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 answer is c

Explanation:

Answer 2

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 = $\frac{\text{ given mass of Na}}{\text{ molar mass of Na}}= \frac{53.976g}{23g/mole}=2.347moles$

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

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

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

For Na = $\frac{2.347}{0.769}=3.05\approx 3$

For B = $\frac{0.769}{0.769}=1$

For O = $\frac{2.347}{0.769}=3.05\approx 3$

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 = $Na_3B_1O_3$  = $Na_3BO_3$