Question

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

Answer 1

Answer: The empirical formula is $NaBO_2$

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 =$\frac{\text{ given mass of Na}}{\text{ molar mass of Na}}= \frac{34.5g}{23g/mole}=1.5moles$

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

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

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

For Na = $\frac{1.5}{1.5}=1$

For B = $\frac{1.5}{1.5}=1$

For O =$\frac{3}{1.5}=2$

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

Hence the empirical formula is $NaBO_2$