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

The empirical formula of the compound is =
`KBrO_3`

The name of the compound is potassium bromate.

Step-by-step explanation:

Mass of potassium = 4.628 g

Moles of potassium =
`(4.628 g)/(39 g/mol)=0.1187 mol`

Mass of bromine = 9.457 g

Moles of bromine =
`(9.457 g)/(80 g/mol)=0.1182 mol`

Mass of oxygen = 5.681 g

Moles of oxygen =
`(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 :

`(0.1187 mol)/(0.1182 mol)=1.0`

Bromine;

`(0.1182 mol)/(0.1182 mol)=1.0`

Oxygen ;

`(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.