Solutions: Mass Percent - Video Tutorials & Practice Problems

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Mass Percent can be examined as the percentage of solute found within a given amount of solution.

Mass Percent in Solutions

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Solutions: Mass Percent Concept 1

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mass or weight percent is the percentage of a given element or compound within a solution. Now we're going to say that mass percent can be seen as mass component, which is usually the mass of our salute divided by the total mass off our solution. Times 100 now, for example, forgiven 23% n a o h. This means that we have 23% which translates to 23 g of N A. O H. Over 100 g of solution. Because we're dealing with the percentage we always assume it's out of 100 g. Now we can further expand on this and say that we have 23 g of N a. O. H. Still on top. And remember, a solution is made up of Salyut plus solvent, so I'll be 23 g still of an A O. H. Plus 100 minus 23 which gives us 77 77 g of our solvent. Within our calculations, it's important that you are able to see that mass percent could be broken like this broken down into this set up, and it can further be expanded where we look at both the solute and solvent components individually. Keep this in mind as we approach more questions dealing with mass percent. Now look, at example. One. Take a look at it once you do attempted on your own. If you get stuck, move onto the next video and see how I approach example. One.

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Solutions: Mass Percent Example 1

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So here it states calculate the amount of water in kilograms that must be added to g of Yuria in the preparation of an 18.3% by mass solution. Here, the molar mass of Yuria has given a 60.55 g per mole. All right, so we need to determine what the kilograms of water are. And we're dealing here with mass percent, so mass percent here would equal the grams of our solid ute so mass off or grams off Salyut divided by grams of solution times 100 plug in what we know. Well, we know that the percentages 18.3% the mass of our salute is this Yuria here and it's still down here is well, we don't know what the mass of our water is, so that's going to be our X. And this is getting multiplied by 100. All right, so what we're gonna do here is we're going to realize that we need to cross multiple. We need toe multiply both sides here by 12 0 plus x. So this cancels out with this. So now we have 18.3 times 12 plus X here in parentheses. This 12 and this 100 are multiplying each other to give us 1200. Next, what we're gonna do is we could solve this in two ways. That we can divide up the 18.3 right now or we can distribute it. It's up to you which method you want to use. So here, when I distributed I get to 19.6 plus 18.3 X equals 1200 so tracked out to 19.6 here. And then I'm gonna get here 18 3 X equals 9 84. Divide both sides here by 18. 3 X equals the mass of our water, which comes out to 53 0. g of H 20 But we don't want the answer at the end of the Ingrams who wanted to be in kilograms. So we do one more conversion for every 1 kg. It's equal to 1000 g. So that gives me 0.0 536 kg of water. So just remember this question. We're talking about mass percent, so use that formula toe isolate are variable, which in this case, is grams of water. Once we do that converted 2 kg to get our final answer. Now that you've seen example, one attempt to do the next example on your own, come back and see if your answer matches up with mine.

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Solutions: Mass Percent Example 2

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Ah, solution was prepared by dissolving 51 g of KBR, potassium bromide and 310 MLS of water. Calculate the mass percent of potassium bromide in the solution. Alright, so that's going to be mass percent equals the grams of our salute, which is KBR divided by grams of solution here. Our solution is made up of our salute KBR and are solvent water. So we need grams of KBR plus grams of water and this gets multiplied by 100. We already know the grams of KBR, so we can immediately plug those in. So that's 51 g of KBR here on top and here on the bottom. Now we need to determine the grams of water, but instead were given. Mill leaders remember that water is roughly about one in terms of its density, So water is 1 g per one millimeter. We use that information, help us get the grams of water. Now, this is a piece of information you just have to remember. The density of water is around roughly one. So that's 310 g that we have of water. So plug that in and this gets multiplied by 100 and it gives us, at the end 14.1%. So this question wasn't that bad. You just have to remember the density of water is one. And use that to find its grams. Once you find all the components, plug it into the formula to get our final answer.