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Name Surname, class
date

CALCIUM CARBONATE (CaCO3) IN VARIOUS MARINE
ORGANISMS
INTRODUCTION
In this practical, we determined the percentage by mass of calcium carbonate (CaCO3) in a
seashell, a coral, a hermit crab and a sea urchin
...
A
known amount of excess acid was added to the sample to ensure that all the calcium
carbonate has reacted
...
Aliquotes of this deluded excess solution were then used for titration with a
standard solution of sodiun hydroxide
...
The
percentage in a shell and hermit crab is fairly similar
...
0001 g
± 0
...
05 cm3

50 cm3 conical flask
10 cm3 pipette

± 0
...
05 cm3

Name Surname, class
date

REACTIONS:
𝐶𝑎𝐶𝑂3 (𝑠) + 2𝐻𝐶𝑙(𝑎𝑞) → 𝐶𝑎𝐶𝑙2 (𝑙) + 𝐶𝑂2 (𝑔) + 𝐻2 𝑂(𝑙)
𝐻𝐶𝑙(𝑎𝑞) + 𝑁𝑎𝑂𝐻(𝑎𝑞) → 𝑁𝑎𝐶𝑙(𝑎𝑞) + 𝐻2 𝑂(𝑙)

PROCEDURE
1
...
The samples were
then grint into small pieces
...
About 0
...
Then, 20
...
00
mol/dm3 hydrochloric acid (HCl) was added into each flask
...

3
...
0 cm3 of distilled water was added to each flask and the contents
of these flasks were then transfered to a 100 cm3 volumetric flask where the diluded
acid was made to be 100 cm3 using more distilled water
...
10
...
100 mol/dm3 sodium hydroxide (NaOH) using phenolphthalein as
an indicator
...
50 cm3 of NaOH was added to a burette and then added drop by drop to solution
aliquotes
...
3 runs
were made for each sample
...

6
...


Name Surname, class
date

DATA PRESENTATION AND CALCULATIONS
RAW DATA:
1 (exoskeleton
of sea urchin)
Run 1
Run 2
Run 3
Average
m (sea urchin) = 0
...
0001
0
...
05 ml)

Vf (±0
...
1 ml)

5
...
30
24
...
30
24
...
80

9
...
7
8
...
0

= 1
...
478415139 × 10−2 % =

0
...
0111111111 = 1
...
0

2 (seashell)
Vi (±0
...
00
Run 2
8
...
50
Average
𝑚 (𝑠𝑒𝑎𝑠ℎ𝑒𝑙𝑙) = 0
...
0001
𝛿(𝑚) =
= 1
...
515840534 ×
0
...
10
𝛿(𝑉𝑢𝑠𝑒𝑑 ) =
= 0
...
21506683 %
8
...
05 ml)
8
...
50
24
...
1 ml)
8
...
3
8
...
23

10−2 %

3 (coral)
Vi (±0
...
05 ml)
Run 1
24
...
50
Run 2
33
...
20
Run 3
8
...
50
Average
m (coral) = 0
...
0001
𝛿(𝑚) =
= 1
...
512859304 × 10−2 %
0
...
1
𝛿(𝑉𝑢𝑠𝑒𝑑 ) =
= 0
...
14547537 %
8
...
1 ml)
8
...
7
8
...
73

4 (hermit crab
Vi (±0
...
50
Run 2
26
...
20
Average
m(hermit crab) = 0
...
0001
𝛿(𝑚) =
= 1
...
542971764 ×
0
...
1
𝛿(𝑉𝑢𝑠𝑒𝑑 ) =
= 0
...
12359551 %
8
...
05 ml)

Vused (±0
...
30
35
...
20

8
...
9
9
...
9

10−2 %

Name Surname, class
date

DATA PROCESSING AND PRESENTATION:
𝑔

M (CaCO3) = 100
...
0 cm3 aliquot:
𝑛(𝑁𝑎𝑂𝐻) = 𝑛(𝐻𝐶𝑙)𝑒𝑥𝑐𝑒𝑠𝑠 = 𝑉 ∗ 𝑐(𝑁𝑎𝑂𝐻) = 8
...
986666667 ∗ 10−3 𝑚𝑜𝑙
∆(𝑛(𝐻𝐶𝑙)𝑒𝑥𝑐𝑒𝑠𝑠 1 ) = 𝑛(𝐻𝐶𝑙)𝑒𝑥𝑐𝑒𝑠𝑠 1 ∗ 𝛿(𝑉𝑢𝑠𝑒𝑑 ) = 9
...
0055066667 𝑚𝑜𝑙
2
∆(𝑛(𝐶𝑎𝐶𝑂3 )) = ∆(𝑛(𝐻𝐶𝑙)) + ∆(𝑛(𝐻𝐶𝑙)𝑒𝑥𝑐𝑒𝑠𝑠 1 ) = 1
...
8148466366 = 𝟖𝟏
...
229473149 %
𝑛(𝐶𝑎𝐶𝑂3 )
𝑤=

2 SEASHELL
In 10
...
233333333 ∗ 10−4 𝑚𝑜𝑙

In 100 cm3:
𝑛(𝐻𝐶𝑙)𝑒𝑥𝑐𝑒𝑠𝑠 1 = 𝑛(𝐻𝐶𝑙)𝑒𝑥𝑐𝑒𝑠𝑠 ∗ 10 = 8
...
000405023 ∗ 10−4 𝑚𝑜𝑙
1
𝑛(𝐶𝑎𝐶𝑂3 ) = ∗ (𝑛(𝐻𝐶𝑙) − 𝑛(𝐻𝐶𝑙)𝑒𝑥𝑐𝑒𝑠𝑠 1 ) = 0
...
500405023 ∗ 10−4 𝑚𝑜𝑙
𝑛(𝐶𝑎𝐶𝑂3 ) ∗ 𝑀(𝐶𝑎𝐶𝑂3 )
= 0
...
𝟐𝟔𝟐𝟐𝟏𝟓𝟏𝟓 %
𝑚
∆(𝑛(𝐶𝑎𝐶𝑂3 )
∆(𝑤) = (
+ 𝛿(𝑚)) ∗ 𝑤 = 2
...
0 cm3 aliquot:
𝑛(𝑁𝑎𝑂𝐻) = 𝑛(𝐻𝐶𝑙)𝑒𝑥𝑐𝑒𝑠𝑠 = 𝑉 ∗ 𝑐(𝑁𝑎𝑂𝐻) = 8
...
733333333 ∗ 10−3 𝑚𝑜𝑙
∆(𝑛(𝐻𝐶𝑙)𝑒𝑥𝑐𝑒𝑠𝑠 1 ) = 𝑛(𝐻𝐶𝑙)𝑒𝑥𝑐𝑒𝑠𝑠 1 ∗ 𝛿(𝑉𝑢𝑠𝑒𝑑 ) = 1
...
0056333334 𝑚𝑜𝑙
2
∆(𝑛(𝐶𝑎𝐶𝑂3 ) = ∆(𝑛(𝐻𝐶𝑙)) + ∆(𝑛(𝐻𝐶𝑙)𝑒𝑥𝑐𝑒𝑠𝑠 1 ) = 1
...
8530110968 = 𝟖𝟓
...
284814318 %
𝑛(𝐶𝑎𝐶𝑂3 )
𝑤=

4 HERMIT CRAB SHELL
In 10
...
900000000 ∗ 10−4 𝑚𝑜𝑙
In 100 cm3:
𝑛(𝐻𝐶𝑙)𝑒𝑥𝑐𝑒𝑠𝑠 1 = 𝑛(𝐻𝐶𝑙)𝑒𝑥𝑐𝑒𝑠𝑠 ∗ 10 = 8
...
000000004 ∗ 10−4 𝑚𝑜𝑙
1
𝑛(𝐶𝑎𝐶𝑂3 ) = ∗ (𝑛(𝐻𝐶𝑙) − 𝑛(𝐻𝐶𝑙)𝑒𝑥𝑐𝑒𝑠𝑠 1 ) = 0
...
500000004 ∗ 10−4 𝑚𝑜𝑙
𝑛(𝐶𝑎𝐶𝑂3 ) ∗ 𝑀(𝐶𝑎𝐶𝑂3 )
𝑤=
= 0
...
𝟕𝟏𝟐𝟎𝟎𝟒𝟑𝟐 %
𝑚
∆(𝑛(𝐶𝑎𝐶𝑂3 )
∆(𝑤) = (
+ 𝛿(𝑚)) ∗ 𝑤 = 2
...
So our hypothesis was not correct, since
we predicted the largest amount of CaCO3 in the coral
...
05 ml)
0
...
10
18
...
05 ml)
9
...
20
27
...
1 ml)
9
...
1
9
...
1

m(Na2CO3*xH2O) = 1
...
0001 g
Relative uncertainties:
0
...
246417013 ∗ 10−5
1
...
1
𝛿(𝑉𝑢𝑠𝑒𝑑 ) =
= 0
...
1

DATA PROCESSING AND PRESENTATION:
Equation of the reaction:
Na2CO3 (aq) + 2 HCl (aq)

2 NaCl (aq) + H2O (l) + CO2 (g)

In 25
...
1 ∗ 10−10 𝑚𝑜𝑙
𝑛(𝐻𝐶𝑙)
𝑛1 (Na2CO3) =
= 4
...
55 ∗ 10−3 𝑚𝑜𝑙
∆(𝑛) = 𝑛(Na2CO3) ∗ 𝛿(𝑉𝑢𝑠𝑒𝑑 ) = 0
...
55 ∗ 10−3 𝑚𝑜𝑙 = 5
...
55 ± 0
...
4823455 𝑔
0
...
4823455 ∗
= 0
...
55
𝑚(Na2CO3) = (0
...
005) 𝑔

Percentage yield:
𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
%𝑦𝑖𝑒𝑙𝑑 =
= 0
...


Name Surname, class
date
To find out the x in Na2CO3*xH2O:
𝑚(Na2CO3*xH2O)
1
...
6923077
𝑛(Na2CO3*xH2O) 4
...
6923077 ∗ (0
...
246417013 ∗ 10−5 ) = 2
...
6923077 − 106
...
6923077
𝑥=

M(xH2O) 131
...
308119184
M(H2O)
18
...
1
2

𝑥 = 7
...
1
Formula of the reactant:

Na2CO3*7H2O

3

2

𝑔
𝑚𝑜𝑙

Name Surname, class
date

DETERMINING THE VALUE OF THE GAS CONSTANT
MATERIALS AND TOOLS:
magnesium ribbon, copper wire, 20 % HCl, 50-Ml, burette, bung with a hole, beakers, thermometer,
barometer, funnel, 250 cm3 cylinder, plastic tray, tap water, distilled water, top pan balance,
analytical balance and sandpaper

PROCEDURE:
•

•

•
•

•
•

•

First an appropriate amount of magnesium ribbon was obtained
...
A 5 cm
long piece of magnesium ribbon was weighted and hence the length of a 0
...
The assumption that the magnesium ribbon is
homogenous was made at this point
...

Then the volume of burette between the bottom graduation and the tap (the “dead
volume”) was measured
...
Using Pasteur pipette the water was filled exactly to the bottom
graduation
...
The volume of water was then calculated from its mass
...

20 % hydrochloric acid was added to near the 40 cm3 mark
...
Acid remained at the bottom as a separate layer
...

The hole in the bung was closed with a finger before the burette was inverted
...

After some time when there was no further reaction (when all the magnesium had dissolved)
the burette was placed in the measuring cylinder so that water levels inside the burette and
the measuring cylinder were the same
...

Room temperature and pressure were measured
...
0472 ± 0
...
1 ± 0
...
2 ± 0
...
8 ± 0
...
0021186441
0
...
0010288066
0
...
0199349800

Name Surname, class
date
Total relative uncertainty = 0
...
0047393365 + 0
...
0120481928 =
= 0
...
2∗0
...
305
=
294
...
0472

8
...
0199349800 * 8
...
1689545976 J⋅K−1⋅mol−1
R = 8
...
2 J⋅K−1⋅mol−1

Taking into consideration the vapour pressure of water:
Not all the pressure inside the burette was a consequence of H2, because there was probably some
water vapour
...
3 kPa
...

(𝑝−2
...
2−2
...
0498∗24
...
3∗0
...
269112869 J⋅K−1⋅mol−1

Absolute uncertainty = 0
...
269112869 = 0
...
3 ± 0
...


CONCLUSION AND EVALUATION:
The final result was quite accurate and precise, specially after taking into consideration the vapour
pressure of water
...
3 ± 0
...
31 J⋅K−1⋅mol−1
...


Name Surname, class
date
This model assumes that the gas is an ideal gas, which means that there are no interactions between
the particles and that the particles do not occupy any space
...
Most heavy gases, like many refrigerants and gases with strong intermolecular forces, like
water vapor also don´t obey the model of ideal gas law
...
Also avoiding the copper spiral would make it
more accurate as some copper still reacted with HCl and produced some gas (H2)
...




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