Answer: The empirical formula is [tex]MgSO_3[/tex]
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 =[tex]\frac{\text{ given mass of Mg}}{\text{ molar mass of Mg}}= \frac{46.6g}{24g/mole}=1.942moles[/tex]
Moles of S =[tex]\frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{61.4g}{32g/mole}=1.918moles[/tex]
Moles of O =[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{92.0g}{16g/mole}=5.75moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For Mg = [tex]\frac{1.942}{1.918}=[/tex]
For S= [tex]\frac{1.918}{1.918}=1[/tex]
For O =[tex]\frac{5.75}{1.918}=3[/tex]
The ratio of Mg: S: O = 1: 1: 3
Hence the empirical formula is [tex]MgSO_3[/tex]
