To weigh the mass of the salt accurately, we use analytical balance and the two-decimal balance to check the accuracy of the mass obtained. Analytical balance is an instrument used to measure mass to a high degree of precision. The two-decimal balance is an instrument used to measure mass at two-decimal point. Along the experiment we weighed the object by using both of the balances, the result obtained shows that the analytical balance is more accurate then the two-decimal balance, because the analytical balance show four decimal result that can detect small change s on the mass of the object. For example, when we reheat and reweigh the dehydrated salt, the result only can show by the analytical balance, but not the two-decimal balance. So, from the above example given, we can conclude that, the analytical balance is more accurate then two-decimal balance.
Calculation:
Mass of Test tube
(Mass of test tube + biker) – (Mass of biker)
Analytical balance:
41.5471 – 36.0436
=5.5035g
Two-decimal balance:
53.49 – 48.89
=5.50g
Mass of Hydrated salt
(Mass of test tube + biker + hydrated salt) – (Mass of test tube + biker)
Analytical balance:
42.0562 – 41.5471
=0.5091g
Two-decimal balance:
54.89 – 54.39
=0.5g
Mass of Dehydrated salt
(Mass of test tube + biker + dehydrated salt) – (Mass of test tube + biker)
Analytical balance:
41.8946 – 41.5471
=0.3475g
Two-decimal balance:
54.74 – 54.39
=0.35g
Mass of Water
(Mass of hydrated salt) – (Mass of dehydrated salt)
Analytical balance:
0.5091 – 0.3475
=0.1616g
Two-decimal balance:
0.50 – 0.35
=0.15g
Question
Q1. H=1.0079 S=32.065 Cu=63.546 O=15.9994
CuSO4=159.6085 H2O=18.0153
Analytical balance:
Weight of water in salt= hydrated salt – dehydrated salt
= 0.5091 – 0.3475
= 0.1616g
Two-decimal balance:
Weight of water in salt= hydrated salt – dehydrated salt
= 0.50 – 0.35
= 0.15g
Mole of dehydrated salt = 0.3475g / 159.6085g/mole
= 0.00218 mole
Mole of water = 0.1616g / 18.0153g/mole
= 0.00897 mole
CuSO4 ∙x H20
X = mole of water / mole of dehydrated salt
= 0.00897mole / 0.00218mole
= 4.1147 Approximate to = 4 Empirical Formula = CuSO4∙4H2O
Q2
Na= 22.9898 S= 32.065 O= 15.9994 H= 1.00794
Na2SO4= 142.0422 H2O= 18.0153
Na2SO4 ∙x H2O= 15g
Water content in Hydrated salt= 7.05g
Dehydrated salt= 7.95g
Mole of dehydrated salt = 7.95g / 142.0422g/mole
= 0.05597mole
Mole of water = 7.05g / 18.0153g/mole
= 0.39133mole
Na2SO4 ∙x H2O
X = mole of water/ mole of dehydrated salt
= 0.39133mole / 0.05597mole
= 6.99178 Approximate to = 7
Empirical Formula = Na2SO4∙7H2O
Q3
Analytical balance is a very expensive and sensitive instrument, so when using it, we must handle it with care to avoid any error or accident occurs during the experiment. To use analytical balance, first, we must ensure that the weighing pan is cleared and make sure it is zeroed. To start weighing, we place the biker as a weighing container on the weighing pan and remember to close the chamber doors after opened it. After putting the weighing container, to clear the readout just simply pressing “TARE” button will do. After that, we add the substance to be weighed, but make sure that the substance did not split on the weighing pan, because this will affect the reading of the readout. If necessary, it is advisable to take out the container from the weighing chamber when adding the substance to be weighed, did not press “TARE” button before weigh the substance that just added in the container. Put back the container on the weighing pan after added the substance, remember to close the chamber doors. The readout will show the net weight of the substance just added. After recorded the readout, take out the container from the weighing chamber and clear the readout by pressing the button “TARE”. Again, remember to close the chamber doors and ensure that the weighing pan is cleared. To turn off the analytical balance, press “off”. Bear in mind, each time using the analytical balance must handle it with care.