A compound has the following composition Mg = 9.76%,S = 13.01%, 0
= 26.01, H2O = 51.22, what is its empirical formula?
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Answered by
103
where the atomic mass of Mg is 24/ O is 16/Su is 32 and for water is 18,using atomic mass v can find the mass ratio
9.76/24=0.406
26.01/16=1.625
13.01/32=0.406
51.22/18=2.846
Then divide the mass ratio with the simple ratio/ where simple ratio value is 0.406 /now v can find the moles using that v can find the empirical formula
the empirical formula for this is MgSO₄.7H₂O
9.76/24=0.406
26.01/16=1.625
13.01/32=0.406
51.22/18=2.846
Then divide the mass ratio with the simple ratio/ where simple ratio value is 0.406 /now v can find the moles using that v can find the empirical formula
the empirical formula for this is MgSO₄.7H₂O
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65
Answer:
Explanation:the required empirical formula for this question is MgSO4.7H2O.
And if you want to calculate molecular formula :-
n×E.F = M.F
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