Symbols, Formula and Equations
Empirical Formula

The empirical formula of a compound is the simplest possible ratio the atoms of the compound can be expressed. It doesn't tell the exact amount of atoms present in the compound.

Molecular formula

The molecular formula of a compound is the actual ratio of moles of atoms in the compound. It is the actual formula of the compound.

The relationship between Empirical formula and Molecular formula is: $$ (E.F)_n = \text{Molar mass} $$ $$ (E.F)_n = M.F $$ $$ n = \text{number of moles} $$ $$ \text{also, R.M.M} = 2 × V.D $$ $$ \text{where, V.D = vapor density} $$ $$\text{R.M.M = Relative molecular mass} $$

Examples

Example 1: An organic compound with relative molecular mass 136 contains 70.75% carbon, 5.09% hydrogen and 23.55% oxygen. Determine its

  1. Empirical formula
  2. Molecular formula
    3. (H = 1, C = 12, O = 16)

Solution

Symbols C H O
Write the mass % 70.75 5.09 23.55
Divide by atomic mass to get mole ratio $$ \frac{70.75}{12} $$ $$= 5.90 $$ $$ \frac{5.09}{1} $$ $$ = 5.09 $$ $$ \frac{23.55}{16} $$ $$ = 1.47 $$
Divide by the smallest mole $$ \frac{5.90}{1.47} $$ $$ = 4 $$ $$ \frac{5.09}{1.47} $$ $$ = 3.5 $$ Express this in fraction because of the decimal part $$ \frac{7}{2} $$ $$ \frac{1.47}{1.47} $$ $$ = 1 $$
Since the mole ratio can't be easily rounded up to a whole number, multiply through by the the den. $$ 4 × 2 $$ $$= 8 $$ $$ 2 × \frac{7}{2} $$ $$ = 7 $$ $$ 2 × 1 $$ $$ = 2 $$

The Empirical formula is then expressed using the final ratio $$ E.F = C_8H_7O_2 $$

  1. $$ (E.F)_n = \text{Molar mass} $$
    2. $$(C_8H_7O_2)_n = 136 $$ $$ (12 × 8) + (7) + (16 × 2)n = 136 $$ $$ 135n = 136 $$ $$ n = \frac{136}{135} $$ $$ n = 1 $$ $$ M.F = {E.F}_n $$ $$ M.F = (C_8H_7O_2)_1 $$ $$ M.F = C_8H_7O_2 $$ The molecular formula = \( C_8H_7O_2 \)

Example 2: A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen. Determine its molecular formula if it's molar mass is 180 [H = 1, C = 12, O = 16]

Solution

Symbols C H O
% of the elements 40 6.7 53.3
divide by the atomic mass to get moles $$ \frac{40}{12}$$ $$ = 3.33 $$ $$ \frac{6.7}{1}$$ $$ = 6.7 $$ $$ \frac{53.3}{16} $$ $$ = 3.33 $$
Divide each by Smallest mole $$ \frac{3.33}{3.33} $$ $$ = 1 $$ $$ \frac{6.7}{3.33} $$ $$ = 2 $$ $$ \frac{3.33}{3.33} $$ $$ = 1 $$

The empirical formula = \( CH_2O \)
To derive the molecular formula; $$ (E.f)_n = \text{Molar mass} $$ $$ (CH_2O)n = 180 $$ $$ [(12 + (1 × 2) + 16)n = 180 $$ $$ 36n = 180 $$ $$ n = \frac{180}{36} $$ $$ n = 6 $$ $$ M.F = (E.F)_n $$ The molecular formula = \( C_6H_{12}O_6 \)


Example 3: A hydrocarbon Z with molecular mass 78 on combustion gave 3.385g of CO2 and 0.692g of H2O. Determine the molecular formula of Z. [H = 1, C = 12, O = 16]

Solution

The combustion of an hydrocarbon produces CO2 and H2O $$ Z + O_2 ——> CO_2 + H_2O $$ To express the molecular formula we need to find the masses of the elements that make up Z which is hydrogen and carbon (Hydrocarbon) $$ \text{For carbon, C} $$ $$ C + O_2 ———> CO_2 $$ Relating masses $$ 12g = 44g $$ $$ x = 3.385g $$ $$\text{mass of carbon} = \frac{12 × 3.385}{44} $$ $$ \text{mass of carbon} = 0.923g $$ Doing same for hydrogen $$ \text{mass of hydrogen} = \frac{2 × 0.692}{18} $$ $$ \text{mass of Hydrogen} = 0.077g $$ Now derive the empirical formula
Symbols C H
reacting mass 0.923 0.077
divide by atomic mass $$ \frac{0.923}{12} $$ $$ = 0.077 $$ $$ \frac{0.077}{0.077} $$ $$ = 0.077 $$
Divide by smallest mole $$ \frac{0.077}{0.077} $$ $$ = 1 $$ $$ \frac{0.077}{0.077} $$ $$ = 1 $$

The Empirical formula = \( CH \)
To get the molecular formula $$ (CH)n = \text{molar mass} $$ $$ (12 + 1)n = 78 $$ $$ 13n = 78 $$ $$ n = \frac{78}{13} $$ $$ M.F = (CH)_6 $$ The molecular formula = \( C_6H_6 \)


Example 4: Determine the molecular formula of an alkane whose vapor density is 15 (H = 1, C = 12)

Solution

An alkane is a hydrocarbon which has a general molecular \( C_nH_{2n+2} \)
Let the empirical formula of the hydrocarbon be CH $$ (CH)_n = 2 × VD $$ $$ 13n = 30 $$ $$ n = 2 $$ Substituting the value for n in the general formula
The molecular formula = \( C_2H_6 \)


Example 5: What is the empirical formula of a hydrocarbon containing 0.08 moles of carbon and 0.32 moles of hydrogen?

Solution

The empirical can be expressed by dividing by the smallest mole
Symbols C H
Reacting moles 0.08 0.32
Divide by smallest mole $$ \frac{0.08}{0.08} $$ $$ \frac{0.32}{0.08} $$
= 1 = 4

Empirical formula = \( {CH}_4 \)

Summary