Mass or weight percent is the percentage of a given element or compound within a solution. Now, we're gonna say that mass percent can be seen as mass component which is usually the mass of our solute divided by the total mass of our solution times 100. Now, for example, if we're given 23% NaOH, this means that we have 23%, which translates to 23 grams of NaOH over 100 grams of solution. Because we're dealing with the percentage, we always assume it's out of 100 grams. Now, we can further expand on this and say that we have 23 grams of NaOH still on top and remember a solution is made up of solute plus solvent. So, that'd be 23 grams still of NaOH plus 100 minus 23, which gives us 77 grams of our solvent. Within our calculations, it's important that you are able to see that mass percent can be broken like this, broken down into this setup 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 1. Take a look at it. Once you do attempt it on your own, if you get stuck, move on to the next video and see how I approach example 1.
Solutions: Mass Percent - Online Tutor, Practice Problems & Exam Prep
Mass Percent can be examined as the percentage of solute found within a given amount of solution.
Mass Percent in Solutions
Solutions: Mass Percent Concept 1
Video transcript
Solutions: Mass Percent Example 1
Video transcript
So here it states calculate the amount of water in kilograms that must be added to 12 grams of urea in the preparation of an 18.3% by mass solution. Here the molar mass of urea is given as 60.055 grams per mole. Alright, 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 solute. So mass of or grams of solute divided by grams of solution times 100. Pulling in what we know. Well, we know that the percentage is 18.3%, the mass of our solute is this urea here and it's still down here as 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. Alright. So what we're going to do here is we're going to realize that we need to cross-multiply, well, we need to multiply both sides here by \(12.0 + x\). So this cancels out with this. So now we have \(18.3 \times (12 + x)\) here in parenthesis. This 12 and this 100 are multiplying each other to give us 1200. Next, what we're gonna do is we can solve this in 2 ways. Either we can divide out 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 distribute it, I get \(219.6 + 18.3x = 1200\). Subtract out 219 here. And then I'm gonna get here, \(18.3x = 980.4\). Divide both sides here by 18.3. \(x = \) the mass of our water which comes out to 53.6 grams of \(H_2O\). But we don't want the answer at the end of being grams, we want it in kilograms, so we do one more conversion. For every 1 kilogram, it's equal to 1,000 grams. So, that gives me 0.0536 kilograms of water. So, just remember in this question we're talking about mass percent so use that formula to isolate our variable which in this case is grams of water. Once we do that, convert it to kilograms to get our final answer. Now that you've seen example 1, attempt to do the next example on your own, come back and see if your answer matches up with mine.
Solutions: Mass Percent Example 2
Video transcript
A solution was prepared by dissolving 51 grams of KBr, Potassium Bromide, in 310 ml 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 solute which is KBr divided by grams of solution. Here our solution is made up of our solute KBr and our 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 grams of KBr here on top and here on the bottom. Now, we need to determine the grams of water, but instead we're given milliliters. Remember that water is roughly about 1 in terms of its density. Water is 1 gram per 1 milliliter. We use that information to 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 1. So that's 310 grams 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 had to remember the density of water is 1 and use that to find its grams. Once you find all the components, plug it into the formula to get our final answer.
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