CHE104
Name of the student
Name of the experiment
Introduction:
Aim of the experiment is...
Equations used (chemical and mathematical):....
Data collection
The measurements in the experiment are shown in the table:
m1
m2
m3
Crucible and lid
Crucible, lid and solid before heating
Crucible, lid and solid after heating
mass [g]
25,10
27,10
26,50
Uncertainty
± 0,01
± 0,01
± 0,01
Here, all observation are stated, changes of colour, state, smell etc.
Colour of solid before heating: Blue
During heating, the blue colour fainted and then the solid turned grey. It started from
the edges…
Few drops of water added to the solid after heating: the solid turns blue again. Heat
is given off and some smoke is formed.
Data processing:
Molar masses:
MWH2O = 18,02 g/mol
MWCuSO4 = 159,6 g/mol
To find x, number of moles of H2O and CuSO4 are needed:
Mass of water in the solid: m2-m3 = 27,10 g – 26,50 g = 0,50 g
Therefore, n H 2O =
m
0,50 g
=
= 0,0277 mol
MW 18,02 g mol
Mass of CuSO4….
nCuSO4 = ….
The number of H2O molecules combined with CuSO4 is therefore,
x=
n H 2O
= .... = 4,67
nCuSO 4
x is a whole number so, 4,67 ≈ 5, the formula is then CuSO4 · 5 H2O
CHE104
Name of the student
Energy + CuSO4 · 5 H2O(s) → CuSO4(s) + 5 H2O(g)
The correct number is 5 (Ref: Chemistry by John Green and Sadru Damji. Year.
Publisher, country).
The percentage difference is then…
% difference =
X correct − X exp erimental
X correct
⋅ 100 =
5 − 4,67
⋅ 100 = 6,6% ≈ 7%
5
Explanation of the observation (if necessary): The blue colour disappeared during
heating. The solid turn blue again when adding water. This indicates that water
causes the colour. Energy was needed to remove the water and when water was
added, heat was given off. The chemical change can then be written as:
Conclusion and evaluation:
(CONCLUSION)
The experimental value of x is 4,7, rounded to 5, and the correct value is 5. The
percentage difference is 7% that is acceptable in this case…. (discuss uncertainty if
necessary).
(EVALUATION OF PROCEDURE)
The process was not successful because some material was lost during heating. The
scale was not accurate enough for such a small amount (± 0,01 g) but the initial mass
was only 2,00 g.
(IMPROVEMENTS)
To improve the procedure, use more accurate scale, a device that heats evenly and is
easy to control. Then, triplicate the sample and use the average masses for
calculation.