Answer:
B) Na3BO3.
Explanation:
Hello!
In this case, since percent compositions are used to identify the empirical formula of an unknown compound, we can assume we have 54.0 g of sodium, 8.50 g of boron and 37.5 g of oxygen, and we compute the moles of each one:
[tex]n_{Na}=54.0g*\frac{1mol}{23g} =2.35mol\\\\n_B=8.50g*\frac{1mol}{11g}=0.773mol\\\\n_O=37.5g*\frac{1mol}{16g}=2.34mol[/tex]
Now, we divide by the moles of boron as those are the fewest:
[tex]Na:\frac{2.35}{0.773}=3\\\\B:\frac{0.773}{0.773}=1\\\\O=\frac{2.34}{0.773}=3[/tex]
Thus, the empirical formula is B) Na3BO3.
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