Vanadium forms four different oxides in which the percent by mass of vanadium is, respectively, (a) 76%, (b) 68%, (c) 61%, and (d) 56%. Determine the formula and the name of each oxide.

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Hey everyone in this example, we need to determine the name or the chemical formula of copper oxide that contains 89% copper by mass. In order to find our chemical formula, we're going to take our moles of our first element in this name, which we can see as copper. So moles of copper. And we're going to divide that by the smallest number of most. And that's going to give us our results of our number of copper in the chemical formula. The next step is to recall that we need to take our moles of our second name named adam. In this compound oxygen. We know it's oxygen because it has the O. X. In the beginning of the name. So we have moles of oxygen that we're going to also want to divide by the smallest number of most to give us our number of oxygen in our chemical formula. So we can just change that end number. So it's clear. So they tell us that we have 89% copper by mass. And we want to know how much oxygen we have in our compound by mass or how much oxygen we have percent by mass. So in order to figure that out, We're going to take 100 And we're going to subtract that from 89% of our copper to give us our percent of oxygen by mass. Which is 11%. So now we're going to just interpret these percentages as grams for mass. And what we're going to want to do is find first moles of our copper Using the 89% as g. So we're going to have 89 g now of copper multiplied by in our denominator. Are sorry, in our denominator, our molar mass of copper where in the numerator, we're going to end up with moles of copper. And we should recall that our Moeller massive copper is 63 . g for one mole of copper. This allows us to cancel our units of grams leaving us with moles. And we're going to get our moles of copper equal to 1.4005. We're sorry about that. That would be 1.4015 moles of our copper. Our next step is to then find moles of our oxygen. We just found that we had 11% of our oxygen by mass. So this is going to be 11 g of our oxygen multiplied by In our denominator are molar mass for oxygen, which we recall is 16 g of oxygen for one mole of oxygen. So we can cancel out grams of oxygen. We're left with moles of oxygen and we're going to get a result equal 2.6875 moles of oxygen. Now we want to figure out which of these moles that we just calculated is the smallest amount. And comparing these two, we can say that oxygen produced the smallest number of most. So to get our chemical formula of our copper oxide. Again, we're going to follow these steps here above. So we're going to get first hour Moles of our copper which according to what we calculated above is 1.4015 moles of copper. And we're dividing that by the smallest number of moles, which we agree is 0.6875 most of our oxygen. So this quotient is going to give us a value Equal to 2.03, which we can round to about two. We just want a single digit here for our results. And so this tells us that we should have a subscript of two in front of our copper and our chemical formula. So we can say therefore we have see you too. Next we want to figure out what our subscript on oxygen is going to be. So we're going to take our molds of our oxygen Which we found to be 0.6875 moles of oxygen. And we're also dividing it by itself because it was the smallest number of moles. And this gives us our results equal to one because it's just divided by itself. And so this allows us to determine that overall, we can say We have a chemical formula of CU two. And then, Oh So our final answer is going to be CU- 20 as our chemical formula of our given adam copper oxide or given molecule copper oxide. 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.

Vanadium forms four different oxides in which the percent by mass of vanadium is, respectively, (a) 76%, (b) 68%, (c) 61%, and (d) 56%. Determine the formula and the name of each oxide.

Verified Solution

Key Concepts

Percent by Mass

Empirical Formula

Oxidation States