Answer:
2)
[tex]\% Mg=20.2\%\\\\\% S=26.6\%\\\\\% O=53.2\%[/tex]
3)
[tex]\% Ag=93.1\%\\\\\% O=6.9\%[/tex]
Explanation:
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2) In this case, since magnesium sulfate is MgSO₄, we can see how magnesium weights 24.305 g/mol, sulfur 32.06 g/mol and oxygen 64.00 g/mol as there is one atom of magnesium as well as sulfur but four oxygen atoms for a total of g/mol; thus the percent compositions are:
[tex]\% Mg=\frac{24.305}{120.36 } *100\%=20.2\%\\\\\% S=\frac{32.06}{120.36 } *100\%=26.6\%\\\\\% O=\frac{64.00}{120.36 } *100\%=53.2\%[/tex]
3) In this case, although the element seems to contain Ag and O, we infer its molecular formula is Ag₂O; thus, since we have two silver atoms weighing 215.74 g/mol and one oxygen atom weighing 16.00 g/mol for a total of 231.74 g/mol, we obtain the following percent compositions:
[tex]\% Ag=\frac{215.74}{231.74} *100\%=93.1\%\\\\\% O=\frac{16.00}{231.74} *100\%=6.9\%[/tex]
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