Question

In a determination of P, an aqueous solution of $NaH_2PO_4$ is treated with a mixture of ammonium magnesium ions to precipitate magnesium ammonium phosphate $Mg(NH_4)PO_4.6H_2O$. This is heated and decomposed to magnesium pyrophosphate, $Mg_2P_2O_7$ which is weighed. A solution of $NaH_2PO_4$ yielded 1.11 g of $Mg_2P_2O_7$. What weight of $NaH_2PO_4$ (in gm) was present originally ? (P=31)

Answer 1

1.20 g

Answer 2

The problem involves a gravimetric determination of phosphorus, where $NaH_2PO_4$ is converted to $Mg_2P_2O_7$ through a series of reactions. The key principle to solve this problem is the Principle of Atom Conservation (POAC) for the phosphorus (P) atom, as all the phosphorus from the initial $NaH_2PO_4$ is ultimately converted into $Mg_2P_2O_7$.

1. Identify the species containing the conserved atom: The phosphorus (P) atom is conserved.

• Initial compound: $NaH_2PO_4$
• Final compound: $Mg_2P_2O_7$

2. Determine the number of P atoms in each molecule:

• One molecule of $NaH_2PO_4$ contains 1 P atom.
• One molecule of $Mg_2P_2O_7$ contains 2 P atoms.

3. Calculate the molar masses of the compounds:

• Molar mass of $NaH_2PO_4$ ($M_1$):
  Na (23) + H (2 × 1) + P (31) + O (4 × 16)
  $M_1 = 23 + 2 + 31 + 64 = 120 \text{ g/mol}$
• Molar mass of $Mg_2P_2O_7$ ($M_2$):
  Using atomic mass of Mg as 24 (consistent with common rounding in such problems and the similar question's molar mass of 222 for $Mg_2P_2O_7$):
  Mg (2 × 24) + P (2 × 31) + O (7 × 16)
  $M_2 = 48 + 62 + 112 = 222 \text{ g/mol}$

4. Apply the Principle of Atom Conservation (POAC) for P:
The total moles of P atoms in $NaH_2PO_4$ must be equal to the total moles of P atoms in $Mg_2P_2O_7$.

Let $w_1$ be the weight of $NaH_2PO_4$ (unknown) and $w_2$ be the weight of $Mg_2P_2O_7$ (given as 1.11 g).

Moles of P in $NaH_2PO_4$ = (Moles of $NaH_2PO_4$) × (Number of P atoms per molecule)
$= \frac{w_1}{M_1} \times 1$

Moles of P in $Mg_2P_2O_7$ = (Moles of $Mg_2P_2O_7$) × (Number of P atoms per molecule)
$= \frac{w_2}{M_2} \times 2$

Equating the moles of P atoms:
$\frac{w_1}{M_1} \times 1 = \frac{w_2}{M_2} \times 2$

5. Substitute the values and solve for $w_1$:
$\frac{w_1}{120} = \frac{1.11}{222} \times 2$
$\frac{w_1}{120} = \frac{2.22}{222}$
$\frac{w_1}{120} = 0.01$
$w_1 = 0.01 \times 120$
$w_1 = 1.20 \text{ g}$

The weight of $NaH_2PO_4$ originally present was 1.20 g.

Explanation of the solution: The principle of atom conservation (POAC) states that the total number of atoms of a specific element remains constant throughout a chemical reaction or series of reactions, even if the compounds change. In this problem, all the phosphorus (P) from the initial $NaH_2PO_4$ is converted into $Mg_2P_2O_7$. By equating the moles of P atoms in the initial and final compounds, considering the number of P atoms per molecule, we can determine the unknown mass. The calculation involves finding the molar masses of $NaH_2PO_4$ (120 g/mol) and $Mg_2P_2O_7$ (222 g/mol). Using the given mass of $Mg_2P_2O_7$ (1.11 g) and the stoichiometric ratio of P atoms (1 P in $NaH_2PO_4$ to 2 P in $Mg_2P_2O_7$), the mass of $NaH_2PO_4$ is calculated as 1.20 g.