An oxide of tungsten (symbol W) is a bright yellow solid. If 5.34 g of the compound contains 4.23 g of tungsten, what is its empirical formula?
Solution:
We will first calculate the moles of tungsten (W) and oxygen (O).
\text{Mol W} = 4.23\,g\text{ W}\times \frac{1\text{mol W}}{183.84\, g\text{ W}}
\text{Mol W} =0.02301\text{mol}
\text{Mol O} = (5.34-4.23)\,g\, \text{O}\times \frac{1\text{mol O}}{16.00\, g\, \text{O}}
\text{Mol O} =0.06938\text{mol}
Now, we will divide each by the smaller number of moles, which will lead us to an integer value for the empirical formula.
\text{W}=\frac{0.02301}{0.02301}=1.0
\text{O}=\frac{0.06938}{0.02301}=3.0
With this result, we can safely say that the empirical formula is \text{WO}_3
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