In a gravimetric determination of phosphorus, 0.248 g of an organic compound was strongly heated in a Carius tube with concentrated `HNO_(3)`. Phosphoric acid so produced was precipitated as `MgNH_(4)PO_(4)` which on ignition yielded 0.444 g of `Mg_(2)P_(2)O_(7)`. The percentage of phosphorus in the compound is

Options

A. 2.5
B. `5.0`
C. 7.5
D. 50

Answer

To determine the percentage of phosphorus in the organic compound, we can follow these steps:

### Step 1: Determine the molar mass of \( \text{Mg}_2\text{P}_2\text{O}_7 \)

The molar mass of \( \text{Mg}_2\text{P}_2\text{O}_7 \) can be calculated using the atomic masses of magnesium (Mg), phosphorus (P), and oxygen (O):

- Atomic mass of Mg = 24 g/mol
- Atomic mass of P = 31 g/mol
- Atomic mass of O = 16 g/mol

The formula for \( \text{Mg}_2\text{P}_2\text{O}_7 \) consists of:
- 2 Mg: \( 2 \times 24 = 48 \) g/mol
- 2 P: \( 2 \times 31 = 62 \) g/mol
- 7 O: \( 7 \times 16 = 112 \) g/mol

Adding these together gives:
\[ \text{Molar mass of } \text{Mg}_2\text{P}_2\text{O}_7 = 48 + 62 + 112 = 222 \text{ g/mol} \]

### Step 2: Calculate the mass of phosphorus in \( \text{Mg}_2\text{P}_2\text{O}_7 \)

From the formula \( \text{Mg}_2\text{P}_2\text{O}_7 \), we see that there are 2 moles of phosphorus in one mole of the compound. Therefore, the mass of phosphorus in one mole of \( \text{Mg}_2\text{P}_2\text{O}_7 \) is:
\[ \text{Mass of phosphorus} = 2 \times 31 = 62 \text{ g} \]

### Step 3: Determine the mass of phosphorus in the sample

We have 0.444 g of \( \text{Mg}_2\text{P}_2\text{O}_7 \). To find the mass of phosphorus in this amount, we use the ratio of the mass of phosphorus to the mass of the compound:
\[ \text{Mass of phosphorus in 0.444 g of } \text{Mg}_2\text{P}_2\text{O}_7 = \frac{62 \text{ g}}{222 \text{ g}} \times 0.444 \text{ g} \]

Calculating this gives:
\[ \text{Mass of phosphorus} = \frac{62}{222} \times 0.444 = 0.124 \text{ g} \]

### Step 4: Calculate the percentage of phosphorus in the original organic compound

To find the percentage of phosphorus in the original organic compound, we use the formula:
\[ \text{Percentage of phosphorus} = \left( \frac{\text{mass of phosphorus}}{\text{mass of organic compound}} \right) \times 100\% \]
Substituting the values:
\[ \text{Percentage of phosphorus} = \left( \frac{0.124 \text{ g}}{0.248 \text{ g}} \right) \times 100\% = 50\% \]

### Final Answer
The percentage of phosphorus in the compound is **50%**.

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