1.615 g of an anhydrous salt was placed in moist air . After a few days it was found to be 2.875 g . Assuming that complete hydration has taken place , calculate the simplest formula of hydrated salt . The composition of anhydrous salt is Zn=40.6%;S=19.8%;O=39.6%.
Asked by Ujjwal Kumar
| 6th Oct, 2010,
10:04: AM
Expert Answer:
symbol
weight
Molecular weight
moles
Ratios of moles
Simplest whole no. ratio
Zn
40.6 g
65.3
40.6/65.3=.621
.62/.61
1
S
19.8 g
32
19.8/32=.618
.61/.61
1
O
39.6
16
39.6/16=2.475
2.475/.61
4
Hence, empirical formula is ZnSO4
Weight of ZnSO4 .XH2O(Hydrated salt)/weight of ZnSO4(Anhydrous salt)=Molecular weight of ZnSO4.XH2O/Molecular weight of ZnSO4
(2.875)/1.615=(161.3+18x)/161.3
x=6.99=7.00
hence, x=7
so we can write it as ZNSO4.7H2O
symbol |
weight |
Molecular weight |
moles |
Ratios of moles |
Simplest whole no. ratio |
Zn |
40.6 g |
65.3 |
40.6/65.3=.621 |
.62/.61 |
1 |
S |
19.8 g |
32 |
19.8/32=.618 |
.61/.61 |
1 |
O |
39.6 |
16 |
39.6/16=2.475 |
2.475/.61 |
4 |
Hence, empirical formula is ZnSO4
Weight of ZnSO4 .XH2O(Hydrated salt)/weight of ZnSO4(Anhydrous salt)=Molecular weight of ZnSO4.XH2O/Molecular weight of ZnSO4
(2.875)/1.615=(161.3+18x)/161.3
x=6.99=7.00
hence, x=7
so we can write it as ZNSO4.7H2O
Answered by
| 6th Oct, 2010,
11:12: PM
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