Application of chemical reactions in chemical analysis

Chemical reactions can be applied for quantitative and qualitative chemical analyses. Reactions produced solids can be used for gravimetric analysis, meaning determination of chemical quantities by measuring the masses of solids. Titrations can be applied for volumetric analysis, quantities determined by measured volumes.

Chemical reactions

Examples

Example 1

How many liters of hydrogen is produced at 300.0 K and 100.0 kPa when 1.00 g of sodium (Na) reacts with water?

solution

Na + H2O = Na(OH) + ½ H2(g)

1.00 g Na 1 mol Na
--------
23 g Na
½ mol H2
----------
1 mol Na
8.3145 kPa L mol-1 K-1 300.0 K
-----------------
100.0 kPa

The last factor is realy R T / P used to convert n moles to volume (V = n R T / P).
= 0.542 L

Example 2

How many liters of hydrogen is produced at 300.0 K and 100.0 kPa when 1.00 g of calcium (Ca) reacts with water?

solution

Ca + H2O = Ca(OH) + H2(g)

1.00 g Ca 1 mol Ca
--------
40.1 g Ca
1 mol H2
----------
1 mol Ca
8.3145 kPa L mol-1 K-1 300.0 K
-----------------
100.0 kPa

= 0.622 L

Example 3

When 2.00 g mixture of Na and Ca reat with water, 1.164 L hydrogen was produced at 300.0 K and 100.0 kPa. What is the percentage of Na in the sample?

solution

2 Na + 2 H2O = 2 Na(OH) + H2(g)
Ca + H2O = Ca(OH) + H2(g)

Let x be the mass of Na, then (2.00-x) is the mass of Ca.

We have the following relationship

x g
-----
23.0 g/mol
1 mol H2
----
2 mol Na
  + (2.0 - x) g Ca
------
40.1 g Ca/mol1 mol
1 mol H2
----------
1 mol Ca
  = 1.164 L H2 * 100.0 kPa
------------------
8.3145 kPa L mol-1 K-1 300.0 K

Simplify to give

x
----
46.0
  + 2
----
40.1
- x
----
40.1
  = 0.0467 all in mol

Multiply all terms by (40.1 * 46.0)

40.1 x + 2 * 46.0 - 46.0 x = 86.1

Simplify

-5.9 x = 86.1 - 92.0 = -5.91

Thus, Mass of Na = x = 1.0 g
Mass of Ca = 2.0 - x = 1.0 g
Mass Percentage of Na = 100* (1 / 2.0) = 50%

Discussion

Mole of Na = 1/23 = 0.0435 mol

Mole percentage = (1/23) / (1/23 + 1/40.1) = 0.635 = 63.5%

Compare this example with gravimetric analyses using the reaction
Ag+(aq) + Cl-(aq) = AgCl (s)
where Cl-(aq) comes from the disolution of two salts such as NaCl and MgCl2.

Also compare with analyses making use of the reaction
Ba2+(aq) + SO42-(aq) = BaSO4 (s)
where the anion SO42-(aq) comes from the disolution of two sulfate salts.

Theoretical Yield and Percent (actual) Yield

When two reactants react, the one that will be use up is called the limiting reactant or limiting reagent. Theoretical and percent yields are usually calculated according to the limiting reagent.

Example 4

Equal volume of CO and O2 at STP are mixed and ignited. Which one is the limiting reactant?

solution

CO(g) + ½ O2(g) = CO2(g)

This indicates that 1.0 L of CO will react with ½ L O2. Assume you have 1.0 L each, the after the reaction, ½ L of O2 will remain. Thus, CO is the limiting reagent.

Example 5

One liter each of CO and O2 at STP are mixed and then ignited. If the products passes through a solution of Ca(OH)2 and 2.0 g of dried CaCO3 is recovered. What is the yield?

solution

CO(g) + ½ O2(g) = CO2(g)
Ca(OH)2 + CO2 = CaCO3 + H2O
The CO gas is the limiting reactant. Therefore:

1.0 L CO 1 mol CO2
------
22.4 L CO
1 mol CO2
------
1 mol CO
100.1 g CaCO3
--------------
1 mol CaCO3

  = 4.47 g (Theoretical yield)

Actual yield = 2.0 g / 4.47 g = 0.45 = 45% yield

Born-Habor Cycle and Lattice Energy Calculations

Example 6

Evaluate the lattice energy for NaCl.

Solution

       IE
Na(g)  =   e-   +    Na+(g)
 |
 |                   +
 |              EA
 |    Cl(g) + e-  =  Cl-(g)
 |Hsub |               |
 |    |               |
 |    |BE             |U
 |    |               |
 |    |               |
Na(s) + ½ Cl2(g) = NaCl(s)
To evaluate the lattice energy, let us make a diagram showing the relationship of various processes involved. This diagram is shown on the right, and the corresponding equations are given below:

ReactionE J
Na(s) = Na(g) 108
½Cl2(g) = Cl(g) ½*240
Na(g) = e- + Na+ 496
Cl(g) + e- = Cl- -349
NaCl(s) = Na(s) + ½ Cl2(g) 411
Sum
NaCl(s) = Na+(g) + Cl-(g) 786

Note that only relevant data have been used.