(b) Nicotine, a component of tobacco, is
composed of C, H, and N....

Question

(b) Nicotine, a component of tobacco, is
composed of C, H, and N. A 5.250-mg sample of nicotine was
combusted, producing 14.242 mg of CO2 and 4.083 mg of
H2O. What is the empirical formula for nicotine?

Accepted Answer

Hey everyone, we're asked to determine the empirical and molecular formula of the following unknown compound. First let's go ahead and determine the number of grams we have for each atom Starting off with carbon, we can take our 0.6581 g of carbon dioxide Using carbon dioxides, Mueller mass of 44.01 g. We can get some moles of carbon dioxide And per one mole of carbon dioxide. We can see that we have one mole of carbon And using carbons atomic mass, we know that per one mole of carbon. We have 12.01 g of carbon. This will get us to a value of 0. g of carbon. Moving on to hydrogen, we can go ahead and take our 0.112, two grams of water. And using waters Mueller mass, we have 18.01 g of water per one mole of water and per one mole of water, we see that we have two mole of hydrogen using hydrogen atomic mass. We know that per one mole we have 1.01 g of hydrogen. So when we calculate this out, we end up with a total of 0.01246g of hydrogen. And now to find our nitrogen We can take our total number of grams of our compound, which is 0.227g. And subtract the values we just calculated which is 0.1794 g of carbon and 0. grams of hydrogen. And when we calculate this out, we end up with a total of 0.03514 g of nitrogen. Now that we have our grams of carbon, hydrogen and nitrogen, let's go ahead and convert this into moles. Starting off with our carbon, we have 0.1794 g of carbon and we know that we have 12.01g of carbon per one mole of carbon. This will get us a total of 0. mole of carbon. Looking at our hydrogen, We have 0.01246g of hydrogen And we know that we have 1.01 g of hydrogen per one mole of hydrogen. So when we calculate this out, we end up with 0. more of hydrogen. Next looking at our nitrogen, we have 0.03514 g of nitrogen. And we know that we have 14.01 g of nitrogen per one mole of nitrogen. So when we calculate this out, we end up with 0.00258 mole of nitrogen. Now, in order to get our empirical formula will have to divide each of our values by the least amount of moles we have present. In this case, it's going to be our nitrogen. So dividing each 1x0. We're going to end up with six carbon five hydrogen and one nitrogen. So our empirical formula Is going to be C6 H five N. Now in order to get the molecular formula will have to calculate the molar mass of our empirical formula. And when we do we find that it's 91.11 g per mole. So to get our molecular formula, We're going to take the molar mass of our unknown compound, which they told us in our questions them as 273. g per mole. And we'll divide this by our empirical formulas Mueller mass which is 91.11 g per mole. This will get us to a total of three. This means our molecular formula is going to be C six H 5 n. Multiplied by three. So we end up with a molecular formula of c. 18, Age 15 N three. And these are our final answers. So I hope this made sense. And let us know if you have any questions

(b) Nicotine, a component of tobacco, is composed of C, H, and N. A 5.250-mg sample of nicotine was combusted, producing 14.242 mg of CO2 and 4.083 mg of H2O. What is the empirical formula for nicotine?

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