The empirical formula of a compound gives the simplest ratio of the number of atoms of different elements present in one molecule of a compound. it will not give the actual number of atomsof different elements present in one molecule of a compound.
example 1 : Empirical formula of ethene is CH2. the ratio of atoms of carbon and hydrogen in ethene is 1:2, which is the simplest ratio
Molecular formula of ethene is C2H4. this formula gives the actualnumber of carbon and hydrogen atoms present in one molecule of a compound.
example 2 : Empirical formula of acetic acid : CH2O
Molecular formula of acetic acid C2H4O2
"Empirical formula of a compound is the simplest formula showing the relative number of atoms of different elements present in one molecule of the compound"
Calculation of the empirical formula.
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Detect elements present in the compound.
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Find out experimentally, the percentage composition by weight of each element present in the compound.
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Divide the percentage of each element by its atomic weight to get the relative number of atoms of each element.
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Divide each number obtained for the respective elements in step (3) by the smallest number among those numbers so as to get the simplest ratio.
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If any number obtained in step (4) is not a whole number then multiply all the numbers by a suitable integer to get whole number ratio. This ratio will be the simplest ratio of the atoms of different elements present in the compound. Empirical formula of the compounds can be written with the help of this ratio.
Model Problem 1: Chemical analysis of a carbon compound gave the following percentage composition by weight of the element present.
carbon= 10.06%, hydrogen=0.84%, chlorine= 89.10%. Calcuate the empirical formula of the compound.
Step 1 : Percentage of the elements present
Carbon Hydrogen Chlorine
10.06 0.84 89.10
Step 2 : Dividing the percentage compositions by the respective atomic weight of elements
10.06 0.84 89.10
12 1 35.5
0.84 0.84 2.51
Step 3 : Dividing each value in step 2 by the smallest number among them to get simple atomic ratio
0.84 0.84 2.51
0.84 0.84 0.84
Step 4 : Ratio of the atoms present in the moecule C : H : Cl
1 : 1 : 3
Therefore The empirical formula of the compound C1H1Cl3 or CHCl
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