Question

What is the empirical formula of a compound found to contain 13.5g Ca, 10.8g O, and 0.675g H? (Molar masses: Ca-40.08g/mol, H-1.008g/mol, O-16.00g/mol)

Answer 1

The empirical formula of a compound Ca_xH_yO_z can be found from the ratios of the number of the moles of calcium, hydrogen and oxygen. The number of the moles of calcium is:

$n(Ca) = \frac{m(Ca)}{M(Ca)} = \frac{13.5 \text{ g}}{40.08 \text{ g/mol}} = 0.337$ mol.

The number of the moles of hydrogen is:

$n(H) = \frac{m(H)}{M(H)} = \frac{0.675 \text{ g}}{1.008 \text{ g/mol}} = 0.670$ mol.

The number of the moles of oxygen is:

$n(O) = \frac{m(O)}{M(O)} = \frac{10.8\text{ g}}{16.00 \text{ g/mol}} = 0.675$ mol.

The smallest value of the number of the moles is that of calcium, 0.337 mol. Therefore, we must divide each value of the number of the moles by 0.337 mol:

0.337/0.337 = 1

0.670/0.337 = 2

0.675/0.337 = 2.

Thus, the simplest empirical formula of the compound is CaO₂H₂, or Ca(OH)₂. This compound really exists, it is calcium hydroxide, traditionally called slaked lime.

Answer: the simplest empirical formula of the compound is Ca(OH)₂.