Gravimetric Analysis - Video Tutorials & Practice Problems

In Gravimetric Analysis the mass of a product in a chemical reaction is used to calculate the amount of the original analyte.

Introduction to Gravimetric Analysis

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Electrogravimetric Analysis Example 1

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in gravimetric analysis. The massive a product within a chemical reaction can be used to determine the amount of the original. An elite. Now, if we take a look here at this example, it states a 25 millimeter solution containing bromide ion was treated with excess lead to sulfate to precipitate 250.7550 g of lead to bromide. What was the polarity of the bromide ion in the unknown? Alright, so we're being asked to determine the polarity of bromide ion. So moller charity of bromide ion would equal the moles of bromide ion divided by leaders of bromide ion solution. Now from the question question given, we can see that we know the middle leaders of bromide ion solution, it's 25 mls. So what I'm gonna do here first is I'm going to convert 25 mls into leaders. Remember one millie is 10 to the negative three leaders. So that comes out to be .025 L. So we've determined the bottom portion of this polarity. Now we need to determine the moles of bromide ion. Now we know the amount of product obtaining the chemical reaction. So we can rely on stoke geometry and use those grams of lead to bromide to help us determine the moles of bromide ion. So we have .7550 g of lead to bromide. It's composed of one lead and two bro means within the compound. When you look up their atomic masses from the periodic table and add them together, you get approximately 367.008g of lead to bromide. And that's for every one mole grams of lead to bro. Might cancel out now, we just have to look at our balanced chemical equation and we see that for every two moles of bromide ion, I have one mole of lead to bromide, So two moles to one mole. So for every one mole of lead to bromine, I have two moles of bromide ion. When we work that out, we'll get 4.114 times 10 to the minus three moles of bromide ion. Take those molds and plug them into our polarity equation. That's negative three. So when we work, when we plug that in, we get 30.1646 Mohler bromide ion. Okay, now realize here that our number at the end has four significant figures because 25.0 has four sig figs and point 7550 g also has four significant figures. Because of that, our answer will have four significant figures. Now that we've seen this example. Move on to the next question. As we delve deeper into gravimetric analysis

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Electrogravimetric Analysis Practice

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So here in this practice question, it states the iron and a 1.1530 g sample of iron ore is precipitated as iron three oxide connected to an unknown amount of water molecules by the addition of ammonia. The residue is ignited. High temperatures to give 30.6310 g of pure iron three 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 gonna 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 gonna plug that in. So that's 1.1530 g of your. What we need to do next though is we need to determine the amount of our iron an elite. 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 gonna take these .6310 g of iron to Iron three oxide. We're gonna change it into moles. So we're gonna say one mole of iron, three oxide. How many grams are involved While it's composed of two irons and three oxygen's. So the combined molar mass of it would come out to 159.687 g Here, the grams of iron three oxide cancel out And we're left with moles of iron three oxide. Now, we need to make the jump from uh moles of Iron three oxide to just moles of iron. If we look at our balanced equation, we can see that it's a 1-1 relationship. We can assume that actually a 1-2. So I wanted to relationship between Iron three Oxide and Iron three. And we can say that that iron three originates from the pure non charged iron form. So we can make that that jump. So we're gonna say here that for every one mole of Iron three oxide, We have two moles of iron three ion. And by extension, two moles of iron. So these moles cancel out. And then we're gonna say here that we need grams of iron. So we're gonna stay here for every one mole of iron. Its atomic mass according to periodic table is 55.845g. Most of iron cancel out and at the end will have .44134 g 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 or sample. So again, with the information from the amount of product obtain were able to gain information on the amount of our original an allied and by extension in this case the weight percent of the an elite. So now that we've seen these two problems, let's move on to our next question dealing with gravimetric analysis.

Electrogravimetric Analysis Calculations

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Electrogravimetric Analysis Calculations Example 1

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So here in this example, it states the reaction between piper zine and acetic acid creates an adult product known as piper scene di acetate. So here we have piper's in one mole piper seen reacting with two moles of acetic acid that produces this larger product here, one mole of it. Now here it says a 7.50 g sample of impure pipe ursine contained 83.1% piper zine. How many grams of piper scene di acetate 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 pipe racine and its mass percent. With this information, we should be able to determine the mass of the product. So up to this point, we've been given the mass of our product and been able to use that to find the original. An elite in this case. Now we're given the an elite 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 soy geometry. So we have 7.50 g of our impure pipe ursine. So that actually represents grams of our solution. Because if it was pure piper seen, then that would be 7.50 g of it. Now they're giving us mass percent of piper seen. What this is really saying is that for every 83.1 g of piper's scene, There are 100 g 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 piper's scene. We can just use geometry to go from grams of piper's in two moles of piper's seen two moles of our product. Alright, so pipe racine itself. Remember that every corner is a carbon. Carbon itself Must make four bonds and those carbons to make four bonds, they have hydrogen that are not visible. Okay, so this formula here has four carbons. It has 10 hydrogen and two nitrogen. When we add all of that up, that will give us the molar mass of piper scene. So when we add all that up, that gives us 86.136 g. And then that's for every one mole of it grams of pipers and cancel out. And now we're gonna figure out the moles of our pipers in di acetate. We can see from the balanced equation up above that, it's a 1 to 1 relationship. So for every one mole of piper's scene, we have one mole of our product and then one mole. And to make things easier, I'll just abbreviated as P. A. Or P. D. Actually for pipers in di acetate 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 g For every one mole moles cancel out. So at the end we're gonna have grams of piper's in di acetate. So when we work that out, we're gonna get 14.9 g. And here my answer has three sig figs because 7.50 has three sig figs. And this percentage here has four sig figs. Right? We always go with the least number of sig figs. So 14 point now will be our best answer. Now, just remember gravimetric analysis were usually given the mass of our product and with that we're able to determine the amount of our original. An elite. In this case, I'm giving us information on the an elite and use that to figure out the amount of product now that we've seen this question. Move on to the next

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Electrogravimetric Analysis Calculations Example 2

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So an example to it says, the amount of iron within an ore sample was determined by an oxidation reduction titrate in using potassium permanganate or came in oh four. As the titans were told that a 40.5600 g sample of the ore was placed into acid and the newly freed iron three ion was then reduced to iron two ions. The dehydration of the solution required 39.82 mL of 0.315 molar potassium permanganate to reach the endpoint determine the mass percent of iron three 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 polarity or known concentration or a standard of potassium permanganate. Alright, so we want to find the mass percent of iron three oxide in the sample. So that would be mass percent of Iron three Oxide equals the g of Iron three 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 .5600 g. So that's gonna go here on the bottom. What we have to do now is we have to use to Akiyama Tree in order to isolate the grams of iron three oxide. So, we're gonna do that by using the only other piece of information given to us that we have 39.82 mls of 0.315 molar potassium permanganate. Remember if we can change these mls into leaders and multiply it by the polarity will get the moles of potassium permanganate. At that point, we'll use geometry to isolate the g of Iron three oxide. So we're gonna say we have 39.82 mls For every one leader it's 1000 ml. So remember that polarity really means .0315 moles Of potassium permanganate over one leader. 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 gonna convert our moles of potassium permanganate into molds of just permanganate. And we're gonna 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 multiple mangan it. Now, we can start and we can keep going to try to get two g of iron three oxide. Iron three oxide is also not in this balanced equation, but iron three oxide has in it. The iron three ion, which is part of this equation. So we're gonna continue by saying that according to my balanced equation and moles of iron through. Go up here According to my balanced equation for every one mole of this, there are five moles of this. So we have that now and because of that, we can establish a relationship with iron three oxide. We're gonna say for every one mole of the entire compound of iron three oxide, we can see that there are two Moles of Iron three Oxide. And then finally we need graham. So for every one mole of iron three oxide, how many grams of iron three oxide do we have? So for that we have to calculate the weight. So if we look at the periodic table, we have two irons and three oxygen's in the compound on the periodic table. Iron weighs 55.845g, oxygen weighs 15.9994 g. So that gives us 111.69 47.9982. Which together is 159.688 g. We take that and plug it in. So we've just isolated our grams of iron. Three oxide. Remember analytical is the chemistry of precision precision, so we cannot round until the very end. So we're gonna write all these numbers down. So there goes the grams of iron three oxide. So take that and plug it into my mass percent formula. Okay, So that's gonna give me a percentage of 89.4,. And we could just do 89.4, which has 3/6 figs. Like this number here has three significant figures. So just remember the wording of a lot 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 tight rations and eventually back tight rations utilize the techniques we've learned to get the answers to those questions as well.