A compound contains equal masses of the elements A, B & C. If the gram atomic masses of A, B & C are 60 g, 20 g & 40 g respectively. Determine the molecular formula of the compound if the molecular mass is 720 g.

A_{3} B_{2} C

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A B_{2} C_{3}

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A_{4} B_{12} C_{6}

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A_{6} B_{3} C_{2}

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Solution

The correct option is C $A_{4} B_{12} C_{6}$
Let individual mass of A, B & C be $x$ gm
Mass Ratio is $A : B : C \Rightarrow x : x : x$
Mole Ratio is $A : B : C \Rightarrow \frac{x}{60} : \frac{x}{20} : \frac{x}{40}$
On simplification it becomes $2 : 6 : 3$
$∴ E m p i r i c a l F o r m u l a b e c o m e s A_{2} B_{6} C_{3}$
Hence empirical formula mass = $2 \times 60 + 6 \times 20 + 3 \times 40 = 360 g$
We know that $M o l e c u l a r F o r m u l a = E m p i r i c a l F o r m u l a \times n$
where, $n = \frac{M o l e c u l a r M a s s}{E m p i r i c a l M a s s} = \frac{720}{360} = 2$
$∴ M o l e c u l a r F o r m u l a = (A_{2} B_{6} C_{3})_{2}$
i.e. $A_{4} B_{12} C_{6}$

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