DISCUSSION

In this experiment was done the determination of heat solubility CuSO4.5H2O and

anhydrate CuSO4. In this experiment using calorimeter that the constant already know from

the preview experiment is about 114 JoC-1. Firstly into calorimeter was entered aquades that

already measured the temperature. When the temperature in calorimeter is constant, added

CuSO4.5H2O is about 5.001 grams, shack it and the temperature was measure until constant.

After the temperature is constant, make a graph between time and temperature. Next, from

that the graph makes extrapolation and obtains the amount of ΔT. Graph relationship of time

and temperature is like this:

0 2 4 6 8 10 1226.7

26.8

26.9

27

27.1

27.2

27.3

27.4

27.5

27.6

Graph of Time vs Temperature

Time (minutes)

Tem

pera

ture

(oC)

The next stage is to use anhydrous CuSO4. Where the calorimeter is used, the same

calorimeter with the same constant is 114 JOC-1. First of all anhydrous CuSO4 made by

heating until the color of CuSO4.5H2O changed to white solids CuSO4.5H2O then left in place

in a desiccator to cool. CuSO4.5H2O crystals are used weighed as much as 5.001 grams. The

crystal of CuSO4.5H2O after heating will be decrease mass. This is because when heating

water contained in CuSO4.5H2O has evaporated so that the weight of CuSO4 will be reduced.

Furthermore distilled water has been known the temperature that added in the calorimeter and

then stirred and the temperature of the calorimeter is measured to obtain a constant

Δt

Picture 1. Graph of the relationship between time and temperature when added

CuSO4.5H2O

temperature. Then continue by entering the anhydrous CuSO4 that obtained from heating of

CuSO4.5H2O. The temperature is measured from the addition of anhydrous CuSO4 until the

temperature in calorimeter is constant. Graph showing the relationship between time and

temperature are made and extrapolated to obtain the value of ΔT as shown in Figure 2.

0 1 2 3 4 5 6 7 8 924

25

26

27

28

29

30

31

Graph of Time vs Temperature

Time (minutes)

Tem

pera

ture

(oC)

Picture 2. Graph relationship among time and temperature when added anhydrate of CuSO4

1. Determination of heat solubility of CuSO4.5H2O

Known

- Crystal mass CuSO4.5H2O = 5,001 grams

- Density of water (c) = 4,18 J/g°C

- Calorimeter capacity (C) = 114 J/°C

- Initial temperature of mixture = 27 °C

- Final temperature of mixture = 27,5 °C

Calculation:

Calculation of mol CuSO4.5H2O:

Mol (n) CuSO4.5H2O =

mass CuSO4 . 5H 2OmassmolarCuSO4 . 5H2O

=

5 ,001 gram249 ,54 g/mol

Δt

= 0,02 mol

Heat calculation:

Q = Q water + Q calorimeter

Q = m.c.∆T + C.∆T

Q = (100 gram x 4,18 J/g°C x 0,5 °C ) + (114 J/°C x 0.5°C)

Q = 209 J + 57 J

Q = 266 J

Calculation of dissolution heat

ΔH =

qmol

ΔH =

266 J0 ,02mol

ΔH = + 13300 J/ mol = + 13.3 kJ/ mol (endotherm)

2. Determination of dissolution heat of CuSO4

Known:

- Mass of CuSO4 crystal = 4.10 gram

- Heat capacity of water (c) = 4,18 J/g°C

- Calorimeter capacity (C) = 114 J/°C

- Initial temperature of mixture = 26.5 °C

- Final temperature of mixture = 30 °C

Calculation:

Mole calculation of CuSO4:

Mol(n) CuSO4 =

mass CuSO4mass molarCuSO4

=

4 . 10 gram154 . 54 g/mol

= 0.026 mol

Heat calculation:

Q = Qwater + Q calorimeter

Q = m.c.∆T + C.∆T

Q = (100 gram x 4,18 J/g°C x 3.5 °C ) + (114 J/°C x 3.5 °C)

Q = 1463 J + 399 J

Q = 1862 J

Calculation of dissolution heat:

ΔH =

qmol

ΔH =

1862 J0 .026mol

ΔH = + 71615 J/ mol = + 71.615 kJ/ mol (endotherm)

Berdasarkan analisis data yang telah dilakukan, diperoleh bahwa nilai kalor pelarutan

pada CuSO4.5H2O adalah +13.3 kJ/ mol sedangkan kalor pelarutan pada CuSO4 adalah

+71.615 kJ/ mol. Dari data yang diperoleh dapat disimpulkan bahwa kalor pelarutan pada

CuSO4.5H2O lebih besar dibandingkan dengan CuSO4. Jadi, dapat diartikan bahwa kalor yang

diserap dalam pelarutan CuSO4.5H2O lebih besar dibandingkan dengan kalor pelarutan

CuSO4.