Answer:
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
