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Problem 54 Medium Difficulty

Nitrogen and oxygen form a series of oxides with the general formula $\mathrm{N}_{x} \mathrm{O}_{y}$. One of them, a blue solid, contains $36.84 \% N .$ What is the empirical formula of this oxide?

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Video Transcript

So for this in this problem were given that we are going to be working with a compound that IHS nitrogen and oxygen. But we don't know the ratio of nitrogen into oxygen because there's a variety of ways that nitrogen and oxygen can combine. But we do know that it is 36.84% nitrogen. Um, and we're trying to figure out what formula would work for that. So what, We're going to start off with this, having, assuming we have 100 grams of this substance? And with that, that would mean that we would have 36.84 grams of nitrogen in that substance and then 100 minus. That would give us 63.16 grams of oxygen. And so again, you know, get this. This number comes from taking 100 minus the 36.84 That's how we got that value. Now we have these given numbers of grams, and we're going to go ahead and convert that into moles. So we're take thes 36.84 grams of nitrogen and multiply by one mole divided by 14.1 grant are yes, grams nitrogen and we get that that is 2.63 malls of nature. And we can do the same thing with oxygen's that we're gonna have 63.16 grams of oxygen times one mole divided by 16 grams, and that gives us 3.95 moles of oxygen. Next, we're going to take those, and we're gonna find the ratio by dividing by this smaller value. So obviously 2.63 moles of nitrogen divided by 2.63 is going to be one. We're gonna take this 3.95 moles of oxygen divided by the 2.63 um and that will give us a about approximately 1.5 or three halves. So that tells us that this is going to be three mole of oxygen to two more of nitrogen so that our empirical formula is going to be end to 03