Basic “chemical arithmetic”

A balanced chemical equation expresses the relative number of moles of each component (product or reactant), but because each formula in the equation implies a definite mass of the substance (its molar mass), the equation also implies that certain weight relations exist between the components. For example, the equation describing the combustion of carbon monoxide to carbon dioxide

2 CO + O₂ → 2 CO₂

The term stoichiometry can refer to any type of calculation that relates the quantities of reactants and products in a chemical reaction.  Stoichiometry is essentially an expression of the principle that atoms are conserved in chemical change.

The term comes from the Greek words στοιχεῖον stoicheion “element” and μέτρον metron “measure”

implies the following relations:

Figure 4.20 Mass in chemical reaction

The relative masses shown in the bottom line establish the stoichiometry of the reaction, that is, the relations between the masses of the various components. Since these masses vary in direct proportion to one another, we can define what amounts to a conversion factor (sometimes referred to as a chemical factor) that relates the mass of any one component to that of any other component.

The stoichiometry of chemical reactions is based on the mole ratios among the reactants and the products.

A general outline summarizes different stoichiometry calculations is given in the figure below:

Figure 4.21 Concept map for Reaction Stoichiometry

Reference: https://pressbooks-dev.oer.hawaii.edu/chemistry/chapter/reaction-stoichiometry/

The video below illustrates the concept of stoichiometry:

Stoichiometry

Example:

2 KClO₃   →     2 KCl    +    3 O₂

The stoichiometry expresses the mole ratios among the reactants and products:

2 moles of KClO₃ / 2moles of KCl and

2 moles of KClO₃ / 3 moles O₂

We can translate this into: 2 moles of KClO₃ will produce 2 moles KCl as well as 3 moles O₂.

Examples:

Number of moles produced from certain amount of moles of one of reactants:

2 KClO₃  …..>2 KCl  +  3 O₂

1. How many moles of O2 will be formed from 1.65 moles of KClO3?

                                                                  = 2.48 mol O₂

2. How many moles of KClO3 are needed to make 3.50 moles of KCl?

                                                                     = 3.50 mol KClO₃

3. How many moles of KCl will be formed from 2.73 moles of KClO3?

= 2.73   mol KCl

Number of grams (mass) produced or reacted from or by certain amount of grams (mass) of one of reactants or products:

4. Consider the reaction below:

ZnSO4    +    SrCl2     …..>      SrSO4    +   ZnCl2

1. Calculate the mass of SrCl2 is consumed when 55.0 g of ZnCl2 is produced?
2. Calculate the mass of SrSO4 is consumed when 88.5 g of ZnSO4 is produced?

1. Mass of SrCl2   = [ 55.0 g of ZnCl2 ] x [mole ZnCl2 / 136.3 g ZnCl2 ] x [1 mol SrCl2 / 1mol ZnCl2] x [158.5 g SrCl2 / mol SrCl2 ] =  63.958 g SrCl2 = 64.0 g SrCl2  (3 sig. figs)

ZnCl2  molar mass = 136.3 g / mol

SrCl2 molar mass =   158.5 g / mol

2. mass of SrSO4 = [ 88.5 g of ZnSO4] x [ mol ZnSO4 / 161.5 g  ZnSO4 ] x [ 1 mol SrSO4 / 1 mol ZnSO4 ] x [ 183.7 g SrSO4 / mol SrSO4 ] = 100.66 g SrSO4 = 0.101 x 10³ g SrSO4

ZnSO4 molar mass = 161.5 g / mol

SrSO4  molar mass = 183.7 g / mol

Consider the reaction below:

5. N₂    +    3 H₂     ….>    2 NH₃    +    Heat

1. How many molecules of NH₃ produced if one has used 0.750 moles of N₂ is used.
2. How many molecules of NH₃ produced if one has used 1.80 moles of H₂ is used.

1. Molecules of NH₃ = [0.750 mole N₂]  x [ 2 mole of  NH₃  / 1 mole  N₂ ] x  [6.022 x 10²³ molecules NH₃ / mole  N₂]   =   9.033 x 10²³ molecules   NH₃

2.  Molecules of NH₃ = [1.80 mole H₂]  x [ 2 mole of  NH₃  / 3 mole  H₂ ] x  [6.022 x 10²³ molecules NH₃ / mole  H₂]   =   7.23 x 10²³ molecules   NH₃

6. Practice problem:

What mass of gallium oxide, Ga₂O₃, can be prepared from 29.0 g of gallium metal? The equation for the reaction is 4Ga+3O2⟶2Ga2O3.4Ga+3O2⟶2Ga2O3.

Ans: 39.0 g

More Practice Problems

Examples:

Number of molecules produced from certain amount of moles of one of reactants:

Consider the reaction below:

N₂    +    3 H₂     ßà    2 NH₃    +    Heat

1. How many molecules of NH₃ produced if one has used 0.750 moles of N₂ is used.
2. How many molecules of NH₃ produced if one has used 1.80 moles of H₂ is used.

1. Molecules of NH₃ = [0.750 mole N₂]  x [ 2 mole of  NH₃  / 1 mole  N₂ ] x  [6.022 x 10²³ molecules NH₃ / mole  N₂]   =   9.033 x 10²³ molecules   NH₃

2.  Molecules of NH₃ = [1.80 mole H₂]  x [ 2 mole of  NH₃  / 3 mole  H₂ ] x  [6.022 x 10²³ molecules NH₃ / mole  H₂]   =   7.23 x 10²³ molecules   NH₃

Examples:

Number of grams (mass) produced or reacted from or by certain amount of grams (mass) of one of reactants or products:

Consider the reaction below:

ZnSO4    +    SrCl2     à      SrSO4    +   ZnCl2

1. Calculate the mass of SrCl2 is consumed when 55.0 g of ZnCl2 is produced?
2. Calculate the mass of SrSO4 is consumed when 88.5 g of ZnSO4 is produced?

1. Mass of SrCl2   = [ 55.0 g of ZnCl2 ] x [mole ZnCl2 / 136.3 g ZnCl2 ] x [1 mol SrCl2 / 1mol ZnCl2] x [158.5 g SrCl2 / mol SrCl2 ] =  63.958 g SrCl2 = 64.0 g SrCl2  (3 sig. figs)

ZnCl2  molar mass = 136.3 g / mol

SrCl2 molar mass =   158.5 g / mol

2. mass of SrSO4 = [ 88.5 g of ZnSO4] x [ mol ZnSO4 / 161.5 g  ZnSO4 ] x [ 1 mol SrSO4 / 1 mol ZnSO4 ] x [ 183.7 g SrSO4 / mol SrSO4 ] = 100.66 g SrSO4 = 0.101 x 10³ g SrSO4

ZnSO4 molar mass = 161.5 g / mol

SrSO4  molar mass = 183.7 g / mol