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How are substances in water measured and analysed?

the principles and stoichiometry of gravimetric analysis to determine the concentration or percentage by mass of an analyte in a sample, including precipitation, filtration, washing, drying to constant mass, and the calculation of the analyte from the mass of the precipitate

A focused VCE Chemistry Unit 2 answer on gravimetric analysis. Covers the choice of precipitating reagent, the lab steps (precipitation, filtration, washing, drying to constant mass), the stoichiometric calculation from precipitate mass to analyte amount, and the common sources of error in a gravimetric determination.

Generated by Claude OpusReviewed by Better Tuition Academy8 min answer

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What this dot point is asking

VCAA wants you to describe and apply gravimetric analysis: a quantitative technique that determines the concentration or percentage by mass of an analyte by precipitating it as an insoluble compound, filtering, washing and drying the precipitate to constant mass, then using stoichiometry to calculate back to the analyte.

The answer

When gravimetric analysis is used

Gravimetric analysis is the right tool when:

  • The analyte forms a highly insoluble, easily filtered, well-defined precipitate (for example sulfate as BaSO4BaSO_4, chloride as AgClAgCl, iron(III) as Fe(OH)3Fe(OH)_3 then ignited to Fe2O3Fe_2O_3).
  • The precipitate composition is known and reproducible.
  • Trace amounts of impurities can be washed out.
  • High accuracy is needed but instruments such as AAS or UV-Vis are not available or are not appropriate.

It is slower than titration or instrumental methods but, done carefully, can be more accurate because every step is direct mass measurement.

The principles

  1. Selective precipitation. Add a reagent that forms an insoluble compound with the analyte but leaves everything else in solution. Use the solubility rules.
  2. Quantitative precipitation. Add the precipitating reagent in excess so essentially every analyte ion is captured in the solid phase.
  3. Constant, known stoichiometry. The precipitate must have a definite formula so you can convert from n(precipitate)n(precipitate) to n(analyte)n(analyte) using a fixed mole ratio.
  4. Direct mass measurement. The result is calculated from the mass of the dried precipitate measured on an analytical balance.

The standard lab procedure

  1. Dissolve the sample in a known mass or volume of water (or other suitable solvent).
  2. Add the precipitating reagent in slight excess. Add slowly and with stirring. Slow addition and gentle warming favour large, well-formed crystals that filter cleanly and trap fewer impurities.
  3. Allow the precipitate to digest (sit in the hot solution for a few minutes). Small particles dissolve and re-deposit on larger ones (Ostwald ripening). The result is a coarser, easier-to-filter solid.
  4. Filter through a pre-weighed sintered glass crucible or a pre-weighed filter paper. Quantitatively transfer all of the solid using a wash bottle.
  5. Wash with small portions of cold distilled water (or sometimes a dilute precipitating-agent solution) to remove soluble impurities while minimising re-dissolution.
  6. Dry to constant mass. Place the crucible in an oven (typically 110 deg C for water removal; higher for ignition to a different stable form, such as Fe(OH)3Fe(OH)_3 to Fe2O3Fe_2O_3). Weigh, dry again, weigh again. Repeat until the mass changes by less than a small tolerance (constant mass).
  7. Calculate using the mass of the dried precipitate and the mole ratio to the analyte.

The calculation framework

The general workflow:

  1. Calculate n(precipitate)=m(precipitate)/M(precipitate)n(precipitate) = m(precipitate) / M(precipitate).
  2. Use the stoichiometric mole ratio to find n(analyte)n(analyte).
  3. Calculate the mass or concentration of analyte:
    • Percentage by mass: %=(m(analyte)/m(sample))Γ—100\% = (m(analyte) / m(sample)) \times 100.
    • Concentration in solution: c=n(analyte)/V(sample)c = n(analyte) / V(sample).

The mole ratio in step 2 comes from the net ionic equation. Most simple gravimetric reactions are 1:11:1 (chloride to AgClAgCl, sulfate to BaSO4BaSO_4). Some are not. Phosphate determined as Mg2P2O7Mg_2P_2O_7 has a 2:12:1 ratio of phosphate to precipitate.

Sources of error

Error Effect on result How to avoid it
Precipitate not fully dried Mass too high (water counted as precipitate) Dry to constant mass
Co-precipitated impurities Mass too high Slow addition, hot solution, careful washing
Loss of precipitate during transfer or washing Mass too low Quantitative transfer, cold wash water
Precipitate partially soluble Mass too low Use a sufficient excess of precipitating reagent; cold wash water
Wrong precipitate formula assumed Result systematically wrong Confirm formula and dry/ignite to the correct stable form

Worked example

A 1.000Β g1.000\ g sample of an alloy is dissolved in nitric acid. All chloride is removed first. Iron is precipitated by adding excess ammonia, giving Fe(OH)3Fe(OH)_3, which is ignited at 800800 deg C to constant mass to give Fe2O3Fe_2O_3. The final mass of Fe2O3Fe_2O_3 is 0.225Β g0.225\ g. Calculate the percentage by mass of iron in the alloy.

M(Fe2O3)=2(55.85)+3(16.0)=159.7Β gΒ molβˆ’1M(Fe_2O_3) = 2(55.85) + 3(16.0) = 159.7\ g\ mol^{-1}.

n(Fe2O3)=0.225/159.7=1.409Γ—10βˆ’3Β moln(Fe_2O_3) = 0.225 / 159.7 = 1.409 \times 10^{-3}\ mol.

Stoichiometry: 1 mol of Fe2O3Fe_2O_3 contains 2 mol of FeFe.

n(Fe)=2Γ—1.409Γ—10βˆ’3=2.818Γ—10βˆ’3Β moln(Fe) = 2 \times 1.409 \times 10^{-3} = 2.818 \times 10^{-3}\ mol.

m(Fe)=2.818Γ—10βˆ’3Γ—55.85=0.1574Β gm(Fe) = 2.818 \times 10^{-3} \times 55.85 = 0.1574\ g.

%Β Fe=(0.1574/1.000)Γ—100=15.7%\%\ Fe = (0.1574 / 1.000) \times 100 = 15.7\%.

Common traps

Forgetting to dry to constant mass. A wet precipitate weighs more than a dry one. The hallmark of a careful gravimetric analysis is the repeat weigh-and-dry cycle.

Using the wrong mole ratio. Always go through the net ionic equation. SO42βˆ’SO_4^{2-} to BaSO4BaSO_4 is 1:11:1. PO43βˆ’PO_4^{3-} to Mg2P2O7Mg_2P_2O_7 is 2:12:1. Read the formula carefully.

Reporting concentration when percentage by mass was asked, or vice versa. Read what the question is asking for and pick the right denominator: mass of sample for %m/m\%m/m, volume of solution for molΒ Lβˆ’1mol\ L^{-1}.

Adding precipitating reagent in stoichiometric amount instead of excess. Equilibrium leaves a small fraction of analyte in solution. The standard practice is to add a moderate excess and check completeness.

Including paper or crucible mass in the precipitate mass. Subtract the pre-recorded empty mass.

In one sentence

Gravimetric analysis converts an analyte to a known, insoluble precipitate, filters, washes and dries it to constant mass, and uses the mass of that precipitate plus the mole ratio in the net ionic equation to calculate the original concentration or percentage of analyte.

Past exam questions, worked

Real questions from past VCAA papers on this dot point, with our answer explainer.

2024 VCE5 marksA 2.500 g sample of a fertiliser is dissolved in water and excess barium chloride solution is added to precipitate all the sulfate as barium sulfate. The filtered, washed and dried precipitate has a mass of 1.846 g. (a) Write the net ionic equation for the precipitation. (b) Calculate the percentage by mass of sulfate (SO4 2-) in the original fertiliser. (c) Identify two precautions during the procedure that improve the accuracy of the final result.
Show worked answer β†’

A 5-mark answer needs the equation, the calculation with units, and two clear precautions.

(a) Net ionic equation:
Ba2+(aq) + SO4 2-(aq) -> BaSO4(s)

(b) Calculation:
M(BaSO4) = 137.3 + 32.1 + 4(16.0) = 233.4 g mol^-1.
n(BaSO4) = 1.846 / 233.4 = 7.909 x 10^-3 mol.
Mole ratio BaSO4 : SO4 2- is 1 : 1, so n(SO4 2-) = 7.909 x 10^-3 mol.
M(SO4 2-) = 32.1 + 4(16.0) = 96.1 g mol^-1.
m(SO4 2-) = 7.909 x 10^-3 x 96.1 = 0.7600 g.
Percentage by mass = (0.7600 / 2.500) x 100 = 30.4%.

(c) Precautions (any two):
Add the barium chloride in excess so all sulfate is precipitated.
Wash the filter cake with distilled water to remove soluble impurities such as chloride and excess barium chloride that would otherwise contribute to the mass.
Dry the precipitate to constant mass to ensure all water is removed; weigh repeatedly until the mass is unchanged.
Use a sintered crucible or pre-weighed filter paper so the mass of the precipitate is measured accurately.

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