Calculate the theoretical yield, in grams, of Eauropium Hydrogen phosphate (1 EuHPOsn formed rom 108.34-grams of Europium nitrate and 122.43-grams of Lithium Hydrogenphosphate. Eu(NO) + 3 Li2HPO4 I Eu2(HPO4)3 + 6 LİNO3 13 pts)One of the compounds iridium and oxygen found to contain 88.90% iridium and 11.10 % What is the empirical formula of compound?
The Correct Answer and Explanation is :
Theoretical Yield of Europium Hydrogenphosphate (EuHPO₄)
Let’s start by calculating the theoretical yield of Europium Hydrogenphosphate (EuHPO₄) from Europium Nitrate (Eu(NO₃)₃) and Lithium Hydrogenphosphate (Li₂HPO₄).
Step 1: Write the balanced chemical equation
The balanced chemical equation is: Eu(NO₃)₃+3Li₂HPO₄→Eu₂(HPO₄)₃+6LiNO₃\text{Eu(NO₃)₃} + 3 \text{Li₂HPO₄} \rightarrow \text{Eu₂(HPO₄)₃} + 6 \text{LiNO₃}
Step 2: Calculate the molar masses of reactants
- Eu(N₃)₃:
- Eu: 151.98 g/mol
- N: 14.01 g/mol (×3)
- O: 16.00 g/mol (×9)
- Molar mass of Eu(NO₃)₃ = 151.98 + (3 × 14.01) + (9 × 16.00) = 151.98 + 42.03 + 144.00 = 338.01 g/mol
- Li₂HPO₄:
- Li: 6.94 g/mol (×2)
- H: 1.008 g/mol (×1)
- P: 30.97 g/mol (×1)
- O: 16.00 g/mol (×4)
- Molar mass of Li₂HPO₄ = (2 × 6.94) + 1.008 + 30.97 + (4 × 16.00) = 13.88 + 1.008 + 30.97 + 64.00 = 109.86 g/mol
Step 3: Convert the mass of the reactants to moles
- Moles of Eu(NO₃)₃: 108.34 g338.01 g/mol=0.3204 mol\frac{108.34 \, \text{g}}{338.01 \, \text{g/mol}} = 0.3204 \, \text{mol}
- Moles of Li₂HPO₄: 122.43 g109.86 g/mol=1.113 mol\frac{122.43 \, \text{g}}{109.86 \, \text{g/mol}} = 1.113 \, \text{mol}
Step 4: Determine the limiting reagent
From the balanced equation, 1 mole of Eu(NO₃)₃ reacts with 3 moles of Li₂HPO₄. Therefore:
- 0.3204 moles of Eu(NO₃)₃ would require: 0.3204 mol×3=0.9612 mol of Li₂HPO₄0.3204 \, \text{mol} \times 3 = 0.9612 \, \text{mol of Li₂HPO₄}
Since we have 1.113 moles of Li₂HPO₄ (which is more than required), Eu(NO₃)₃ is the limiting reagent.
Step 5: Calculate the moles of Eu₂(HPO₄)₃ formed
From the balanced equation, 1 mole of Eu(NO₃)₃ produces 1 mole of Eu₂(HPO₄)₃. Therefore, the moles of Eu₂(HPO₄)₃ formed are 0.3204 mol.
Step 6: Calculate the theoretical yield of Eu₂(HPO₄)₃
- Molar mass of Eu₂(HPO₄)₃:
- Eu: 151.98 g/mol (×2)
- H: 1.008 g/mol (×3)
- P: 30.97 g/mol (×3)
- O: 16.00 g/mol (×12)
- Molar mass of Eu₂(HPO₄)₃ = (2 × 151.98) + (3 × 1.008) + (3 × 30.97) + (12 × 16.00) = 303.96 + 3.024 + 92.91 + 192.00 = 591.89 g/mol
The theoretical yield of Eu₂(HPO₄)₃ is: 0.3204 mol×591.89 g/mol=189.78 g0.3204 \, \text{mol} \times 591.89 \, \text{g/mol} = 189.78 \, \text{g}
So, the theoretical yield of Eu₂(HPO₄)₃ is 189.78 g.
Empirical Formula of the Compound Containing Iridium and Oxygen
Given that the compound contains 88.90% iridium (Ir) and 11.10% oxygen (O), let’s calculate the empirical formula.
Step 1: Assume 100 grams of the compound
- Mass of Iridium = 88.90 g
- Mass of Oxygen = 11.10 g
Step 2: Convert masses to moles
- Moles of Iridium: 88.90 g192.22 g/mol=0.463 mol of Ir\frac{88.90 \, \text{g}}{192.22 \, \text{g/mol}} = 0.463 \, \text{mol of Ir}
- Moles of Oxygen: 11.10 g16.00 g/mol=0.694 mol of O\frac{11.10 \, \text{g}}{16.00 \, \text{g/mol}} = 0.694 \, \text{mol of O}
Step 3: Determine the simplest ratio
To find the ratio of Ir to O, divide each by the smaller number of moles (0.463):
- Ratio of Iridium: 0.4630.463=1\frac{0.463}{0.463} = 1
- Ratio of Oxygen: 0.6940.463≈1.5\frac{0.694}{0.463} \approx 1.5
Since the ratio of O is approximately 1.5, we multiply both by 2 to get whole numbers:
- Iridium: 1×2=21 \times 2 = 2
- Oxygen: 1.5×2=31.5 \times 2 = 3
Thus, the empirical formula of the compound is Ir₂O₃.