Determine the moles of water in copper sulcate
Data Collecting and Processing
Raw data
Table of raw data
Qualitative observation: At the start of the experiment, the crystals were blue. During the heating, the crystals turned white which means that they lost their water molecules.
The source of hydrated copper sulphate consists of both small grains and large grains of the salt. Therefore during the heating process, the smaller grains of the hydrated salt turned from blue to white in just 1 minute. In other words, there is an uneven heat distribution, which give rises to inconsistent colour of the hydrated salt (i.e. there were a few white speck and the rest is still blue). Moreover, some of the hydrated salt also spattered. In addition, at the end of the heating process, the finer grains were also turned to light yellowish-green.
Processed Data
Calculation needed:
In order to find ‘’ in the formula CuSO4.H2O, we need to find the ratio of mole between anhydrous copper II sulfate and the water vapour by using the formula:
In this case, the molar mass of:
Copper II sulfate (CuSO4) is: 63.55 + 32.06 + 15.99 x 4 = 159.57 g/mol
Water (H2O) is: 1.01 x 2 + 15.99 = 18.01 g/mol
(e.g. 0.37/18.01= 0.02054 mol.)
Then, is obtained by:
Calculation 1st attempt 2nd attempt 3rd attempt
Mole of water (mol)
= 0.06163
Mole of anhydrous copper sulfate (CuSO4) (mol)
= 0.00394
= 0.00789
= 0.01184
Gradient =
Examples: Gradient = = 5.21
Uncertainties
Uncertainty % = x 100
Uncertainties in raw data
Analytical balance: ±0.001 g
Absolute uncertainties
A. Mass of crucible (g)
±0.01 g
B. Mass of hydrated copper II sulfate (g) (C-A)
(±0.01) g + (±0.01) g = ±0.002 g
C. Mass of crucible + hydrated copper II sulfate (g)
±0.01 g
D. Mass of crucible + anhydrous copper sulfate (g)
±0.01 g
E. Mass of anhydrous copper sulfate (g) (D-A)
(±0.01) g + (±0.01) g = ±0.02 g
F. Mass of water lost (g) (C-D)
(±0.01) g + (±0.01) g = ±0.02 g
Mass (g)
Raw data
Table of raw data
Qualitative observation: At the start of the experiment, the crystals were blue. During the heating, the crystals turned white which means that they lost their water molecules.
The source of hydrated copper sulphate consists of both small grains and large grains of the salt. Therefore during the heating process, the smaller grains of the hydrated salt turned from blue to white in just 1 minute. In other words, there is an uneven heat distribution, which give rises to inconsistent colour of the hydrated salt (i.e. there were a few white speck and the rest is still blue). Moreover, some of the hydrated salt also spattered. In addition, at the end of the heating process, the finer grains were also turned to light yellowish-green.
Processed Data
Calculation needed:
In order to find ‘’ in the formula CuSO4.H2O, we need to find the ratio of mole between anhydrous copper II sulfate and the water vapour by using the formula:
In this case, the molar mass of:
Copper II sulfate (CuSO4) is: 63.55 + 32.06 + 15.99 x 4 = 159.57 g/mol
Water (H2O) is: 1.01 x 2 + 15.99 = 18.01 g/mol
(e.g. 0.37/18.01= 0.02054 mol.)
Then, is obtained by:
Calculation 1st attempt 2nd attempt 3rd attempt
Mole of water (mol)
= 0.06163
Mole of anhydrous copper sulfate (CuSO4) (mol)
= 0.00394
= 0.00789
= 0.01184
Gradient =
Examples: Gradient = = 5.21
Uncertainties
Uncertainty % = x 100
Uncertainties in raw data
Analytical balance: ±0.001 g
Absolute uncertainties
A. Mass of crucible (g)
±0.01 g
B. Mass of hydrated copper II sulfate (g) (C-A)
(±0.01) g + (±0.01) g = ±0.002 g
C. Mass of crucible + hydrated copper II sulfate (g)
±0.01 g
D. Mass of crucible + anhydrous copper sulfate (g)
±0.01 g
E. Mass of anhydrous copper sulfate (g) (D-A)
(±0.01) g + (±0.01) g = ±0.02 g
F. Mass of water lost (g) (C-D)
(±0.01) g + (±0.01) g = ±0.02 g
Mass (g)