2.128 g Be, 7.557 g S, 15.107 g O

We need to get the moles of each element in the compound.

The atomic masses are 9.012 g/mol Be, 32.06 g/mol S, and 16.00 g/mol O.

$\mathbf{2}\mathbf{.}\mathbf{128}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{Be}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}\mathbf{Be}}{\mathbf{9}\mathbf{.}\mathbf{012}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{Be}}}$ **= 0.2361 mol Be**

$\mathbf{7}\mathbf{.}\mathbf{557}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{S}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}\mathbf{S}}{\mathbf{32}\mathbf{.}\mathbf{06}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{S}}}$** = 0.2357 mol S**

$\mathbf{15}\mathbf{.}\mathbf{107}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{O}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}\mathbf{O}}{\mathbf{16}\mathbf{.}\mathbf{00}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{O}}}$ **= 0.9442 mol O**

Divide the number of moles of each by the smallest value.

A chemist decomposes samples of several compounds; the masses of their constituent elements are listed. Calculate the empirical formula for each compound.

2.128 g Be, 7.557 g S, 15.107 g O

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