NAME: HIEP TON
October 10, 2014
LAB 3 DETERMINATION OF THE EMPIRICAL FORMULA OF A HYDRATE
I.
OBJECTIVE :
• To determine the formula of an unknown by using the molar ratio from the mass and mole of hydrate.
• To establish the formula of a hydrate using gravimetric analysis.
• To understand how constant-mass and careful measurement really matter in the experiment.
II.
DATA and REPORT :
Gravimetric Analysis of an Unknown Hydrate
Sample #
34
.
Table 1: Weight Data
Measurement (g)
Trial 1
Trial 2
Trial 3
7.5521
Mass of Test Tube (heated)
7.5507
7.5504
Mass of Test Tube (heated constant) + Hydrated
Salt
Mass of Test Tube (heated constant) + Anhydrous
Salt (heated constant)
8.0545
8.1740
7.8398
7.875
7.9722
7.7886
7.873
7.9721
7.7877
7.8726
III.
DATA and CALCULATION :
Calculating the mass of hydrated salt, anhydrous salt, water, moles of water, moles of salt and
molar ratio in order to figure out and determine the salt formula.
Table 2: Naming Variables
MTT
Mass of Test Tube (heated constant at third time)
MTT&H
Mass of Test Tube (heated constant) + Hydrated Salt
Mass of Test Tube (heated constant) + Anhydrous Salt (heated constant at third time)
MTT&A
MH
MA
MW
Mol W
Mol A
Mass of the hydrate
Mass of the anhydrous
Mass of the water
Mole of water
Mole of Anhydrous or Salt
Page 1 of 4
NAME: HIEP TON
October 10, 2014
Three Hydrate possibilities (instructor provided):
 MgSO₄ ∙ 7 H2O
 ZnSO₄ ∙ 7 H2O
 CaSO₄ ∙ 2 H2O
Taking Trial 1 (to show calculations):
Calculate the mass of Hydrate:
 MH = ( MTT&H ) – ( MTT ) = ( 8.0545 ) – ( 7.5504 ) = 0.5041 (g)
Calculate the mass of Anhydrous:
 MA = ( MTT&A ) – ( MTT ) = ( 7.8726 ) – ( 7.5504 ) = 0.3222 (g)
Calculate the mass of Water:
 Mw = ( MH ) – ( MA ) = ( 0.5041 ) – ( 0.3222 ) = 0.1819 (g)
Calculate the mole of Water:
 Mol W = Mw (g) ÷ 18.02 (mol/g) = 0.1819 (g ) ÷ 18.02 (mol/g) = 0.01009 (mol)
NOTES: Since there are three possibilities of anhydrous/salt, the molar mass of anhydrous
are depended on the anhydrous/salt formula (only use none water)
Calculate that molar mass of three different possible anhydrous/salt:
 MgSO₄ = (24.305) + (32.065) + (15.999 × 2) = 88.368 (mol/g)
 ZnSO₄ = (65.38) + (32.065) + (15.999 × 2) = 129.443 (mol/g)
 CaSO₄ =(40.078) + (32.065) + (15.999 × 2) = 104.141 (mol/g)
Calculate the mole of Anhydrous(salt):
Mol A = MA (g) ÷ MolarMassOfAnhydrous (mol/g)
 MgSO₄ : Mol A = 0.3222 (g ) ÷ 88.368 (mol/g) = 3.6461 × 10-03 (mol)
 ZnSO₄ : Mol A = 0.3222 (g ) ÷ 129.443 (mol/g) = 2.4891 × 10-03 (mol)
 CaSO₄ : Mol A = 0.3222 (g ) ÷ 104.141 (mol/g) = 3.0939 × 10-03 (mol)
Page 2 of 4
NAME: HIEP TON
October 10, 2014
Calculate the Molar Ratio (water/Anhydrous(salt)):
MgSO₄ : Molar Ratio =
ZnSO₄ : Molar Ratio =
CaSO₄ : Molar Ratio =
Mol W
Mol A
Mol W
Mol A
Mol W
Mol A
=
0.01009
3.6461 × 10−03
0.01009
=
2.4891 × 10−03
=
= 2.7673
= 4.0537
0.01009
3.0939 × 10−03
= 3.2612
Table 3: Calculation Data
Trial 1
Trial 2
Trial 3
Mass of the hydrate (g)
0.5041
0.6236
0.2894
Mass of the anhydrous (g)
0.3222
0.4217
0.2373
Mass of the water (g)
0.1819
0.2019
0.0521
Moles of the the water (mol)
0.01009
0.0112
2.8912 × 10-03
MgSO₄
3.6461 × 10-03
4.7721 × 10-03
2.6854 × 10-03
ZnSO₄
2.4891 × 10-03
3.2578 × 10-03
1.8332 × 10-03
CaSO₄
3.0939 × 10-03
4.0493 × 10-03
2.2786 × 10-03
MgSO₄
2.7673
2.3470
1.0766
ZnSO₄
4.0537
3.4379
1.5771
CaSO₄
3.2612
2.7659
1.2688
Mole of salt (mol)
Molar ratio (water/salt)
Calculate the Average Molar Ratio :
 MgSO₄ : (2.7673 + 2.3470 + 1.0766) ÷ 3 = 2.0637
 ZnSO₄ : (4.0537 + 3.4379 + 1.5771) ÷ 3 = 3.0229
 CaSO₄ : (3.2612 +2.7659 + 1.2688 ) ÷ 3 = 2.4319
IV.
CALCULATION and RESULT :
As the result shown from calculation section above, the hydrate sample (#34) can
be determined as CaSO₄ ∙ 2 H2O because the average Molar Ratio is calculated as 2.4319,
which can be rounded off as 2.
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NAME: HIEP TON
V.
October 10, 2014
CONCLUSION :
 By using molar ratio, if given an unknown sample and many different possibilities of
hydrates, it is possible to figure out, determine and establish the possible compound sample.
 During the heating process, if not letting the test-tube or sample being heated long enough,
the test-tube or sample can still contain a small amount of water. And it will take even more
times to re-heat to approach to the constant mass.
 Furthermore, in the weighting process, if no letting the test-tube’s temperature is totally
cooled down, the outcome weight might be the invalid data since the air pressure in the
balance scale is changed by the heat that still in the test-tube and sample.
 As just a small mistake in any process throughout the experiment, the result outcome might
be totally wrong. Even though the data might be recorded correctly with a minimized
tolerance of error, the calculation process might be a huge impact on the invalid result(s)
without writing out all the needed formulas and checking every single step carefully.
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