(b) If the molecular weight is 326.26, what is the molecular formula?

Video transcript

Hey everyone, we're told that sample X has a molecular weight of 378. g and only contains carbon, hydrogen and chlorine. If 1.12 g of our sample X produces 1. g of carbon dioxide and 0.56 g of water. After combustion analysis, what is the molecular formula first? We need to go ahead and determine our mole and our grams of each element. Starting off with carbon, We can take our 1.57 g of carbon dioxide and we can go ahead and use our dimensional analysis here. So, using carbon dioxide smaller mass, we know that we have 44.01 g of carbon dioxide per one mole of carbon dioxide. Looking at our multiple ratios, we see that one mole of carbon dioxide contains one mole of carbon. Now, when we calculate this out and cancel out our units, we end up with 0.03567 mole of carbon. Now we can use our mole of carbon to determine our grams of carbon. So we had 0.03567 mole of carbon. Again, using our dimensional analysis, We know that one mole of carbon contains 12.01 g of carbon. And when we calculate this out, we end up with a total of 0.4284 g of carbon. Now let's go ahead and determine our mole of hydrogen. Starting with 0.56 g of water. We're going to use our dimensional analysis again And using water's molar mass, we have 18.016g of water per one mole of water. Now, looking at our multiple ratios, We see that one mole of water contains two mole of hydrogen. Now, when we calculate this out and cancel out our units, we end up with 0.06217 mole of hydrogen. Now let's go ahead and determine our g of hydrogen. Again, using our mole of hydrogen, we have 0.06217 mole of hydrogen. Using hydrogen solar mass, we know that one mole of hydrogen contains 1.8 g of hydrogen. So when we calculate this out, We end up with 0.06267 g of hydrogen. Now we can go ahead and use our grams of carbon and our grams of hydrogen to calculate our grams of chlorine. Now we were told that we had 1.12g of our sample X. So we can go ahead and subtract the sum of 0.4284 grams of carbon And 0.06267 g of hydrogen. And this will get us 0.6- grams of chlorine. Now, let's go ahead and determine our mole of chlorine. Again. Using dimensional analysis, we had 0.62893g of chlorine Using Chlorine Moller Mass, we know that we have 35.5 g of chlorine per one mole of chlorine. So when we calculate this out, we end up with 0. mol of chlorine. Now let's go ahead and look at our multiple ratios in order to determine our formula. So we had 0. mole of carbon. Now we also had 0.06217 mole of hydrogen. And for a chlorine we had 0. mole of chlorine. Now, comparing these values, we want to divide these values by the smallest mole. In this case it will be our mole of chlorine. So dividing each one of these by 0.01772. We end up with about two of carbon 3.5 of hydrogen and one of chlorine. Now when we write this into a formula, we end up with two of carbon 3.5 of hydrogen and one of chlorine. But since we need whole numbers for a formula, we have to multiply each subscript by two. So this gets us to a formula of four of carbon seven of hydrogen And two of Chlorine. So right now we have our empirical formula in order to get our molecular formula which they asked of us. We need to get the molar mass of our empirical formula. So we have four of carbon And we're going to multiply this by 12.01 g per mole. This gets us to 48.04 g per mole. Next looking at our hydrogen, we have seven of hydrogen. Now we're going to multiply this by its molar mass of 1. g per mole. This will get us to a total of 7. g per mole. Now looking at our chlorine, we have two of chlorine. So we're going to multiply this by 35.5 g per mole. This will get us to a total of 71 g per mole. Now, when we add this all up, we end up with a molar mass of 126.096 g per mole. Now in our questions them, they told us that our molecular weight was 378.29 g per mole. So to find our end we're going to take our 378. and divide it by our empirical formula's molar mass which was 126.096 g per mole. This gets us to a value of three. So we're going to take our empirical formula which was C four H seven cl two. And we're going to multiply our subscript by three. So our molecular formula comes up to C 12 H 21 c L six. And this is going to be our final answer. Now, I hope that made sense. And let us know if you have any questions.