How to Calculate the Empirical Formula

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The empirical formula in chemistry provides the relative numbers of each type of atom in a particular molecule. It does not provide the exact number of each type of atom in the molecule, nor does it provide any information on the arrangement of those atoms. Stoichiometry, a branch of analytical chemistry which studies the composition of reactants and products in chemical reactions, uses the empirical formula. Calculate the empirical formula of a compound from the amount of each element that is in a given sample of the compound.

TL;DR (Too Long; Didn't Read)

The empirical formula of a compound provides the proportions of each element in the compound but not the actual numbers or arrangement of atoms.

  1. Determine Mass of Each Element

  2. Determine the mass of each element in a compound. For this example, assume that you have 13.5 grams (g) of calcium (Ca), 10.8 g of oxygen (O) and 0.675 g of hydrogen (H).

  3. Find Atomic Weight of Each Element

  4. Determine the number of grams in a mole (mol) of each element. This is known as the atomic weight of the element and is available from a periodic table. In this example, the atomic weight of Ca is 40.1, the atomic weight of O is 16.0 and the atomic weight of H is 1.01.

  5. Calculate Number of Moles

  6. Calculate the number of moles of each element in the compound. For example, 13.5 g Ca ÷ (40.1 g/mol Ca) = 0.337 mol Ca, 10.8 g O ÷ (16.0 g/mol O) = 0.675 mol O and 0.675 g H ÷ (1.01 g/mol H) = 0.668 mol H.

  7. Find Ratio of Elements

  8. Determine the ratio of the elements in the compound. Divide the molar amount of each element by the smallest quantity. In this case, the smallest quantity is for calcium at 0.337 mol. By dividing each molar amount by 0.337 mol, we get 0.337 ÷ 0.337 = 1 for calcium, 0.675 ÷ 0.337 = 2 for oxygen and 0.668 ÷ 0.337 = 2 for hydrogen.

  9. Express Empirical Formula

  10. Express the empirical formula for the sample. From Step 4, we know there are two atoms of oxygen and two atoms of hydrogen for each atom of calcium. The empirical formula for the sample compound is therefore CaO2H2.

References

About the Author

Allan Robinson has written numerous articles for various health and fitness sites. Robinson also has 15 years of experience as a software engineer and has extensive accreditation in software engineering. He holds a bachelor's degree with majors in biology and mathematics.

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