determine the initial and final mass of cobalt chloride compound, the mass of the water that was driven off,the # of moles of dry CoCl2 there were?Dry Lab-involves a sample of hydrated cobalt...
determine the initial and final mass of cobalt chloride compound, the mass of the water that was driven off,the # of moles of dry CoCl2 there were?
Dry Lab-involves a sample of hydrated cobalt choride consisting of molecules of COCl2 joined to molecules of H2O. Each molecule of CoCl2 has the same number of water molecules bonded to it. The formula is written as CoCl2 (XH2O for now. X is the number of water molecules.
Since there won't be fractions of an H2O molecule, X must be a whole number. You are to determine the value of X and write the correct formula.
The method is to heat the hydrated cobalt chloride and drive off the water. Knowing the weight of water and dry CoCl2 you can determine the number of moles of each.
mass of test tube=10.41g
mass of test tube and the hydrated cobalt chloride =12.43g
mass of test tube and contents after heating=11.52g
Mass of test tube=10.42g
mass of test tube and the hydrated cobalt chloride=13.06g
Mass of test tube and contents after heating=11.86g
I have to do a chart for trial 1 and 2
mass of test tube and Hydrated cobalt cholride=
mass of test tube=
mass of hydrated cobalt chloride=
mass of test tube and cobalt chloride after heating=
mass of cobalt chloride=
mass of water=
moles of cobalt chloride (not hydrated)=
moles of water=
mole ratio-water/cobalt chloride=
Mass of water predicted from mole ratio=
Start by writing a chemical equation so you can follow what is happening.
CoCl2-XHOH --> CoCl2 + X HOH
Now determine the molar mass of each product.
CoCl2 = 129.83 g/mole
HOH = 18.02 g/mole
You are now ready to do the calculations.
12.43 g - 10.41 g = 2.02 g of hydrate
11.53-10.41 = 1.11 g of CoCl2
2.02 - 1.11 = 0.91 g of water removed by heating
13.06 - 10.42 = 2.64 g of hydrate
11.86-10.42 = 1.46 g of CoCl2
2.64-1.46 = 1.18 g of water removed by heating.
1.11 g CoCl2/129.83 g/mol = .00855 moles CoCl2
0.91 g HOH/18.02 g/mol = 0.0505 mol of water.
Mole ratio for trial one is: .0505/.00855 = 5.91 moles water per mole of CoCl2. This would round to an actual whole number ratio of 6:1
1.46 g CoCl2/129.83 g/mole = 0.01125 moles CoCl2
1.18 g HOH/18.02 g/mole = 0.0655 moles of HOH
mole ratio for trial two is: 0.0655/0.01125 = 5.82 moles of water to one mole of CoCl2. Again round to whole number ratio of 6:1
Average determined ratio is:
(5.91 + 5.82)/2 =5.865
% error: (6 - 5.865)/6 * 100 =5.25%
Now what should you have obtained from your starting material?
Assuming you have 6 moles of water per mole of CoCl2 in the hydrate, the molar mass of the hydrate is 237.95 g/mole.
Trial 1: 2.02/237.95 = 0.0085 moles
So you should get 0.0085 moles of CoCl2 = 1.10 g of CoCl2
and 0.919 g of water released.
2.64/237.95 = 0.011 moles of hydrate, producing 0.011 moles of CoCl2 = 1.44 g CoCl2 and 1.19 g HOH
As a post note: you got very good results with the two trials. Well Done?