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What is the molecular formula of a compound if its empirical formula is CFBrO and its molar mass is 381.01 g/mol?

A.CFBrO

B.C2F2Br2O2

C.C3F3Br3O3

D.C2F3Br2O3

E.C4F4Br4O HELP ME PLEASE

Verified Answer :

the answer would be C3F3Br3O3

Answer:  C. \(C_3F_3Br_3O_3\)

Explanation:

Molecular formula is the chemical formula which depicts the actual number of atoms of each element present in the compound.

Empirical formula is the simplest formula which depicts the whole number of atoms of each element present in the compound.

Given empirical formula: \(CFBrO\)

Thus empirical mass= 12+19+80+16= 127 g

Given : Molar mass= 381.01 g

Now we have to calculate the molecular formula.

\(n=\frac{\text{Molecular weight of metal}}{\text{Equivalent of metal}}\)

\(n=\frac{381.01g/mole}{127g/eq}=3\)

Thus Molecular formula is :\(3\times CFBrO=C_3F_3Br_3O_3\)

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