I need to calculate the actual perce t in unknow hydrate. the combination of the unknow is CuSO4•2h2o and FeSO4•7h2o.the intial mass of crucible and unknow was 25.43g.After it dehydrate was 23 .75 g and the crucible mass by itself was 21.11g.

Initial mass of crucible plus hydrate = 25.43g

Mass of crucible alone = 21.11g

Thus, Mass of hydrate only = 25.43g - 21.11g = 4.32g

Dehydrated mass of crucible plus hydrate = 23.75g

Thus, dehydrated mass of crucible plus hydrate = 23.75g - 21.11g = 2.64g

Thus, weight of water was = 4.32g - 2.64g = 1.68g

Thus water as a % of mass was = 1.68/4.32 = 38.889%

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