Question

Find the empirical formula of 58.8% Barium, 13.7% Sulfur, 27.5% Oxygen

Answer 1

Answer:

BaSO4.

Explanation:

What is given?

If we assume that the total mass of the compound is 100 g:

Mass of barium (Ba) = 58.8 g.

Mass of sulfur (S) = 13.7 g.

Mass of oxygen (O) = 27.5 g.

Molar mass of Ba = 137 g/mol.

Molar mass of S = 32 g/mol.

Molar mass of O = 16 g/mol.

Step-by-step solution:

The first step is to calculate the number of moles of each element using their respective molar mass. Let's start with Ba:

$58.8\text{ g Ba}\cdot\frac{1\text{ mol Ba}}{137\text{ g Ba}}=0.429\text{ moles Ba.}$

For S:

$13.7\text{ g S}\cdot\frac{1\text{ mol S}}{32\text{ g S}}=0.428\text{ moles S.}$

And for O:

$27.5\text{ g O}\cdot\frac{1\text{ mol O}}{16\text{ g O}}=1.72\text{ moles O.}$

The next step is to divide each number of moles by the least number of moles obtained. In this case, the least number of moles obtained was 0.428 moles:

$\begin{gathered} 0.429\text{ moles Ba/0.428=1.002 moles Ba}\approx1\text{ mol Ba,} \\ 0.428\text{ moles S/0.428=1 mol S,} \\ 1.72\text{ moles O/0.428=4.01 moles O}\approx4\text{ moles O.} \end{gathered}$

As we obtained 1 mol of Ba, 1 mol of S, and 4 moles of O, the empirical formula would be BaSO4.