Question

A compound is found to contain 2.270 % hydrogen, 34.80 % phosphorus, and 62.93 % oxygen by mass. What is the empirical formula for this compound

Answer 1

Answer: The empirical formula is
`H_4P_2O_7`

Step-by-step explanation:

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 H= 2.270 g

Mass of P = 34.80 g

Mass of O = 62.93 g

Step 1 : convert given masses into moles

Moles of H =
`\frac{\text {given mass}}{\text {Molar mass}}=(2.270g)/(1g/mol)=2.270`

Moles of P =
`\frac{\text {given mass}}{\text {Molar mass}}=(34.80g)/(31g/mol)=1.122`

Moles of O =
`\frac{\text {given mass}}{\text {Molar mass}}=(62.93g)/(16g/mol)=3.933`

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

For H =
`(2.270)/(1.122)=2`

For P=
`(1.122)/(1.122)=1`

For O =
`(3.933)/(1.122)=3.5`

The simples ratio will be = H: P : O= 4: 2: 7

Hence the empirical formula is
`H_4P_2O_7`