Researchers obtained the following data from experiments to find the molecular formula of benzocaine, a local anesthetic, which contains only carbon, hydrogen, nitrogen, and oxygen. Complete combustion of a 3.54-g sample of benzocaine with excess O₂ forms 8.49 g of CO₂ and 2.14 g H₂O. Another 2.35-g sample contains 0.199 g of N. The molar mass of benzocaine is 165 g/mol. Find the molar formula of benzocaine.

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Hi everyone today, we have a question telling us that the following data were obtained from experiments to determine the molecular formula of aniline, a precursor of urethane, which is made up of only carbon, hydrogen and nitrogen combustion analysis of 11. g gives 7.96 g of water and 1.75 g of nitrogen. And Alan has a molar mass of 93 g per mole. What is the molecular formula of aniline? So our first step here is going to be calculate our mass of hydrogen and nitrogen. So we're going to start off with our 7.96 grams of water. And we're going to multiply that by one mole of water time over water's molar mass which is 18. and then multiply that by our multiple ratio. So two moles of hydrogen over one mole of water. And then we're going to multiply that by Hydrogen Molar mass, which is 1. And that is over one mole and that cancels out. Our grams of water are moles of water And our moles of hydrogen leaving us with g of hydrogen. So our massive hydrogen is 0. 907 g. Next we're going to do the same for nitrogen. So we start out with 1.75 grams of nitrogen And then we're going to multiply that by one mole of nitrogen over its molar mass Which is 28.02 g of into. And then we're gonna multiply by our multiple ratio. So two moles of nitrogen For every one mole of into. And then we're going to multiply by our molar mass of nitrogen. So 14 .01g of nitrogen over one mole of nitrogen. And our grams of and two are canceling out our moles of into are canceling out our moles of nitrogen are canceling out, leaving us with grams of nitrogen And that equals 1.75 grams of nitrogen. Now we're going to calculate the mass of carbon. And to do that, we're just going to subtract our masses of water and into from our total mass. So our massive carbon is going to equal 11.65 g -0.8907g of hydrogen -1. g of nitrogen. And that gives us 9. grams of carbon. Now, our next step is going to be to calculate our moles. So we're going to start off with our carbon. So we have 9. grams of carbon. Then we're going to multiply it by one mole Time over its molar mass, which is 12. and our grams of carbon are going to cancel out. And that's going to give us 0.7501 moles of carbon. Next we're going to our hydrogen. So we have 0. grams of hydrogen. And we're going to multiply it by one mole of hydrogen Over hydrogen smaller mass, which is 1.008g. And our grams are going to cancel out and that equals 0.88 moles of hydrogen. And lastly we're going to do our nitrogen. So we have 1.75 g of nitrogen. And we're going to apply that by one mole of nitrogen Over its molar mass, which is 14.01 g of nitrogen. And our grams of nitrogen are canceling out here And that equals 0.124, nine moles of nitrogen. And now we need to divide all of these by our smallest, which is 0.1249. So 0.7501 divided by 0.1249 equals six carbons. 0.8836 divided by 0.1249 equals seven hydrogen And 0.1249, divided by 02, 0.1249 of course equals one. So our empirical formula is going to be C six H n. And to figure out our molecular formula, we need to add up our molar mass of that. So we have 12.01, which is the Mueller massive carbon times six Plus 1.008, which is the more mass of hydrogen times seven Plus 14.01. And that equals 93 g per mole, which is the same as our molar mass given to us in the problem. So our empirical formula is going to be our molecular formula, So C six H 7 N is our final answer. Thank you for watching. Bye.

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Key Concepts

Combustion Analysis

Empirical Formula

Molecular Formula