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The mass percentage of each constituent elements present in any compound is called its Percentage composition
$$Mass\; percentage\; of element=\frac{Total\; mass\; of\; particular\; element}{Molecular\; mass\; of\; compound}\times\;100\% $$
For example % composition of potassium nitrate (KNO_{3}) can be calculated as,
1 mole of KNO_{3} contains 1 mole of K, 1-mole of N and 3-mole of O .
The molecular mass of KNO_{3} = 39 + 14 + 98 = 101 amu
Now,
% of K =\( \frac {39}{101}\) x 100% = 38.6%
% of N = \(\frac {14}{101}\) x 100% = 13.9%
And % of O = \(\frac{98}{101}\) x 100% = 47.5%
Therefore, the % composition of KNO_{3} is K = 38.6%, N = 13.9% and O = 47.5%
The empirical formula of a compound is defined as the simplest formula that shows the simplest whole number ratio of the different atoms present in one molecule of the compound.
Benzene has a molecular formula C_{6}H_{6}. The simplest whole no. the ratio of C and H is 1:1 so empirical formula of C_{6}H_{6} is CH.
Molecular formula | Empirical formula | Common integer(n) |
H_{2}O_{2} | HO | 2 |
C_{6}H_{12}O_{6} | CH_{2}O | 6 |
C_{6}H_{6} | CH | 6 |
C_{2}H_{4} | CH_{2} | 2 |
C_{3}H_{6} | CH_{2} | 3 |
H_{2}O | H_{2}O | 1 |
Na_{2}S_{2}O_{3} | Na_{2}S_{2}O_{3} | 1 |
It is defined as the symbolic representation of a molecule that shows the actual number of atoms present in that molecule of the compound.
The molecular formula may be the same or an integral multiple of its empirical formula. Therefore the relation between empirical and molecular formula is given as.
Molecular Formula = ( empirical formula ) x n
Where n = 1, 2, 3, 4 ......... (Interger)
And the value of 'n' can be obtained by the following relation,
n = \(\frac{Molecular formula mass(mole. wt.)}{Empirical formula mass(emp. wt.)}\)
This process involves the following rules:
II. For molecular formula:
This process involves the following rules :
Solved Example 1
3.0-gram sample of copper chloride on analysis gave 1.42 gm copper and 1.5 gm chlorine. What is the Empirical formula of copper chloride?
Solution :
i. calculation of % of elements
wt. of copper chloride = 3.0 gm
wt. of copper = 1.42 gm
wt. of chlorine = 1.58 gm
Now, % of Cu = \(\frac{wt.\ of \ copper}{wt.\ of\ copper\ chloride}\) x 100% = \(\frac{1.42}{3.0}\) x 100% = 47.33%
And % of Cl = \(\frac{wt. of chlorine}{wt. of copper cholride}\) x 100% = \(\frac{1.58}{3.0}\) x 100% = 52.66%
ii. Claculation of empirical formula :
Element | % | At. Wt. | Relative mole of atom = \(\frac{\%}{at.\ wt.}\) | Simplest atomic ratio | Simplest whole no. rato of atom |
Cu | 47.33 | 63.5 | \(\frac{47.33}{63.5}\) = 0745 | \(\frac{0.722}{0.722}\) = 1 | 1 |
Cl | 52.66 | 35.5 | \(\frac{52.66}{35.5}\) = 1.483 | \(\frac{1.483}{0745}\) = 1.99 | 2 |
Therefore, empirical formula of the compound is CuCl_{2}.
Solved question 2
An inorganic compound contains 28.16 % potassium and 25.63% chlorine. Determine its empirical and molecular formula, if its molecular mass is 138.5 amu.
Solution:(i) Calculation of % of elements
% of K = 28.16 %
% of Cl = 25.63 %
And % of O = 100 - (28.16 + 25.63) = 43.21 %
(ii) Calculation of empirical formula :
Element | % | At. Wt. | Relative mole of atom = \(\frac{\%}{at.\ wt.}\) | Simplest atomic Ratio | Simplest whole no. ratio of atom |
K | 28.16 | 39 | \(\frac{28.16}{39}\) = 0.722 | \(\frac{0.722}{0.722}\) = 1 | 1 |
Cl | 25.63 | 35.5 | \(\frac{25.63}{35.5}\) = 0.722 | \(\frac{0.722}{0.722}\) = 1 | 1 |
O | 46.21 | 16.0 | \(\frac{46.21}{16}\) = 2.89 | \(\frac{2.89}{0.722}\) = 4 | 4 |
So, Empirical formula = KCLO_{4}
(iii) Calculation of molecular formula
Now, empirical formula (KCLO_{4}) = 39 + 35.5 + 64 = 138.5 amu
Given, molecular mass = 138.5 amu
We have, n = \(\frac{molecular\ mass}{empirical\ formula\ mass} = \frac{138.5}{138.5}\) = 1
Hence, molecular formula = (empirical formula ) x n
= (KCLO_{4}) x 1
= KCLO_{4}
Therefore, molecular formula of compound = KCLO_{4}
The reactant in accordance with the stoichiometry indicated by the balanced equation. Generally, one of the reaction mixtures is present in the lesser amount than other reactants as required by the balanced chemical equation. So the amount of product formed during the reaction depends on the mass of reactant which is consumed completely.
Limiting reagents is defined as the chemical reagent present in the reaction mixture which is in deficit and completely consumed at first during the chemical reaction.
Limiting reagent controls or limits the formation of the product.
Limiting reagent is essential in stoichiometric calculation because of the following reasons:
The general method of calculations for all the problems of reactions consists of the following steps:
To tackle the problems on chemical equations, we should be able to relate the given quantities in terms of,
References :-
Ghosh, A.K. Chemical Calculations. 15th Edition. India: Scientific Book Company, 1991.
M.L Sharma & P.N. Chaudhary. Advanced Level Chemistry. 2nd Edition . Vol. I. Kathmandu, Nepal: Ekta Books, 2011.
Palak, K.R. Fundamentals of Chemistry. Kathmandu, Nepal: Ratna Pustak Bhandar, 2000.
Pathak, Prof. Dr Tulsi Prasad. Rectified Chemistry. Kalikasthan, Kathmandu : Airawati Prakashan (P.) Ltd., 2014.
ASK ANY QUESTION ON Empirical, Molecular Formula And Limiting Reagents
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