Question

1. The composition of a compound is 34.6% carbon, 3.9% hydrogen, and 61.5% oxygen . What is the empirical formula of the compound?

2. A mixture consisting of 15.0 g of iron(II) sulfate and 15.0 g of sodium phosphate is reacted in the following reaction:

3 F e S O 4 ( a q ) + 2 N a 3 P O 4 ( a q ) ⟶ F e 3 ( P O 4 ) 2 ( s ) + 3 N a 2 S O 4 ( a q )

Formula weight of FeSO₄ = 151.91 g/mol

Formula weight of Na₃PO₄ = 163.94 g/mol

Formula weight of Na₂SO₄ = 119.05 g/mol

Answer the following two questions:

• What is the maximum number of grams of sodium sulfate that can be obtained?

• Calculate the grams remaining of the excess reactant.

Answer 1

1)

Let mass of compound be 100gm

Mass of carbon = 34.6 gm

Mole of carbon = 34.6 / 12 = 2.883

Mass of H = 3.9 gm

Mole of H = 3.9 / 1.00784 = 3.87

Mass of O = 61.5 gm

Mole of O = 61.5 / 16 = 3.84

C2.88H3.87O3.84

Emperical formula :- C3H4O4

2)

Mole of FeSO4 = mass / molar mass

= 15 / 151.91 = 0.0987

Mole of Na3PO4 = mass / molar mass

= 15 / 163.94 = 0.0915

(Mole/coefficient)FeSo4 is less than that of Na3PO4

So, FeSO4 is limiting reagent.

(Mole/coefficient)FeSO4 = (Mole/coefficient)Na2SO4

Mole of FeSO4 / 3 = mole of Na2SO4 / 3

Mole of Na2SO4 = mole of FeSO4 = 0.0987

Mass of Na2SO4 formed = 0.0987 * 119.05

= 11.75 gm

Remaining mole of excess Na3PO4

= 0.0915 - 2* 0.0987/3

= 0.0915 - 0.0658 mole

= 0.0257 mole

Mass of Na3PO4 remains = remaining mole * molar mass

= 0.0257 * 163.94

= 4.213 gm