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ChemistryEdu Logo Mole Concept | Empirical and Molecular Formula#

Mass Percentage or Percentage by Mass#

  • Mass percentage represents the relative quantities of each element in a compound.
  • Mass percentage of an element in the compound can be calculated by:
\[Mass\ Percentage = {Mass\ of\ element \times 100 \over Mass\ of\ compound}\]

Example: Let us calculate mass percentage of CaCO3

Please note that any amount of CaCO3, be it 1 molecule or 100 g or 1 mole or 1000 moles, will contain elements having same mass percentages.

Let us take 1 mole of CaCO3.

\[Mass\ of\ 1\ mole\ CaCO_3 = 100\ g\]

Since, 1 mole CaCO3 contains 1 mole Ca, 1 mole C and 3 moles O:

\[Mass\ of\ 1\ mole\ Ca = 40\ g\]
\[Mass\ of\ 1\ mole\ C = 12\ g\]
\[Mass\ of\ 3\ moles\ O = 48\ g\]
\[Mass\ percentage\ of\ Ca = {40 \times 100 \over 100} = 40\]
\[Mass\ percentage\ of\ C = {12 \times 100 \over 100} = 12\]
\[Mass\ percentage\ of\ O = {48 \times 100 \over 100} = 48\]

Molecular and Empirical Formula#

  • Molecular formula gives the number of each atom in a single molecule of a compound whereas empirical formula gives the simplest ratio of each atom in a single molecule of the compound.
  • Relation between molecular formula and empirical formula can be described as:
\[Molecular\ formula = (Empirical\ formula)_n\]

Example

Let us consider a compound whose molecular formula is C4H10O2. Then, its empirical formula will be C2H5O.

  • Mass percentage of each element in the compound will be same in both empirical and molecular formula.

Question#

An organic compound having mass percentage of C = 53.33%, mass percentage of H = 11.11%, mass percentage of O = 35.56%. Molecular weight of the organic compound is 90 g/mole. Find its empirical formula and molecular formula.

Let 100 g of that organic compound is taken.

\[Moles\ of\ C = {53.33 \over 12}\]
\[Moles\ of\ H = {11.11 \over 1}\]
\[Moles\ of\ O = {35.56 \over 16}\]
\[Simplest\ ratio\ of\ moles\ of\ C,\ H,\ and\ O = {53.33 \over 12} : {11.11 \over 1} : {35.56 \over 16}\]
\[∴ Simplest\ ratio = 2:5:1\]

Hence, the empirical formula is C2H5O.

Let the molecular mass of compound be (C2H5O)n.

\[Molecular\ mass = 12 \times 2n + 1 \times 5n + 16 \times n\]
\[90 = 45n\]
\[∴n = 2\]
\[Hence,\ Molecular\ formula = (C_2H_5O)_n = (C_2H_5O)_2 = C_4H_{10}O_2\]