Experiment 4
"Water of hydration"
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I. Observations and results:
Unknown number
Molar mass of anhydrous salt
Mass of crucible
Mass of crucible and hydrate
Mass of crucible and residue after heating
Mass of hydrate
Mass of anhydrous salt
Mass of water of anhydrous salt
Moles of anhydrous salt
Moles of water
Moles of water per 1 mol hydrate
% water in the hydrate
Molar mass of hydrate
Na2CO3010H2O
106 g/mol
21.321 g
21.321 g
123.80
1.43 g (1)
0.28 g
1.17 g (3)
0.00175 mol (4)
0.065 mol (5)
3.887 mol
81% (6)
176 g/mol (7)
II. Calculation:
(1) Mass of hydrate = 124.95-123.52=1.43 g
(2) Mass of anhydrous salt =123.80-21.321 =0.28 g
(3) Mass of water of anhydrous salt =21.321 g -123.80=1.17 g
(4) n =m/M=0.28 /106= 0.00175 mol
(5) n= m/M=1.17 /18=0.065 mol
(6) %water= (mwater/msample)*100%= (1.17 /1.43)*100%=81%
(7) In hydrate (anhydrous salt : water) ,that’s mean the ratio between
anhydrous salt and water in hydrate as (0.00175: 0.065) or (1:38.23) ,from
the ratio the hydrate is Na2CO304H2O ,and molar mass is 176 g/mol
III. Answer the following questions:
1. What percent by weight of water is in alum KAI(SO4)2.12H2O
Take sample of hydrate
,nhydrate =m/M =100\474=0.211 mol
1 mole hydrate === 12 mole water
0.211 mol
=== ??
Moles of water =2.532 mol
Mass of water =n*M =2.532*18 =45.576 g
%water = (mwater/msample)*100% = (45.576/100)*100%=45.576%
2. A solid hydrate weighting 2.691 g was heated to drive off the water .A solid
anhydrate residue remained which weighed 2.259 g /Calculate the percent by
mass of water in the hydrate .If the anhydrous residue has a molar mass of 282
g/mol how many moles of water contained in one mole of hydrous salt?
,mwater = mhydrate-manhydrate=2.692-2.259=0.432 g
%water =(mwater/msample)*100% =(0.432/2.691)*100% =16%
,nanhydrate = m/M=2.259/282 =0.008 mol
, nwater = 0.432/18=0.024 mol
0.008 mol ===== 0.024 mol
1 mol
===== ??
, nwater =3 mol in one mole of hydrous
Not: For experiment number , For calculation number , For the solution of
answers .