Gravimetric Analysis via Precipitation

Gravimetric analysis turns a bathroom-scale idea into a precision tool. Suppose you can force every atom of an analyte into an insoluble solid. Then you filter that solid out, dry it, and weigh it. From there, you can work backward through the mole ratio. This tells you exactly how much analyte you started with. The entire method rests on two things. Both are already familiar from Precipitation Reactions and Solubility Rules and Solution Stoichiometry. First, a reaction that goes essentially to completion. Second, a balanced equation that converts grams of product into moles of the original species. What makes gravimetric analysis a distinct skill? It is not just another stoichiometry problem. Instead, it involves a chain of practical constraints layered on top of the chemistry. The precipitate must be pure. It must be insoluble enough to leave nothing behind in solution. It must also be stable enough to survive drying without changing composition. Get any link in that chain wrong, and the final mass lies to you. Get every link right, though, and gravimetric analysis remains one of the most accurate ways known. It can determine a substance's identity or purity. With careful technique, it is sensitive to a part in ten thousand or better.

Key takeaways

A gravimetric determination converts mass of dried precipitate → moles of precipitate → moles of analyte → mass or % of analyte, using the balanced precipitation equation.
The precipitating reagent is added in excess to drive the reaction toward completion, minimizing residual analyte left dissolved.
A gravimetric factor (moles analyte per mole precipitate, times molar mass ratio) converts precipitate mass directly to analyte mass in one multiplication.
Sources of error include coprecipitation, occlusion, incomplete drying, and solubility losses during washing — each biases the mass, and hence the answer, in a predictable direction.
The technique demands the precipitate be filterable, of known and constant composition, and essentially insoluble under the wash/dry conditions used.
Key idea: mass is a stoichiometric messenger — weigh a pure, known solid, and the mole ratio tells you everything about the invisible species that made it.

The core insight

Every gram of dried precipitate carries a fixed number of moles of analyte locked inside it, set by the reaction's stoichiometry — weighing the solid is really weighing the analyte in disguise.

The physical picture

Picture a solution containing an unknown mass of chloride ion, $\text{Cl}^-$ , perhaps from a dissolved salt of unknown purity. Add a soluble silver salt, $\text{AgNO}_3(aq)$ , and immediately a white curdy solid forms:

\text{Ag}^+(aq) + \text{Cl}^-(aq) \rightarrow \text{AgCl}(s)

This reaction is chosen for gravimetric work precisely because $\text{AgCl}$ has an extremely small solubility product ( $K_{sp} \approx 1.8\times10^{-10}$ ), so adding excess $\text{Ag}^+$ pushes essentially all the chloride out of solution and into the solid. The solid is collected on a fine filter, washed to remove adsorbed impurities without dissolving the precipitate itself, then dried (sometimes ignited at high temperature) to a constant, reproducible mass. That final mass is the experimental data point; everything else is bookkeeping through the mole ratio. The logic is a closed loop. You know the balanced equation. You measure the final solid's mass. Molar mass converts that mass to moles of precipitate. The equation's coefficients convert moles of precipitate to moles of analyte. Molar mass converts moles of analyte back to a mass, or a percentage, of the original sample.

Figure

The gravimetric analysis pipeline: sample → precipitate → dried solid → back-calculated analyte mass

The core relations

The heart of the method is a single conversion chain, run in reverse from what you might expect: you measure the product and infer the reactant.

n_{\text{precipitate}} = \frac{m_{\text{precipitate}}}{M_{\text{precipitate}}}

where: $m_{\text{precipitate}}$ is the dried, weighed mass of the solid (g); $M_{\text{precipitate}}$ is its molar mass (g/mol); $n_{\text{precipitate}}$ is moles of precipitate formed.

n_{\text{analyte}} = n_{\text{precipitate}} \times \frac{a}{b}

where: $a$ and $b$ are the stoichiometric coefficients of analyte and precipitate respectively in the balanced equation, so $a/b$ is the mole ratio (analyte : precipitate).

m_{\text{analyte}} = n_{\text{analyte}} \times M_{\text{analyte}}

where: $M_{\text{analyte}}$ is the molar mass of the species being determined (g/mol). These three steps collapse into one multiplicative shortcut called the gravimetric factor (GF):

\text{GF} = \frac{a}{b}\times\frac{M_{\text{analyte}}}{M_{\text{precipitate}}} \qquad\Longrightarrow\qquad m_{\text{analyte}} = m_{\text{precipitate}}\times \text{GF}

where: GF has units of (g analyte / g precipitate) and is a fixed number for any given analyte–precipitate pair, computable once and reused for every sample of that type. Finally, the mass fraction (percent composition) of analyte in the original sample is

\%\,\text{analyte} = \frac{m_{\text{analyte}}}{m_{\text{sample}}}\times 100\%

where: $m_{\text{sample}}$ is the mass of the original, undissolved sample weighed out before the experiment began.

Derivation ABuilding the gravimetric factor from the balanced equation

Goal. Show why a single number, the gravimetric factor, correctly converts any measured precipitate mass into the corresponding analyte mass, for a general reaction with unequal coefficients.

Step 1. Write the balanced precipitation reaction in general form, with $a$ moles of analyte $X$ producing $b$ moles of precipitate $P$ :

a\,X(aq) + \ldots \rightarrow b\,P(s) + \ldots

The coefficients $a$ and $b$ need not be equal — for instance, two moles of an analyte might combine with one mole of reagent to give one mole of a 2:1 precipitate, or vice versa.

Step 2. Convert the measured precipitate mass to moles, since mass is what the balance reports but moles are what the equation relates:

n_P = \frac{m_P}{M_P}

Step 3. Use the mole ratio from the balanced equation to step across to moles of analyte. Because $a$ moles of $X$ always accompany $b$ moles of $P$ in this reaction, the ratio $a/b$ is fixed regardless of how much material is present:

n_X = n_P \times \frac{a}{b} = \frac{m_P}{M_P}\times\frac{a}{b}

Step 4. Convert moles of analyte back into a mass, since the final answer is usually wanted in grams (or as a percentage of sample mass):

m_X = n_X \times M_X = \frac{m_P}{M_P}\times\frac{a}{b}\times M_X

Step 5. Group every term that does not depend on the measured mass $m_P$ into a single constant — the gravimetric factor:

\text{GF} = \frac{a}{b}\times\frac{M_X}{M_P} \quad\text{so that}\quad m_X = m_P \times \text{GF}

Because $a$ , $b$ , $M_X$ , and $M_P$ are all fixed by the chemistry of the reaction, GF can be computed once, in advance, for a given analyte–precipitate pair, and reused for any sample size.

Step 6. Sanity check the limiting case $a = b$ (a 1:1 precipitation, as with $\text{Cl}^-/\text{AgCl}$ or $\text{SO}_4^{2-}/\text{BaSO}_4$ ). Then $\text{GF} = M_X/M_P$ , simply the ratio of molar masses — exactly what naive intuition would guess when one mole of analyte becomes one mole of precipitate, confirming the general formula reduces correctly in the simplest case.

Worked examples

Example 1 — Chloride content of an impure salt (1:1 stoichiometry). A 0.5416 g sample of an impure chloride salt is dissolved in water and treated with excess $\text{AgNO}_3(aq)$ . The dried precipitate of $\text{AgCl}$ has a mass of 0.8845 g. Find the mass and percentage of $\text{Cl}^-$ in the original sample. The reaction is

\text{Ag}^+(aq) + \text{Cl}^-(aq) \rightarrow \text{AgCl}(s)

a 1:1 mole ratio between $\text{Cl}^-$ and $\text{AgCl}$ . Step 1 — moles of precipitate. With $M_{\text{AgCl}} = 107.87 + 35.45 = 143.32\ \text{g/mol}$ ,

n_{\text{AgCl}} = \frac{0.8845\ \text{g}}{143.32\ \text{g/mol}} = 6.play{-}

Carrying the division through carefully:

n_{\text{AgCl}} = \frac{0.8845}{143.32} = 6.172\times10^{-3}\ \text{mol}

Step 2 — moles of analyte. The mole ratio is 1:1, so

n_{\text{Cl}^-} = 6.172\times10^{-3}\ \text{mol}

Step 3 — mass of analyte. With $M_{\text{Cl}} = 35.45\ \text{g/mol}$ ,

m_{\text{Cl}^-} = 6.172\times10^{-3}\ \text{mol}\times 35.45\ \text{g/mol} = 0.2188\ \text{g}

Step 4 — percentage in the sample.

\%\,\text{Cl}^- = \frac{0.2188}{0.5416}\times100\% = 40.40\%

Equivalently, using the gravimetric factor directly: $\text{GF} = M_{\text{Cl}}/M_{\text{AgCl}} = 35.45/143.32 = 0.24735$ , so $m_{\text{Cl}^-} = 0.8845\times0.24735 = 0.2188$ g — the same result, reached in one line once GF is known.

Example 2 — Phosphate content via a non-1:1 precipitate. A fertilizer sample of mass 1.245 g is dissolved and all phosphate is precipitated as magnesium ammonium phosphate, which on ignition converts cleanly to magnesium pyrophosphate, $\text{Mg}_2\text{P}_2\text{O}_7(s)$ , a standard gravimetric form for phosphate determination. The relevant net stoichiometry is

2\,\text{PO}_4^{3-}(aq) \rightarrow \text{Mg}_2\text{P}_2\text{O}_7(s)

so two moles of analyte correspond to one mole of precipitate — a genuinely asymmetric 2:1 ratio, unlike Example 1. The ignited precipitate weighs 0.3092 g. Find the mass and percentage of phosphate, $\text{PO}_4^{3-}$ , in the fertilizer. Step 1 — molar masses.

M_{\text{PO}_4^{3-}} = 30.97 + 4(16.00) = 94.97\ \text{g/mol}

M_{\text{Mg}_2\text{P}_2\text{O}_7} = 2(24.31) + 2(30.97) + 7(16.00) = 48.62 + 61.94 + 112.00 = 222.56\ \text{g/mol}

Step 2 — moles of precipitate.

n_{\text{Mg}_2\text{P}_2\text{O}_7} = \frac{0.3092}{222.56} = 1.3894\times10^{-3}\ \text{mol}

Step 3 — moles of analyte, using the 2:1 ratio.

n_{\text{PO}_4^{3-}} = 2 \times 1.3894\times10^{-3} = 2.7788\times10^{-3}\ \text{mol}

Step 4 — mass of analyte.

m_{\text{PO}_4^{3-}} = 2.7788\times10^{-3}\ \text{mol}\times 94.97\ \text{g/mol} = 0.2639\ \text{g}

Step 5 — percentage.

\%\,\text{PO}_4^{3-} = \frac{0.2639}{1.245}\times100\% = 21.20\%

Checked against the gravimetric factor shortcut:

\text{GF} = \frac{2}{1}\times\frac{94.97}{222.56} = 2\times0.42679 = 0.85358

m_{\text{PO}_4^{3-}} = 0.3092\times0.85358 = 0.2639\ \text{g}

matching Step 4 exactly, and confirming that ignoring the factor of 2 in the mole ratio — a common slip — would have understated the phosphate content by exactly half.

Limiting cases & common traps

Situation	What happens to the calculation	Consequence
Mole ratio treated as 1:1 when it is not (e.g., Example 2 above)	GF misses the factor $a/b$	Analyte mass off by that exact factor
Precipitate not fully dried (retains wash water)	$m_{\text{precipitate}}$ measured too high	Reported analyte mass and % are falsely high
Precipitate partly soluble in wash liquid	Some product mass lost down the drain	Reported analyte mass and % are falsely low
Coprecipitation of a second ion (adsorbed impurity, e.g. NO₃⁻ on AgCl)	Extra mass included that isn't the true precipitate	Reported analyte mass and % are falsely high
Excess precipitating reagent insufficient	Equilibrium leaves analyte dissolved	Reported analyte mass and % are falsely low
$a = b$ (simplest 1:1 case, as in Example 1)	GF reduces to plain molar-mass ratio $M_X/M_P$	Matches intuitive "molar mass swap," a useful sanity check

The chemist behind it

Gravimetric analysis as a rigorous discipline owes its refinement largely to the German chemist Karl Remigius Fresenius, whose mid-nineteenth-century textbooks on qualitative and quantitative analysis systematized precipitation, filtration, and weighing into a reproducible protocol taught in laboratories for the following century. Fresenius founded one of the first dedicated schools of analytical chemistry, in Wiesbaden, and his procedures — precise reagent excess, controlled digestion, careful washing to remove soluble contaminants, and ignition to a constant mass — remain, in essence, the same steps used in a modern gravimetric determination. The method's reliability also depends on accurate atomic and molecular masses, a project advanced enormously by the careful gravimetric redeterminations of atomic weights carried out later by Theodore William Richards, whose exhaustive precipitation-based measurements of silver and chlorine helped fix the atomic mass scale used in calculations like Example 1.

Origins · 1 minKarl Remigius Fresenius and how this was discovered

Karl Remigius Fresenius

1818–1897 · Systematized precipitation-based quantitative and qualitative analysis, founding a school of analytical chemistry in Wiesbaden and authoring the standard reference texts of his era

Ask the tutor: How would the calculation change if the precipitate itself contains water of hydration that must be accounted for separately from the ignited, anhydrous form?

Why must the precipitating reagent be added in excess?

Precipitation reactions are governed by an equilibrium solubility product, not an infinite driving force. Adding excess reagent shifts that equilibrium (via Le Châtelier's principle, familiar from Precipitation Reactions and Solubility Rules) so that the concentration of remaining dissolved analyte becomes negligible. Without excess reagent, some analyte stays in solution, is lost during filtration, and never gets weighed — making the result too low.

Why is the precipitate washed instead of just filtered and dried directly?

Washing removes soluble impurities and adsorbed ions trapped in or on the precipitate's surface, especially from the mother liquor. Without washing, extra dissolved salts crystallize onto the solid as it dries, adding mass that is not the true precipitate and biasing the analyte result upward. The wash liquid is chosen so that it does not appreciably dissolve the precipitate itself, avoiding the opposite error.

Can gravimetric analysis be used if the precipitate is somewhat soluble in water?

It is far less reliable in that case. Any precipitate solubility means some analyte remains dissolved, is not filtered, and is never weighed, systematically lowering the answer. This is exactly why classic gravimetric reactions are chosen for their very small solubility products — AgCl and BaSO₄ are standard examples precisely because they leave almost nothing behind in the filtrate.

How is the gravimetric factor different from ordinary stoichiometric mole-ratio calculations?

It is not conceptually different — it is the same Solution Stoichiometry mole-ratio logic, pre-multiplied by the molar-mass ratio so the whole calculation collapses to one multiplication step. The gravimetric factor is a convenience constant, computed once for a given analyte–precipitate pair, that lets a chemist convert precipitate mass to analyte mass directly without re-deriving moles each time.

What happens to the calculation if the precipitate is heated so strongly that it decomposes into a different compound?

The molar mass used in the calculation must match whatever solid is actually being weighed. If ignition converts the precipitate into a new, well-defined compound (as with magnesium ammonium phosphate converting to magnesium pyrophosphate in Example 2), the calculation simply uses the molar mass and mole ratio of that final ignited form, not the original precipitate. Using the wrong molar mass for the actual weighed solid is a frequent source of gross error.