(b) When a brass sample with a mass of 0.544 g was sub-jected to the preceding analysis, 10.82 mL of 0.1220 M sodium thiosulfate was required for the reaction with iodine. What is the mass percent copper in the brass?

Video transcript

Welcome back everyone. A bronze sample was subjected to proceeding analysis. The reactions involved have the following unbalanced equations 10 2 plus reacting with two for my Dean and 10 4 plus and Aydin reacting with dia sulfate to form our diet an ion and our tetra an ion. If 3.6 to 18 g of sample needs 11.26 mL of 0.11 67 molar sodium sulfate. What is the mass percent of tin in the sample? Our first step is to recognize that the prompt states we have unbalanced equations. So looking at equation one versus equation to sorry, equation to here, we want to balance out equation one and recognize that we're gonna use redox. So we're going to break it up into half reactions. So our first half reaction is going to be 10 to plus county. On Which forums are 10, 4 plus Cascione balancing out our atoms. We see we have one mole of 10 on both sides. So now we can move on to balancing charges. We have a net charge of plus two on the reactant side and net charge of plus four on the product side. We're going to mend this by adding two electrons to the product side so that we now have a net charge of plus two on both sides of the reaction. And our charges are balanced. Now moving on to the second half reaction, we have I date which forms our iodine compound on the product side. We will begin by bouncing on our atoms. We have two moles of iodine on the product side and just one on the reacting side. So we'll place a coefficient of two in front of my date. Now, we're going to need to bounce out oxygen here, we have a total of six moles of oxygen. Now, with that coefficient of two, so we're going to need oxygen on the product side, specifically six moles. And we're going to recall that we bounce oxygen using water. So we're going to add six moles of water to the product side. But now with adding water, we've introduced 12 moles of hydrogen to the equation. And so we're going to need to recall that we balance hydrogen using protons, so H Plus, and so we're going to expand our react inside. And sorry, let's do this better. And we're going to add 12 moles of hydrogen. So that's 12 moles here of H Plus. Now we need to balance out charges now that all of our atoms are balanced. So we have a net charge of plus 12 minus two on the reactant side. That gives us a net charge of plus 10 and we have a neutral charge on the product side. Recall that we add electrons to the more positive side. And so we're going to cancel out that charge of plus 10 by expanding our reactant side again and we're going to add 10 electrons to cancel out that plus 10 charge. And so now we have a net charge of zero on both sides of our reaction. And now we just need to bounce out our electrons in both of the half reactions. So we have only two electrons in the first half reaction. And the second half reaction has 10. So we're going to take this entire first half reaction and multiply it by a coefficient of five. And so now going back to the first half reaction, We would have five moles of 10, 2 plus, forming five moles of 10, 4 plus Plus 10 electrons. And now we can straightforwardly add both of our half reactions. So in doing so, we would be able to cancel out our 10 electrons on the product side here and on the reactant side here and now we'll just carry everything down. So what we have overall is two moles of I date plus five moles of 10, 2 plus plus 12 protons produces, We have one mole of iodine Plus five moles of 10, 2 plus, sorry, five moles of SN two plus Plus six moles of water. And just to make a correction. The 10 2 plus should be 10 4 plus on the product side. So apologies for that area there. And this is our overall equation for the first half worth For the first equation. Now we're going to move on to our second equation where we have our first half reaction. So, using the second equation, we have iodine which produces our I died an ion Beginning by bouncing out our atoms we have two moles of iodine on the product side, sorry, on the reactant side we need two moles on the product side. So we'll place a coefficient of two On the product side. Now bouncing out net charges, we have a net charge of -2 on the product side. And so we're going to expand our react inside Because it's neutral. And we're going to add two electrons as a second reactant, which gives us a net charge of -2 on both sides of the reaction. Now, with Adams and charges bounds, we can move on to our second half reaction. And so that would be so sulfate S 2032 minus, which forms tetra tino S 406 to minus. And so beginning by bouncing out our non oxygen atoms and non hydrogen atoms. We have sulfur to bounce. We have four moles on the product side, and only two on the reactant side. So we'll place a coefficient of two in front of our sulfate on the reactant side. That now gives us four moles of sulfur and six moles of oxygen. And as we did earlier, we need to bounce oxygen using water. But looking at our product side, it does check out that we have also six moles of oxygen. So it looks like our oxygen oxygen's are already bounced. Now moving onto bouncing charges, we have a net charge of -4 on the reactant side and net charge of -2 on the product side. And so we're going to need to expand our product side and add two electrons as a product to give us a net charge of -4 on both sides of the equation. So now our charges are balanced and now looking at both half reactions, we do have matching number of electrons. So we can actually simply add up both of these half reactions. So just moving this over for room, we can place a plus sign here and now add up these together, canceling our electrons on the product side and on the reactant side here and carrying everything down we have idea Plus two moles of hi Nate Forms two moles of I died And one mole of Tetra dining. So now using info from the prompt, we want to find moles of our cyan eight. An eye on So moles of as 2032 -. And we're doing so because the prompt mentions the salt sodium dye in it. So we are told the volume of this salt needed as 11.26 ml of sodium diet. And sorry there should be a two A subscript of two next to N. A. So beginning with this volume we recall that we are also given the molar concentration of this salt in polarity and recall that form polarity. We can interpret it as moles per liter. So we need to go ahead and convert this volume from male leaders to leaders. So multiplying by our conversion factor with male leaders in the denominator and leaders in the numerator recalled that are prefix milli tells us that one leader is equivalent to 10 to the third power milliliters. So canceling out milliliters. We are at leaders and now can incorporate our concentration given in the prompt for the salt as point 1300 moles of sodium sulfate, S 203 over one more of sodium sulfate Or sorry over one leader. So with that we can cancel out our units of leaders. And now we have moles of sodium sulfate or salt where we can now go into the ratio using the formula of sodium sulfate between sodium sulfate. So N. A two S 203 and the ion that it has S 2032 minus. And so using the formula we see we have one mole of co sulfate for one mole of sodium sulfate. So that's a 1 to 1 molar ratio and we can just cancel out moles of sodium sulfate the salt. And we're left with molds of our an ion which we want. So using our calculators we should find that we get .001414638 moles of our bio sulfate an ion from our salt. And now we can go ahead and find our massive tin by using our molar ratios from our bounced equations that we came up with above. So here we would find again the mass of tin. So beginning with our molds of our an ion 0.14638 moles of sulfate as 2032 minus. We will begin by looking at the ratio between moles of sulfate as 203 to minus. And going back to our equation, we see that we can compare it to our iodine from the second bounce equation. So moles of iodine would be in the numerator and we can see that from the second balanced reaction, we have a we have one mole of iodine for two moles of sulfate. And so that's a 1-2 molar ratio. So canceling out now moles of sulfate, we can go from moles of iodine, so moles of I 22 moles of the caddy on in our original sample. And so going back up to our equation, we see that our Catalan of tin that is in the original sample would be on the reactant side and that would be our 10 2 plus catalon. So let's make note of that that are S. N two Plus is a part of the original sample. So the bronze sample and that is why we look at the molds of the 10 2 plus catalon and not the 10 4 plus cat ion, even though when we look at the reaction between or in the reaction that involves Aydin and tin. We see that the moles of 10 4 plus match the moles of tin two plus. So we can it wouldn't make a difference. But just so it's understood. We would want to look at the ratio between 10 to plus and I dine and we see that for five moles of 10 2 plus, we have one mole of iodine. So, plugging in that molar ratio, we have again five moles of tin two plus for one mole of iodine. So canceling out moles of iodine. We next want to get two moles of tin. And so we're going to let's follow on the bottom here, we have moles of 10 in the denominator and we want our final unit to be are massive. 10 so grams of 10. So we're called that from the periodic table, Tim Tinh has a molar mass of 118.71 g per mole of 10. So per one mole of 10. So now canceling out moles of 10 with our ion of 10. We're left with grams of 10 as our final unit. And this could this is going to yield in our calculators a mass of .4344 g of 10. So now with this massive tin, we want to recall our formula for mass percent. And so what we would do is take our mass Of our compound, which is 10. So mass of sn over the mass of our sample. And that is going to give us our massive tin above. In our numerator, we calculated as .4344 g of 10. The mass of the sample given in the prompt is 3.6 - 18 g of our bronze sample. And in our calculators, we also want to multiply by 100 since this is a mass percent. So by 100% g will cancel out. And this is going to yield on our calculators, a mass% 11.99 of 10. And so for our final answer, we've successfully determined the mass percent of tin as 11.99% of tin. In the bronze sample, this corresponds to choice a in the multiple choice. I hope that everything I reviewed was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video.