Question

What is the empirical formula for 22.1% Al, 25.4% P, 52.5% O​

Answer 1

Answer:
`Al_(1)P_(1)O_(4)`

Solution : Given,

If percentage are given then we are taking total mass is 100 grams.

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

Mass of Al = 22.1 g

Mass of P = 25.4 g

Mass of O = 52.5 g

Step 1 : convert given masses into moles.

Moles of Al =
`\frac{\text{ given mass of Al}text{ molar mass of Al}}= (22.1g)/(27g/mole)=0.8moles`

Moles of P = \frac{\text{ given mass of P}}{\text{ molar mass of P}}= \frac{25.4g}{31g/mole}=0.8moles [/tex]

Moles of O = \frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{52.5g}{16g/mole}=3.2moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Al =
`(0.8)/(0.8)=1`

For P =
`(0.8)/(0.8)=1`

For O =
`(3.2)/(0.8)=4`

The ratio of Al: P : O= 1 : 1 : 4

Hence the empirical formula is
`Al_(1)P_(1)O_(4)`