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
Answer:
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
