A sample was contaminated with sodium phosphate. The sample was dissolved in water and added to an excess of acidified (NH₄)₂MoO₄ to produce a precipitate of (NH₄)₃PO₄·12MoO₃ (MM = 1877 g mol⁻¹). If 24.21 g of dry (NH₄)₃PO₄·12MoO₃ was obtained, what was the mass of sodium phosphate in the original sample?

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A sample was contaminated with sodium phosphate - HSC - SSCE Chemistry - Question 17 - 2021 - Paper 1

Question 17

A sample was contaminated with sodium phosphate. The sample was dissolved in water and added to an excess of acidified (NH₄)₂MoO₄ to produce a precipitate of (NH₄)₃P... show full transcript

Worked Solution & Example Answer:A sample was contaminated with sodium phosphate - HSC - SSCE Chemistry - Question 17 - 2021 - Paper 1

Step 1

Calculate the number of moles of (NH₄)₃PO₄·12MoO₃

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Answer

To find the number of moles of the precipitate formed, use the formula:

$n = \frac{mass}{molar\ mass}$

Given mass = 24.21 g and molar mass = 1877 g mol⁻¹:

$n = \frac{24.21}{1877} \approx 0.0129 \text{ mol}$

Step 2

Determine the relation between sodium phosphate and (NH₄)₃PO₄·12MoO₃

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Answer

The reaction indicates that 1 mole of sodium phosphate produces 1 mole of (NH₄)₃PO₄·12MoO₃. Thus, the moles of sodium phosphate will also be 0.0129 mol.

Step 3

Calculate the mass of sodium phosphate

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Answer

To find the mass of sodium phosphate, the molar mass of sodium phosphate (Na₃PO₄) needs to be calculated:

Na: 22.99 g/mol × 3 = 68.97 g/mol
P: 30.97 g/mol
O: 16.00 g/mol × 4 = 64.00 g/mol

Molar mass of Na₃PO₄ = 68.97 + 30.97 + 64.00 = 163.94 g/mol.

Using the number of moles calculated:

$mass = n \times molar\ mass = 0.0129 \times 163.94 \approx 2.115\ g$

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