Answer:
[tex]HgO[/tex]
Explanation:
Hello,
In this case, by knowing the mass of the oxide and the mass of the element mercury, one obtains its percentage as follows:
[tex]\% Hg=\frac{0.189g}{0.204g}*100\%=92.6\%\\[/tex]
Consequently, oxygen's percentage is:
[tex]\% O=100\%-92.6\%=7.4\%[/tex]
Now, the moles of each element for a 100-g sample:
[tex]n_{Hg}=92.6gHg*\frac{1molHg}{200.59gHg} =0.462molHg\\n_{O}=7.4gO*\frac{1molO}{16gO}=0.462molO[/tex]
In such a way, the subscripts in the formula are given by:
[tex]Hg=\frac{0.462mol}{0.462mol}=1\\ O=\frac{0.462mol}{0.462mol}=1[/tex]
Therefore, the empirical formula turns out:
[tex]HgO[/tex]
Best regards.
