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Ch.2 - Atoms, Molecules, and Ions

Chapter 2, Problem 11c

A 1.0-g sample of carbon dioxide (CO2) is fully decomposed into its elements, yielding 0.273 g of carbon and 0.727 g of oxygen. If a sample of a different compound decomposes into 0.429 g of carbon and 0.571 g of oxygen, what is its ratio of the mass of O to C? (c) According to Dalton's atomic theory, what is the empirical formula of the second compound?

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Welcome back everyone in this example, we're told a sample of water weighs one g and decomposes into .111g of hydrogen and .889 g of oxygen. Another sample of a different compound decomposes into 0.59 g of hydrogen and 0.941 g of oxygen. And we need to identify our empirical formula of the other compound based on Dalton's atomic theory. So our first step is to recall what Dalton's atomic theory tells us. And specifically, we're recalling our fourth postulate of Dalton's theory. So we should recall that. That tells us that the composition of our chemical compound should be consistent with the ratio of the second compound. And by ratio we are referring to our masses of our atoms and where it states composition. That is referring to the quantities of the atoms that we have in our compound, which is what will give us our empirical formula. And so we first want to write out our mass ratios of both of our compounds. So for our first mass ratio, according to the prompt, we have the comparison of our mass of oxygen to our mass of hydrogen and according to the prompt, we have a massive oxygen being 0.889 g of oxygen, whereas our massive hydrogen from the prompt is given as 0. g of hydrogen. So this is going to give us a ratio in our calculators equal to a value of eight. And then moving on to our second mass ratio for the second compound, we are again comparing our mass of oxygen to our mass of hydrogen. And according to the prompt for our massive oxygen for compound two we have a mass of 0.9 41 g of oxygen for a compound. Or sorry for our adam hydrogen. We have a mass given from the prompt as 0.59 g of hydrogen. And this is going to give us a ratio equal to a value of 16. So, if we know our ratio of our second compound is this value here, we want to see if our Ratio of our atomic masses of oxygen to hydrogen is also going to equal this value here, 16. Which is again our ratio for our second compound given in the prompt. And so we want to recall our atomic mass of oxygen and hydrogen in our periodic tables. And we would see that for oxygen we have an atomic mass equal to a value of 16.0 g. And then for our atomic mass of hydrogen, we would see on a periodic tables that we have a value of zero point or 1.8 g. And so calculating for our atomic mass ratio between these two atoms, we would get a value equal to 16. And because we have these two consistent values here, we would say that therefore our empirical formula is going to be just H. And then, oh and again, this was due to the fact that Dalton's atomic theory applied, because the ratio of our masses of oxygen and hydrogen was equal to the ratio of our second compound mentioned in the prompt. And so for our final answer, we have our empirical formula being H. O. So what's highlighted in yellow is our final answer. I hope that everything I reviewed was clear. If you have any questions, leave them down below, and I will see everyone in the next practice video.
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Textbook Question

A 1.0-g sample of carbon dioxide (CO2) is fully decomposed into its elements, yielding 0.273 g of carbon and 0.727 g of oxygen. (a) What is the ratio of the mass of O to C?

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A 1.0-g sample of carbon dioxide (CO2) is fully decomposed into its elements, yielding 0.273 g of carbon and 0.727 g of oxygen. (b) If a sample of a different compound decomposes into 0.429 g of carbon and 0.571 g of oxygen, what is its ratio of the mass of O to C?

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