Problem

A sample of $\mathrm{HgCl}_{2}$ weighing 9.41 $\m…

01:37

Problem 68 Hard Difficulty

An organic compound has a composition of 93.46$\% \mathrm{C}$ and 6.54$\% \mathrm{H}$ by mass. A solution of 0.090 $\mathrm{g}$ of this compound in 1.10 $\mathrm{g}$ of camphor melts at $158.4^{\circ} \mathrm{C}$ . The metting point of pure camphor is $178.4^{\circ} \mathrm{C}$ . $K_{\mathrm{f}}$ for camphor is $37.7^{\circ} \mathrm{C} / \mathrm{m}$ . What is the molecular formula of the solute? Show your calculations.

Answer

the coefficient of $\mathrm{C}$ is 12 and that of $\mathrm{H}$ is $10 .$ The empirical formula is $\mathrm{C}_{12} \mathrm{H}_{10}$ .

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Video Transcript

Well, everyone, this is Ricky. And today we're working on Problems 68 from chapter 11 and we will be working from the freezing point formula all the way to calculate the molecular formula of thes Salyut. All right, so some things that we know it kf for camp for is 37.7. Um, we know the composition of our organic compound by percent. Um, you know, we used 0.90 grams of it and 1.10 grams of camphor, and that campers melting point is 1 to 58.4 degrees Celsius. Oh, this is of the new Social of Solution. Yeah, than melting points of cam for is 1 78.4 degrees Celsius. All right, so here we see a depression of the freezing point. So we'll be using that formula to calculate our morality. And so the way we're going to be getting this is first rather, let's calculate are delta t f is equal to our Is he with the 1 78.4 minus 1 58.4 Celsius? Um is it is 20 degrees Celsius Is theme magnitude of the Depression. Off are melting point. All right, so now we don't know the molar mass of our organic compounds of religious state. Mueller Mass of organic is equal to x, And whenever you see X, just know that we're referring to the more massive are organic compound of our organic compound. All right, so to get the Malala t of the solution we have morality is equal to the number of moles over the mass of the solvent. So we have 1.10 grams of organic compound and that should be divided by the molar mass. But which is just X and that will be multiplied by are the weight of the solvent used in kilograms. And we're using points. We're using 1.10 grams of camp, for which which translates to point zero 011 kilograms. Hi. So now we can solve through or now that we have morality, we can plug things into our freezing point equation or freezing depression. Freezing point of depression equation. So delta T f is equal to kf that morality is equal to, um 20 degrees Celsius equals 37 17 centigrade from allow times 1.10 grams over 0.11 kilogram, times X. All right. After this, we will solve for X. So cross multiply. We have 20 times point. 011 because 0.22 it's a point. Oh, team to X is equal to 37.7 times 1.1. Whoa! We have 41.5. Divide by point out. You two. And the molar mass is 8 1885 Huh, Grams? Ramon. All right, so now that we know the molar mass of organic compound, we can go ahead and start trying to solve for the empirical formula. All right, so we have all of our we have our carbon and our hydrogen percentages and or amounts of percentage. So we'll say we have 100 grams organic, which means we have 93.46 grams of carbon in 6.54 grams of hydrogen. And then we want to divide this by the molar mass of carbon. No, my boy. 12.1, remember? And this is equal to 7.78 Same thing for hydrogen. Ah, this is equal to 6.48 My was hydrogen moles of carbon. And, um so now that we have our approximates Moeller numbers, we can divide to get our ratios. So it's about 1.2 carbon for one hydrogen, and then we have to convert this into an imager, so multiply both sides by 10. So we have 12 carbon per 10 and hydrogen. So the empirical formula is C 12 each 10. And then to get the molecular formula, we need to get the Moler massive C 12 each 10. And so our molar masses 154 courage from And then we want to divide that, uh, so we want to put 1885 over 1 54 is equal to 12. So that means C 12 age, 10 times 12 is tthe e molecular formula for organic compound. So we have see 1 44 h one. All right, so I hope this video was helpful, and I will see you in the next one

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