Question

What is the empirical formula of a compound that is 14 g calcium, 11 g oxygen, and 0.7 g hydrogen?
A. Ca2OH2
B. Ca(OH)2
C. CaO2H
D. CaOH

Answer 1

We are given with the mass of the elements. Taking a 100-gram basis, we convert the mass of the reactants to moles. Calcium is equal to 0.35 moles , oxygen is equal to 0.6875 and hydrogen is equal to 0.7 moles. We divide each number to the less amount. Hence Ca is 1, O is 1.96 and H is 2. Answer is B. Ca(OH)2, calcium hydroxide.

Answer 2

Answer : The correct option is, (B) $Ca(OH)_2$

Solution : Given,

Mass of calcium = 14 g

Mass of oxygen = 11 g

Mass of hydrogen = 0.7 g

Molar mass of calcium = 40 g/mole

Molar mass of oxygen = 16 g/mole

Molar mass of hydrogen = 1 g/mole

First we have to calculate the moles of calcium oxygen and hydrogen.

$\text{Moles of Ca}=\frac{\text{ Given mass of Ca}}{\text{ Molar mass of Ca}}=\frac{14g}{40g/mole}=0.35moles$

$\text{Moles of O}=\frac{\text{ Given mass of O}}{\text{ Molar mass of O}}=\frac{11g}{16g/mole}=0.68moles$

$\text{Moles of H}=\frac{\text{ Given mass of H}}{\text{ Molar mass of H}}=\frac{0.7g}{1g/mole}=0.70moles$

Now for the mole ratio, divide each value of moles by the smallest number of moles calculated.

$\text{For Ca}=\frac{0.35}{0.35}=1$

$\text{For O}=\frac{0.68}{0.35}=1.9\approx 2$

$\text{For Ca}=\frac{0.70}{0.35}=2$

The ratio of C : O : H = 1 : 2 : 2

The mole ratio of the element is represented by the subscripts in empirical formula.

The Empirical formula = $Ca_1O_2H_2$ or $Ca(OH)_2$

Therefore, the empirical formula of a compound is, $Ca(OH)_2$