Determining empirical formula for hydrated cobalt chloride compound. RESULTS, DATA CHART, QUESTION LISTED BELOWResults:Trial 1Mass of test...

Mass of
H2O

You simply deduce the mass of hydrated
chloride to the mass of cobalt chloride which is after heating
process.

#1= 2.02- 1.11=
0.91g

#2= 2.64-1.44=
1.2g

mole of unhydrated CoCl2

You need to determine first the molar mass
of CoCl2

CoCl2= 129.83
g/mol

#1.) 1.11 g CoCl2 ( 1mol/ 129.83 g CoCl2) =0.0085 mol
CoCl2

#2.) 1.44 ( 1mol/ 129.83 g CoCl2)= 0.0110 mol
CoCl2

mole of
H2O

from the mass of water we can get the
mole by using the molar mass of water which is

H2O= 18. 016
g/mol

#1.) 0.91 g H2O ( mol/ 18.016g) = 0.0505 mol
H2O

#2) 1.2 g ( mol/ 18.016g) = 0.0666 mol
H2O

mole ratio of H2O to CoCl2

you simply use the mole of water divided by
the moles of CoCl2

#1.) 0.0505 mol H2O/0.0085 mol CoCl2=
5.94 or 6

formula of hydrated CoCl2= CoCl2
6H20

#2.) 0.0666 mol H2O/0.0110 mol
CoCl2=6.05

formula of hydrated CoCl2= CoCl2
6H20

therefore there are 6 moles of water per mole of CoCl2
based from the mole ratio acquired which is
6.

finding the actual % of water from the
CoCl2 6H20

you will use the actual mass of
the sample. use a littel stoichiometry to derive the mass of water. here it is
how:

you need to determine first the molar mass
of CoCl2 6H20

#1.) 2.02 g
CoCl2 6H2O ( mol CoCl26H2O/ 237.926 g) = 0.0084 mol

0.0084
mol CoCl2 6 H2O ( 6 mol H2O/ 1 mol CoCl2) (18.016 gH2O/ 1mol H2O) = 0.908 g
H2O

#2.) 2.64 g CoCl2 6H2O ( mol CoCl26H2O/ 237.926 g)
=0.0110 mol

0.0110 molCoCl2 6 H2O ( 6 mol H2O/ 1 mol CoCl2)
(18.016 gH2O/ 1mol H2O) = 1.189 g
H2O

determining your percent error from the
theoretical yield and actual yield

% error =   | your result -
accepted value | x  100 %
accepted
value

#1.) % error= 0.91-0.908/
0.908 X 100 = 0.22 %

#.2) % error= 1.2- 1.189/1.189 X 100=
0.92 %