Compound X contains only the elements C, H, O, and S. A 5.00 g sample undergoes complete combustion to give 4.83 g of CO2, 1.48 g of H2O, and a certain amount of SO2 that is further oxidized to SO3 and dissolved in water to form sulfuric acid, H2SO4. On titration of the H2SO4, 109.8 mL of 1.00 M NaOH is needed for complete reaction. (Both H atoms in sulfuric acid are acidic and react with NaOH.) (a) What is the empirical formula of X?

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everyone in this example, we have the reaction of tungsten chloride with aluminum. We're told that it produces a compound with just tungsten and chlorine. So it says that this compound is then reduced with hydrogen and they want us to find the empirical formula of our tungsten chloride crystal. So let's go ahead and write this reaction out or this reduction out. So we have a compound with just tungsten and chlorine. And we're putting these X and Y subscript because we do not know the empirical formula yet. We're reducing it with hydrogen. So we're going to have a church to gas reacting and our products are hydrochloric acid and we produced tungsten metal. Now again we want to find empirical formula. So we want to recall that to find the empirical formula for each atom in tungsten chloride. We're going to recall that we take the moles of our atom and divided by the smallest number of moles. So beginning with finding our moles of tungsten, we're gonna go ahead and utilize that mask that. The question gives us, it says that we have 0.89 16 g of tungsten metal produced as a product so we can write that out. And we want to recall from our periodic tables that molar mass is given in grams per mole. So because we want to find moles of tungsten, we can use the molar mass of tungsten as a conversion factor. So we'll have grams of tungsten in the denominator. So it can cancel out And when we refer to our periodic tables, we see that for Tungsten, we have a molar mass of 183.84 g of tungsten for one mole of Tungsten. This allows us to cancel out grams of tungsten leaving us with moles. And this gives us 0.00485 moles of tungsten. Our next step is to find our moles of chlorine. So our product hydrochloric acid is tight traded by our base sodium hydroxide and they give us some information on this hydration. So we're going to want to find the moles of hcl to find the moles of chlorine. So In order to find this according to the problem, we need .2102 moller sodium hydroxide or base And a volume of 92.4 ml of the space to neutralize our hcl in a tight situation. So we should interpret that because they give us more clarity for the base. We want to recall that polarity is equal to moles divided by leaders. And they give us the volume in ml. So we're going to want to go ahead and first begin by converting 92. ml of our base sodium hydroxide to the proper units leaders. So we want male leaders in the denominator and we're going to recall that our prefix milli tells us that we have 10 to the negative third power of our base unit leader. This allows us to cancel out middle leaders leaving us with the proper unit for volume Leaders and we get a volume of 0.09-4 leaders of sodium hydroxide that are required to neutralize our acid hydrochloric acid. So our next step is to then end up with our moles of Hcl. And in order to do so we're going to take that volume that we just found and multiply it by the concentration given of our base to find how much molds of acid it neutralizes. So what we're going to get To find moles of Hcl. Sorry, I don't need to rewrite that is 0.09-4 liters of our sodium hydroxide base. We're going to multiply it by its polarity given in the problem as . moller sodium hydroxide. And this is going to give us our molds of hcl that is going to be neutralized As 0.19422. Most of Hcl that are going to be neutralized. So now that we have this information we want to recognize so far we've confirmed our moles of our tungsten as well as molds of our hcl. And we're going to use this information to find our empirical formula. So what we should have, we have our W. X. C. L. Y compound that we're trying to find our empirical formula for our tungsten chloride. So first we'll find most of ceo And in order to do so we're going to take the moles that we found of Hydrochloric acid. So above that was 0.19422 moles of hcl. And we want to divide that by the smallest number of moles. And so comparing these two more calculations we found we have 0.19422 moles of hcl versus 0.4 85 moles of tungsten. So we would say that tungsten produced the smallest number of most. And so this is what we can use to divide each of our atoms to get our empirical formula. So in our denominator we're going to divide by 0.00485 moles of tungsten. And this is going to give us a value Equal to 4.0054. Which we can round to about four. So this tells us that we should have the subscript for next to R C. L. Moving on to our moles of our tungsten. We're going to go ahead and do the same thing. We're going to take the molds that we found above of tungsten. So that was 0.00485 moles of Tungsten. And we're dividing that by our smallest number of moles. Which is the exact amount of Tungsten 0.004 moles. So this gives us a value for our empirical formula subscript equal to one because it's just divided by itself. And so we can say that therefore we have the empirical formula W cl four for our tungsten chloride. And so this is actually going to complete this example as our final answer for the empirical formula of tungsten chloride. So I hope that everything we did was clear. But if you have any questions, please leave them down below, and I will see everyone in the next practice video.

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Key Concepts

Combustion Analysis

Empirical Formula

Titration and Stoichiometry