The empirical formula of a boron compound is ${BH}_{3}$ . What will be the molecular formula of the compound if its measured molecular mass is 27.66 g?

B_{3} H_{9}

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B_{2} H_{4}

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B_{2} H_{6}

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B_{3} H_{6}

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Solution

The correct option is C $B_{2} H_{6}$
Given:
Empirical formula of the compound = BH3
Molecular formula
= n x Empirical formula
Thus, n =
Molecular massEmpirical mass
Hence, Empirical mass
= 10.81 + 3 x 1 = 13.81 g
Therefore, n is almost equal 2.
Hence, Molecular formula would be 2( ${BH}_{3}$ ) = $B_{2} H_{6}$

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The empirical formula of a boron compound is BH3. What will be the molecular formula of the compound if its measured molecular mass is 27.66 g?

The correct option is C B2H6Given: Empirical formula of the compound = BH3 Molecular formula = n x Empirical formula Thus, n = Molecular massEmpirical mass Hence, Empirical mass = 10.81 + 3 x 1 = 13.81 g Therefore, n is almost equal 2. Hence, Molecular formula would be 2(BH3) = B2H6

The empirical formula of a boron compound is ${BH}_{3}$ . What will be the molecular formula of the compound if its measured molecular mass is 27.66 g?