1. The composition of a compou
1. The composition of a compound is 34.6% carbon, 3.9% hydrogen,and 61.5% oxygen . What is the empirical formula of thecompound?
2. A mixture consisting of 15.0 g of iron(II) sulfate and 15.0 gof 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 = 163.94g/mol
Formula weight of Na2SO4 = 119.05g/mol
Answer the following two questions:
- What is the maximum number of grams of sodium sulfate that canbe obtained?
- Calculate the grams remaining of the excess reactant.
Answer:
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
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