Combustion analysis of a 31.472 mg sample of the widely used flam...

Question

Combustion analysis of a 31.472 mg sample of the widely used flame retardant Decabrom gave 1.444 mg of CO2. Is the molecular formula of Decabrom C12Br10 or C12Br10O?

Accepted Answer

Hey everyone, we're told that per floral methyl Declan is widely regarded as a blood replacement. The combustion analysis of 31.468 mg of per floral methyl Declan produced 29.751 mg of carbon dioxide. Identify whether the molecular formula of Per floral method Declan is C 11 F 20 or C 11 F 20 oh. To answer this question, we need to determine the percent carbon of each of our formulas and compare it to the percent carbon from our analysis and as we've learned, our mass percent is going to be the mass of our element divided by the mass of our compound. So first let's do this for our first molecular formula, which is C 11 f 20. To calculate percent carbon, we're going to take the mass of our element. So we have 12 point oh one g per mole, which is the molar mass of carbon. And we're going to multiply this by 11 since we have 11 of carbon. Next, we will divide this by the mass of our compound. So we have 12 point oh one g per mole times 11 plus 19.0 g per mole times 20. Since the molar mass of flooring is 19.0 g per mole and we have 20 of flooring. And since we want this in percentage form, we're going to multiply this by 100. So when we calculate this out, we end up with a percent carbon of 25.80%. Now let's go ahead and do this for our second molecular formula of C 11 f 20. Oh again, we have our percent carbon, We're going to take 12.01 g per mole, which is carbons Mueller mass and multiply it by 11 to get the mass of our compound. We're going to take the same steps as we did before. So we have 12 point oh one g from old times 11 plus 19.0 g per mole times 20. And lastly, we're going to add in that oxygen, which is 16.0 g per mole times one of oxygen. And we're going to multiply this by 100 in order to get that percentage. So we end up with a percent carbon of 25.02%. Now let's go ahead and get the mass of carbon from our analysis. So we had 29.751 mg of our carbon dioxide. We first want to convert this into grams. So we know we have 10 to 30 mg per one g Using carbon dioxides molar mass. We know that we have 44.01 g of carbon dioxide per one mole of carbon dioxide. Using our multiple ratios, we know that one mole of carbon dioxide contains one mole of carbon using carbons Mueller mass. We know that we have 12 point oh one g of carbon per one mole of carbon. And lastly we can convert this into milligrams again and we know that we have 10 to 30 mg per one g. So when we calculate this out and cancel out all of our units, we end up with a mass of carbon Of 8.119 mg. So taking the percent carbon In our sample, we're going to take 8.119 mg and divide that by Our 31.468 mg that they provided to us in our questions, then. And we're going to multiply this by 100 to get that percentage, And this gets us to 25.80%. Now, comparing this to our values that we calculated in the beginning, It looks like this matches our molecular formula of C11 F20, which is going to be our final answer. Now, I hope that made sense. And let us know if you have any questions.

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