STOICHIOMETRY

by clever academyJanuary 19, 20180 Comments

Stoichiometry is the study of the qualitative and quantitative relationship between substances undergoing chemical changes. It is the calculations of the quantities of reactants and products involved in a chemical change.

       CHEMICAL FORMULA AND                                                  EQUATIONS


A chemical formula indicates or specifies the number of atoms in a molecule. These chemical symbols and formulars can be combined to form a statement called chemical equation, which describes a chemical reactions. For example, combustion of carbon or ethene in oxygen can be represented in form of chemical equations. Balancing of chemical equations in stoichiometry is very paramount because it gives us an insight on the type of reaction and the state at which each of the species involve. To get a proper understanding on how to balance a chemical equation click here

       EMPIRICAL AND MOLECULAR                                           FORMULAR


An emperical formula is a chemical formula that gives only the relative number of each type in a molecule. It can be evaluated from the percentage composition of a substance or from the masses of each atom of a substance.
Molecular formulars are chemical formulars which indicate the actual composition of a molecule of the compound. It additionally provides structural information. It may be obtained from the emperical formulae if the molecular mass of the substance is known.

ELEMENTAL ANALYSIS: determination of percentage composition of carbon, hydrogen and oxygen


For a newly discovered compound whose formula is to be determined, the following steps are employed in obtaining it's percentage composition  As an example let us consider the determination of the percentages of carbon, hydrogen and oxygen in a compound containing only these three element.

  1. Heat a small, accurately weighed sample of the unknown compound to 700 degree Celsius in a steam of pure and dry oxygen, and in the presence of pure copper (ii) oxide, to ensure complete combustion of the vapours. The hydrogen and carbon are oxidizer to steam and carbon (iv) oxide respectively. The mass of steam present is determined by passing it through a previously weighed tube of calcium (ii) chloride,  and carbon  (iv) oxide by passing it through a previously weighed wash - bottle of concentrated potassium hydroxide.
  2. Relate the masses of carbon (iv) oxide and steam to the masses of carbon and hydrogen. Once the masses of carbon and hydrogen in the sample are known, the mass percentage of carbon and hydrogen can then be calculated.
  3. The mass percentage of oxygen can be obtained by subtracting the mass percentage of carbon and hydrogen from 100.

                  WORKED EXAMPLES


Question 1:

14.28grams of hydrogen react with 85.72 grams of carbon. What is the emperical formular of the compound? [C=12.0; H= 1.0 g/mol].


Solution: In this example, the actual masses of the reacting elements are given.
                                                C                                H
Mass of each atom           85.72                         14.28

Ratio of the atom               85.72/12                 14.28/1
(Divided by atomic mass)    7.14                       14.28

Divide by the smallest to get whole number:                                                             7.14/7.14                  14.28/7.14
                                                                                                                                                        1                :               2

Therefore,emperical formula is CH2


Question 2:

Caculeate the emperical formula of a compound containing 40.2% K, 26.9% Cr and the remaining being oxygen [K= 39.0, Cr= 52.0; O= 16.0 g/mol].


Solution: In this example,  the percentage composition of the compound are given:
                                                                                                                                       K                     Cr                     O
% Composition                                                                                                           40.2               26.9                 32.9

Ratio of the atom  
(Divided by atomic mass)                                                                                  40.2/39             26.9/52          32.9/16

                                                                                                                                 1.03                     0.52                  2.06

 Divide by the smallest to get whole number               Errand boy, stitchAstyle.com                                                                                1.03/0.52        0.52/0.52           2.06/0.52
                                                                                                                                 2             :            1            :           4
Therefore, emperical formula is K2CrO4.


                   EXERCISE QUESTIONS

  1. In a laboratory, 1.55 grams of an organic compound containing carbon, hydrogen and oxygen is combusted for analysis. Combustion resulted in 1.45 grams of carbon (iv) oxide and 0.890 grams of steam.What is the emperical formula of the organic compound? 

  2. 110 grams of hydrated calcium tetraoxosulphate (vi) (CaSO4. XH2O) weigh 7.9 grams after heating and cooling. Calculate the emperical formula of the compound  [Ca=40.0, S=32.0, O= 16.0, H= 1.0, g/mol]. 

  3. 0.07 grams of an hydride of carbon occupies 56.0 cm3 at STP when vapourised and contains 14.29% by mass of hydrogen. What is the formula of the hydrocarbon.

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