A sample that weighs 25.12 g contains 6.022 * 10^23 particles. If...

Question

A sample that weighs 25.12 g contains 6.022 * 10^23 particles. If 25.00% of the total number of particles are argon atoms and 75.00% are another element, what is the chemical identity of the other constituent?

Accepted Answer

Hello. Everyone in this video, we're trying to identify the other element in our alloy. We know that 80% of the sample of the alloy is containing nickel and 20% will contain the other element that we're trying to solve for. So given the percent by mass of each Or the percentages of the alloy, the most of Nickel, then would be equal to 0.80 moles. And then for the molds of our other unknown is we'll just put X is equal to 0.20 moles. Alright, let's go ahead and first saw for the mass of nickel. So, we're starting off with the 0.80 moles of Nickel. Then we'll go ahead and use the molar mass of nickel which for every one mole of nickel we get 58. g of nickel. You see here that the most Nikola council. All right now give us the g being 46.9547 g of nickel. Next, we can solve for the mass of our unknown element And that's equal to 57. g. And that's just the mass of our sample that's given to us. Here, we subtract this with how much nickel there is, which is 46.9547 g. And that will give us the mass of our unknown which is equal to 10.3953 g. All right now for our moller mass moller mass of X. We know the components of any molar masses grams from all We have both components. So we can go ahead and just divide those numbers and get the molar mass. So for my grams we just solve for is 10.3953 g earlier. With the ratios we can see that my moles of our X are unknown as 0.20 moles. Once you put these numerical values into the calculator, we can see that we get out 51. units being of course g per mole. So now just taking a look at the pr table and seeing which one has a similar molar mass to this calculated value, we can see that our element X is going to be chromium. Alright folks, this is going to be my final answer for this problem. Thank you all so much for watching.

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