Question

A sample of an ionic compound that is often used as a dough conditioner is analyzed and found to contain 4.628 g of potassium, 9.457 g of bromine, and 5.681 g of oxygen.
What is the empirical formula for this compound?

What is it chemical name?

Answer 1

Answer:

The empirical formula of the compound is = $KBrO_3$

The name of the compound is potassium bromate.

Explanation:

Mass of potassium = 4.628 g

Moles of potassium = $\frac{4.628 g}{39 g/mol}=0.1187 mol$

Mass of bromine = 9.457 g

Moles of bromine = $\frac{9.457 g}{80 g/mol}=0.1182 mol$

Mass of oxygen = 5.681 g

Moles of oxygen = $\frac{5.681 g}{16 g/mol}0.3551$

For empirical formula of the compound, divide the least number of moles from all the moles of elements present in the compound:

Potassium :

$\frac{0.1187 mol}{0.1182 mol}=1.0$

Bromine;

$\frac{0.1182 mol}{0.1182 mol}=1.0$

Oxygen ;

$\frac{0.3551 mol}{0.1182 mol}=3.0$

The empirical formula of the compound is = $KBrO_3$

The name of the compound is potassium bromate.