Combustion Analysis Example 2

Jules Bruno
Was this helpful?
in this example question. It says, Ah, solve aging agent that contains carbon, hydrogen and chlorine is used for spectroscopic processes. Work told also that it has a molar mass of 1 47 g per mole. Now here, it says determined its molecular formula when a 0.250 g sample contains 0.451 g of carbon dioxide and 0.617 g of water upon combustion. Now here we're dealing with a non hydrocarbon, so we're gonna have an element that's different from just carbon and hydrogen. Here, the element is chlorine. This is gonna be our third element. Some of the steps you will be similar to what we've seen in the past. But because we're dealing with chlorine is just gonna be a little bit different at one point now, if we take a look at Step One, it says, in present, convert the grams of carbon dioxide, 2 g of carbon. So this is what we've done before with combustion analysis. We take our grams of carbon dioxide. We change those grants into moles. So one mole off carbon dioxide, remember, it's possesses one carbon and two oxygen when you look up there, masses on the periodic table and add them together. That gives you 44.1 g of carbon dioxide grams Here, cancel out. Next, we're gonna convert moles of carbon dioxide to just moles of carbon. For every one mole of carbon dioxide we see in the formula that there's just exactly one carbon. So that's one more of carbon. Then finally, we're going to convert. Finally, we're going to convert our moles of carbon into grams of carbon. So one mole of carbon weighs 12.1 g. When we multiply everything out together, we're gonna get here. 0.1231 grams of carbon. Step two, we're gonna convert the grams of water into grams of just hydrogen. So here we're gonna take our 0.617 g of water. Water possesses two. Hydrogen is and one oxygen. When you add their masses together from the periodic table, you'll see that waterways 18.16 g of water for every one mole of water. Then we're gonna stay here for every one mole of water we can see in the formula. We have two hydrogen. So for every one mole of water. There are two moles of just H and then for everyone, Mol of H. It weighs 1.8 g h. Moles cancel out and then we'll get the mass of hydrogen as 0.69 g of H now for step three. We're accustomed to subtracting out the amount of carbon and hydrogen from our sample to get our oxygen. But here we're dealing with a non hydrocarbon. So our third element is not oxygen, but instead chlorine. So step three, if necessary, to attract the grams of steps one and two from the grams of the sample to determine the third element, we're told that our sample is 20.250 g that contains carbon, hydrogen and glory. Subtract out the grams off the carbon and the grams off the hydrogen, and that will give us our grams of chlorine. So when we do that, we're going to get as our grams of chlorine 0.1 to g of chlorine. So at this point we have the grams of carbon, grams of hydrogen and grams of chlorine. So Step four tells us to convert all the masses into moles. So we're gonna convert each of those grams that we got into moles. So remember to do that, you just divide them by their atomic masses from the periodic table. So one mole of carbon, one mole of H one mole of chlorine Here are their atomic masses. From the periodic table grams he will cancel out, and we'll be left with moles of each one of these elements. So when we get that, we're gonna write their moles here, down here. So 0.102 moles of C, we're gonna have here 0.68 Moles of H. And then finally, we're gonna have our moles of chlorine, which is 0.0 34 Moles of chlorine. Now, Step five divide each Milan to buy the smallest more value in order to obtain whole numbers for each element. So we're gonna divide all of them by the smallest mall. Answer that we got, which is 0.34 and again 0.34 So that's going to give me three carbons, two hydrogen and then here, one CEO. That means that my empirical formula at this point is C three H two C L Now, step six. If you get a value, that was 60.1 or 0.9. We could just round to a whole number here. We don't have to worry about that because all of these numbers are whole numbers. Now, if we go back up here, notice that they're asking what is the molecular formula? All we've done at this point is determined the empirical formula. So we come back down here now, remember, it is R N factor R n factor Times are empirical formula equals are molecular formula. Right. So here we need to determine what our end factor is. So n equals R. Moller Mass divided by our empirical mass. Remember, your empirical mass comes from the empirical formula the C three h to seal that we isolated. So here, if we add up the three carbons, the two hydrogen and one cl, it will give us the empirical mass. So when we do that so we're gonna say we have three carbons. Two hydrogen is one cl. You multiply them by their atomic masses on the periodic table. Okay, so then that's gonna give us 36.3 g 2. g and 35.45 g. Add them all together that's going to give us our empirical mass are empirical mass. When we add them all together, get comes out to 73.496 g per mole. And on top, we just have the molar mass that give us in the very beginning, which is 1 47 g per mole. When you divide, those two gives us approximately two. So that tells me that my in factor is too. So I'd say to timed my empirical formula of C three h two c l all the subsequent get multiplied by two. So now come out to C six h four c l two so that there would represent our molecular formula. So realize that this is slightly different from our previous calculations in combustion analysis when we're dealing with typical carbon compounds. But now we're dealing with non hydrocarbons, so we're gonna have a different element that we're not normally going to see such as chlorine. In this case,