Sodium hypochlorite (Na xCl yO z) is the active ingredient in household bleach. A 250.0 g sample of sodium hypochlorite contains 77.2 g sodium, 119.1 g chlorine, and the remainder is oxygen. Calculate the empirical formula of sodium hypochlorite.

The empirical formula is NaClO

Explanation:

Step 1: Data given

Mass of the sample sodium hypochlorite = 250.0 grams

Mass of sodium = 77.2 grams

Atomic mass sodium = 22.99 g/mol

Mass of chlorine = 119.1 grams

Atomic mass of chlorine = 35.45 g/mol

Step 2: Calculate moles Na

Moles Na = 77.2 grams / 22.99 g/mol

Moles Na = 3.36 moles

Step 3: Calculate moles chlorine

Moles Cl = 119.1 grams / 35.45 g/mol

Moles Cl = 3.36 moles

Step 4: Calculate mass oxygen

Mass oxygen = 250.0 grams - 77.2 grams - 119.1 grams

Mass oxygen = 53.7 grams

Step 5: Calculate moles O

Moles O = 53.7 grams / 16.0 g/mol

Moles O = 3.36 moles

Step 6: Calculate mol ratio

We divide by the smallest amount of moles

Na: 3.36 moles / 3.36 = 1

Cl: 3.36 moles / 3.36 moles = 1

O: 3.36 moles / 3.36 moles = 1

The empirical formula is NaClO

Empyrical formula → NaClO

Explanation:

To determine the empirical formula of sodium hypochlorite we need the centesimal composition:

Grams of an element in 100 g of compound.

77.1 g of Na in 250 g of compound

119.1 g of Cl in 250 g of compound

(250g - 119.1g - 77.1g) = 53.8 g of O in 250 g of compound

We make this rules of three:

In 250 g of compound we have 77.1 g of Na, 119.1 g of Cl and 53.8 g of O

In 100 g of compound we must have:

(77.1. 100) / 250 = 30.84 g of Na

(119.1. 100) / 250 = 47.64 g of Cl

(53.8. 100) / 250 = 21.52 g of O

We divide the mass by the molar mass of each element:

30.84 g / 23 g/mol = 1.34 mol of Na

47.64 g / 35.45 g/mol = 1.34 moles of Cl

21.52 g / 16 g/mol = 1.34 mol of O

We divide by the lowest value of obtained mol, but in this case, it's the same number for the three of them.

In conclussion, the empyrical formula for the sodium hypochlorite is: NaClO