0.248 g of an organic compound was strongly heated in a Carius tube with fuming HNO3. Phosphoric acid so produced was precipitated as MgNH4PO4. This on ignition gave 0.444 g of Mg2P2O7. The percentage of phosphorus in the compound is (Atomic weight: Mg = 24 u, P = 31 u, O = 16 u)

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q138 – AP-EAMCET 2026

← Back

Home Chemistry Basic ChemistryQuestion #18

0.248 g of an organic compound was strongly heated in a Carius tube with fuming HNO

_{3}

. Phosphoric acid so produced was precipitated as MgNH

_{4}

_{4}

. This on ignition gave 0.444 g of Mg

_{2}

_{2}

_{7}

. The percentage of phosphorus in the compound is (Atomic weight: Mg = 24 u, P = 31 u, O = 16 u)

Clue 1 of 3: Gravimetric Analysis Principle

The problem involves gravimetric analysis, where an element (phosphorus) is converted into a stable compound (Mg $_{2}$ P $_{2}$ O $_{7}$ ) of known composition, and its mass is used to determine the amount of the element in the original sample.

🧠 Full Solution Path

Step 1: Identify the stoichiometry of phosphorus in the final product✦ Active

The final product obtained after ignition is magnesium pyrophosphate, Mg $_{2}$ P $_{2}$ O $_{7}$ . From its chemical formula, one molecule of Mg $_{2}$ P $_{2}$ O $_{7}$ contains two atoms of phosphorus (P).

Step 2: Calculate the molar mass of Mg

_{2}

_{2}

_{7}

and the mass of phosphorus it contains○ Expand

The atomic weights are given as: Mg = $24 u$ , P = $31 u$ , O = $16 u$ . The molar mass of Mg $_{2}$ P $_{2}$ O $_{7}$ is calculated as:

(2 \times 24) + (2 \times 31) + (7 \times 16) = 48 + 62 + 112 = 222 g/mol

The mass of phosphorus in one mole of Mg $_{2}$ P $_{2}$ O $_{7}$ is:

2 \times 31 = 62 g

💡 Teacher's Secret Hint

Ensure to count all atoms of each element in the formula when calculating molar mass, especially for phosphorus, which appears twice.

Step 3: Calculate the mass of phosphorus in the precipitate obtained○ Expand

The mass of Mg $_{2}$ P $_{2}$ O $_{7}$ obtained is $0.444 g$ . We can find the mass of phosphorus (P) in this precipitate using the ratio of phosphorus mass to the molar mass of Mg $_{2}$ P $_{2}$ O $_{7}$ :

Mass of P = \frac{{Mass of P in Mg}_{2} P_{2} O_{7}}{{Molar mass of Mg}_{2} P_{2} O_{7}} \times {Mass of Mg}_{2} P_{2} O_{7}

Mass of P = \frac{62 g}{222 g} \times 0.444 g = 0.124 g

💡 Teacher's Secret Hint

This step applies the principle of stoichiometry: the mass of an element in a compound is proportional to its contribution to the compound's total molar mass.

Step 4: Calculate the percentage of phosphorus in the organic compound○ Expand

The mass of the original organic compound is $0.248 g$ . Now, we can calculate the percentage of phosphorus in the organic compound:

Percentage of P = \frac{Mass of P}{Mass of organic compound} \times 100 %

Percentage of P = \frac{0.124 g}{0.248 g} \times 100 % = 0.5 \times 100 % = 50 %

💡 Teacher's Secret Hint

Always double-check your arithmetic, especially when dealing with percentages, to avoid small calculation errors.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Chemistry Question 138 – AP-EAMCET 2026

Clue 1 of 3: Gravimetric Analysis Principle

🧠 Full Solution Path