Sunday 31 January 2021

Counting Atoms (Determine the Empirical Formula) slide presentation ppt



KKCS International Program

Chemistry IGCSE 1

Mr. Markus

3 kinds of e chemical formulae:

  • The molecular formula of a compound shows the actual number and kinds of atoms present.

ex: the molecular formula of ethane is C2H6

  • The empirical formula of a compound shows the simplest ratio of the atom present.

ex: the empirical formula of ethane is CH3

  • The structural formula of a compound shows how the atoms are joined in the molecules.

Ex: the structural formula of ethane

H     H

|      |

    H     C     C      H

|      |

H      H


Suppose a compound is analyzed to contain 48.8 g of cadmium, 20.8 g of carbon, 2.62 g of hygrogen, and 27.8 g of oxygen. Determine the empirical formula of this compound.

To determine the empirical formula of a compound for which the amounts of each element are

given you convert the amounts to moles using the elements atomic weights from the periodic table

as follows:

This gives you the mole ratio of the elements in the formula and theoretically the formula for the

compound could be written as: Cd0.434C1.67H2.62O1.74

But we know that atoms are not fractional in nature so we must find the smallest whole number ratio for the elements making up the compound. To do this we divide each of the numbers by the smallest number in the set. In our case the smallest number is 0.434 moles. *

These whole numbers become the subscripts for the empirical formula: CdC4H6O4


  • Suppose a compound whose molecular mass is 695 is analyzed to contain 26.7% phosphorus, 12.1% nitrogen, and 61.2% chlorine.

Therefore the Empirical formula is PNCl2

To determine the molecular formula you set the EF mass taken X times and set it equal to the

Molecular Weight of the compound. Like this:

(PNCl2) X = 695

Now substitute the atomic weight of the elements into the equation to get this:

[30.97 +14 +2(35.45)] X = 695

[115.87] X = 695

X = 695/115.87

X = 6

To get the Molecular Formula we must multiply each of the subscripts in the Empirical Formual by

6 to give us this:


As a check, you can find the formula mass as usual from the sum of the atomic weights:

P = 6 x 30.97 = 185.82

N = 6 x 14 = 84

Cl = 12 x 35.45 = 425.4

Total = 695.22

You can see from this example that the MF is a multiple of the EF and represents the actual number of each type of atom that makes up the compound.

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