determine empirical formula of a compound with O=24 percent and water molecule=45.4percent and fe=20 percent and s=11.5percent.
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Firstly, Divide the constituent by their molecular masses or atomic masses
O=24/16=1.5
H20=45.4/18=2.5
Fe=20/56=0.3
S=11.5/32=0.3
Divide these values by the minimum of the above obtained value
And round up to nearest unit places
O=1
H2O=8
Fe=1
S=1
So the empirical formula would be FeOS.8H2O
Hope this helps
O=24/16=1.5
H20=45.4/18=2.5
Fe=20/56=0.3
S=11.5/32=0.3
Divide these values by the minimum of the above obtained value
And round up to nearest unit places
O=1
H2O=8
Fe=1
S=1
So the empirical formula would be FeOS.8H2O
Hope this helps
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