Salicylic acid, used in the manufacture of aspirin, contains only...

Question

Salicylic acid, used in the manufacture of aspirin, contains only the elements C, H, and O and has only one acidic hydrogen that reacts with NaOH. When 1.00 g of salicylic acid undergoes complete combustion, 2.23 g CO2 and 0.39 g H2O are obtained. When 1.00 g of salicylic acid is titrated with 0.100 M NaOH, 72.4 mL of base is needed for complete reaction. What are the empirical and molecular formulas of salicylic acid?

Accepted Answer

Hello everyone today. We have the following problem sucks in the acid only contains carbon, hydrogen and oxygen elements. It has two acidic hydrogen. Is that react with sodium hydroxide. 2.24 g of carbon dioxide and 0.685 g of water are obtained after the complete combustion of 1.5 g of socks acid. The complete reaction of that 1.5 g of socks acid with 0.892 molar sodium hydroxide needed 28.5 mL of the base, determine the empirical and molecular formulas of success acid. So the very first thing that we must do is we must find how many moles we have of each of our elements. So we'll start off with the moles of carbon. So with the moles of carbon, we're going to start and solve that using our mass that we were given. So we were given 2.24 g of co two or carbon dioxide. We can simply multiply by the molar mass and that one mole of carbon dioxide is equal to 44. g of carbon dioxide and then multiply the multiple ratio. And by that we can say that in one mole of C. 02, We only have one carbon. So we can see that we have one mole of carbon. And once our unit's canceled out, we're left with 0. 98 moles of carbon. We're gonna do the same for hydrogen thermals of hydrogen. We're gonna take our mass of water .685g of water and multiplied by its molar mass, which we can say that one mole of our water is equal to 18.2 g of H 20. And then we'll apply the same multiple ratio and that in one mole of water we have two hydrogen. So we have two moles of hydrogen. And what's happening is canceled for that. We're left with 0.076 moles of hydrogen. Now for oxygen, we don't have anything that we can use for that. So what we can do is we can first find a massive oxygen and then and then find the most from that. So too, in finding the mass of our oxygen that's going to be taking the total mass which is 1.50 g. And that is subtracting from the masses that we have of carbon and hydrogen. So we're going to subtract that by 0.6113g and then subtract that from our 0.07 seven g of hydrogen. And this is going to give us 0. g of oxygen. We are then going to take that 0.812 g of oxygen and use that To solve for our moles. So we can simply just take the mass here and multiply by our molar mass. So one mole of oxygen is equal to 16g. And once our units cancel out, it's going to be 0.05074 moles of oxygen. And so now we have our number of moles for each of our components. And so the next thing I wanna do is we want to find the mole ratio and finding the mole ratio, we simply take our number of moles that we sold for each one and divide by the smallest number. So, for example, for carbon, what we can do, you can take the number of moles we have of carbon, which is . 0898 moles. We can divide by the smallest number of moles that we sold for and that's going to be the oxygen. So it's going to be dividing by 0. moles. And that's gonna give us one for hydrogen will do the same principle, 0.076 moles of hydrogen divided by 0.5074 moles. And that's going to give us 1.5. And then for oxygen That is going to give us one because we're just dividing by the same number, we're moles. And that's going to give us one. So for our subscript, we're gonna have to multiply each one by two to get a whole number. So when we do that, We end up with two, three and 2. So our empirical formula is going to be, our empirical formula is going to be C two H 302. So this is the first step they're finding the empirical formula. And now what we have to do is find the empirical mass. So the empirical mass essentially going to be finding the molar mass of our empirical formula. So we have two carbons And the molar mass of carbon is 12 g per mole. We're gonna add that to our three hydrogen and the Mueller massive hydrogen is about one g per mole. And then we're gonna add that to our two oxygen's, which has a molar mass of 16 g per mole. And in total, that's gonna give us 59 . g per mole. That's the molar mass of our empirical formula. And so with that we can find our moles of success acid that we will eventually need to use. So what we're going to start off with is we're going to take our Volume are 28.5 ml convert that to leaders by using the conversion factor, that one leader is equal to 1000 ml. Any, we're going to multiply by the polarity of our base, which is in units of moles per liter. So we're gonna say 0.892 moles of our sodium hydroxide Is equal to one leader. And then we are going to multiply by our mole ratio that in one mole of our sodium hydroxide. We have one hydrogen. So we have one mole of hydrogen. And then lastly we'll draw this down here, we're going to multiply by another conversion factor that in one mole of our Sucks in acid. We have two moles of hydrogen. And when we add those numbers up, we're going to get 0. 2711 moles. And so what we're gonna do with that, we're going to find a molar mass. Using that. So over here we're gonna put a molar mass Is going to be able to the g over moles. We have 1.5 g according to the question and our moles is 0.0127 moles. And then we put that together and divide that. We're gonna get 118 g per mole. So this is the molecular mass of success acid. And so next we're going to take our bowler mass and divide that by our empirical mass. We're gonna take our molar mass and divide that by our empirical mass. So that's gonna be our 1 18 g per mole, Divided by our 59 .05 g per mole. And that is going to give us about two. And so that tells us that we need to multiply our empirical formula by two to give us the molecular formula. So the molecular formula molecular formula is going to be as follows C four H 604. And so with that we have answered the question and found the empirical formula and the molecular formula overall, I hope this helped. And until next time

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