Question

A substance contains 36.1 percent calcium and 63.9 percent chlorine by weight. What is the empirical formula of the given compound?

Answer 1

A(Ca)=40.08
A(Cl)=35.45
w(Ca)=0.361
w(Cl)=0.639

CaxCly
Solve the system of two equations:
40.08x/(40.08x+35.45y)=0.361
35.45y/((40.08x+35.45y)=0.639

x=1
y=2
CaCl₂ - calcium chloride

Answer 2

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

Step-by-step explanation:

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

Step 1: Converting the given percentages into mass

Total mass of the compound is taken as 100 g.

So, the mass of each element is equal to the percentage given.

Mass of Calcium = 36.1 g

Mass of Chlorine = 63.9

Step 2: Converting the given masses into moles.

Moles of Ca =
`\frac{\text{Given mass of Ca}}{\text{Molar mass of Ca}}=(36.1g)/(40.08g/mole)=0.9moles`

Moles of Cl =
`\frac{\text{Given mass of Cl}}{\text{Molar mass of Cl}}=(63.9g)/(35.45g/mole)=1.80moles`

Step 3: 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 0.9

For Ca =
`(0.9)/(0.9)=1`

For Cl =
`(1.8)/(0.9)=2`

Step 4: Taking the mole ratio as their subscripts.

The ratio of Ca : Cl = 1 : 2

Hence, the empirical formula of the given compound is
`CaCl_2`