(d) Estimation of phosphorous:
A known mass of compound is heated with fuming `HNO_(3)` or sodium peroxide `(Na_(2)O_(2))` in Carius tube which converts phosphorrous to `H_(3)PO_(4)` . Magnesia mixture `(MgCL_(2) + NH_(4)Cl)` is then added, which gives the precipitate of magnesium ammonium phosphate `(MgNH_(4).PO_(4))` which on heating gives magnesium pyrophosphate `(Mg_(2)P_(2)O_(7))` , which is weighed.
An organic compound has 6.2 % of phosphorus. On sequence of reaction, the phosphorous present in 10 gm of organic compound is converted of `Mg_(2)P_(2)O_(7)` . Find the weight of `Mg_(2)P_(2)O_(7)` formed.

Options

A. 2.22 gm
B. 10.0 gm
C. 4.44 gm
D. 1.11 gm

Answer

To find the weight of magnesium pyrophosphate (Mg₂P₂O₇) formed from the phosphorus present in a given organic compound, we can follow these steps:

### Step 1: Calculate the amount of phosphorus in the organic compound
Given:
- Mass of the organic compound = 10 g
- Percentage of phosphorus = 6.2%

To find the mass of phosphorus:
\[ \text{Mass of phosphorus} = \left(\frac{6.2}{100}\right) \times 10 \, \text{g} = 0.62 \, \text{g} \]

### Step 2: Determine the molar mass of magnesium pyrophosphate (Mg₂P₂O₇)
The molar mass of Mg₂P₂O₇ can be calculated as follows:
- Molar mass of Mg = 24.31 g/mol
- Molar mass of P = 31.00 g/mol
- Molar mass of O = 16.00 g/mol

Calculating the molar mass:
\[ \text{Molar mass of Mg}_2\text{P}_2\text{O}_7 = (2 \times 24.31) + (2 \times 31.00) + (7 \times 16.00) \]
\[ = 48.62 + 62.00 + 112.00 = 222.62 \, \text{g/mol} \approx 222 \, \text{g/mol} \]

### Step 3: Relate the moles of phosphorus to moles of magnesium pyrophosphate
From the formula of magnesium pyrophosphate (Mg₂P₂O₇), we see that:
- 1 mole of Mg₂P₂O₇ contains 2 moles of phosphorus (P).

### Step 4: Calculate the moles of phosphorus present
Using the molar mass of phosphorus:
\[ \text{Moles of phosphorus} = \frac{\text{mass of phosphorus}}{\text{molar mass of phosphorus}} = \frac{0.62 \, \text{g}}{31.00 \, \text{g/mol}} \approx 0.02 \, \text{mol} \]

### Step 5: Calculate the moles of magnesium pyrophosphate formed
Since 2 moles of phosphorus correspond to 1 mole of magnesium pyrophosphate:
\[ \text{Moles of Mg}_2\text{P}_2\text{O}_7 = \frac{\text{Moles of phosphorus}}{2} = \frac{0.02}{2} = 0.01 \, \text{mol} \]

### Step 6: Calculate the weight of magnesium pyrophosphate formed
Using the moles of magnesium pyrophosphate and its molar mass:
\[ \text{Weight of Mg}_2\text{P}_2\text{O}_7 = \text{moles} \times \text{molar mass} = 0.01 \, \text{mol} \times 222 \, \text{g/mol} = 2.22 \, \text{g} \]

### Final Answer
The weight of magnesium pyrophosphate (Mg₂P₂O₇) formed is **2.22 g**.

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