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

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

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RomeliaThurston RomeliaThurston · Answer 2 · 2019-05-14T08:59:17+02:00

Answer: The empirical formula for the given compound is [tex]CaCl_2[/tex]

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 = [tex]\frac{\text{Given mass of Ca}}{\text{Molar mass of Ca}}=\frac{3.609g}{40.08g/mole}=0.09moles[/tex]

Moles of Cl = [tex]\frac{\text{Given mass of Cl}}{\text{Molar mass of Cl}}=\frac{6.384g}{35.45g/mole}=0.18moles[/tex]

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 = [tex]\frac{0.09}{0.09}=1[/tex]

For Cl = [tex]\frac{0.18}{0.9}=2[/tex]

Step 3: Taking the mole ratio as their subscripts.

The ratio of Ca : Cl = 1 : 2

Hence, the empirical formula is [tex]CaCl_2[/tex]