Question

find the empirical formula of a compound formed when 6.75 g of aluminium reacts with 26.63g of chlorine [A:AL,27.0,CL,35.5 ]​

Answer 1

The empirical formula would be AlCl₃

Step-by-step explanation:

The ratio of number of moles of each reacted specie can help to determine the empirical formula

moles of Al

`mole = (given mass (g))/(molecular mass (g/mole))`

`mole = (6.75 g)/(27.0 g /mole)`

`moles of aluminium = 0.250`

moles of Cl

`mole = (given mass (g))/(molecular mass (g/mole))`

`mole = (26.63 g)/(35.5 g /mole)`

`moles of chlorine = 0.750`

Ratio of moles

`ratio = (moles of Cl)/(moles of Al)`

`ratio = (0.750)/(0.250)`

`ratio = 3`

Empirical formula

Calculation shows that the number of moles of chlorine are three times higher than aluminium, hence empirical formula would be AlCl₃