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}}=(14g)/(40g/mole)=0.35moles`

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

`\text{Moles of H}=\frac{\text{ Given mass of H}}{\text{ Molar mass of H}}=(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}=(0.35)/(0.35)=1`

`\text{For O}=(0.68)/(0.35)=1.9\approx 2`

`\text{For Ca}=(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`