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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!

## Example: Determining the purity of a mixture containing ${\text{MgCl}}_{2}$‍  and ${\text{NaNO}}_{3}$‍

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 $0.7209\phantom{\rule{0.167em}{0ex}}\text{g}$ of a mysterious mixture containing ${\text{MgCl}}_{2}$ and ${\text{NaNO}}_{3}$. 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, ${\text{AgNO}}_{3}\left(aq\right)$, and observe the formation of a precipitate, $\text{AgCl}\left(s\right)$. Once the precipitate is filtered and dried, we find that the mass of the solid is $1.032\phantom{\rule{0.167em}{0ex}}\text{g}$.
What is the mass percent of ${\text{MgCl}}_{2}$ 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 ${\text{MgCl}}_{2}\left(aq\right)$ and ${\text{AgNO}}_{3}\left(aq\right)$ to make $\text{AgCl}\left(s\right)$, when ${\text{AgNO}}_{3}\left(aq\right)$ 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, $\text{AgCl}\left(s\right)$. That means we just have to identify the other product, ${\text{Mg(NO}}_{3}{\right)}_{2}\left(aq\right)$, and make sure the overall reaction is balanced. The resulting balanced chemical equation is:
${\text{MgCl}}_{2}\left(aq\right)+2{\text{AgNO}}_{3}\left(aq\right)\to 2\text{AgCl}\left(s\right)+{\text{Mg(NO}}_{3}{\right)}_{2}\left(aq\right)$
The balanced equation tells us that for every $1\phantom{\rule{0.167em}{0ex}}{\text{mol MgCl}}_{2}\left(aq\right)$, which is the compound we are interested in quantifying, we expect to make $2\phantom{\rule{0.167em}{0ex}}\text{mol AgCl}\left(s\right)$, our precipitate. We will use this molar ratio to convert moles of $\text{AgCl}\left(s\right)$ to moles of ${\text{MgCl}}_{2}\left(aq\right)$. We are also going to make the following assumptions:
• All of the precipitate is $\text{AgCl}\left(s\right)$. We don't have to worry about any precipitate forming from the ${\text{NaNO}}_{3}$.
• All of the ${\text{Cl}}^{-}\left(aq\right)$ has reacted to form $\text{AgCl}\left(s\right)$. In terms of the stoichiometry, we need to make sure we add an excess of the precipitating agent ${\text{AgNO}}_{3}\left(aq\right)$ so all of the ${\text{Cl}}^{-}\left(aq\right)$ from ${\text{MgCl}}_{2}\left(aq\right)$ reacts.
Now let's go through the full calculation step-by-step!

### Step $1$‍ : Convert mass of precipitate, $\text{AgCl}\left(s\right),$‍  to moles

Since we are assuming that the mass of the precipitate is all $\text{AgCl}\left(s\right)$, we can use the molecular weight of $\text{AgCl}$ to convert the mass of precipitate to moles.
$\text{mol of AgCl}\left(s\right)=1.032\phantom{\rule{0.167em}{0ex}}\overline{)\text{g AgCl}}×\frac{1\phantom{\rule{0.167em}{0ex}}\text{mol AgCl}}{143.32\phantom{\rule{0.167em}{0ex}}\overline{)\text{g AgCl}}}=0.007201\phantom{\rule{0.167em}{0ex}}\text{mol AgCl}=7.201×{10}^{-3}\phantom{\rule{0.167em}{0ex}}\text{mol AgCl}$

### Step $2$‍ : Convert moles of precipitate to moles of ${\text{MgCl}}_{2}$‍

We can convert the moles of $\text{AgCl}\left(s\right)$, the precipitate, to moles of ${\text{MgCl}}_{2}\left(aq\right)$ using the molar ratio from the balanced equation.
${\text{mol of MgCl}}_{2}\left(aq\right)=7.201×{10}^{-3}\phantom{\rule{0.167em}{0ex}}\overline{)\text{mol AgCl}}×\frac{1\phantom{\rule{0.167em}{0ex}}{\text{mol MgCl}}_{2}}{2\phantom{\rule{0.167em}{0ex}}\overline{)\text{mol AgCl}}}=3.600×{10}^{-3}\phantom{\rule{0.167em}{0ex}}{\text{mol MgCl}}_{2}$

### Step $3$‍ : Convert moles of ${\text{MgCl}}_{2}$‍  to mass in grams

Since we are interested in calculating the mass percent of ${\text{MgCl}}_{2}$ in the original mixture, we will need to convert moles of ${\text{MgCl}}_{2}$ into grams using the molecular weight.
${\text{Mass of MgCl}}_{2}=3.600×{10}^{-3}\phantom{\rule{0.167em}{0ex}}\overline{){\text{mol MgCl}}_{2}}×\frac{95.20\phantom{\rule{0.167em}{0ex}}{\text{g MgCl}}_{2}}{1\phantom{\rule{0.167em}{0ex}}\overline{){\text{mol MgCl}}_{2}}}=0.3427\phantom{\rule{0.167em}{0ex}}{\text{g MgCl}}_{2}$

### Step $4$‍ : Calculate mass percent of ${\text{MgCl}}_{2}$‍  in the original mixture

The mass percent of ${\text{MgCl}}_{2}$ in the original mixture can be calculated using the ratio of the mass of ${\text{MgCl}}_{2}$ from Step $3$ and the mass of the mixture.
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:
${\text{Mass of MgCl}}_{2}=\underset{⏟}{1.032\phantom{\rule{0.167em}{0ex}}\overline{)\text{g AgCl}}×\frac{1\phantom{\rule{0.167em}{0ex}}\overline{)\text{mol AgCl}}}{143.32\phantom{\rule{0.167em}{0ex}}\overline{)\text{g AgCl}}}}×\underset{⏟}{\frac{1\phantom{\rule{0.167em}{0ex}}\overline{){\text{mol MgCl}}_{2}}}{2\phantom{\rule{0.167em}{0ex}}\overline{)\text{mol AgCl}}}}×\underset{⏟}{\frac{95.20\phantom{\rule{0.167em}{0ex}}\overline{){\text{g MgCl}}_{2}}}{1\phantom{\rule{0.167em}{0ex}}\overline{){\text{mol MgCl}}_{2}}}}=0.3427\phantom{\rule{0.167em}{0ex}}{\text{g MgCl}}_{2}$

## 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 ${\text{AgNO}}_{3}\left(aq\right)$ in excess
What would happen to our results in the above situations?

### 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 $\text{AgCl}\left(s\right)$ than we actually do (since we are now measuring the mass of $\text{AgCl}\left(s\right)$ plus the residual water). A higher mass of $\text{AgCl}\left(s\right)$ will result in calculating more moles of $\text{AgCl}\left(s\right)$ in Step $1$, which will be converted into more moles of ${\text{MgCl}}_{2}\left(s\right)$ in our mixture. In the last step, we will end up calculating that the mass percent of ${\text{MgCl}}_{2}\left(s\right)$ 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:
This equation tells us (incorrectly!) that for every mole of $\text{AgCl}\left(s\right)$ we make, we can infer that we started with $1$ mole of ${\text{MgCl}}_{2}$ in the original mixture. When we use that stoichiometric ratio to calculate the mass of ${\text{MgCl}}_{2}$, we will get:
${\text{Mass of MgCl}}_{2}=1.032\phantom{\rule{0.167em}{0ex}}\overline{)\text{g AgCl}}×\frac{1\phantom{\rule{0.167em}{0ex}}\overline{)\text{mol AgCl}}}{143.32\phantom{\rule{0.167em}{0ex}}\overline{)\text{g AgCl}}}×\underset{⏟}{\frac{1\phantom{\rule{0.167em}{0ex}}\overline{){\text{mol MgCl}}_{2}}}{1\phantom{\rule{0.167em}{0ex}}\overline{)\text{mol AgCl}}}}×\frac{95.20\phantom{\rule{0.167em}{0ex}}\overline{){\text{g MgCl}}_{2}}}{1\phantom{\rule{0.167em}{0ex}}\overline{){\text{mol MgCl}}_{2}}}=0.6854\phantom{\rule{0.167em}{0ex}}{\text{g MgCl}}_{2}$
We just calculated that the mass of ${\text{MgCl}}_{2}$ in our mixture is double the correct amount! This will result in overestimating the mass percent of ${\text{MgCl}}_{2}$ by a factor of $2$:

### Situation $3$‍ : Adding ${\text{AgNO}}_{3}\left(aq\right)$‍  in excess

In the last scenario we wonder what would happen if we didn't add ${\text{AgNO}}_{3}\left(aq\right)$ in excess. We know this would be bad because if ${\text{AgNO}}_{3}\left(aq\right)$ is not in excess, we will have unreacted ${\text{Cl}}^{-}$ in solution. That means the mass of $\text{AgCl}\left(s\right)$ will no longer be a measure of the mass of ${\text{MgCl}}_{2}$ in the original mixture since we won't be accounting for the ${\text{Cl}}^{-}$ still in solution. Therefore, we will underestimate the mass percent of ${\text{MgCl}}_{2}$ 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 ${\text{AgNO}}_{3}\left(aq\right)$ in excess?
If we knew the answer to that question, we could be extra confident in our calculations! In this problem:
• We have $0.7209\phantom{\rule{0.167em}{0ex}}\text{g}$ of a mixture that contains some percentage of ${\text{MgCl}}_{2}$.
• We also know from our balanced equation that for each mole of ${\text{MgCl}}_{2}$, we will need $2$ moles of ${\text{AgNO}}_{3}\left(aq\right)$ at a minimum.
It is okay if we have extra ${\text{AgNO}}_{3}\left(aq\right)$, since once all the ${\text{Cl}}^{-}$ has reacted, the rest of the ${\text{AgNO}}_{3}$ will simply stay part of the solution which we will be able to filter away.
If we don't know how many moles of ${\text{MgCl}}_{2}$ are in our original mixture, how do we calculate the number of moles of ${\text{AgNO}}_{3}$ necessary to add? We know that the more moles of ${\text{MgCl}}_{2}$ we have in our original mixture, the more moles of ${\text{AgNO}}_{3}$ we need. Luckily, we have enough information to prepare for the worst case scenario, which is when our mixture is $100\mathrm{%}\phantom{\rule{0.167em}{0ex}}{\text{MgCl}}_{2}$. This is the maximum amount of ${\text{MgCl}}_{2}$ we can possibly have, which means this is when we will need the most ${\text{AgNO}}_{3}$.
Let's pretend that we have $100\mathrm{%}\phantom{\rule{0.167em}{0ex}}{\text{MgCl}}_{2}$. How many moles of ${\text{AgNO}}_{3}$ will we need? This is another stoichiometry problem! We can calculate the number of moles of ${\text{AgNO}}_{3}$ by converting the mass of the sample to moles of ${\text{MgCl}}_{2}$ using the molecular weight, and then converting to the moles of ${\text{AgNO}}_{3}$ using the molar ratio:
${\text{mol of AgNO}}_{3}=0.7209\phantom{\rule{0.167em}{0ex}}\overline{){\text{g MgCl}}_{2}}×\frac{1\phantom{\rule{0.167em}{0ex}}\overline{){\text{mol MgCl}}_{2}}}{95.20\phantom{\rule{0.167em}{0ex}}\overline{){\text{g MgCl}}_{2}}}×\frac{2\phantom{\rule{0.167em}{0ex}}{\text{mol AgNO}}_{3}}{1\phantom{\rule{0.167em}{0ex}}\overline{){\text{mol MgCl}}_{2}}}=1.514×{10}^{-2}\phantom{\rule{0.167em}{0ex}}{\text{mol AgNO}}_{3}$
This result tells us that even if we don't know exactly how much ${\text{MgCl}}_{2}$ we have in our mixture, as long as we add at least $1.514×{10}^{-2}\phantom{\rule{0.167em}{0ex}}{\text{mol AgNO}}_{3}$ we should be good to go!

## Summary

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 ${\text{Ag}}^{+}$ to quantify the amount of a halide ion such as ${\text{Br}}^{-}\left(aq\right)$. 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 $0.4015\phantom{\rule{0.167em}{0ex}}\text{g}$ of a mixture of ${\text{MgCl}}_{2}$ and $\text{NaCl}$. We add an excess of ${\text{AgNO}}_{3}\left(aq\right)$ and find that we have $1.032\phantom{\rule{0.167em}{0ex}}\text{g}$ of the precipitate, $\text{AgCl}\left(s\right)$.
How many moles of ${\text{MgCl}}_{2}$ and $\text{NaCl}$ did we have in our original mixture?
Express your answers with $4$ significant digits.
${\text{mol MgCl}}_{2}$
$\text{mol NaCl}$

## Want to join the conversation?

• 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?
• 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.
• 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?
• 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.
• 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)
• 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.
• In the example, why is NaNO3 mentioned in the question but not used in the equation?
• why does sodium nitrate not take part in the reaction? Doesn't chemical (all) react once it is mixed?
• 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
• 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?
• 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.
• The "just for fun" answer is totally unclear. How do you reach the first equation of Hint 1?
• What does (aq) stAnd for?
• (aq) stands for aqueous. This means that the compound has been dissolved in water and is now in solution.
For example, if you dissolve (solid) table salt in water you have: NaCl(s) --> NaCl(aq)
• 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?
• 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?