DETERMINING CHEMICAL FORMULAS

Chapter 7.4

Objectives:1. Define empirical formula, and explain how

the term applies to ionic and molecular compounds

2. Determine an empirical formula from either a percentage or a mass composition

3. Explain the relationship between the empirical formula and the molecular formula of a given compound

4. Determine a molecular formula from an empirical formula

Empirical formula Defined as: consists of the symbols for

the elements combined in a compound, with subscripts showing the smallest whole-number mole ratio of the different atoms in the compound For ionic compounds – formula unit is usually

the compounds empirical formula For molecular compounds – empirical formula

does not necessarily indicate the actual numbers of atoms Example:

BH3 (diborane) empirical formula

B2H6 molecular formula

Calculation of empirical formulas Sample problem 1:

Quantitative analysis shows that a compound contains 32.38 % sodium, 22.65 % sulfur, and 44.99% oxygen. Find the empirical formula of this compound.1. Assume 100 g

samples32.38 g Na

22.65 g S

44.99 g O

2. Convert to moles

x 1 mol Na 22.99 g Nax 1 mol S 32.07 g S

x 1 mol O 16.00 g O

= 1.408 mol Na

= 0.7063 mol S

= 2.812 mol O

3. Divide by the smallest number÷

o.7063

÷ o.7063

÷ o.7063

= 1.993 mol Na

= 1 mol S

= 3.981 mol O

4. Round

2

4

1

5. Use as subscripts

Na2SO4

Sample Problem 2 Sample problem 2:

Analysis of a 10.150 g sample of a compound known to contain only phosphorus and oxygen indicates a phosphorus content of 4.433 g. What is the empirical formula of this compound.

4.433 g P

5.717 g O

1. Convert to moles

x 1 mol P 30.97 g P

x 1 mol O 16.00 g O

= 0.1431 mol P

= 0.3573 mol O

3. Divide by the smallest number = 1 mol P

= 2.497 mol O

4. Round

1

2.5

5. Use as subscripts

P2O5

÷ 0.1431

÷ 0.1431

5. Multiply by 2 2

5

Practice Problems1. A compound is found to contain 63.52 % iron

and 36.48 % sulfur. Find its empirical formula.

2. Find the empirical formula of a compound found to contain 26.56 % potassium, 35.41 % chromium, and the remainder oxygen.

3. Analysis of 20.0 g of a compound containing only calcium and bromine indicates that 4.00 g of calcium are present. What is the empirical formula of the compound formed?

Answer: FeS

Answer: K2Cr2O7

Answer: CaBr2

Calculation of Molecular formulas

Remember: empirical formula is smallest possible whole number ratio of atoms in a compound Molecular formula is the ACTUAL formula

Relationship can be written as: x(empirical formula mass) = molecular

formula massx = whole number multipleFactor which the subscripts are

multiplied to give empirical formula.

Sample Problem 1: Molecular formula In the last sample problem, the empirical formula of a

compound of phosphorus and oxygen was found to be P2O5. Experimentation shows that the molar mass of this compound is 283. 89 g/mol. What is the compound’s molecular formula?

x(empirical formula mass) = molecular formula massmolecular formula

massempirical formula mass

x =

1. Empirical formula mass

P 2 x 30.97 amu = 61.94 amu O 5 x 16.00 amu = 80.00 amu

= 141.94 amu

x =283.89 amu141.94 amu

= 2.0001

So molecular formula would be

2 x (P2O5)

= P4O10

Practice problems1. A sample of a compound with a formula

mass of 34.00 amu is found to consist of 0.44 g H and 6.92 g O. Find its molecular formula.

First: find the empirical formula

0.44 g H

6.92 g O

1. Convert to moles

x 1 mol H 1.01 g H

x 1 mol O 16.00 g O

= 0.436 mol H

= 0.433 mol O

3. Divide by the smallest number = 0.929mol

H

= 1 mol O

4. Round

1

1

5. Use as subscripts

HO

÷ 0.433

÷ 0..433

H 1 x 1.01 amu = 1.01 amu O 1 x 16.00 amu = 16.00 amu

= 17.01 amu

x =molecular formula massempirical formula massx =34.00 amu17.01 amu

= 22 x

(HO)= H2O2

Practice Problem 2 Determine the molecular formula of the

compound with an empirical formula of CH and a formula mass of 78.110 amu.x(empirical formula mass) = molecular

formula massmolecular formula massempirical formula mass

x =

1. Empirical formula mass

C 1 x 12.o1 amu = 12.01 amu H 1 x 1.01 amu = 1.01 amu

= 13.02 amu

x =78.110 amu 13.02 amu

= 5.999

So molecular formula would be

6 x (CH) = C6H6