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

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 $BaSO_4$, chloride as $AgCl$, iron(III) as $Fe(OH)_3$ then ignited to $Fe_2O_3$).
• 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 $n(precipitate)$ to $n(analyte)$ 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 $Fe(OH)_3$ to $Fe_2O_3$). 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 $n(precipitate) = m(precipitate) / M(precipitate)$.
2. Use the stoichiometric mole ratio to find $n(analyte)$.
3. Calculate the mass or concentration of analyte:
   • Percentage by mass: $\% = (m(analyte) / m(sample)) \times 100$.
   • Concentration in solution: $c = n(analyte) / V(sample)$.

The mole ratio in step 2 comes from the net ionic equation. Most simple gravimetric reactions are $1:1$ (chloride to $AgCl$, sulfate to $BaSO_4$). Some are not. Phosphate determined as $Mg_2P_2O_7$ has a $2:1$ ratio of phosphate to precipitate.

Sources of error

Worked example

A $1.000\ g$ sample of an alloy is dissolved in nitric acid. All chloride is removed first. Iron is precipitated by adding excess ammonia, giving $Fe(OH)_3$, which is ignited at $800$ deg C to constant mass to give $Fe_2O_3$. The final mass of $Fe_2O_3$ is $0.225\ g$. Calculate the percentage by mass of iron in the alloy.

$M(Fe_2O_3) = 2(55.85) + 3(16.0) = 159.7\ g\ mol^{-1}$.

$n(Fe_2O_3) = 0.225 / 159.7 = 1.409 \times 10^{-3}\ mol$.

Stoichiometry: 1 mol of $Fe_2O_3$ contains 2 mol of $Fe$.

$n(Fe) = 2 \times 1.409 \times 10^{-3} = 2.818 \times 10^{-3}\ mol$.

$m(Fe) = 2.818 \times 10^{-3} \times 55.85 = 0.1574\ g$.

$\%\ Fe = (0.1574 / 1.000) \times 100 = 15.7\%$.

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. $SO_4^{2-}$ to $BaSO_4$ is $1:1$. $PO_4^{3-}$ to $Mg_2P_2O_7$ is $2:1$. 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 $\%m/m$, volume of solution for $mol\ L^{-1}$.

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.

Examples in context

Example 1. Sulfate in Latrobe Valley brown coal ash. Chemists at AGL Loy Yang determine sulfate in coal-combustion ash before disposal. A $5.000 \, \text{g}$ sample of ash is dissolved in dilute $\text{HCl}$, filtered, and an excess of $\text{BaCl}_2$ added. The reaction $\text{Ba}^{2+} + \text{SO}_4^{2-} \to \text{BaSO}_4 \downarrow$ produces a fine white precipitate of barium sulfate ($K_{sp}$ near $10^{-10}$). After digesting at $90^{\circ}\text{C}$ to grow the crystals, filtering through pre-weighed ashless paper and igniting in a crucible at $800^{\circ}\text{C}$, the residue weighs $0.4583 \, \text{g}$. Moles of $\text{BaSO}_4 = 0.4583 / 233.4 = 1.964 \times 10^{-3}$ mol; mass of sulfate $= 1.964 \times 10^{-3} \times 96.1 = 0.1888 \, \text{g}$. Percentage sulfate $= 3.78\%$, informing whether the ash can be sold as roadbase or must be land-filled.

Example 2. Chloride in Mornington Peninsula irrigation water. Wineries on the Mornington Peninsula test recycled-water chloride to avoid leaf damage in chardonnay vines. A $250 \, \text{mL}$ sample is acidified and excess $0.100 \, \text{mol/L}$ silver nitrate added. The reaction $\text{Ag}^{+} + \text{Cl}^{-} \to \text{AgCl} \downarrow$ produces a white precipitate ($K_{sp} = 1.8 \times 10^{-10}$). After filtering on a sintered-glass crucible and drying at $110^{\circ}\text{C}$ to constant mass, $0.0717 \, \text{g}$ of $\text{AgCl}$ is collected. Moles of $\text{AgCl} = 0.0717 / 143.3 = 5.00 \times 10^{-4}$ mol; mass of $\text{Cl}^{-} = 5.00 \times 10^{-4} \times 35.5 = 0.01775 \, \text{g}$. Chloride concentration $= 71.0 \, \text{mg/L}$, well below the $250 \, \text{mg/L}$ ANZECC livestock guideline.

Try this

Q1. Outline the four essential steps in a gravimetric analysis. [4 marks]

• Cue. (1) Dissolve sample. (2) Add excess precipitating reagent. (3) Filter, wash, dry to constant mass. (4) Calculate from precipitate mass and stoichiometry.

Q2. A water sample of $100 \, \text{mL}$ gives $0.0537 \, \text{g}$ of dried $\text{BaSO}_4$ when treated with excess $\text{BaCl}_2$. Calculate the concentration of sulfate in $\text{mg/L}$. [3 marks]

• Cue. $n(\text{BaSO}_4) = 0.0537 / 233.4 = 2.30 \times 10^{-4}$ mol; mass $\text{SO}_4^{2-} = 2.30 \times 10^{-4} \times 96.1 = 0.02211 \, \text{g} = 22.1 \, \text{mg}$; concentration $= 221 \, \text{mg/L}$.

Q3. A $1.000 \, \text{g}$ sample of an unknown sulfate salt produces $0.6645 \, \text{g}$ of $\text{BaSO}_4$. (a) Calculate the percentage of sulfate. (b) Determine the moles of sulfate. (c) Identify the salt if the cation is $\text{Na}^+$. [2+2+2 marks]

• Cue. (a) $0.6645 / 233.4 \times 96.1 / 1.000 \times 100 = 27.4\%$. (b) $n = 2.85 \times 10^{-3}$ mol. (c) For $\text{Na}_2 \text{SO}_4$, $M_r = 142.0$; $0.4046 \, \text{g}$ expected vs $0.274 \, \text{g}$. Suggests $\text{Na}_2 \text{SO}_4 \cdot 10 \text{H}_2 \text{O}$ (Glauber's salt, $M_r = 322$).