If a 1.271-g sample of aluminum metal is heated in a chlorine gas atmosphere,
the mass of aluminum chloride produced is 6.280 g. calculate the empirical
formula of the aluminum chloride.

AlCl₃

Explanation:

Data Given:

Mass of aluminum metal = 1.271 g

Mass of aluminum chloride = 6.280 g

Empirical formula of aluminum chloride = ?

Solution:

First find mass of Chlorine

As 6.280 g of aluminum chloride produced by 1.271 g so

the mass of chlorine in 6.280 g will be 6.280 g -1.271 g)

Mass of chlorine = 5.009 g

Now

Find the number of moles of Al and chlorine in aluminum chloride.

Molar Mass of Al = 26.98 g/mole

Molar mass of Cl =  35.5 g/mol

Mole of Al

Formula Used

                no. of moles = mass in grams / molar mass . . . . . . . . (1)

Put values in equation 1

             no. of moles of Al = 1.271 g / 26.98 g/ mole

             no. of moles of Al = 0.0471

Mole of Chlorine

Formula used

                no. of moles = mass in grams / molar mass . . . . . . . . (1)

Put values in equation 1

             no. of moles of Cl = 5.009 g / 35.5 g/ mole

             no. of moles of Cl = 0.1411

Now we have

Al = 0.0471 moles

Cl = 0.1411 moles

As we Know

Empirical formula shows the simplest ratio of atoms in the molecule but not whole numbers of atoms in a compound.

So,

The ratio of moles of Al to chlorine is

                          Al     :   Cl

                      0.0471     0.1411

Divide the ratio by smallest number to get simplest whole number ratio

                              Al                    :           Cl

                      0.0471 / 0.0471           0.1411/ 0.0471  

                               Al     :   Cl

                                1      :   3

The simplest ratio of Al to cl is 1 to 3, so the formula will be

Emperical formula of  aluminum chloride  = AlCl₃

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