Calculate the percentage composition in Fe2(SO4)3 (3M)
(Fe=56, S=32, O=16)
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Answer:
Explanation:
To find the percentage of oxygen in ferric sulphate, we need to find molar mass of it. Then the percentage can be found.
Molar mass of
= 56 × 2 + (32 + 16 ×4) × 3
(since, molar mass of Fe = 56 g , S = 32 g and O = 16 g)
= 400 g
Now, total molar mass of O in
= 16 × 4 × 3
= 192 g
So, percentage of oxygen in ferric sulphate will be
⇒ 192/400 × 100
⇒ 192/4
= 48 %
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