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Kurs: Chemia - program rozszerzony > Rozdział 2

Lekcja 3: Bilans reakcji w przypadku niestechiometrycznej ilości reagentów

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.

A precipitation reaction is when two aqueous ionic compounds react to form a new ionic compound that is insoluble in water. The insoluble compound is called the precipitate.

In real life, precipitation reactions can appear quite magical! This beautiful video shows a variety of precipitation reactions close up.

For more on precipitation reactions, which are a type of double replacement reaction, see this article on double replacement reactions.

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!

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 ${MgCl}_{2}$ ‍ and ${NaNO}_{3}$ ‍

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 g

of a mysterious mixture containing

{MgCl}_{2}

and

{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,

{AgNO}_{3} (a q)

, 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

1.032 g

What is the mass percent of ${MgCl}_{2}$ ‍ 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

{MgCl}_{2} (a q)

and

{AgNO}_{3} (a q)

to make

AgCl (s)

, when

{AgNO}_{3} (a q)

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(NO}_{3})_{2} (a q)

, and make sure the overall reaction is balanced. The resulting balanced chemical equation is:

${MgCl}_{2} (a q) + 2 {AgNO}_{3} (a q) \to 2 AgCl (s) + {Mg(NO}_{3})_{2} (a q)$ ‍

The balanced equation tells us that for every

1 {mol MgCl}_{2} (a q)

, which is the compound we are interested in quantifying, we expect to make

2 mol AgCl (s)

, our precipitate. We will use this molar ratio to convert moles of

AgCl (s)

to moles of

{MgCl}_{2} (a q)

. 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 ${NaNO}_{3}$ ‍.
All of the ${Cl}^{-} (a q)$ ‍ has reacted to form $AgCl (s)$ ‍. In terms of the stoichiometry, we need to make sure we add an excess of the precipitating agent ${AgNO}_{3} (a q)$ ‍ so all of the ${Cl}^{-} (a q)$ ‍ from ${MgCl}_{2} (a q)$ ‍ 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.

$mol of AgCl (s) = 1.032 g AgCl \times \frac{1 mol AgCl}{143.32 g AgCl} = 0.007201 mol AgCl = 7.201 \times 10^{- 3} mol AgCl$ ‍

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

We can convert the moles of

AgCl (s)

, the precipitate, to moles of

{MgCl}_{2} (a q)

using the molar ratio from the balanced equation.

${mol of MgCl}_{2} (a q) = 7.201 \times 10^{- 3} mol AgCl \times \frac{1 {mol MgCl}_{2}}{2 mol AgCl} = 3.600 \times 10^{- 3} {mol MgCl}_{2}$ ‍

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

Since we are interested in calculating the mass percent of

{MgCl}_{2}

in the original mixture, we will need to convert moles of

{MgCl}_{2}

into grams using the molecular weight.

${Mass of MgCl}_{2} = 3.600 \times 10^{- 3} {mol MgCl}_{2} \times \frac{95.20 {g MgCl}_{2}}{1 {mol MgCl}_{2}} = 0.3427 {g MgCl}_{2}$ ‍

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

The mass percent of

{MgCl}_{2}

in the original mixture can be calculated using the ratio of the mass of

{MgCl}_{2}

from Krok

3

and the mass of the mixture.

${Mass% MgCl}_{2} = \frac{0.3427 {g MgCl}_{2}}{0.7209 g mixture} \times 100 % = 47.54 % {MgCl}_{2} in mixture (Thanks Igor!)$ ‍

Shortcut: We could also combine Kroks

1

through

3

into a single calculation which will involve careful checking of units to make sure everything cancels out properly:

${Mass of MgCl}_{2} = \underset{⏟}{1.032 g AgCl \times \frac{1 mol AgCl}{143.32 g AgCl}} \times \underset{⏟}{\frac{1 {mol MgCl}_{2}}{2 mol AgCl}} \times \underset{⏟}{\frac{95.20 {g MgCl}_{2}}{1 {mol MgCl}_{2}}} = 0.3427 {g MgCl}_{2}$ ‍

$Step 1: Step 2: Step 3:$ ‍
$find mol AgCl use mole ratio {find g MgCl}_{2}$ ‍

Potencjalne źródła błędu

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 ${AgNO}_{3} (a q)$ ‍ 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

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 Krok

1

, which will be converted into more moles of

{MgCl}_{2} (s)

in our mixture. In the last step, we will end up calculating that the mass percent of

{MgCl}_{2} (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:

${MgCl}_{2} (a q) + {AgNO}_{3} (a q) \to AgCl (s) + {Mg(NO}_{3})_{2} (a q) (Warning : Not balanced!)$ ‍

This equation tells us (incorrectly!) that for every mole of

AgCl (s)

we make, we can infer that we started with

1

mole of

{MgCl}_{2}

in the original mixture. When we use that stoichiometric ratio to calculate the mass of

{MgCl}_{2}

, we will get:

{Mass of MgCl}_{2} = 1.032 g AgCl \times \frac{1 mol AgCl}{143.32 g AgCl} \times \underset{⏟}{\frac{1 {mol MgCl}_{2}}{1 mol AgCl}} \times \frac{95.20 {g MgCl}_{2}}{1 {mol MgCl}_{2}} = 0.6854 {g MgCl}_{2}

wrong molar ratio!

We just calculated that the mass of

{MgCl}_{2}

in our mixture is double the correct amount! This will result in overestimating the mass percent of

{MgCl}_{2}

by a factor of

2

${Mass% MgCl}_{2} = \frac{0.6854 {g MgCl}_{2}}{0.7209 g mixture} \times 100 % = 95.08 % {MgCl}_{2} in mixture (Compare to 47.54% !!)$ ‍

Situation $3$ ‍: Adding ${AgNO}_{3} (a q)$ ‍ in excess

In the last scenario we wonder what would happen if we didn't add

{AgNO}_{3} (a q)

in excess. We know this would be bad because if

{AgNO}_{3} (a q)

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

{MgCl}_{2}

in the original mixture since we won't be accounting for the

{Cl}^{-}

still in solution. Therefore, we will underestimate the mass percent of

{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 ${AgNO}_{3} (a q)$ ‍ in excess?

If we knew the answer to that question, we could be extra confident in our calculations! In this problem:

We have $0.7209 g$ ‍ of a mixture that contains some percentage of ${MgCl}_{2}$ ‍.
We also know from our balanced equation that for each mole of ${MgCl}_{2}$ ‍, we will need $2$ ‍ moles of ${AgNO}_{3} (a q)$ ‍ at a minimum.

It is okay if we have extra

{AgNO}_{3} (a q)

, since once all the

{Cl}^{-}

has reacted, the rest of the

{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

{MgCl}_{2}

are in our original mixture, how do we calculate the number of moles of

{AgNO}_{3}

necessary to add? We know that the more moles of

{MgCl}_{2}

we have in our original mixture, the more moles of

{AgNO}_{3}

we need. Luckily, we have enough information to prepare for the worst case scenario, which is when our mixture is

100 % {MgCl}_{2}

. This is the maximum amount of

{MgCl}_{2}

we can possibly have, which means this is when we will need the most

{AgNO}_{3}

Let's pretend that we have

100 % {MgCl}_{2}

. How many moles of

{AgNO}_{3}

will we need? This is another stoichiometry problem! We can calculate the number of moles of

{AgNO}_{3}

by converting the mass of the sample to moles of

{MgCl}_{2}

using the molecular weight, and then converting to the moles of

{AgNO}_{3}

using the molar ratio:

${mol of AgNO}_{3} = 0.7209 {g MgCl}_{2} \times \frac{1 {mol MgCl}_{2}}{95.20 {g MgCl}_{2}} \times \frac{2 {mol AgNO}_{3}}{1 {mol MgCl}_{2}} = 1.514 \times 10^{- 2} {mol AgNO}_{3}$ ‍

This result tells us that even if we don't know exactly how much

{MgCl}_{2}

we have in our mixture, as long as we add at least

1.514 \times 10^{- 2} {mol AgNO}_{3}

we should be good to go!

It turns out that due to something called the common-ion effect, having an excess of

{Ag}^{+} (a q)

helps maximize the formation of precipitate. The reason for this is that the reaction to from solid

AgCl

is technically reversible due to a small fraction of the

AgCl

being able to dissolve back in the solution, which is represented in the following reaction:

$AgCl (s) ⇋ {Ag}^{+} (a q) + {Cl}^{-} (a q)$ ‍

As long as there is some solid

AgCl

around, there will be an equilibrium between the very very very small amount of dissociated

{Ag}^{+} (a q)

and

{Cl}^{-} (a q)

and the solid

AgCl (s)

. While this particular equilibrium favors the reactant side (or else this would not be very useful for gravimetric analysis!), we can drive the equilibrium even further to the reactant side by adding more

{Ag}^{+}

. This is an application of Le Châtelier’s principle!

Podsumowanie

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}^{-} (a q)

. 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 g

of a mixture of

{MgCl}_{2}

and

NaCl

. We add an excess of

{AgNO}_{3} (a q)

and find that we have

1.032 g

of the precipitate,

AgCl (s)

How many moles of ${MgCl}_{2}$ ‍ and $NaCl$ ‍ did we have in our original mixture?

Express your answers with

4

significant digits.

{mol MgCl}_{2}

mol NaCl

This time we have

2

sources of

{Cl}^{-}

, so the mass of the precipitate will reflect the sum of the

{Cl}^{-}

ions from

{MgCl}_{2} (a q)

and

NaCl (a q)

That means the following relationship is true:

${mols of Cl}^{-} = moles of NaCl + ({moles of MgCl}_{2} \times \frac{2 {mol Cl}^{-}}{1 {mol MgCl}_{2}}$ ‍)

We can make this a little tidier by renaming some variables. Let's say:

$m = {moles of MgCl}_{2}$ ‍
$n = moles of NaCl$ ‍

Now our equation looks like this:

${mols of Cl}^{-} = n + 2 m$ ‍

We can calculate

{mols of Cl}^{-}

based on the moles of precipitate (in fact, we already did this above in Krok 1 of our article example). Therefore, we know:

${mol Cl}^{-} = 1.032 g AgCl \times \frac{1 mol AgCl}{143.32 g AgCl} \times \frac{1 {mol Cl}^{-}}{1 mol AgCl} = 7.201 \times 10^{- 3} {mol Cl}^{-}$ ‍

We can use this value in our equation from Hint 1:

${mols of Cl}^{-} = 7.201 \times 10^{- 3} mol = n + 2 m$ ‍

We now have an equation with

2

variables,

m

and

n

. In order to solve for the

2

unknowns, we need another equation describing a relationship between them. Luckily, we know the total mass of our mixture, which is also related to the

{moles of MgCl}_{2} (m)

and

moles of NaCl (n)

$mass of mixture = (MW of NaCl \times mol of NaCl) + ({MW of MgCl}_{2} \times {mol of MgCl}_{2})$ ‍

$0.4015 = 58.44 n + 95.20 m$ ‍

Since we have

2

different equations for our

2

unknowns, we can now solve for

n

and

m

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Chemia - program rozszerzony

Kurs: Chemia - program rozszerzony > Rozdział 2

What is precipitation gravimetry?

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

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

Step 2‍ : Convert moles of precipitate to moles of MgCl2‍

Step 3‍ : Convert moles of MgCl2‍ to mass in grams

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