A compound contains 21.6% Na, 33.3% Cl, and 45.1% O. Write the empirical formula and name the compound that is formed

The empirical formula for the given compound is and the name of compound is sodium chlorate.

Explanation : Given,

Percentage of Na = 21.6 %

Percentage of Cl = 33.3 %

Percentage of O = 45.1 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of Na = 21.6 g

Mass of Cl = 33.3 g

Mass of O = 45.1 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Na =

Moles of Cl =

Moles of O =

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.938 moles.

For Na =

For Cl =

For O =

Step 3: Taking the mole ratio as their subscripts.

The ratio of Na : Cl : O = 1 : 1 : 3

Hence, the empirical formula for the given compound is and the name of compound is sodium chlorate.

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answer is c