10.000g of boron (B) combines with hydrogento form 11.554g of a pure compound. What is the empirical formula of this compound?

B3H5

Explanation:

The law of conservation of mass states that matter in an closed system is neither created nor destroyed by physical transformations or chemical reactions but changes from one form to the other.

That is, the sum of masses of the reactants = The sum of masses of the product

10.00g of Boron + x grams of Hydrogen = 11.55g of the product

Mass of hydrogen = 11.55 - 10.00 = 1.55g

Molar mass of Boron = 10.811g

Molar mass of Hydrogen = 1.00784g

Number of moles of Boron = (mass of Boron) / (molar mass of Boron) = 10/10.811 = 0.9249 mols

Number of moles of Hydrogen = (mass of Hydrogen) / (molar mass of Hydrogen) = 1.55/1.00784 = 1.5379mols

0.9249 mols of Boron combines with 1.5379mols of Hydrogen

Dividing both sides mols by 0.9249 gives

1 mole of Boron combines with 1.66266 mols of Hydrogen

Converting 1.66266 to fractions we have 1.66266 approximately 5/3

or 1 mole of Boron combines with 5/3 moles of Hydrogen

Multiplying both sides by 3 we have

3 moles of Boron combines with 5 moles of Hydrogen

Molecular formula of the compound is

B3H5