A 67.2 g sample of a gold and palladium alloy contains 2.49 * 1023 atoms. What is the composition (by mass) of the alloy?

Video transcript

Hey everyone in this example, we need to determine the mass composition of a copper and zinc alloy with the given mass and containing 5.13 times 10 to the 23rd power atoms. Our first step is to recall our molar mass for copper, which from our periodic tables, we're going to see that it's 63.5 46 g per mole. And we're going to recall our Mueller mass for zinc, Which is going to be equal to according to our periodic tables 65. g per mole. The next step is to recognize that made up of our copper zinc alloy. We have individual copper and zinc atoms. So what we're going to use is a substitution so that X. Is equal to our atoms of copper. And why is going to equal our atoms of zinc. And so That allows us to understand that based on our alloy containing 5.13 times 10 to the 23rd power atoms. We can say that X plus Y is going to equal 5.13 times 10 to the 23rd power atoms. And so we can go ahead and sulfur Y. By saying that Y is equal to 5.13 times 10 to the 23rd atoms minus our X value. And so that means we're going to want to solve for X in order to figure out our atoms of Y. Or zinc and our atoms of copper. Now again, this question wants us to find mass composition. So we need to write an expression where we will have our units X and Y for our zinc and copper represented in mass. And so they give us our units. Or we've expressed our units in atoms based on what we are given in the question. However, what we're going to do is we're going to convert from Adams, two malls and then from moles to grams. So beginning with our X atoms of copper, what we should have is that is going to be multiplied by our conversion factor. Where we should recall that for one mole of copper. That gives us 6.022 Times 10 to the 23rd power atoms. And so this way this conversion factor makes sense to use because we can go ahead and now cancel out atoms and we're left with moles of copper. And so what we should have is that this is going to equal our X term for our copper over 6.02, 2 times 10 - 23rd power atoms. We're sorry we canceled out atoms. And so this would be in parenthesis and we would say times moles of copper. And so again we want to go from moles to g. So how do we do that? We recall our conversion factor for molar mass. We're going to multiply by that conversion factor and that's why above we said that the molar mass for copper is going to be 63.5, grams of copper for one mole of copper. And so that way we're able to cancel our units moles of copper leaving us with grams of copper as the final unit. And so this allows us to get 63. x over 6.02, 2 times 10 to the 23rd power. And then this is going to be now in grams of copper. So now we've successfully made an expression with our X term for atoms of copper in units of grams. And so now we can do the same thing for our y atoms of zinc. And so we're going to multiply again to get into molds. We're going to multiply by the conversion factor that One mole of zinc is equal to 6.02, 2 times 10-23rd power atoms. Because we want to cancel out the atom term here or unit. And so we're left with moles of zinc just like we were left with most of copper in the first part. And so this is going to equal in our numerator we're going to have instead of why we should be writing our expression for why? Which we said is right down here. So we can go ahead and write in 5.13 times 10 to the 23rd Power -X. This is going to be over six point oh 22 times 10 to the 23rd power, which doesn't have units anymore because now our units are moles of zinc for the numerator and the denominator. And now we are at moles. We want to go to grams. So we're going to multiply this by our molar mass conversion for our zinc adam. And according to our periodic tables, we said that we have 65.39 g of our zinc for one mole of zinc. And so this way we can cancel out moles of zinc with molds of zinc leaving us with grams of zinc. And this gives us our second expression, which is going to be 65. multiplied by our Y substitution, which is 5.13 times 10 to the 23rd power minus X Over 6.02, 2 times 10 to the 23rd power. And I'm just going to scoot this over a little bit so that it's all visible. This is times grams of zinc. And I'll just fix this here. Sorry about that. Okay, so now we can go ahead and take our expressions and what we should recall is that according to this question, it's telling us that our mass of copper within our alloy, we have our massive copper plus our mass of zinc which is going to equal the given 54.7 g that are alloy consists of. And so we can take these two expressions that we've just made here and plug this into this equation here to solve for our value for X to ultimately get our atoms of copper and atoms of sync. And so let's go ahead and do that. What we should have is our grams of copper. We're substituting with that first expression we formed 63. x. And I'll just make sure we keep the colors consistent X Over 6.022 times 10 to the 23rd power grams of copper Plus our expression 4g of zinc Which we determined above as 65.39. Multiplied by 5.13 times 10 to the 23rd power minus X. Which is our substitution for ry Over 6.022 times 10 to the 23rd power. And our units are grams of zinc And then this is equal to 54.7 g which is our alloy mass. And so now we want to simplify this to solve for our x term. So the first thing to do to simplify this is to multiply both sides by the denominator. So multiply both sides by 6.02, 2 times 10 to the 23rd power which we recall is our avocados number. So we cancel that out on both sides. And then Here we're going to have 6.022 times 10 to the 23rd power again. So this will simplify to 63. X. Because we have our numerator alone. Now they're not infractions no any longer. This is still going to be times grams of copper and let's just make that blue so it's not confusing. Plus we should have 65.39 our numerator times our wide substitution. 5.13 times 10 to the 23rd power minus X. And then this is still multiplied by grams of sink. And so this is set equal to the product of this multiplication here. Equal to 3.29403 times 10 to the 25th Power g. Our next step is to go ahead and simplify further by substituting this value here to both of our terms and the parentheses. So what we're going to get when we do so Is we still have that 63.546 times x. And then this is four g of copper Added to what we should get in. Our substitution is 3. times 10 to the 25th power -65. X. And this is still going to be grams of sink Units equal to we still have 3.29403 times 10 to the 25th power gramps. Now, our next step is going to be combining our like terms. So we'll just underline those so that it's clear what we're combining. So we're going to combine the 63.5 46 X times or with the negative 65.39 X term there. And what we will have when doing so in our next line is negative 1.844 X. Because we're going to add This value to this value here. So we're going to get negative 1.844 X. And this is going to be grams Added to our 3.35451 times 10 to the 25th power set equal to what we have on the right hand side, which has remained consistent 3.29403 times to the 25th power gramps. And so now it's a lot easier to simplify for X. We can just focus on our next step which is to subtract 3.35451 times 10 to the 25th power from both sides. So what we're going to get when we do so is negative 1. x grams. Set equal to the difference on the right hand side. Which is going to give us a result equal to negative 6.48 times 10. 10 to the 23rd power grams. And so what we can do next to finally isolate for X is divide both sides by negative 1.844 g. So this allows that to cancel out on the left hand side. And then we're going to have grams cancel out on the right hand side. And so ultimately we will get that our x term is equal to a value of 3.279 83 times 10 to the 23rd power. And we should recall that we said above X. Is our atoms of copper here. That's what we said above here. X is atoms of copper. So we're going to say that this value is going to be atoms And we're going to round this to about 3.279 times 10 to the 23rd power atoms of copper. So now we have our X term. We can go ahead and figure out our Y term. So we said that our y value for our zinc should be equal to 5.13 times 10 to the 23rd power atoms of our alloy Subtracted from our x value from above which will plug in as 3.279 times 10 to the 23rd power atoms of our copper. And this is going to equal A value of 1.85017 times 10 to the 23rd power atoms of zinc. And we're going to go ahead and say that we're going to round this. So that we can say that why is about one point 850 times 10 to the 23rd power atoms of zinc. So now we finally have our units for our individual atoms of copper and our individual atoms of zinc within our alloy. We can go ahead now and find our mass of copper and massive zinc. And so for the massive copper, we're going to take our expression for our copper um individually as an atom Which above we formed as 63.546 grams per mole which was the molar mass of our copper. Multiplied by our X. Value from above, which we just found as 3. times 10 to the 23rd power atoms of copper divided by Our 6.02, 2 times 10 to the 23rd power atoms per mole. Because we should recall that this is avocados number in our denominator and it's always in units of atoms per mole. So now we can go ahead and cancel out our units. So we can get rid of moles with moles, we can get rid of atoms and this leaves us with grams for our mass of copper. And so this gives us a massive copper equal to 34.6 g. We're gonna follow the same steps for the mass of zinc so that we should have Our molar mass of zinc from above which we recall as 39 g per mole, Multiplied by what we found above for our y value of our zinc as 1.850 times 10 to the 23rd power atoms of our zinc. And then this is going to be divided by our six point oh 22 times 10 to the 23rd power Adams Permal, which is our avocados number. And so now we can go ahead and cancel our units again. We can get rid of moles with moles because it's in the numerator and denominator we can get rid of atoms with atoms and we're left with grams as our final unit giving us a value equal to 20.1 g for our massive zinc. And so finally we can find our mass percent for each of these values. So we would have our mass percent first for copper equal to our numerator. We should recall for mass percent. We need to plug in our mass of our atoms. So 34.6 g for our copper atom. And then in our denominator from mass percent. We want to recall, we plug in the mass of the alloy. And so that's going to be given in the problem as 54.7 g. And what we're going to do is also, sorry, multiply this by 100 and this gives us 63.3% of copper as our first answer for the mass percent. Making up copper in our alloy. Next we can go ahead and do the same thing for our mass percent of zinc. So we're going to again plug in our mass for individual zinc atoms equal to 20.1 g of zinc. And then our denominator we should still have our 54.7 g of the alloy Multiplied by 100%. And this gives us a value equal to 36.7% of zinc. And so this would be our second and final answer to complete our mass percent of zinc that makes up our alloy and as a whole, this will complete this example as our final answers. So I hope that everything we reviewed was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.