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Chemistry library
Course: Chemistry library > Unit 5
Lesson 3: Limiting reagent stoichiometry- Limiting reactant and reaction yields
- Worked example: Calculating the amount of product formed from a limiting reactant
- Introduction to gravimetric analysis: Volatilization gravimetry
- Gravimetric analysis and precipitation gravimetry
- 2015 AP Chemistry free response 2a (part 1 of 2)
- 2015 AP Chemistry free response 2a (part 2/2) and b
- Limiting reagent stoichiometry
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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 start text, M, g, C, l, end text, start subscript, 2, end subscript and start text, N, a, N, O, end text, start subscript, 3, end subscript
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, point, 7209, start text, g, end text of a mysterious mixture containing start text, M, g, C, l, end text, start subscript, 2, end subscript and start text, N, a, N, O, end text, start subscript, 3, end subscript. 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, start text, A, g, N, O, end text, start subscript, 3, end subscript, left parenthesis, a, q, right parenthesis, and observe the formation of a precipitate, start text, A, g, C, l, end text, left parenthesis, s, right parenthesis. Once the precipitate is filtered and dried, we find that the mass of the solid is 1, point, 032, start text, g, end text.
What is the mass percent of start text, M, g, C, l, end text, start subscript, 2, end subscript 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 start text, M, g, C, l, end text, start subscript, 2, end subscript, left parenthesis, a, q, right parenthesis and start text, A, g, N, O, end text, start subscript, 3, end subscript, left parenthesis, a, q, right parenthesis to make start text, A, g, C, l, end text, left parenthesis, s, right parenthesis, when start text, A, g, N, O, end text, start subscript, 3, end subscript, left parenthesis, a, q, right parenthesis 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, start text, A, g, C, l, end text, left parenthesis, s, right parenthesis. That means we just have to identify the other product, start text, M, g, left parenthesis, N, O, end text, start subscript, 3, end subscript, right parenthesis, start subscript, 2, end subscript, left parenthesis, a, q, right parenthesis, and make sure the overall reaction is balanced. The resulting balanced chemical equation is:
The balanced equation tells us that for every 1, start text, m, o, l, space, M, g, C, l, end text, start subscript, 2, end subscript, left parenthesis, a, q, right parenthesis, which is the compound we are interested in quantifying, we expect to make 2, start text, m, o, l, space, A, g, C, l, end text, left parenthesis, s, right parenthesis, our precipitate. We will use this molar ratio to convert moles of start text, A, g, C, l, end text, left parenthesis, s, right parenthesis to moles of start text, M, g, C, l, end text, start subscript, 2, end subscript, left parenthesis, a, q, right parenthesis. We are also going to make the following assumptions:
- All of the precipitate is start text, A, g, C, l, end text, left parenthesis, s, right parenthesis. We don't have to worry about any precipitate forming from the start text, N, a, N, O, end text, start subscript, 3, end subscript.
- All of the start text, C, l, end text, start superscript, minus, end superscript, left parenthesis, a, q, right parenthesis has reacted to form start text, A, g, C, l, end text, left parenthesis, s, right parenthesis. In terms of the stoichiometry, we need to make sure we add an excess of the precipitating agent start text, A, g, N, O, end text, start subscript, 3, end subscript, left parenthesis, a, q, right parenthesis so all of the start text, C, l, end text, start superscript, minus, end superscript, left parenthesis, a, q, right parenthesis from start text, M, g, C, l, end text, start subscript, 2, end subscript, left parenthesis, a, q, right parenthesis reacts.
Now let's go through the full calculation step-by-step!
Step 1: Convert mass of precipitate, start text, A, g, C, l, end text, left parenthesis, s, right parenthesis, comma to moles
Since we are assuming that the mass of the precipitate is all start text, A, g, C, l, end text, left parenthesis, s, right parenthesis, we can use the molecular weight of start text, A, g, C, l, end text to convert the mass of precipitate to moles.
Step 2: Convert moles of precipitate to moles of start text, M, g, C, l, end text, start subscript, 2, end subscript
We can convert the moles of start text, A, g, C, l, end text, left parenthesis, s, right parenthesis, the precipitate, to moles of start text, M, g, C, l, end text, start subscript, 2, end subscript, left parenthesis, a, q, right parenthesis using the molar ratio from the balanced equation.
Step 3: Convert moles of start text, M, g, C, l, end text, start subscript, 2, end subscript to mass in grams
Since we are interested in calculating the mass percent of start text, M, g, C, l, end text, start subscript, 2, end subscript in the original mixture, we will need to convert moles of start text, M, g, C, l, end text, start subscript, 2, end subscript into grams using the molecular weight.
Step 4: Calculate mass percent of start text, M, g, C, l, end text, start subscript, 2, end subscript in the original mixture
The mass percent of start text, M, g, C, l, end text, start subscript, 2, end subscript in the original mixture can be calculated using the ratio of the mass of start text, M, g, C, l, end text, start subscript, 2, end subscript 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:
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space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, start text, f, i, n, d, space, m, o, l, space, A, g, C, l, end text, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, start text, u, s, e, space, m, o, l, e, space, r, a, t, i, o, end text, space, space, space, space, space, space, start text, f, i, n, d, space, g, space, M, g, C, l, end text, start subscript, 2, end subscript, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space
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 start text, A, g, N, O, end text, start subscript, 3, end subscript, left parenthesis, a, q, right parenthesis 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 start text, A, g, C, l, end text, left parenthesis, s, right parenthesis than we actually do (since we are now measuring the mass of start text, A, g, C, l, end text, left parenthesis, s, right parenthesis plus the residual water). A higher mass of start text, A, g, C, l, end text, left parenthesis, s, right parenthesis will result in calculating more moles of start text, A, g, C, l, end text, left parenthesis, s, right parenthesis in Step 1, which will be converted into more moles of start text, M, g, C, l, end text, start subscript, 2, end subscript, left parenthesis, s, right parenthesis in our mixture. In the last step, we will end up calculating that the mass percent of start text, M, g, C, l, end text, start subscript, 2, end subscript, left parenthesis, s, right parenthesis 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 start text, A, g, C, l, end text, left parenthesis, s, right parenthesis we make, we can infer that we started with 1 mole of start text, M, g, C, l, end text, start subscript, 2, end subscript in the original mixture. When we use that stoichiometric ratio to calculate the mass of start text, M, g, C, l, end text, start subscript, 2, end subscript, we will get:
We just calculated that the mass of start text, M, g, C, l, end text, start subscript, 2, end subscript in our mixture is double the correct amount! This will result in overestimating the mass percent of start text, M, g, C, l, end text, start subscript, 2, end subscript by a factor of 2:
Situation 3: Adding start text, A, g, N, O, end text, start subscript, 3, end subscript, left parenthesis, a, q, right parenthesis in excess
In the last scenario we wonder what would happen if we didn't add start text, A, g, N, O, end text, start subscript, 3, end subscript, left parenthesis, a, q, right parenthesis in excess. We know this would be bad because if start text, A, g, N, O, end text, start subscript, 3, end subscript, left parenthesis, a, q, right parenthesis is not in excess, we will have unreacted start text, C, l, end text, start superscript, minus, end superscript in solution. That means the mass of start text, A, g, C, l, end text, left parenthesis, s, right parenthesis will no longer be a measure of the mass of start text, M, g, C, l, end text, start subscript, 2, end subscript in the original mixture since we won't be accounting for the start text, C, l, end text, start superscript, minus, end superscript still in solution. Therefore, we will underestimate the mass percent of start text, M, g, C, l, end text, start subscript, 2, end subscript 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 start text, A, g, N, O, end text, start subscript, 3, end subscript, left parenthesis, a, q, right parenthesis in excess?
If we knew the answer to that question, we could be extra confident in our calculations! In this problem:
- We have 0, point, 7209, start text, g, end text of a mixture that contains some percentage of start text, M, g, C, l, end text, start subscript, 2, end subscript.
- We also know from our balanced equation that for each mole of start text, M, g, C, l, end text, start subscript, 2, end subscript, we will need 2 moles of start text, A, g, N, O, end text, start subscript, 3, end subscript, left parenthesis, a, q, right parenthesis at a minimum.
It is okay if we have extra start text, A, g, N, O, end text, start subscript, 3, end subscript, left parenthesis, a, q, right parenthesis, since once all the start text, C, l, end text, start superscript, minus, end superscript has reacted, the rest of the start text, A, g, N, O, end text, start subscript, 3, end subscript will simply stay part of the solution which we will be able to filter away.
If we don't know how many moles of start text, M, g, C, l, end text, start subscript, 2, end subscript are in our original mixture, how do we calculate the number of moles of start text, A, g, N, O, end text, start subscript, 3, end subscript necessary to add? We know that the more moles of start text, M, g, C, l, end text, start subscript, 2, end subscript we have in our original mixture, the more moles of start text, A, g, N, O, end text, start subscript, 3, end subscript we need. Luckily, we have enough information to prepare for the worst case scenario, which is when our mixture is 100, percent, start text, M, g, C, l, end text, start subscript, 2, end subscript. This is the maximum amount of start text, M, g, C, l, end text, start subscript, 2, end subscript we can possibly have, which means this is when we will need the most start text, A, g, N, O, end text, start subscript, 3, end subscript.
Let's pretend that we have 100, percent, start text, M, g, C, l, end text, start subscript, 2, end subscript. How many moles of start text, A, g, N, O, end text, start subscript, 3, end subscript will we need? This is another stoichiometry problem! We can calculate the number of moles of start text, A, g, N, O, end text, start subscript, 3, end subscript by converting the mass of the sample to moles of start text, M, g, C, l, end text, start subscript, 2, end subscript using the molecular weight, and then converting to the moles of start text, A, g, N, O, end text, start subscript, 3, end subscript using the molar ratio:
This result tells us that even if we don't know exactly how much start text, M, g, C, l, end text, start subscript, 2, end subscript we have in our mixture, as long as we add at least 1, point, 514, times, 10, start superscript, minus, 2, end superscript, start text, m, o, l, space, A, g, N, O, end text, start subscript, 3, end subscript 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 start text, A, g, end text, start superscript, plus, end superscript to quantify the amount of a halide ion such as start text, B, r, end text, start superscript, minus, end superscript, left parenthesis, a, q, right parenthesis. 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, point, 4015, start text, g, end text of a mixture of start text, M, g, C, l, end text, start subscript, 2, end subscript and start text, N, a, C, l, end text. We add an excess of start text, A, g, N, O, end text, start subscript, 3, end subscript, left parenthesis, a, q, right parenthesis and find that we have 1, point, 032, start text, g, end text of the precipitate, start text, A, g, C, l, end text, left parenthesis, s, right parenthesis.
How many moles of start text, M, g, C, l, end text, start subscript, 2, end subscript and start text, N, a, C, l, end text did we have in our original mixture?
Express your answers with 4 significant digits.
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?(42 votes)
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.(2 votes)
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?(17 votes)
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.(21 votes)
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)
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.(15 votes)
- In the example, why is NaNO3 mentioned in the question but not used in the equation?(7 votes)
- why does sodium nitrate not take part in the reaction? Doesn't chemical (all) react once it is mixed?(5 votes)
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(7 votes)
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)
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)
- The "just for fun" answer is totally unclear. How do you reach the first equation of Hint 1?(6 votes)
What does (aq) stAnd for?(4 votes)
(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)(5 votes)
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)- What is le chatelier's principle ?(2 votes)
- You can watch this video to learn more about le Chatelier's principle:
https://www.khanacademy.org/science/chemistry/chemical-equilibrium/factors-that-affect-chemical-equilibrium/v/le-chatelier-s-principle(4 votes)
