Question

Please help

Answer 1

Answer: The empirical formula is $MgSO_3$

Explanation:

Mass of Mg= 46.6 g

Mass of S= 61.4 g

Mass of O = 92.0 g

Step 1 : convert given masses into moles.

Moles of Mg =$\frac{\text{ given mass of Mg}}{\text{ molar mass of Mg}}= \frac{46.6g}{24g/mole}=1.942moles$

Moles of S =$\frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{61.4g}{32g/mole}=1.918moles$

Moles of O =$\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{92.0g}{16g/mole}=5.75moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Mg = $\frac{1.942}{1.918}=$

For S= $\frac{1.918}{1.918}=1$

For O =$\frac{5.75}{1.918}=3$

The ratio of Mg: S: O = 1: 1: 3

Hence the empirical formula is $MgSO_3$