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Ch.4 - Reactions in Aqueous Solution

Chapter 4, Problem 149b

(b) When 5.00 g of X is titrated with NaOH, it is found that X has two acidic hydrogens that react with NaOH and that 54.9 mL of 1.00 M NaOH is required to completely neu-tralize the sample. What is the molecular formula of X?

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Hello everyone today. We have the following problem. X. Is a compound that has two acidic hydrogen 6.27 g of X consumes 57.2 mL of 1.5 molar sodium hydroxide to complete to be completely neutralized, determine the molecular formula for X. If its empirical formula is C three H 502. So the very first thing in this problem we want to do is we want to find the moles of sodium hydroxide. So we want to find the number of moles for any O. H. And we do that by starting off what we're given from any O. H. Which is its volume. So we take our 57.2 millions of sodium hydroxide And we want to convert this into leaders. So we see that one middle leaders equal to 10 to the negative third leaders. And then lastly we multiplied by the polarity that were given similarities in terms of moles per liter, which means that this is 1.5 moles over one liter. Our units of leaders canceled out and our units for middle leaders canceled out. We're left with 0.0858 moles of sodium hydroxide. And so we'll just keep this number. It's also important to note that we need two moles of sodium hydroxide to titrate, one mole of X because it contains too acidic hydrogen. So essentially for every amount of acid we need to base. So we have one parts acid. We needed two parts base. Our second step is to find our moles of X. So we found our moles of sodium hydroxide. Now we need to find our moles of X. And so to do that, we're gonna take what we're given of X. Which we're going to take our Moles of sodium hydroxide that we just found our 0.0858 moles of sodium hydroxide. And we're going to multiply by a multiple ratio. And so we already noted that for every one asset we have, we need two parts based so we have one mole of X, which means that it's going to use two moles of sodium hydroxide to completely neutralize it. And so our units of moles for sodium hydroxide canceled out. We're left with 0.0429 moles of X. And we will go ahead and save that for later. And so thirdly, we want to ultimately find the molar mass for X. So you want to find the molar mass of X. And how do we go about doing that? Well, first we start with our mass, Our 6.27 g of x. And we divide by the moles of X that we have since molar masses and units of g per mole. So we have 6.27g divided by 0.0429 moles of X. And this ultimately gives us a molar mass of 146. g per mole. And so now we have the molar mass of our compound X. Next we want to find the empirical mass of this formula that we have. We want to see how does it compare to our molar mass. And so how do we find the empirical mass? We simply take however many moles or of our elements that we have. And we multiply by the molar mass. So for example, we have three carbons and the molar mass for carbon is 12.01 g per mole. And then we add that to that of hydrogen. So we have five hydrogen atoms. The molar mass of hydrogen is 1.8 g per mole. And then lastly we add that to our oxygen and we have two oxygen atoms. We multiply that by the molar mass of oxygen which is 16 g per mole per the periodic table. And so once we do that, we end up with 73.08 g per mole. Our 5th step is going to be finding the ratio between our Mueller mass. So you want the ratio between our solar mass and our empirical mass that we just found it. So our molar mass divided by our empirical mass. We said our Molar mass was 146.15, g per mole. And our empirical mask that we just sold for was 73.08 g per mole. Our units of g per mole canceled out. And since this is a ratio it's going to be a whole number. This actually comes to 1.99. But we round this for two, we rounded to two. And so the very last step is going to be finding our molecular formula. And so in finding our molecular formula, we simply take our empirical mass formula and we multiply by the ratio that we have, for example, we have three carbons, so we take that three, we multiply it by two, we have five hydrogen, we take that five, we multiply it by two and we have two oxygen's. So we take that two and we multiply that by two. And this ultimately gives us C six or six carbons, H 10 or 10 hydrogen oh four or four oxygen's as our final answer. Overall, I hope this helped. And until next time.
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A mixture of FeCl2 and NaCl is dissolved in water, and addi-tion of aqueous silver nitrate then yields 7.0149 g of a pre-cipitate. When an identical amount of the mixture is titrated with MnO4 -, 14.28 mL of 0.198 M KMnO4 is needed for complete reaction. What are the mass percents of the two compounds in the mixture? (Na+ and Cl-do not react with MnO4 -. The equation for the reaction of Fe2+ with MnO4 was given in Problem 4.146.)
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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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A 1.268 g sample of a metal carbonate (MCO3) was treated with 100.00 mL of 0.1083 M sulfuric acid (H2SO4), yielding CO2 gas and an aqueous solution of the metal sulfate (MSO4). The solution was boiled to remove all the dissolved CO2 and was then titrated with 0.1241 M NaOH. A 71.02 mL volume of NaOH was required to neutralize the excess H2SO4. (a) What is the identity of the metal M?

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Element M is prepared industrially by a two-step procedure according to the following (unbalanced) equations:

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