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A 0.025-g sample of a compound composed of Boron and hydrogen with the molecular mass of 28 AMU burns spontaneously when exposed to air, produced 0.063 g of B2O3. What are the empirical and the molecular formulas of the compound.

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Answer:

The empirical formula is BH₃ and molecular formula is B₂H₆

Explanation:

To know the number of borons you must obtain the moles number of the initial and B₂O₃ compound. So:

Moles of initial compound:

0,025 g of X × ( 1 mol / 28 g) = 8,9 × 10⁻⁴ moles

Moles of B₂O₃ compound:

AMU of B₂O₃ = (2 × 10,8 g/mol + 3 × 16 g/mol) = 69,6 g/mol

0,063 g  ( 1 mol B₂O₃ / 69,6 g) = 9,1 × 10⁻⁴ moles

As moles number of initial and final compounds are the same, the number of borons must be equals. So, our compound has two borons.

These two borons weight: 2 × 10,8 g/mol = 21,6 g/mol

If UMA number of our compound is 28 g/mol we need, yet,

28 g/mol - 21,6 g/mol = 6,4 g/mol

These 6,4 g/mol comes from hydrogen that weights 1 g/mol. So, we have 6 hydrogens.

Thus, the molecular formula is B₂H₆

The empirical formula is the simplest way to represent the atoms of a chemical compound. If we divide the molecular formula in two, we will obtain the empirical formula: BH₃

I hope it helps!