This problem is providing us with the mass of a sample containing osmium and oxygen and the mass of former, so the empirical formula of this substance is required and found to be OsO₄ after the following.
Empirical formulas
In chemistry, chemicals have specific formulas showing the number and type of atoms present, via the elements' symbols and subscripts in the formula. In this case, since this compound contains 2.16 g of osmium, one can calculate the mass of oxygen with:
[tex]m_O=2.89g-2.16g=0.73g[/tex]
In such a way, one can then calculate the moles of the both of the elements:
[tex]n_{Os}=2.16g/(190.23g/mol)=0.011355mol\\\\n_O=0.73g/(16.0g/mol)=0.045625mol[/tex]
Then, we divide by the fewest moles to get their subscripts in the formula:
[tex]Os:\frac{0.011355}{0.011355}=1 \\\\O=\frac{0.045625}{0.011355}=4[/tex]
Which means the oxide has a formula of OsO₄.
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