A chemist finds that 30.82 g of nitrogen will react with 17.60,
...

Question

A chemist finds that 30.82 g of nitrogen will react with 17.60, 
35.20, 70.40, or 88.00 g of oxygen to form four different 
compounds. (b) How do the numbers in 
part (a) support Dalton’s atomic theory?

Accepted Answer

Welcome back everyone in this example. We need to identify if the given scenario supports Dalton's atomic theory and briefly explain why. So we're told that it's determined that in an experiment a 32.31 g sample of carbon reacts with 43.04g and all these other masses of oxygen to make different compounds. So what we should first recognize is that this scenario is dealing with elemental oxygen because typically we should recall that oxygen exists as a di atomic molecule. But in this case because this is dealing with elemental oxygen, we would just write it as the formula. Oh, so we're going to list each of our different masses as different letters. So we'll say this first massive oxygen is A. The second mass of oxygen is B. We have C. And then we have Masti. Now these different masses of our oxygen are going to combine with our carbon to form different compounds. And this is why it's important for us to recall what Dalton's atomic theory is referring to, which we would recognize as the law of multiple proportions. And we want to recall that our law of multiple proportions tells us that our elements can combine in different ways to form different compounds. And these different compounds are going to have mass ratios of these elements. A simple whole number multiples of one another. So we want to find what these mass ratios are for each of our masses of carbon reacting with our oxygen masses. So beginning with part or mass a we have Our massive oxygen given as 43.04 g. And we're going to divide this by our massive carbon, which is given in the prompt as 32.31 g. So this gives us a ratio of these two elements equal to a value of 1.33. Now we want to see what ratios we get for the rest of our masses of oxygen reacting with carbon. So moving on to scenario B, we have a mass of 86.08 g. We're going to divide this by our massive carbon given as 32.31 g of carbon. So we have our mass of oxygen to carbon. And this is going to yield a ratio equal to a value of 2.66. Moving on to scenario C, we have a massive oxygen equal to 129.12 g of oxygen. We're going to divide this by our massive carbon 32. g of carbon. And this will yield a ratio equal to a value of 3.99. And lastly, we have scenario D Where we have a mass of oxygen given as 172.16 g of oxygen. We're going to divide this by our massive carbon, 32.31 g of carbon. And this will yield a ratio equal to 5.33. So now that we've calculated all of our mass ratios of our combinations of oxygen and carbon. We want to find out of all of these mass ratios which is the smallest mass ratio. And looking at our data, we can see that we have the smallest value being 1.33. So this is going to be our smallest mass ratio. And with this information we can now use this value to divide by all of our mass ratios to get whole numbers hopefully. And so we're going to begin with scenario A. We divide by 1.33. This is going to yield a whole number equal to one, dividing our scenario B by 1.33. We get a whole number equal to a value of two, dividing scenario C by 1.33. We get a whole number equal to three. And then dividing scenario D by 1.33. We get a whole number equal to a value of four. Now, because we have these simple whole number results as our whole number ratios for our mass ratios. We can say that therefore our experiment follows the or we can say Dalton's theory, for short. So the law of multiple proportions and so thus carbon and oxygen will form different compounds by combining in a ratio of whole numbers. And so for our final answer, we're going to say yes, the experiment follows Dalton's atomic theory. So we will highlight yes. And we'll also highlight our explanation here. And so what's highlighted in yellow represents our final answer. I hope that everything I explained was clear. If you have any questions, please lead them down below, and I will see everyone in the next practice video.

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