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.

(b) What is the molecular weight and molecular formula of methyl tert-butyl ether?

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.

(b) What is the molecular weight and molecular formula of methyl tert-butyl ether?

Accepted Answer

Hi everyone. This problem reads towline is used as a solvent and paint thinner and permanent markers. Elemental analysis of towline shows that it is composed of 91.25% carbon and 8.75% hydrogen a point 250 g sample of towline is placed in a 500 millimeter flask at 25 degrees Celsius where the pressure of the gas was measured to be 250.13 to 8 atmospheres, determined the molecular weight and molecular formula of towline. So we have two things that we want to answer for this problem. The first is the molecular weight and the second is the molecular formula. Okay, so let's go ahead and assume that we have 100 g sample. If we assume that we have 100 g sample, the percentages that we see, we can convert that to grams. So that means that we're going to have 91.25 g of carbon and we're going to have 8.75 g of hydrogen. So, because we know the amount of grams we have, we want to figure out the moles of each. Okay, so we want to go from grams of carbon to moles of carbon. And we want to do the same thing for hydrogen. We want to go from grams of hydrogen, two moles of hydrogen. So let's go ahead and do that. And we're going to start off with what we're given. So four carbon, we have 91.25 g of carbon to go from grams to moles. We need the molar mass of carbon. Okay, so in one mole of carbon, The mass is 12.0107 g of carbon, We'll make sure our units cancel grams of carbon cancel and we're left with moles of carbon. So once we calculate this, we get 7.59 seven moles of carbon. So let's go ahead and do the same thing for hydrogen. We have 8.75 g of hydrogen. We want to go from grams of hydrogen, two moles of hydrogen And one mole of hydrogen. There is 1.00794 g of hydrogen Moles of our g of hydrogen cancel. And we're left with moles of hydrogen. When we do this calculation, we get 8.681 moles of hydrogen. So now we know the moles of carbon and the moles of hydrogen. We want to get the lowest ratio of moles. So we're going to divide each by the lowest ratio of moles. Here we see that out of the two are lowest moles ratio. Or yes, our lowest ratio of moles is 7.597. So we're going to take both of the moles. We just calculated and divide it by the lowest mole ratio. Okay, or the lowest ratio of moles. Excuse me. So we're going to go ahead and take for carbon, It's 7.597 moles of carbon. And when we divide it by the lowest ratio of moles 7.597 we get one carbon and four hydrogen when we take the moles that we calculated which is 8.681 moles of hydrogen and were divided by the lowest ratio of moles which is 7.597. We get 1.143 hydrogen. Okay, so We have a decimal here and we don't want that. So we're going to multiply by a number that's going to give us closest to a whole number and the number that we're going to multiply by seven. Multiplying by seven will give us, Let me write this in a different color. Multiplying by seven gives us the empirical formula Yeah, C seven H eight. Okay, so now that we know the empirical formula, we can calculate the molar mass. Okay, so that is going to be step two. Step one was for us to find the empirical formula. So now let's go ahead and calculate the molar mass. Alright, so the molar mass is going to equal the mass times gas constant R times temperature over pressure, times volume. So let's go ahead and plug in these values. Okay, the mass of the sample we're told in the problem is 0.250 g. Okay, so 0.250 g times gas constant R. Which is 0.8206 leaders, atmosphere over mole kelvin times temperature. We're told the temperature is 25°C. But we need this in Kelvin because our unit for temperature in our gas constant is Kelvin. So we're going to add 273.15. And this is all over pressure times volume. Okay. And the problem we're told the pressure is .13-8 atmospheres. So we're going to write that here, .13-8 atmospheres. And this is multiplied by the volume. So we have 500 ml but we need this in leaders. So we're going to multiply by 10 to the -3. Okay, so once we calculate the molar mass we get mass is equal to 92.117 g per mole. So now we can go ahead and calculate the molecular formula. Okay, so our molecular formula we know that the empirical formula is C seven H eight. So we have seven carbons and eight hydrogen. So we're going to take seven and multiply it by the molar mass of carbon, which is 12.107 g per mole of carbon plus we have eight Hydrogen. So we're going to take eight times the molar mass of hydrogen. 1.00794 grams Permal of hydrogen. And this is equal to 92.138 g per mole. Okay, so now what we're going to do is calculate the ratio which is Z. So Z is going to equal the empirical mass over the molar mass. Mhm And this is going to give us a number and that number is what we're going to multiply our our our units in our empirical mass by so C seven H eight. Whatever we calculate for Z is what we're going to multiply those sub scripts by. Okay, so our empirical mass we calculated is 92.1 38 and our molar mass is 92.117. So the value we get for this is 0.9998, which is about one. So we're going to multiply our empirical mass by one. Our empirical formula by one. So that means then that our final answer for our molecular formula is going to equal the same as our empirical formula. So our molecular formula is C seven H eight. So that is one of our answers and the question also asked us for the molecular weight. Okay. And our molecular weight was 92. Okay, so our molecular weight is 92.117 g per. Okay, so these are our final answers and that is it for this problem. I hope this was helpful

Verified Solution

Key Concepts

Combustion Analysis

Ideal Gas Law

Molecular Weight and Molecular Formula