Respuesta :
Empirical Formula is Fe1S1O4
(the numbers are suppose to be subscript)
Answer: [tex]FeSO_4[/tex]
Solution : Given,
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of Fe = 36.76 g
Mass of S = 21.11 g
Mass of O = 42.13 g
Step 1 : convert given masses into moles.
Moles of Fe =[tex] \frac{\text{ given mass of Fe}}{\text{ molar mass of Fe}}= \frac{36.76g}{56g/mole}=0.65moles[/tex]
Moles of S = \frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{21.11g}{32g/mole}=0.65moles
[/tex]
Moles of O = \frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{42.13g}{16g/mole}=2.63[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For Fe = [tex]\frac{0.65}{0.65}=1[/tex]
For S = [tex]\frac{0.65}{0.65}=1[/tex]
For O =[tex]\frac{2.63}{0.65}=4[/tex]
The ratio of Fe : S : O= 1 : 1 : 4
Hence the empirical formula is [tex]Fe_{1}S_{1}O_4[/tex]