Question

Write the empirical formula for at least four ionic compounds that could be formed from the following ions: Fe2+, Fe3+, CroC,H,02

Can anyone explain why it's FeCrO4 instead of Fe2(CrO4)2?

please explain more details as possible, i will give upvote.

Answer 1

Answer:-

As we know that the empirical formula is the formula of a compound in which simple whole number ratio of atoms of various elements contained whereas molecular formula is the formula of a compound which gives the actual number of the atoms of various elements present in one molecule of the compound.

Also we know that there is following relationship between empirical formula and molecular formula of the compound :-

molecular formula mass of the compound = n $\times$ empirical formula mass of the compound

or

n = molecular formula mass of the compound / empirical formula mass of the compound

where

n = whole integers such as 1, 2, 3,4 ..........................

the e.g. empirical formula and molecular formula of some compounds are given below :-

empirical formula of hydrogen peroxide is HO which represents the hydrogen (H) and oxygen (O) ratio is 1:1 present in hydrogen peroxide whereas molecular formula of hydrogen peroxide is H₂O₂ which represents actual number of hydrogen (H) and oxygen (O) atoms present in one molecule of the hydrogen peroxide.

molecular formula mass of the hydrogen peroxide = n $\times$ empirical formula mass of the hydrogen peroxide

n = molecular formula mass of the hydrogen peroxide / empirical formula mass of the hydrogen peroxide

n = molecular formula mass of the hydrogen peroxide (H₂O₂) / empirical formula mass of the hydrogen peroxide(HO)

n = 34 / 17

n = 2

similarly

empirical formula of benzene is CH which represents the carbon (C) and hydrogen (H) ratio is 1:1 present in benzene whereas molecular formula of benzene is C₆H₆ which represents actual number of carbon (C) and hydrogen (H) atoms present in one molecule of the benzene.

molecular formula mass of the benzene = n $\times$ empirical formula mass of the benzene

n = molecular formula mass of the benzene / empirical formula mass of the benzene

n = molecular formula mass of the benzene (C₆H₆) / empirical formula mass of the benzene(CH)

n = 78 / 13

n = 6

similarly

empirical formula of glucose is CH₂O which represents the carbon (C) , hydrogen (H) and oxygen (O) ratio is 1:2:1 present in glucose whereas molecular formula of benzene is C₆H₁₂O₆ which represents actual number of carbon (C) hydrogen (H) and oxygen (O) atoms present in one molecule of the glucose.

molecular formula mass of the glucose = n $\times$ empirical formula mass of the glucose

n = molecular formula mass of the glucose / empirical formula mass of the glucose

n = molecular formula mass of the glucose (C₆H₁₂O₆) / empirical formula mass of the glucose (CH₂O)

n = 180 / 30

n = 6

Therefore the empirical formula of iron (II) chromate is FeCrO₄ which represents the iron (Fe) , chromium (Cr) and oxygen (O) ratio is 1:1:4 present in iron (II) chromate whereas molecular formula of iron (II) chromate is FeCrO₄  which represents actual number of carbon (C) hydrogen (H) and oxygen (O) atoms present in one molecule of the iron (II) chromate.

molecular formula mass of the iron (II) chromate = n $\times$ empirical formula mass of the iron (II) chromate

n = molecular formula mass of the iron (II) chromate / empirical formula mass of the iron (II) chromate

n = molecular formula mass of the iron (II) chromate (FeCrO₄) / empirical formula mass of the iron (II) chromate (FeCrO₄)

n = 171.8 / 171.8

n = 1

So we can say that empirical formula of iron (II) chromate is FeCrO₄ instead of Fe₂(CrO₄)₂ in which minimum simple whole number ratio of iron (Fe) , chromium (Cr) and oxygen (O) is 1:1:4 whereas in Fe₂(CrO₄)₂ in which simple whole number ratio of iron (Fe) , chromium (Cr) and oxygen (O) is 2:2:8 which is not correct as per definition of empirical formula of compound.therefore empirical formula of iron (II) chromate is written as FeCrO₄ .

Another reason for correct empirical formula of iron (II) chromate ( FeCrO₄ ) is that in this formula there is complete balance of opposite charge in compound and there is no need to equate the charge in cation and anion of the iron (II) chromate ( FeCrO₄ ) as mentioned in other empirical formula of iron.

FeCrO₄ = Fe²⁺ +  CrO₄^--

	C2H3O2-	CrO42-
Fe3+	Fe(C2H3O2)3	Fe2(CrO4)3
Fe2+	Fe(C2H3O2)2	FeCrO4