 # Gravimetric Analysis Tutorial

## Key Concepts

• Gravimetric analysis is the quantitative isolation of a substance by precipitation and weighing of the precipitate.1
• An analyte is the substance to be analysed.
• A precipitating reagent is the reactant used to precipitate the analyte.2
• The precipitate must be a pure substance of definite chemical composition.
• One advantage of gravimetric analysis compared to volumetric analysis (titrimetric analysis) is that there is greater likelihood of any impurities being seen, and therefore a correction can be applied.
• One disadvantage of gravimetric analysis is that it is generally more time-consuming.
• Gravimetric analysis can be used to determine the:
(i) percentage of sulfate in lawn fertiliser

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## General Procedure:

1. Weigh the sample to be analysed
2. Dissolve the sample in a suitable solvent, eg, water
3. Add an excess of the precipitating reagent to precipitate the analyte
4. Filter the mixture to separate the precipitate from the solution3
5. Wash the precipitate to remove any impurities4
6. Dry the precipitate by heating to remove water
7. Cool the precipitate in a dessicator to prevent the precipitate absorbing moisture from the air
8. Weigh the cooled precipitate
9. Repeat the drying and weighing process until a constant mass for the precipitate is achieved
10. Calculate the percent by mass of analyte in the sample

## General calculation of the percent by mass of analyte in a sample:

1. Write the balanced chemical equation for the precipitation reaction
2. Calculate the moles of precipitate: moles = mass ÷ molar mass
3. Calculate moles of analyte from the balanced chemical equation using the mole ratio of analyte : precipitate
(also known as the stoichiometric ratio of analyte to precipitate)
4. Calculate mass of analyte: mass = moles × molar mass
5. Calculate percent by mass of analyte in sample: (mass analyte ÷ mass sample) × 100

## Sources of Error:

• Incomplete precipitation results in a value for the percentage of analyte in the sample that is too low.
• Incomplete drying of the sample results in a value for the percentage of analyte in the sample that is too high
• Other ions in the sample may also be precipitated resulting in a value for the percentage of analyte in the sample that is too high

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## Worked Example

Question:

A 2.00 g sample of limestone was dissolved in hydrochloric acid and all the calcium present in the sample was converted to Ca2+(aq).

Excess ammonium oxalate solution, (NH4)2C2O4(aq), was added to the solution to precipitate the calcium ions as calcium oxalate, CaC2O4(s).

The precipitate was filtered, dried and weighed to a constant mass of 2.43 g.

Determine the percentage by mass of calcium in the limestone sample.

1. Wite the balanced chemical equation for the precipitation reaction:

Ca2+(aq) + C2O42-(aq) → CaC2O4(s)

2. Calculate the moles of calcium oxalate precipitated.

moles(CaC2O4(s)) = mass ÷ molar mass

moles(CaC2O4(s)) = 2.43 ÷ (40.08 + 2 x 12.01 + 4 x 16.00)

moles(CaC2O4(s)) = 2.43 ÷ 128.10

moles(CaC2O4(s)) = 0.019 mol

3. Find the moles of Ca2+(aq).

From the balanced chemical equation, the mole ratio of Ca2+ : CaC2O4(s) is 1 : 1

So, moles(Ca2+(aq)) = moles(CaC2O4(s)) = 0.019 mol

4. Calculate the mass of calcium in grams

mass (Ca) = moles × molar mass

mass (Ca) = 0.019 × 40.08 = 0.76 g

5. Calculate the percentage by mass of calcium in the original sample:

%Ca = (mass Ca ÷ mass sample) × 100

%Ca = (0.76 ÷ 2.00) x 100 = 38%

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## Problem Solving using StoPGoPS method

 Question: A particular water soluble fertiliser contains phosphorus in the form of phosphate ions, PO43-. A student used the following procedure to determine the percentage of phosphorus in a sample of soluble fertiliser. 5.17 g of fertiliser was added to a 250.0 mL volumetric flask and water was added to make it up to the mark. 20.00 mL of this solution was pipetted into a conical flask. A slight excess of precipitating agent was added to precipitate the phosphate ions as MgNH4PO4. The precipitate was filtered, washed with water and then converted by heating into Mg2P2O7 The mass of Mg2P2O7 was 0.0312 g Calculate the percentage by mass of phosphate in the fertiliser.

 STOP STOP! State the Question. What is the question asking you to do? Calculate the percentage by mass of phosphate in the fertiliser. %(PO43-) = ? % PAUSE PAUSE to Prepare a Game Plan (1) What information (data) have you been given in the question? mass of fertiliser = m(fertiliser) = 5.17 g initial volume of solution = Vi = 250.0 mL volume of solution used in experiment = Vf = 20.00 mL mass of Mg2P2O7 = m(Mg2P2O7) = 0.0312 g (2) What is the relationship between what you know and what you need to find out? (i) moles(Mg2P2O7) = mass(Mg2P2O7) ÷ molar mass(Mg2P2O7) (ii) 2P (from PO43-) → Mg2P2O7 (iii) In 20.00 mL sample solution: moles(P) = 2 × moles(Mg2P2O7) (iv) moles(P) in 250.0 mL solution = moles(P) in 20.00 mL sample × 250.0/20.00 (v) each mole PO43- contains 1 mole of P (vi) mass(PO43-) = moles(PO43-) × molar mass(PO43-) (vii) %(PO43-) = 100 × mass(PO43-)/mass(fertiliser) GO GO with the Game Plan (i) moles(Mg2P2O7) = mass(Mg2P2O7) ÷ molar mass(Mg2P2O7) mass(Mg2P2O7) = 0.0312 g (given) molar mass(Mg2P2O7) = (2 × 24.31) + (2 × 30.97) + (7 × 16.00) = 222.56 g mol-1 moles(Mg2P2O7) = 0.0312 ÷ 222.56 = 1.40 × 10-4 mol (ii) 2P (from PO43-) → Mg2P2O7 mole ratio P : Mg2P2O7 is 2:1 (iii) In 20.00 mL sample solution: moles(P) = 2 × moles(Mg2P2O7) moles(P) = 2 × 1.40 × 10-4 = 2.80 × 10-4 mol (iv) moles(P) in 250.0 mL solution = moles(P) in 20.00 mL sample × 250.0/20.00 moles(P) = 2.80 × 10-4 × 250.0/20.00 = 3.50 × 10-3 mol (v) each mole PO43- contains 1 mole of P moles(PO43-) = moles(P) = 3.50 × 10-3 mol (vi) mass(PO43-) = moles(PO43-) × molar mass(PO43-) moles(PO43-) = 3.50 × 10-3 mol molar mass(PO43-) = 30.97 + 4 × 16.00 = 94.97 g mol-1 mass(PO43-) = 3.50 × 10-3 × 94.97 = 0.332 g (vii) %(PO43-) = 100 × mass(PO43-)/mass(fertiliser) %(PO43-) = (0.332/5.17) × 100 = 6.42 % PAUSE PAUSE to Ponder Plausibility Is your answer plausible? Approximate by rounding off the numbers to the nearest 10 (or 5): n(Mg2P2O7) ≈ (5 × 10-2 g) ÷ (2 × 102) g mol-1 ≈ 2.5 × 10-4 mol n(P) ≈ 2 × 2.5 × 10-4 mol = 5 × 10-4 mol n(PO43-) ≈ 5 × 10-4 mol m(PO43-) ≈ 5 × 10-4 mol × (30 + 60 g mol-1) ≈ 5 × 10-4 × 100 ≈ 5 × 10-2 g %(PO43-) ≈ (0.05/5) × 100 ≈ 1% Our answer of 6.42% is in the same region as the approximated 1% so we are reasonably confident out answer is correct. STOP STOP! State the Solution Percentage by mass of phosphate ion in fertiliser is 6.42 %

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1. The separation of the element or the compound containing it may done a number of different ways, precipitation is one of the most common. Other methods include volatilisation and electro-analytical methods.

2. The precipitate formed must be stable and must be so slightly soluble that no appreciable loss occurs when the precipitate is separated by filtration and weighed.

3. The particles making up the precipitate must be large enough so that they do not pass through the filtering medium, and, the must remain larger than the pore size of the filter medium during the washing process.
Ordinary quantitative filter paper will retain particles up to a diameter of about 10 μm.

4. Washing removes impurities from the surface of the precipitate particles but will not remove occluded foreign substances.