Question

What is the empirical formula of a phosphoric acid that contains 0.3086 g of hydrogen, 3.161 g of phosphorus, and 6.531 g of oxygen?

Answer 1

H:P:O = 0,3086/1 : 3,161/31 : 6,531/16 = 0,3086 : 0,10197 : 0,4082
H:P:O ≈ 3 : 1 : 4

H₃PO₄

Answer 2

Answer: The empirical formula is
`H_3PO_4`

Step-by-step explanation:

Given:

Mass of H = 0.3086 g

Mass of P = 3.161 g

Mass of O = 6.531 g

Step 1 : convert given masses into moles.

Moles of H=
`\frac{\text{ given mass of H}}{\text{ molar mass of H}}= (0.3086g)/(1g/mole)=0.309moles`

Moles of P=
`\frac{\text{ given mass of P}}{\text{ molar mass of P}}= (3.161g)/(31g/mole)=0.102moles`

Moles of O =
`\frac{\text{ given mass of O}}{\text{ molar mass of O}}= (6.531g)/(16g/mole)=0.408moles`

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

For H=
`(0.309)/(0.102)=3`

For P=
`(0.102)/(0.102)=1`

For O =
`(0.408)/(0.102)=4`

The ratio of H:P:O= 3: 1: 4

Hence the empirical formula is
`H_3PO_4`