Question

A compound is found to consist of 34.5% sodium, 16.4% boron, and 48.6% oxygen. What is the empirical formula?

A. NaBO3
B. NaBO2
C. Na2B4O7
D. Na3BO3

Answer 1

Answer: The empirical formula is
`NaBO_2`

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 Na= 34.5 g

Mass of B= 16.4 g

Mass of O = 48.6 g

Step 1 : convert given masses into moles.

Moles of Na =
`\frac{\text{ given mass of Na}}{\text{ molar mass of Na}}= (34.5g)/(23g/mole)=1.5moles`

Moles of B =
`\frac{\text{ given mass of B}}{\text{ molar mass of B}}= (16.4g)/(11g/mole)=1.5moles`

Moles of O =
`\frac{\text{ given mass of O}}{\text{ molar mass of O}}= (48.6g)/(16g/mole)=3moles`

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

For Na =
`(1.5)/(1.5)=1`

For B =
`(1.5)/(1.5)=1`

For O =
`(3)/(1.5)=2`

The ratio of Na: B: O= 1: 1: 2

Hence the empirical formula is
`NaBO_2`