Applications to Chemistry

 

 

 

 

 

 

Application 1 It takes three different ingredients A, B, and C, to produce a certain chemical substance. A, B, and C have to be dissolved in water separately before they interact to form the chemical. Suppose that the solution containing A at 1.5 g/cm3 combined with the solution containing A at 3.6 g/cm3 combined with the solution containing C at 5.3 g/cm3 makes 25.07 g of the chemical. If the proportion for A, B, C in these solutions are changed to 2.5, 4.3, and 2.4 g/cm3, respectively (while the volumes remain the same), then 22.36 g of the chemical is produced. Finally, if the proportions are 2.7, 5.5, and 3.2 g/cm3, respectively, then 28.14 g of the chemical is produced. What are the volumes (in cubic centimeters) of the solutions containing A, B, and C?

 

Solution Let x, y, z be the corresponding volumes (in cubic centimeters) of the solutions containing A, B, and C. Then 1.5x is the mass of A in the first case, 3.6y is the mass of B, and 5.3z is the mass of C. Added together, the three masses should give 25.07 g. So

 

 

 

      The same reasoning applies to the other two cases. This gives the linear system

 

 

     

 

       The augmented matrix of this system is

 

 

 

  

         Reducing the above matrix would give the solution

 

 

Application 2 Another typical application of linear systems to chemistry is balancing a chemical equation.  The rationale behind this is the Law of conservation of mass which states the following:

 

“mass is neither created nor destroyed in any chemical reaction. Therefore balancing of equations requires the same number of atoms on both sides of a chemical reaction. The mass of all the reactants (the substances going into a reaction) must equal the mass of the products (the substances produced by the reaction).”

 

As an example consider the following chemical equation

 

C2H  6 +  O2 → CO2 +  H2O.

 

Balancing this chemical reaction means finding values of x, y, z  and t so that the number of atoms of each element is the same on both sides of the equation:

 

 

xC2H  6 +  yO2 → zCO2 +  tH2O.

 

 This gives the following linear system:

 

 

The general solution of the above system is

 

 

Since we are looking for whole values of the variables x, y z, and t, choose x=2 and get y=7, z= 4 and t=6. The balanced equation is then:

 

2C2H  6 +  7O2 → 4CO2 +  6H2O.