Gravimetric Analysis: Videos & Practice Problems

In gravimetric analysis, the mass of a product from a chemical reaction helps determine the original analyte's amount. For example, a 25 mL solution of bromide ion precipitated 0.7550 grams of lead(II) bromide. The molarity of bromide ion is calculated using the formula: $M=\frac{\mathrm{moles}}{\mathrm{liters}}$. Converting 25 mL to liters gives 0.025 L, and stoichiometry reveals 4.114 x 10^-3 moles of bromide ion, resulting in a molarity of 0.1646 M.

Electrogravimetric Analysis Example 1

In gravimetric analysis, the mass of a product within a chemical reaction can be used to determine the amount of the original analyte. Now, if we take a look here at this example, it states a 25 milliliter solution containing a bromide ion was treated with excess lead (II) sulfate to precipitate 0.7550 grams of lead (II) bromide. What was the molarity of the bromide ion in the unknown? Alright. So, we're being asked to determine the molarity of the bromide ion.

The molarity of bromide ion would equal the moles of bromide ion divided by liters of bromide ion solution. Now, from the question, we can see that we know the milliliters of the bromide ion solution; it's 25 mL. So, what I'm going to do here first is to convert 25 mL into liters. Remember, 1 milli is ${10}^{-3}$ liters.

So, that comes out to be 0.025 liters. We've determined the bottom portion of this molarity. Now, we need to determine the moles of bromide ion. We know the amount of product obtained in the chemical reaction, so we can rely on stoichiometry and use those grams of lead (II) bromide to help us determine the moles of bromide ion. We have 0.7550 grams of lead (II) bromide.

It's composed of 1 lead and 2 bromines within the compound. When you look up their atomic masses from the periodic table and add them together, you'll get approximately 367.008 grams of lead (II) bromide, and that's for every 1 mole. Grams of lead (II) bromide cancel out. Now, we just have to look at our balanced chemical equation and we see that for every 2 moles of bromide ion, I have 1 mole of lead (II) bromide. So, 2 moles to 1 mole.

So, for every 1 mole of lead (II) bromide, I have 2 moles of bromide ion. When we work that out, we'll get $4.114\times {10}^{-3}$ moles of bromide ion. Take those moles and plug them into our molarity equation. That's ${10}^{-3}$. So when we plug that in, we get 0.1646 molar bromide ion.

Now realize here that our number at the end has 4 significant figures because 25.00 has 4 sig figs and 0.7550 grams also has 4 significant figures. Because of that, our answer will have 4 significant figures. Now that we've seen this example, let's move on to the next question as we delve deeper into gravimetric analysis.

Electrogravimetric Analysis Practice

So here in this practice question, it states, the iron in a 1.1530 gram sample of iron ore is precipitated as iron (III) oxide connected to an unknown amount of water molecules by the addition of ammonia. The residue is ignited at high temperatures to give 0.6310 grams of pure iron (III) oxide. Calculate the weight percent of iron in the ore. Alright. So they want us to determine the mass percent or weight percent of iron within our sample of iron ore.

We're going to say here that the weight percent or mass percent of iron equals the grams of iron divided by the mass of the ore, so grams of ore. Now we already know how much of our iron ore we have. We're told that in the very first sentence, so we're going to plug that in. So that's 1.1530 grams of ore. What we need to do next though is we need to determine the amount of our iron analyte, and the way we're able to determine that is by using the mass of the product that we obtained from the chemical reaction.

Alright. So we're going to take these 0.6310 grams of iron (III) oxide, we're going to change it into moles. So we're going to say 1 mole of iron (III) oxide. How many grams are involved? Well, it's composed of 2 irons and 3 oxygens, so the combined molar mass of it would come out to 159.687 grams.

Here, the grams of iron (III) oxide cancel out and we're left with moles of iron (III) oxide. Now we need to make the jump from moles of iron (III) oxide to just moles of iron. If we look at our balanced equation, we can see that it's a 1 to 2 relationship. We can assume that, oh actually a 1 to 2 relationship between iron (III) oxide and iron ion, and we can say that that iron originates from the pure non-charged iron form. So we can make that jump.

So we're going to say here that for every one mole of iron (III) oxide, we have 2 moles of iron. So these moles cancel out, and then we're going to say here that we need grams of iron, so we're going to say here for every 1 mole of iron, its atomic mass according to the periodic table is 55.845 grams. Moles of iron cancel out and at the end we'll have 0.44134 grams of iron. Take those grams of iron and plug them in, multiply this by 100 to get our percent, and when we do that we get approximately 38.3% as the weight percent of our iron within this ore sample. So, again, with the information from the amount product obtained, we're able to gain information on the amount of our original analyte and by extension, in this case, the weight percent of that analyte.

So now that we've seen these problems, let's move on to our next question dealing with gravimetric analysis.

Electrogravimetric Analysis Calculations Example 1

So here in this example, it states, the reaction between Piperazine and acetic acid creates an adduct product known as Piperazine diacetate. So here we have Piperazine, 1 mole of Piperazine, reacting with 2 moles of acetic acid that produces this larger product here, 1 mole of it. Now here it says a 7.50 gram sample of impure Piperazine contained 83.01% Piperazine. How many grams of Piperazine diacetate would be formed in the process? Alright.

So here, they're actually asking us to determine the amount of our product, the grams of this, and they're giving us basically the grams of the impure Piperazine and its mass percent. With this information, we should be able to determine the mass of the product. Up to this point, we've been given the mass of our product and been able to use that to find the original analyte. In this case, now, we're given the analyte and we're just trying to figure out how much product we have. So you can approach this in the same way as any dimensional analysis question coupled with stoichiometry.

So we have 7.50 grams of our impure Piperazine, so that actually represents grams of our solution. Because if it were pure Piperazine, then that would be 7.50 grams of it. Now they're giving us mass percent of Piperazine. What this is really saying is that for every 83.01 grams of Piperazine, there are 100 grams of our solution. And by lining it up and setting it up this way, we can see that the grams of solution will cancel out.

Now that we have our grams of Piperazine, we can just use stoichiometry to go from grams of Piperazine to moles of Piperazine to moles of our product. Alright. So Piperazine itself, remember that every corner is a carbon. Carbon itself must make 4 bonds, and those carbons to make 4 bonds, they have hydrogens that are not visible. K.

So this formula here has 4 carbons, it has 10 hydrogens and 2 nitrogens. When we add all of that up, that'll give us the molar mass of Piperazine. So when we add all that up, that gives us 86.136 grams, and then that's for every 1 mole of it. Grams of Piperazine cancel out and now we're gonna figure out, the moles of our Piperazine diacetate. We can see from the balanced equation up above that it's a one-to-one relationship.

So for every 1 mole of Piperazine, we have 1 mole of our product and then 1 mole, and to make things easier, I'll just abbreviate it as PA or PD, actually, for Piperazine diacetate. Moles cancel out. I now have moles of my product, so all we have to do now is just convert those moles into grams. When we add up all the portions of the structure, its combined mass comes out to 206.236 grams for every 1 mole. Moles cancel out, so at the end, we're gonna have grams of Piperazine.

So when we work that out, we're gonna get 14.9 grams. And here, my answer has 3 sig figs because 7.50 has 3 sig figs and this percentage here has 4 sig figs. Right? We always go with the least number of sig figs, so 14.9 will be our best answer. Now just remember gravimetric analysis, we're usually given the mass of our product and with that, we're able to determine the amount of our original analyte.

In this case, I'm giving us information on the analyte and use that to figure out the amount of product. Now that we've seen this question, move on to the next.

Electrogravimetric Analysis Calculations Example 2

So in example 2, it says the amount of iron within an ore sample was determined by an oxidation-reduction titration using potassium permanganate or KMnO₄ as the titrant. We're told that a 0.5600 gram sample of the ore was placed into acid, and the newly freed iron (III) ion was then reduced to iron (II) ion. The titration of the solution required 39.82 milliliters of 0.0315 molar potassium permanganate to reach the endpoint. Determine the mass percent of iron oxide in the sample. Alright, so in this question, we're dealing with volumetric determination.

We're trying to figure out the content of iron within this sample by using the known molarity or known concentration of a standard of potassium permanganate. Alright, so we want to find the mass percent of iron (III) oxide in the sample. So that would be mass percent of iron oxide equals the grams of iron oxide divided by grams of sample times 100. We're already given one of these right off the bat. We're told that our sample weighs 0.5600 grams.

So that's going to go here on the bottom. What we have to do now is we have to use stoichiometry in order to isolate the grams of iron oxide. So, we're going to do that by using the only other piece of information given to us that we have 39.82 ml of 0.0315 molar potassium permanganate. Remember, if we can change these ml's into liters and multiply it by the molarity, we'll get the moles of potassium permanganate. At that point, we'll use stoichiometry to isolate the grams of iron (III) oxide.

So, we're going to say we have 39.82 ml's. For every 1 liter, it's 1000 milliliters. So remember, that molarity really means 0.0315 moles of potassium permanganate over 1 liter. Now at this point, we should realize that we have a balanced equation before us, but it doesn't have exactly potassium permanganate within it. What it has closest to that is just permanganate ion.

So, we're going to convert our moles of potassium permanganate into moles of just permanganate. And we're going to say here, according to the relationship, for every one mole of my entire compound of potassium permanganate, we can see that there's exactly one mole of just permanganate. Now that we have moles of permanganate, now we can start and we can keep going to try to get to grams of iron (III) oxide. Iron (III) oxide is also not in this balanced equation, but iron (III) oxide has in it the iron (III) ion, which is part of this equation. So we're going to continue by saying that according to my balanced equation, and moles of iron (III) go up here.

According to my balanced equation, for every one mole of this, there are 5 moles of this. So, we have that now and because of that we can establish a relationship with iron (III) oxide. We're going to say for every one mole of the entire compound of iron (III) oxide, we can see that there are 2 moles of iron (III) oxide. And then finally, we need grams. So for every 1 mole of iron oxide, how many grams of iron oxide do we have?

So for that, we have to calculate the weight. So if we look at our periodic table, we have 2 irons and 3 oxygens in the compound. On the periodic table, iron weighs 55.845 grams. Oxygen weighs 15.99994 grams, so that gives us 111.69+47.9982, which together is 159.688 grams. We take that and plug it in.

So we've just isolated our grams of iron (III) oxide. Remember, analytical chemistry is the chemistry of precision, so we cannot round until the very end. So, we're going to write all these numbers down. So, there goes the grams of iron (III) oxide. So, take that and plug it into my mass percent formula.

So that's going to give me a percentage of 89.42%. And we could just do 89.4, which has 3 significant figures like this number here has 3 significant figures. So just remember, the wording of many of these questions can be complicated at times, but what we've always done is we've written down first what they're asking me to find, then from that write down all the given information and try to decipher what parts can cancel out to get my desired units at the end. Take this to heart as we approach further questions dealing with titrations and eventually back titrations. Utilize the techniques we've learned to get the answers to those questions as well.

Here’s what students ask on this topic:

What is gravimetric analysis and how is it used to determine the amount of an analyte?

Gravimetric analysis is a method in analytical chemistry where the mass of a product from a chemical reaction is used to determine the amount of the original analyte. This technique involves precipitating the analyte as a solid, filtering, drying, and weighing the precipitate. The mass of the precipitate is then used to calculate the amount of the analyte based on stoichiometry. For example, if a solution containing bromide ions is treated with excess lead(II) sulfate to form lead(II) bromide, the mass of the lead(II) bromide precipitate can be used to determine the concentration of bromide ions in the original solution.

How do you calculate the molarity of an ion in a solution using gravimetric analysis?

To calculate the molarity of an ion in a solution using gravimetric analysis, follow these steps: First, convert the volume of the solution from milliliters to liters. Then, determine the mass of the precipitate formed. Use the molar mass of the precipitate to convert the mass to moles. Next, use the stoichiometry of the reaction to find the moles of the ion in the solution. Finally, divide the moles of the ion by the volume of the solution in liters to find the molarity. For example, if 0.7550 grams of lead(II) bromide is formed from a 25 mL solution, the molarity of bromide ions can be calculated as follows:

$\frac{4.114\times {10}^{-3} \mathrm{mol}}{0.025 L}=0.1646 M$

What are the steps involved in performing a gravimetric analysis?

The steps involved in performing a gravimetric analysis are as follows: 1) Prepare the sample solution. 2) Add a reagent to precipitate the analyte. 3) Filter the precipitate from the solution. 4) Wash the precipitate to remove impurities. 5) Dry the precipitate to a constant mass. 6) Weigh the dried precipitate. 7) Calculate the amount of analyte based on the mass of the precipitate and the stoichiometry of the reaction. These steps ensure accurate determination of the analyte's amount in the original sample.

What are some common applications of gravimetric analysis in real-world scenarios?

Gravimetric analysis is widely used in various real-world applications, including: 1) Determining the purity of a substance by measuring the amount of a specific component. 2) Analyzing the composition of ores and minerals in the mining industry. 3) Measuring the concentration of pollutants in environmental samples, such as water and air. 4) Quality control in the pharmaceutical industry to ensure the correct dosage of active ingredients. 5) Food industry for analyzing the content of specific nutrients or contaminants. These applications highlight the importance of gravimetric analysis in different fields.

What are the advantages and limitations of gravimetric analysis?

Gravimetric analysis offers several advantages, including high accuracy and precision, simplicity of the procedure, and the ability to determine the analyte without the need for expensive instrumentation. However, it also has limitations, such as being time-consuming, requiring careful handling to avoid loss of the precipitate, and being less suitable for very low concentrations of analytes. Additionally, the method relies on the formation of a stable and pure precipitate, which may not always be achievable for all analytes.