empirical formula

Empirical Formulas and Molecular Formulas

CLE.3221.3.3 Explore the Mathematics of Chemical Equations

Empirical FormulasObjective

Define empirical formula

Determine the empirical formula when given percentage composition or mass composition

Determine the molecular formula from empirical formula

Definition

Consists of the symbols for the elements combined in compound

with subscripts showing the smallest whole number ratio of the atoms in compound

Ionic Compounds

ionic compounds

formula unit is usually the empirical formula

NaCl is in the smallest whole number ratio

CaCO3 is in the smallest whole number ratio

Molecular Compounds

empirical formula does not necessarily indicate the actual numbers of atoms present in the molecule

B2H6 can be reduced to BH3

C6H12O6 can be reduced to CH2O

Empirical Formula from Percent Composition

Quantitative analysis shows that a compound contains 32.38% Na, 22.65% S, and 44.99% O. Find the Empirical Formula (EF)

Empirical Formula from Percent Composition

1. assume 100 g sample of substance since percentages are given

32.38 % = 32.38g ; 22.65% = 22.65g; 44.99% = 44.99g

Empirical Formula from Percent Composition

2. Find the moles of each element in the fomula

Na = 32.38 g/ 23.0g = 1.408 mol

S = 22.65 g/ 32.1g = 0.7056 mol

O = 44.99 g/ 16.0g = 2.812 mol

Empirical Formula from Percent Composition

3. Develop small whole number mole ratio by dividing through by smallest number of mols

Na = 1.408 mol/0.7056 mol = 1.99 ~ 2

S = 0.7056 mol/0.7056 mol = 1

O = 2.812 mol/0.7056 mol = 3.98 ~ 4

Empirical Formula from Percent Composition

Mole ratio is 2:1:4

Empirical Formula is Na2SO4

Empirical Formula from Percent Composition

Find the EF of a compound that is 26.56% K, 35.41% chromium, and the remainder oxygen.

Empirical Formula from Percent Composition

K = 26.56g/39.1g = 0.6793 mol/0.6793mol =1

Cr = 35.41 g/52.0g = 0.6810 mol/0.6793mol=1

O = 100g - (26.56 g + 35.41g)/16.0g = 2.377mol/0.6793mol= 3.5

Empirical Formula from Percent Composition

K = 1 Cr = 1 O = 3.5

the number of oxygens is not close to a whole number, multiply by a number that will make 3.5 a whole number

K = 1 x2 = 2

Cr = 1 x 2 = 2

O = 3.5 x 2 = 7 EF = K2Cr2O7

Empirical Formula from Mass Composition

Analysis of a 10.150 g sample contains 4.433g phosphorous and 5.717g oxygen.

P = 4.433g/31.0g = 0.1430 mol/0.1430 mol = 1

O = 5.717g/16.0g = 0.3573mol/0.1430mol=2.5

P 1 x 2 = 2 O = 2.5 x 2 = 5 EF= P2O5

Molecular Formula

The actual formula of the compound

relationship between EF and molecular formula

x(empirical formula mass ) = (molecular formula mass)

x is the whole-number multiple indicating the factor by which subscripts in EF are multiplied to obtain molecular formula

Molecular Formula

In the previous problem EF was P2O5. The molar mass of the compound is 283.89g/mol. What is the molecular formula of the compound?

Molecular Formula

x(EF mass) = (Molar Mass)

EF mass is (2x 31.0g + 5 x 16.0g) = 142.0g

x(142.0g) = 283.89g/mol

x = 2

MF = (P2O5)2 = P4O10