Question

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

Answer 1

Solution:

Mass of Boron = 10.000 g

Atomic mass of Boron = 10.811 g mol-1

Thus,

Number of moles = mass./atomic mass

= 10.000 g / 10.811 g mol-1 = 0.925 mol

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

Atomic mass of hydrogen = 1.00784 g mol-1

Number of moles = 1.554 g / 1.00784 g mol-1 = 1.542 mol

Therefore, divide the number of moles by lowest number of mole will give the number of atoms.

Number of Boron = 0.925 / 0.925 = 1

Number of H = 1.542 / 0.925 = 1.67

For converting to whole number, multiply by 3, we get,

Number of Boron = 1 x 3 = 3

Number of Hydrogen = 1.67 x 3 = 5

Hence, empirical formula of compound = B3H5