Answer:
[tex]Sn_2O[/tex]
Explanation:
Hello,
In this case, given the mass of the sample and mass of tin we can compute the mass of oxygen via:
[tex]m_O=0.534g-0.500g=0.034g[/tex]
Thus, by using the atomic bas of tin and oxygen we can compute their moles:
[tex]n_{Sn}=0.500gSn*\frac{1molSn}{118.8gSn} =0.00421mol\\\\n_O=0.034gO*\frac{1molO}{16gO}=0.002125mol[/tex]
Next, we need to divide both moles by the moles of oxygen as those are the smallest in order to compute the subscript in the chemical reaction:
[tex]Sn=\frac{0.00421}{0.002125}=2\\ \\O=\frac{0.002125}{0.002125}= 1[/tex]
Therefore, empirical formula of the oxide should be:
[tex]Sn_2O[/tex]
Best regards.
