Lab Overview Investigation #26: "A Cobalt Complex"

The study of metal complexes has gained a lot of attention lately because of the realization of its importance in biological aspects. One example is hemoglobin which is a Fe complex. Much like the cobalt complex in this lab, it combines with oxygen. When it does so, it is a red color. When it supplies its oxygen atom to your muscles it changes back to a blue color. That is why blood coming from your lungs is red and is considered to be in your arteries on the way to your muscles. Blood that has delivered the oxygen and on its way back to the heart and lungs is blue and is said to be in your veins. In this lab you will look for color changes in the cobalt complex just like those in the hemoglobin.

This lab will involve different methods of product preparation to include: heating, filtering, and crystal formation.

Concepts of the Experiment

The following balanced equations are important in determining amounts and yields of products and reactants.

step 1: 2CoCl₂^.6H₂O + 4C₄H₈N₂O₂ ® 2CoH₁₄C₈N₄O₄ + 4HCl +12H₂O

step 2: 2CoH₁₄C₈N₄O₄ + 4HCl + 4C₅H₅N + 1/2 O₂® 2CoH₁₉C₁₃N₅O₄Cl + 2C₅H₆NCl + H₂O

overall reaction:

FW of cobalt complex:

Co 58.9332g

H 1.00794*19=19.15086 g

C 12.011*13=156.143 g

N 5*14.0067=70.0335 g

O 4*15.9994=63.9976 g

Cl 35.4527g

CoH₁₉C₁₃N₅O₄Cl = 403.71086 g/mole
 
 

FW of CoCl₂^.6H₂O

Co 58.9332 g

Cl 35.4527*2=70.854 g

H 1.00794*12=12.09528 g

O 6*15.9994=95.9964 g

CoCl₂^.6H₂O = 237.87888 g/mole
 
 

FW of C₄H₈N₂O₂

H 1.00794*8=8.06352 g

C 12.011*4=48.044 g

N 2*14.0067=28.0134 g

O 2*15.9994=31.9988 g

C₄H₈N₂O₂ = 116.11972 g/mole
 
 

Example Calculation: Calculate the mass of CoCl₂^.6H₂O required for the preparation of 0.70 g of the complex. Assume a 60 percent yield in this experiment and consider CoCl₂^.6H₂O to be the limiting reactant.

0.70 g CoH₁₉C₁₃N₅O₄Cl = ? g * 0.60 (60 percent of the total theoretical amount is 0.70g)

? g = 1.167 g CoH₁₉C₁₃N₅O₄Cl

1.167 g CoH₁₉C₁₃N₅O₄Cl * 1 mole/403.71086g * 2 moles/2 moles * 237.87888 g/mole =

0.687 g CoCl₂^.6H₂O (now using stoichiometric calculations we can find the CoCl₂^.6H₂O mass)

Example Calculation: The mass of CoCl₂^.6H₂O calculated above should be reacted with 110% of the stoichiometrically required amount of dimethylgyoxime. How many grams of dimethylglyoxime are needed?

0.687 g CoCl₂^.6H₂O * 1 mole/ 237.87888 g * 4 moles/ 2 moles * 1.10 * 116.11972 g/mole = 0.738 g C₄H₈N₂O₂

The rest of the calculations, I will leave to y'all. Good luck.