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Gravimetric analysis and precipitation gravimetry

Definition of precipitation gravimetry, and an example of using precipitation gravimetry to determine the purity of a mixture containing two salts. 

What is precipitation gravimetry?

Precipitation gravimetry is an analytical technique that uses a precipitation reaction to separate ions from a solution. The chemical that is added to cause the precipitation is called the precipitant or precipitating agent. The solid precipitate can be separated from the liquid components using filtration, and the mass of the solid can be used along with the balanced chemical equation to calculate the amount or concentration of ionic compounds in solution. Sometimes you might hear people referring to precipitation gravimetry simply as gravimetric analysis, which is a broader class of analytical techniques that includes precipitation gravimetry and volatilization gravimetry. If you want to read more about gravimetric analysis in general, see this article on gravimetric analysis and volatilization gravimetry.
In this article, we will go through an example of finding the amount of an aqueous ionic compound using precipitation gravimetry. We will also discuss some common sources of error in our experiment, because sometimes in lab things don't go quite as expected and it can help to be extra prepared!
From left to right, 3 different insoluble silver salts as precipitates in test-tubes. The left tube contains yellowish silver(I) iodide, the middle tube contains cream-colored silver(I) bromide, and the right tube contains white silver(I) chloride.
Soluble silver salts such as silver(I) nitrate can be used as precipitating agents to determine the amount of halide ions present in a sample. Not only does the mass of the precipitate tell you about the concentration of the halide ions in solution, the color is also distinctive for different silver salts. This picture shows test tubes containing yellowish AgI (left), cream-colored AgBr (middle), and white AgCl (right) Photo of silver precipitates by Cychr from Wikipedia, CC BY 3.0

Example: Determining the purity of a mixture containing MgCl2 and NaNO3

Oh no! Our sometimes less-than-helpful lab assistant Igor mixed up the bottles of chemicals again. (In his defense, many white crystalline solids look interchangeable, but that is why reading labels is important!)
As a result of the mishap, we have 0.7209g of a mysterious mixture containing MgCl2 and NaNO3. We would like to know the relative amount of each compound in our mixture, which is fully dissolved in water. We add an excess of our precipitating agent silver(I) nitrate, AgNO3(aq), and observe the formation of a precipitate, AgCl(s). Once the precipitate is filtered and dried, we find that the mass of the solid is 1.032g.
What is the mass percent of MgCl2 in the original mixture?
Any gravimetric analysis calculation is really just a stoichiometry problem plus some extra steps. Since this is a stoichiometry problem, we will want to start with a balanced chemical equation. Here we are interested in the precipitation reaction between MgCl2(aq) and AgNO3(aq) to make AgCl(s), when AgNO3(aq) is in excess.
You might remember that precipitation reactions are a type of double replacement reaction, which means we can predict the products by swapping the anions (or cations) of the reactants. We might check our solubility rules if necessary, and then balance the reaction. In this problem we are already given the identity of the precipitate, AgCl(s). That means we just have to identify the other product, Mg(NO3)2(aq), and make sure the overall reaction is balanced. The resulting balanced chemical equation is:
The balanced equation tells us that for every 1mol MgCl2(aq), which is the compound we are interested in quantifying, we expect to make 2mol AgCl(s), our precipitate. We will use this molar ratio to convert moles of AgCl(s) to moles of MgCl2(aq). We are also going to make the following assumptions:
  • All of the precipitate is AgCl(s). We don't have to worry about any precipitate forming from the NaNO3.
  • All of the Cl(aq) has reacted to form AgCl(s). In terms of the stoichiometry, we need to make sure we add an excess of the precipitating agent AgNO3(aq) so all of the Cl(aq) from MgCl2(aq) reacts.
Now let's go through the full calculation step-by-step!

Step 1: Convert mass of precipitate, AgCl(s), to moles

Since we are assuming that the mass of the precipitate is all AgCl(s), we can use the molecular weight of AgCl to convert the mass of precipitate to moles.
mol of AgCl(s)=1.032g AgCl×1mol AgCl143.32g AgCl=0.007201mol AgCl=7.201×103mol AgCl

Step 2: Convert moles of precipitate to moles of MgCl2

We can convert the moles of AgCl(s), the precipitate, to moles of MgCl2(aq) using the molar ratio from the balanced equation.
mol of MgCl2(aq)=7.201×103mol AgCl×1mol MgCl22mol AgCl=3.600×103mol MgCl2

Step 3: Convert moles of MgCl2 to mass in grams

Since we are interested in calculating the mass percent of MgCl2 in the original mixture, we will need to convert moles of MgCl2 into grams using the molecular weight.
Mass of MgCl2=3.600×103mol MgCl2×95.20g MgCl21mol MgCl2=0.3427g MgCl2

Step 4: Calculate mass percent of MgCl2 in the original mixture

The mass percent of MgCl2 in the original mixture can be calculated using the ratio of the mass of MgCl2 from Step 3 and the mass of the mixture.
Mass% MgCl2=0.3427g MgCl20.7209g mixture×100%=47.54%MgCl2in mixture      (Thanks Igor!)
Shortcut: We could also combine Steps 1 through 3 into a single calculation which will involve careful checking of units to make sure everything cancels out properly:
Mass of MgCl2=1.032g AgCl×1mol AgCl143.32g AgCl×1mol MgCl22mol AgCl×95.20g MgCl21mol MgCl2=0.3427g MgCl2
                                                 Step 1:                              Step 2:                  Step 3:
                                          find mol AgCl                   use mole ratio      find g MgCl2                                  

Potential sources of error

We now know how to use stoichiometry to analyze the results of a precipitation gravimetry experiment. If you are doing gravimetric analysis in lab, however, you might find that there are various factors than can affect the accuracy of your experimental results (and therefore also your calculations). Some common complications include:
  • Lab errors, such as not fully drying the precipitate
  • Stoichiometry errors, such as not balancing the equation for the precipitation reaction or not adding AgNO3(aq) in excess
What would happen to our results in the above situations?
Vacuum-adapted Erlenmeyer flask with glass frit containing orange-yellow solid. The Erlenmeyer is about one third full of slightly cloudy yellowish supernatant.
We might use a vacuum filtration set-up such as this one to separate the precipitate from the supernatant in a precipitation gravimetry experiment. Image from OpenStax Chemistry, CC BY 4.0

Situation 1: The precipitate is not fully dried

Maybe you ran out of time during the lab period, or the vacuum filtration set-up was not producing sufficient vacuum. It probably doesn't help that water is notoriously difficult to fully remove compared to typical organic solvents because it has a relatively high boiling point as well as a tendency to hang on with hydrogen-bonds whenever possible. Let's think about how residual water would affect our calculations.
If our precipitate is not completely dry when we measure the mass, we will think we have a higher mass of AgCl(s) than we actually do (since we are now measuring the mass of AgCl(s) plus the residual water). A higher mass of AgCl(s) will result in calculating more moles of AgCl(s) in Step 1, which will be converted into more moles of MgCl2(s) in our mixture. In the last step, we will end up calculating that the mass percent of MgCl2(s) is higher than it really is.
Lab tip: If you have time, one way to check for water in the sample is to recheck the mass a few times during the end of the drying process to make sure the mass is not changing even if you dry it longer. This is called drying to constant mass, and while it does not guarantee that your sample is completely dry, it certainly helps! You can also try stirring up your sample during the drying process to break up clumps and increase surface area. Make sure you don't tear holes in the filter paper, though!

Situation 2: We forgot to balance the equation!

Remember how we said earlier that gravimetric analysis is really just another stoichiometry problem? That means that working from an unbalanced equation can mess up our calculations. For this scenario, we would be using stoichiometric coefficients from the following unbalanced equation:
MgCl2(aq)+AgNO3(aq)AgCl(s)+Mg(NO3)2(aq)           (Warning: Not balanced!)
This equation tells us (incorrectly!) that for every mole of AgCl(s) we make, we can infer that we started with 1 mole of MgCl2 in the original mixture. When we use that stoichiometric ratio to calculate the mass of MgCl2, we will get:
Mass of MgCl2=1.032g AgCl×1mol AgCl143.32g AgCl×1mol MgCl21mol AgCl×95.20g MgCl21mol MgCl2=0.6854g MgCl2
                                                                                 wrong molar ratio!                                            
We just calculated that the mass of MgCl2 in our mixture is double the correct amount! This will result in overestimating the mass percent of MgCl2 by a factor of 2:
Mass% MgCl2=0.6854g MgCl20.7209g mixture×100%=95.08%MgCl2in mixture   (Compare to 47.54% !!)

Situation 3: Adding AgNO3(aq) in excess

In the last scenario we wonder what would happen if we didn't add AgNO3(aq) in excess. We know this would be bad because if AgNO3(aq) is not in excess, we will have unreacted Cl in solution. That means the mass of AgCl(s) will no longer be a measure of the mass of MgCl2 in the original mixture since we won't be accounting for the Cl still in solution. Therefore, we will underestimate the mass percent of MgCl2 in the original mixture.
A related and perhaps more important question we might want to answer is:
How do we make sure that we are adding AgNO3(aq) in excess?
If we knew the answer to that question, we could be extra confident in our calculations! In this problem:
  • We have 0.7209g of a mixture that contains some percentage of MgCl2.
  • We also know from our balanced equation that for each mole of MgCl2, we will need 2 moles of AgNO3(aq) at a minimum.
It is okay if we have extra AgNO3(aq), since once all the Cl has reacted, the rest of the AgNO3 will simply stay part of the solution which we will be able to filter away.
If we don't know how many moles of MgCl2 are in our original mixture, how do we calculate the number of moles of AgNO3 necessary to add? We know that the more moles of MgCl2 we have in our original mixture, the more moles of AgNO3 we need. Luckily, we have enough information to prepare for the worst case scenario, which is when our mixture is 100%MgCl2. This is the maximum amount of MgCl2 we can possibly have, which means this is when we will need the most AgNO3.
Let's pretend that we have 100%MgCl2. How many moles of AgNO3 will we need? This is another stoichiometry problem! We can calculate the number of moles of AgNO3 by converting the mass of the sample to moles of MgCl2 using the molecular weight, and then converting to the moles of AgNO3 using the molar ratio:
mol of AgNO3=0.7209g MgCl2×1mol MgCl295.20g MgCl2×2mol AgNO31mol MgCl2=1.514×102mol AgNO3
This result tells us that even if we don't know exactly how much MgCl2 we have in our mixture, as long as we add at least 1.514×102mol AgNO3 we should be good to go!


Precipitation gravimetry is a gravimetric analysis technique that uses a precipitation reaction to calculate the amount or concentration of an ionic compound. For example, we could add a solution containing Ag+ to quantify the amount of a halide ion such as Br(aq). Some useful tips for precipitation gravimetry experiments and calculations include:
  • Double check stoichiometry and make sure equations are balanced.
  • Make sure that the precipitate is dried to constant mass.
  • Add an excess of the precipitating agent.

Just for fun!

Let's say we started with 0.4015g of a mixture of MgCl2 and NaCl. We add an excess of AgNO3(aq) and find that we have 1.032g of the precipitate, AgCl(s).
How many moles of MgCl2 and NaCl did we have in our original mixture?
Express your answers with 4 significant digits.
mol MgCl2
mol NaCl

Want to join the conversation?

  • leaf blue style avatar for user Angela Connor
    I am very confused about this just for fun question - how do we go from knowing the mass of mixture to figuring out n and m?
    (45 votes)
    Default Khan Academy avatar avatar for user
    • aqualine ultimate style avatar for user Stefan van der Waal
      We have 1.032 grams of AgCl. We can use some stoichiometry to calculate this includes 7.201*10^-3 moles of chlorine atoms. Since all chlorine atoms come from the MgCl_2 and NaCl, we get our first equation:
      7.201*10^-3 = n + 2m

      Next up we can look at the mass of the original mixture (0.4015 g) together with the molar masses of MaCl_2 (58.44) and NaCl (95.20). This gives us our second equation:
      0.4015 = 58.44n + 95.2m

      Now it's required to do some algebra. One way is to first multiply the first equation by 58.44: 0.42082644 = 58.44n + 116.88m
      Next we subtract the second equation from that equation:
      0.42082644 - 0.4015 = 58.44n + 116.88m - 58.44n - 95.2m
      0.01932644 = 21.68m
      By dividing both sides by 21.68 we get:
      m = 8.914 * 10^-4

      Plugging that into the first equation we get:
      7.201*10^-3 = n + 2 * 8.914 * 10*-4
      7.201*10^-3 = n + 1.7828*10^-3
      n = 5.418*10^-3

      This answer is 6 years late, but better late than never I suppose.
      (8 votes)
  • leaf green style avatar for user khjhzw
    What is the chemical equation for "just for fun" problem? Don't we have to work out stoichiometry (to make sure the equation is balanced) before doing other calcuations?
    (18 votes)
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    • blobby green style avatar for user SnazzyAlpaca
      If there was only one chemical reaction occurring here, yes! That would be the right way to approach a problem like the example above (Determining the purity of a mixture containing MgCl2 and NaNO3), because the precipitating agent, AGNO3, only reacts with MgCl2 (it does NOT react with NaNO3). However, because our precipitating agent in the "just for fun" problem reacts with both MgCl2 and NaCl, a different approach must be taken. Even if we did write out the chemical equations for both reactions, it wouldn't do us much good, because AgCl, the precipitate, is produced in two separate ways. That's why this new, clever approach that uses two mathematical equations needs to be applied here. This question is a bit unfair at this stage - you would need to look at the types of chemical reactions section (I would think the info would be there) to know that both MgCl2 and NaCl react with AgNO3.
      (24 votes)
  • male robot donald style avatar for user johnny
    I am kinda lost, can someone explain to me where I went wrong and why ?
    step 1 - set up the equation
    MgCl2 + NaCl + AgNO3 = AgCl + Na + NO3 + Mg
    step 2 - balance the equation
    MgCl2 + NaCl + 3AgNO3 = 3AgCl + Na + 3NO3 + Mg
    step 3 - cross out the spectators
    MgCl2 + NaCl + 3AgNO3 = 3AgCl
    step 4 - map the solution
    g(AgCl) > mol(AgCl) > mol(MgCl2)
    step 5 - plug in numbers
    1.032g(AgCl) x ( 1mol(AgCl)/143.32g ) x ( 1mol(MgCl2)/3mol(AgCl)
    results = 2.400 x 10^-3 mol(MgCl2)
    (11 votes)
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    • spunky sam blue style avatar for user Ernest Zinck
      You have two separate equations.
      Step 1. Set up the equations.
      MgCl₂ + AgNO₃ → Mg(NO₃)₂ + AgCl
      NaCl + AgNO₃ → NaNO₃ + AgCl
      You don't get Na and Mg as products.
      Step 2. Balance the equations
      MgCl₂ + 2AgNO₃ → Mg(NO₃)₂ + 2AgCl
      NaCl + AgNO₃ → NaNO₃ + AgCl
      Step 3. Cross out spectators.
      Cl⁻ + Ag⁺ → AgCl
      Step 4. Map the solution.
      g AgCl → moles AgCl
      We can't go further without more information, like mass of MgCl₂ and AgCl.
      (17 votes)
  • blobby green style avatar for user Ellie Blight
    In the example, why is NaNO3 mentioned in the question but not used in the equation?
    (7 votes)
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  • old spice man blue style avatar for user Ronald Ramon
    I find the extra example VERY different from what was taught in the previous example. Is there any way to apply the steps taught before to that example?Please i am very confused
    (8 votes)
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  • orange juice squid orange style avatar for user Flo Tawns
    Hi - it may be that my maths is so poor but - how did he find the value of m and n from: 0.4015=58.44n+95.20m ??

    The other equation is 7.201×10​−3 mol=n+2m?
    (4 votes)
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    • piceratops seed style avatar for user RogerP
      The first equation has two unknowns, n and m, so this equation cannot be solved without using the second equation, which also has the same two unknowns, n and m.

      By rearranging the second equation, you can get that n = 7.201×10​−3 - 2m (this is done by subtracting 2m from both sides).

      This equation for n can then be substituted into the first equation in place of n that is in that equation:

      0.4015 = 58.44 (7.201×10​−3 - 2m) +95.20m

      After multiplying out and collecting together like terms you can solve for m. Once you have m, then you substitute that into either of the equations to get n.
      (7 votes)
  • leaf green style avatar for user Nechama Sara Cohen
    The "just for fun" answer is totally unclear. How do you reach the first equation of Hint 1?
    (6 votes)
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  • leafers ultimate style avatar for user Vygandas Razhas
    What does (aq) stAnd for?
    (4 votes)
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  • male robot hal style avatar for user RN
    could someone explain the just-for-fun question at the end? like how do we find the moles of MgCl2 and NaCl in the original mixtures?And why we have to figure out the moles of Cl?
    (4 votes)
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  • stelly blue style avatar for user EHAN
    Well it's kind of nice to the entire comment section is also flabbergasted by the 'just for fun question'
    In all seriousness, what kind of chemical reaction is this where there's two reactions going on at once?
    (2 votes)
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    • leaf red style avatar for user Richard
      Well it’s two chemical reactions happening alongside each other in solution.

      Mg^(+2)(aq) + 2Cl^(-)(aq) + 2Ag^(+)(aq) + 2NO3^(-)(aq) → 2AgCl (s) + Mg^(+2)(aq) + 2NO3^(-)(aq)

      Na^(+)(aq) + Cl(-)(aq) + Ag^(+)(aq) + NO3^(-)(aq) → AgCl(s) + Na^(+)(aq) + NO3^(-)(aq)

      In both cases the net ionic equation is: Ag^(+)(aq) + Cl(-)(aq) → AgCl(s)
      So in that sense there’s only one reaction happening here producing a new product.

      Hope that helps.
      (4 votes)