A hydrated salt contains Cu = 25.50% , S =12.90%, O = 25.60% and the remaining water of crystallization. Calculate the emperical formula of the salt.
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Percentage of water of crystallization = 100 – (25.5+12.9+25.6) = 36 %
We can consider these percentages as masses
Moles of Cu = 25.5/63.5 = 0.4
Moles of S = 12.9/32 = 0.4
Moles of O = 25.6/16 = 1.6
Moles of water = 36/18 = 2
Mole ratios (We divide by the smallest)
Moles of Cu = 25.5/63.5 = 0.4/0.4 = 1
Moles of S = 12.9/32 = 0.4/0.4 = 1
Moles of O = 25.6/16 = 1.6/0.4 = 4
Moles of water = 36/18 = 2/0.4 = 5
The empirical formula is:
CuSO4.5H2O
We can consider these percentages as masses
Moles of Cu = 25.5/63.5 = 0.4
Moles of S = 12.9/32 = 0.4
Moles of O = 25.6/16 = 1.6
Moles of water = 36/18 = 2
Mole ratios (We divide by the smallest)
Moles of Cu = 25.5/63.5 = 0.4/0.4 = 1
Moles of S = 12.9/32 = 0.4/0.4 = 1
Moles of O = 25.6/16 = 1.6/0.4 = 4
Moles of water = 36/18 = 2/0.4 = 5
The empirical formula is:
CuSO4.5H2O
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