Question

A compound contains sodium, boron, and oxygen. An experimental analysis gave values of 53.976 % sodium and 8.461 % boron, by weight; the remainder is oxygen. What is the empirical formula of the compound?

Answer 1

Answer : The empirical formula of a compound is,
`Na_3BO_3`

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 Na = 53.976 g

Mass of B = 8.461 g

Mass of O = [100 - (53.976 + 8.461)] = 37.563 g

Molar mass of Na = 23 g/mole

Molar mass of B = 11 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of Na =
`\frac{\text{ given mass of Na}}{\text{ molar mass of Na}}= (53.976g)/(23g/mole)=2.347moles`

Moles of B =
`\frac{\text{ given mass of B}}{\text{ molar mass of B}}= (8.461g)/(11g/mole)=0.769moles`

Moles of O =
`\frac{\text{ given mass of O}}{\text{ molar mass of O}}= (37.563g)/(16g/mole)=2.347moles`

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

For Na =
`(2.347)/(0.769)=3.05\approx 3`

For B =
`(0.769)/(0.769)=1`

For O =
`(2.347)/(0.769)=3.05\approx 3`

The ratio of Na : B : O = 3 : 1 : 3

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula =
`Na_3B_1O_3` =
`Na_3BO_3`