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Lab Day You nemust show the entire calculation in order to receive full marks. 1

ID: 477541 • Letter: L

Question

Lab Day You nemust show the entire calculation in order to receive full marks. 1t is not sumeient record the answer only. Remember to use significant figures and units. Note that you may seed to find your answers in the introduction of the lab Determination of Phosphate Concentration in Water Sample Unknown Sample Number (1 mark) Date of Reading Mass (g) Trial 1 Mass (g) Trial 2 Pre weighed weigh boat and filter paper Preweighed weigh boat, filter paper and Preweighed weigh boat, filter paper and product... 2nd weighing Product (a the last mass of weigh boat, filter paper and product the preweighed weigh boat and filter paper)

Explanation / Answer

Date of reading

Mass (g) Trial 1

Mass (g) Trial 2

Preweighed weigh boat and filter paper

25/01

14.7969

14.6040

Preweighed weigh boat, filter paper and product….1st weighing

25/01

15.8936

15.1912

Preweighed weigh boat, filter paper and product….2nd reading

27/01

15.8785

15.1871

Product (= the last mass of weigh boat, filter paper and product – the preweighed weigh boat and filter paper)

15.8785 – 14.7969 = 1.0816

15.1871 – 14.6040 = 0.5831

1. Struvite has the chemical formula NH4MgPO4.6H2O; molar mass of struvite = [14 + 4*1 + 24.3 + 31 + 4*16 + 6*(2*1 + 16)] g/mol = 245.3 g/mol.

Molar mass of PO4 = (31 + 4*16) g/mol = 95 g/mol.

As per the molecular formula, 1 mole struvite = 1mole PO4.

Mass of PO4 obtained from 0.5700 g struvite = (0.5700 g struvite)*(1 mole struvite/245.3 g struvite)*(1 mole PO4/1 mole struvite)*(95 g PO4/1 mole PO4) = 0.22075 g.

Mass of PO4 obtained from 0.6154 g struvite = (0.6154 g struvite)*(1 mole struvite/245.3 g struvite)*(1 mole PO4/1 mole struvite)*(95 g PO4/1 mole PO4) = 0.23833 g.

Volume of solution taken =10.0 mL.

Average mass of PO4 obtained from the two samples of struvite = (0.22075 + 0.23833)/2 g = 0.22954 g.

Mass percent (m/v) = (0.22954 g)/(10.0 mL)*100 = 2.2954% (ans).

Date of reading

Mass (g) Trial 1

Mass (g) Trial 2

Preweighed weigh boat and filter paper

25/01

14.7969

14.6040

Preweighed weigh boat, filter paper and product….1st weighing

25/01

15.8936

15.1912

Preweighed weigh boat, filter paper and product….2nd reading

27/01

15.8785

15.1871

Product (= the last mass of weigh boat, filter paper and product – the preweighed weigh boat and filter paper)

15.8785 – 14.7969 = 1.0816

15.1871 – 14.6040 = 0.5831