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. (a) What is the empirical formula of methyl tert-butyl ether?

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Hey everyone, we're asked to determine the empirical formula of methyl paraben. If it is only composed of carbon, hydrogen and oxygen and were given the amount of grams of each compound in our methyl paraben. So let's go ahead and start off by determining the grams of carbon. We have. We can do this by taking our 1.570 g of carbon dioxide and we can use carbon dioxides molar mass, which is 44.01 g of carbon dioxide to get two moles of carbon dioxide. And we know that one mole of carbon dioxide has one mole of carbon. And looking at our periodic table, One mole of carbon contains 12.01 g of carbon. So when we calculate this out, we end up with 0.4284 g of carbon. Moving on to Hydrogen, we can take our 0.3233 g of water. And using water's molar mass, which is 18.01 g of water Per one mole of water and per one mole of water, We have two moles of hydrogen. And looking at our periodic table, we know that one mole of hydrogen contains one point oh one g of hydrogen. This will get us a total of 0.0359 g of hydrogen. To find our oxygen. We can simply take our 0.678g of our methyl paraben and subtract the values that we just calculated for. So 0.4284 g of carbon plus our 0.0359 g of hydrogen. This will get us to a value of 0.2137 g of oxygen. Now we can go ahead and calculate for the mole of each atom. Starting off with our carbon, we have 0.4284 g of carbon. And we know that we have 12.01 g of carbon per one mole of carbon. This will get us to a value of 0. mole of carbon. Moving on to our hydrogen, we have 0.0359 g of hydrogen and we know that we have 1.01 g per one mole. And when we calculate this out, we end up with 0.0359 mole of hydrogen. And lastly looking at our oxygen, we have 0.2137 g of oxygen. And we know that we have 16 g of oxygen Per one mole of oxygen. This will get us to a value of 0. mole of oxygen. Now, in order to get our empirical formula, we're going to need to divide all of our values by the least amount of moles. In this case it's going to be our mole of oxygen, which is 0.01336. So we end up with about 2.67 of carbon, 2.67 of hydrogen and one of oxygen. Since we have these decimal points, we can't round up or round down so we need to multiply all of our values in order to get a whole number, so if we multiply each one by three, We end up with about eight of carbon eight of hydrogen and three of oxygen. So our empirical formula is going to be C 8 H803, and this is our final answer. So I hope this made sense and let us know if you have any questions.

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Key Concepts

Combustion Analysis

Empirical Formula

Ideal Gas Law