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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?

1 Answer

3 votes

Answer:

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.

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