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!
