Convert the following per cent composition into molecular formula.
Diborane $-$ $H : 21.86 %, B : 78.14 %$ .

B_{2} H_{6}

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B_{8} H_{24}

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B_{4} H_{12}

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All of the above

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Solution

The correct option is D All of the above
According to given conditions,
% % $\frac{w t .}{m o l . w t .}$ simple ratio empirical formula
$B$ $78.14$ $\frac{78.14}{11}$ $7$
$H$ $21.86$ $\frac{21.86}{1}$ $21$ $B H_{3} = 14 g$
As molar mass is $28$ g, the compound is $B_{2} H_{6}$

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Convert the following per cent composition into molecular formula.Diborane−H:21.86%,B:78.14%.

The correct option is D All of the aboveAccording to given conditions, % %wt.mol.wt. simple ratio empirical formulaB 78.14 78.1411 7H 21.86 21.861 21 BH3=14gAs molar mass is 28 g, the compound is B2H6

Convert the following per cent composition into molecular formula.
Diborane $-$ $H : 21.86 %, B : 78.14 %$ .

The correct option is D All of the above
According to given conditions,
% % $\frac{w t .}{m o l . w t .}$ simple ratio empirical formula
$B$ $78.14$ $\frac{78.14}{11}$ $7$
$H$ $21.86$ $\frac{21.86}{1}$ $21$ $B H_{3} = 14 g$
As molar mass is $28$ g, the compound is $B_{2} H_{6}$