Combustion analysis of 0.1500 g of methyl tert-butyl
ether, an o...

Question

Combustion analysis of 0.1500 g of methyl tert-butyl 
ether, an octane booster used in gasoline, gave 0.3744 g of 
CO2 and 0.1838 g of H2O. When a flask having a volume 
of 1.00 L was evacuated and then filled with methyl tertbutyl 
ether vapor at a pressure of 100.0 kPa and a temperature 
of 54.8 °C, the mass of the flask increased by 3.233 g. 
 
(d) The enthalpy of combustion for methyl tert-butyl ether 
is ΔH° combustion = -3368.7 kJ>mol. What is its standard 
enthalpy of enthalpy of formation, ΔH°f?

Accepted Answer

Hello everyone today. We have the following problem. Diet isopropyl, either or either is a compound used as a solvent in paint thinners and standard movers. A 0.21 50 g sample of dice april either was burned in excess oxygen which produced 500.55 57 g of carbon dioxide and 570.26 54 g of water. A separate sample of dice approval. Either vapor was placed in a one liter vessel at 70.5 degrees Celsius and had a final pressure of 98. Kill a pascal's. The mass of the vessel increased by 3.504 g. If the entropy of combustion of the isopropyl either is as follows -4010.1 killers from all calculated standard entropy of formation. So the first thing I need to do is we need to find out how many moles of this compound that we have. So die. Isopropyl contains carbons, hydrogen and oxygen based on this given diagram or either either contain carbon, hydrogen and oxygen as well. So first we need to find the moles of carbon that we have. So first we're gonna start With our 0.55 57 g of our carbon dioxide. We are then going to multiply by its molar mass, which one roll of C. 02 is equal to 44. g of carbon dioxide. We're then going to multiply by the multiple ratio and we're going to say that for every one mole of carbon, we're going to have one mole of C. 02 or in other words in carbon dioxide, we have one mole of CO two. And so our units of g and moles of CO2 will cancel out. And we are left with 0. moles of carbon. We're going to do the same process and find the number of moles for our hydrogen. We're going to take our water which is 0. g of water. We're gonna multiply the molar mass of water, which is going to be one mole of water is equal to 18.16 g of water. And they were gonna multiply by the multiple ratio. So in one mole of H 20. We have two moles Of our hydrogen. And so our units for our moles and grams of water cancel out. We were left with 0. moles of hydrogen. We have these two values here and last but not least. We're going to calculate the number of moles in our oxygen. And so to do this. What we're gonna do is we're essentially going to take our total masks of our sample which was 0.21 50g. And then we're going to subtract by our mass of carbon and our massive carbon was zero point 1516 g an hour mass of hydrogen was .02970. And that's gonna give us 0.03370 g of oxygen. So we have those values there. The next thing we're gonna do is we're gonna take that mask and we're gonna find out how many moles of oxygen we have. So we're going to take our 0. g of oxygen and simply multiplied by its molar mass. That's gonna be one mole of oxygen is equal to 16 g of oxygen. And when our units cancel out, we're left with 0.00- moles of oxygen. Next, we need to divide these number of moles by the smallest number of moles. And so what I mean by that is we're simply going to take, for example, for carbon, We're going to take the moles for carbon, which is 0.012 moles. And divide that by the smallest number of moles that we solved for. It's going to be for our oxygen, which is going to be 0.0021 of six moles. And that's gonna give us roughly six. We'll do the same for hydrogen and oxygen. So for hydrogen that's 0.2946 moles Divided by 0.00 21 06. And that's gonna give us 14. And then for oxygen, we're going to do the same thing. 0.00 21 06 moles Divided by 0 2106 moles. That's gonna give us one. And so therefore we can determine what the empirical formula is going to be. So our empirical formula is going to be for carbons six, hydrogen is 14 and oxygen's one. So this is our empirical form that we can work with. Next we then can calculate the molar mass of what we just found. So we can calculate the molar mass for our new empirical formula, which is going to be For carbons. We have six and then times the molar mass according to the periodic table is 12.1 g per mole Plus we have 14 hydrogen and that's just going to be times 1.008 g per mole. And lastly for oxygen, we only have one. So we're just gonna add 16 grams per mole. And that is gonna give us 0.172 g per mole for molar mass. Now we need to find the number of moles for our di isopropyl either and so to do that, we can simply use our ideal gas law with the values that we have. So first we need to find out our pressure in terms of atmospheres and our temperature in terms of kelvin. That way we can plug in our values and be set. So for pressure We started with 98 killer Pascal's and to convert that to atmospheres, we need to do several things first, you need to convert this to regular pascal's and use the conversion factor that one kg pascal is equal to 10 to the third pascal's. And then we can finally multiply that one atmosphere is equal to 101,325 pascal's, our units for pascal's and pascal's cancel out. And we're left with 0.9672 atmospheres. Our volume is normal. So then we just have to find out for our temperature Our temperature has to be in terms of Kelvin. So we're gonna take our 70.5°C and add it to 73.15 to that. That gives us 343.65 Kelvin. And now we can plug in our values. So the number of moles are gonna be equal to the pressure, which we calculated was 0.9672 atmospheres Times are volume, which was one leader Over our gas constant, which was 0.08206 leaders times, atmospheres over most Times Kelvin. And then lastly multiply that by 343. Kelvin. This gives us a total number of moles of 0.03430 moles of our di isopropyl. Either Next we're gonna use the moles mammograms that we have to plug in our molar mass. So right here molar masses and units of grams moles. So for our grams we're gonna plug in our 3.504 g over the multi to salt for which was 0.3430 moles. That's gonna give us 102.16 g per mole. So this is known as the molecular molar mass. And next we have to find the ratio between the molecular molar mass and the empirical molar mass. And so to simply do that, We take our molar mass of our empirical, which is 102.172 g per mole. And we divide that by our molecular Mueller bass. And this is going to give us roughly one. So our molecular mass, our molecular mass is going to be the same as our or molecular formula is going to be the same as our empirical formula C6 H 14 and then one oxygen. The next thing you must do is you must write a balanced chemical reaction for this. So we have our di isopropyl either, which is going to be our C 16 H 14, oxygen plus oxygen gas. And that is going to form and give us carbon dioxide and water. Now to balance this, we simply have to add a nine in front of the oxygen's so that we have 19 on the left. And what we're gonna do is we're gonna add a six here on the right and a seven here as well. And so we see that we have an equal number of carbons, hydrogen and oxygen on the left and the right. And so as the question stated, we must calculate the change in our entropy standard entropy of formation. So this is going to be our delta H of entropy formation is equal to that of our products subtracted by our reactant. So expanding this formula out, we're going to have the entropy change here. It's equal to Our information of our products which is going to be c. 0 2 plus water and that is going to be subtracted our entropy information of our gas plus our di isopropyl either. And these values can be found in your reference text, How we're going to combine this is we're gonna take that of C. 02 which is negative 393.5 and multiply how many we have, which is six moles. We're gonna add that to our negative 85.8, multiply the number of moles we have for that which is seven moles. And then we're gonna take those values and we're going to subtract it by the one more that we have for our isopropyl either because oxygen does not have it since we don't have that value, we're simply going to solve for it And this is actually going to be equal to negative 4,010.1. And so when we solve for our entropy of formation for our diet isopropyl either it's going to give us a value of negative 351.5 kg joules per mole. And with that we have solved the problem overall, I hope this helped ahead until next time.

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