Question

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

Answer 1

To determine the empirical formula of the boron-hydrogen compound, the mass of hydrogen is found by subtraction and then both elements are converted to moles. The empirical formula is found by creating a ratio of the moles of boron to hydrogen, which simplifies to BH2.

Step-by-step explanation:

To determine the empirical formula of the boron and hydrogen compound, let's first find the mass of hydrogen in the compound by subtracting the mass of boron from the total mass of the compound:

Mass of hydrogen = Total mass of compound - Mass of boron

Mass of hydrogen = 11.554g - 10.000g = 1.554g

Now, we will convert the masses of boron and hydrogen to moles using their atomic masses:

Atomic mass of boron (B) = 10.8 amu (approximately)

Atomic mass of hydrogen (H) = 1.008 amu (approximately)

Moles of boron = Mass of boron / Atomic mass of boron

Moles of boron = 10.000g / 10.8g/mol ≈ 0.926 moles

Moles of hydrogen = Mass of hydrogen / Atomic mass of hydrogen

Moles of hydrogen = 1.554g / 1.008g/mol ≈ 1.542 moles

To find the simplest ratio of boron to hydrogen, we divide the moles of each element by the smallest number of moles:

Ratio of boron to hydrogen = Moles of boron : Moles of hydrogen

Ratio of boron to hydrogen ≈ 0.926 : 1.542

Since dividing both numbers by 0.926 gives approximately 1 : 1.666, this ratio simplifies to about 1:2 when rounded to the nearest whole number. Therefore, the empirical formula of the compound is BH2.

Answer 2

B3H5

Step-by-step 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