mol4. Write the simplest or empirical formula.

CH2Cl

Learning Check

Aspirin is 60.0% C, 4.5 % H and 35.5% O. Calculate the empirical formula.

nC = 60.0 g = 5.00 mol

12.01 g/mol

nH = 4.5 g = 4.5 mol

1.01 g/mol

nO = 35.5 g = 2.22 mol

16.00 g/mol

5.00 mol C = 2.25 mol

2.22

4.5 mol H = 2.0 mol

2.22

2.22 mol O = 1.00 mol

2.22

Therefore, the Empirical Formula (EF) = C9H8O4

X 4 = 9 mol C

X 4 = 8 mol H

X 4 = 4 mol O

a) A compound is Cl 71.65%, C 24.27%, and H 4.07%. What is the empirical formula? (from yesterday, CH2Cl)

b) The molar mass is known to be 99.0 g/mol. What is the molecular formula?

1. Calculate EFM (empirical formula mass)

1(12.01g/mol) + 2(1.01 g/mol) + 1(35.45g/mol) = 49.48 g/mol

2. Calculate Multiplier: Molar mass (M) = 99.0 g/mol = 2.00 EFM 49.48 g/mol

3. Multiply the empirical formula subscripts by the multiplier

(CH2Cl) x 2 = C2H4Cl2

Finding the molecular formula

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?

nS = 27.4 g = 0.855 mol

32.06 g/mol

nN = 12.0 g = 0.857 mol

14.01 g/mol

nCl = 60.6 g = 1.71 mol

35.45 g/mol

Solution

0.855 mol S = 1.00 mol

0.855

0.857 mol N = 1.00 mol

0.855

1.71 mol Cl = 2.00 mol

0.855

Therefore, the Empirical Formula (EF) = SNCl2

Solution

Solution

empirical formula mass (EFM)= 32.06 g/mol + 14.01 g/mol + (2)(35.45 g/mol)= 116.97 g/molMolar Mass = 351 g/mol

Multiplier = 351 g/mol = 3.00 116.97 g/mol

so MF is S3N3Cl6