(b) Menthol, the substance we can smell in mentholated cough drops, is composed of C, H, and O. A 0.1005-g sample of menthol is combusted, producing 0.2829 g of CO₂ and 0.1159 g of H₂O. What is the empirical formula for menthol? If menthol has a molar mass of 156 g/mol, what is its molecular formula?

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Hey everyone, we're told that an organic compound only consists of carbon, hydrogen and oxygen. What is its empirical formula? If 0.263 g of sample is combusted and produces 0.750 g of carbon dioxide and 0. g of water. What is its molecular formula? If its molar mass is 462 g per mole first, let's go ahead and determine the mass of carbon and hydrogen To determine carbons mass, we're going to take 0.750 g of carbon dioxide. And using carbon dioxide Molar mass, we know that we have 44.01 g of carbon dioxide per one mole of carbon dioxide. Now, looking at our multiple ratios, we know that per one mole of carbon dioxide, we have one mole of carbon. And lastly using Carbons Moller mass, we know that we have 12.01 g of carbon per one mole of carbon. Now, when we calculate this out and cancel out all of our units, we end up with a total of 0.205 g of carbon. Now, let's go ahead and calculate our grams of hydrogen. Taking that value of 0.276 g of water. Again, we're going to use dimensional analysis here and using water's molar mass, we know that we have 18 point oh one g of water Per one mole of water. Looking at our multiple ratios, we know that one mole of water contains two mole of hydrogen. And lastly using hydrogen solar mass, we know that one mole of hydrogen contains 1.01 g of hydrogen. So when we calculate this out and cancel out all of our units, we end up with a total of 0.0309 g of hydrogen. Now to calculate our mass of oxygen, we can calculate our mass of oxygen by taking our total mass of our compound and subtracting the sum of our mass of carbon and our mass of hydrogen plugging in our values. We know our total mass of our sample is 0.263 g. And we're going to subtract the sum of 0.205 g and 0.309 g, This will get us to a total of 0.0271 g of oxygen. Now let's go ahead and calculate our moles of carbon, hydrogen and oxygen for carbon. We're going to take 0.205 g of carbon and we're going to convert this into moles using carbons Mueller mass. We know that we have 12.1 g of carbon Per one mole of carbon. When we calculate this out and cancel out our units, we get 0.017069 mole of carbon. Next looking at hydrogen, we're going to take 0.309 g of hydrogen and we know that we have one point oh one g of hydrogen per one mole of hydrogen. When we calculate this out, we end up with 0.03065 mole of hydrogen. Next looking at oxygen, we have 0.0271g of oxygen And we know that we have 16.0 g of oxygen Per one mole of oxygen. And when we calculate this out, we end up with 0. mole of oxygen. To get our empirical formula will need to divide by the smallest mole to get our ratios. In this case it will be 0.001693. So dividing each of our values by 0.001693 mol, We end up with 10 of carbon, 18 of hydrogen and one of oxygen. So our empirical formula comes up to c 10 h 18 0. To find a molecular formula will have to determine the molar mass of our empirical formula. We can do so by taking 12.01 g per mole of carbon. And multiplying this by 10 And adding 1. g per mole from hydrogen and multiplying this by 18. And lastly adding 16.0 g per mole of oxygen. And multiplying that by one. This will get us to a molar mass of 154.244 g per mole of our empirical formula. Now to determine our molecular formula, We can do so by taking the molar mass that they provided us, with which was 462 g per mole And dividing this by 154. g per mole. This gives us a value of three. So now we're going to take that three and multiply it into our empirical formula. So this means our molecular formula is going to come up to see 30, Age 54 03, and this is our molecular formula. Now, I hope this made sense and let us know if you have any questions.

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Key Concepts

Combustion Analysis

Empirical Formula

Molecular Formula