Question

What is the empirical formula of a compound that contains 3.609 g calcium and 6.384 g chlorine by mass?

Answer 1

Answer: The empirical formula for the given compound is
`CaCl_2`

Step-by-step explanation:

We are given:

Mass of calcium = 3.609 g

Mass of chlorine = 6.384 g

To find the empirical formula of the compound, we must follow some steps:

Step 1: Converting the given masses into moles.

Moles of Ca =
`\frac{\text{Given mass of Ca}}{\text{Molar mass of Ca}}=(3.609g)/(40.08g/mole)=0.09moles`

Moles of Cl =
`\frac{\text{Given mass of Cl}}{\text{Molar mass of Cl}}=(6.384g)/(35.45g/mole)=0.18moles`

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, divide each value of moles by the smallest number of moles calculated that is 0.09

For Ca =
`(0.09)/(0.09)=1`

For Cl =
`(0.18)/(0.9)=2`

Step 3: Taking the mole ratio as their subscripts.

The ratio of Ca : Cl = 1 : 2

Hence, the empirical formula is
`CaCl_2`

Answer 2

For a compound that contains 3.609g Ca; we shall calculate the number of moles as follows
No of moles = mass/molar mass = 3.609/40 = 0.090
Also for chlorine we have 6.384/35.4 = 0.19
So we divide by the smallest ratio (0.090)
I. e 0.090/0.090 and 0.19/0.090 this gives a ratio of 1 : 2. Hence the formula is Cacl2