The osmotic pressure of a solution containing 2.10 g of an unknown compound dissolved in 175.0 mL of solution at 25 °C is 1.93 atm. The combustion of 24.02 g of the unknown compound produced 28.16 g CO₂ and 8.64 g H₂O. What is the molecular formula of the compound (which contains only carbon, hydrogen, and oxygen)?

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So we asked the molecular formula of an unknown compound. Given the following data below The combustion of 22.98 g of the unknown compound. And it produced 56.10 g of carbon dioxide 7. g of H 20. and osmotic pressure is 1.59 atmosphere of a solution containing 0.703g of an unknown compound result in 200 ml of the solution At 25°C. First calculate the similarity. You know that the osmotic pressure similarity. How is the gas constant times the temperature. And I had to rearrange the equation to solve for similarity to get clarity because the osmotic pressure about about our times T and I'm going to convert the temperature into kelvin 25°C. If you add 273 15, Get 298, 115 kelvin and put their bodies into the pleasure. We're going to get the clarity Because 1. atmosphere By 0. 26. Later, some atmosphere. Bye bye. Malls times, Calvin Times, 15 Calvin. We get 0.06 five mahler and we need to convert the volume into leaders and calculate the moles of the unknown. You have 200 l Under one leader we have 1000 ml. We get 0.2 L. We know that polarity equals the malls of the song divide by the leaders of a solution for the malls of the unknown. 0.06 five malls in one liter We have 0.2 liters of a solution. 0.013 balls. And I'm catholic mass is in the formula. And it was a mask. What about the Mueller mats? Now we can rearrange this to solve for them on the mask and get mass. What about N 0.70. three grams. About a number of malls which is 0.013 malls. Can we get 54 08g from all the nominees catholic the malls of each element in the compound using a mass of carbon dioxide and H 20. Produce 56 10 grams from the oxide. And in one month carbon dioxide of the Mueller mass Get 12.011 g plus two 15.999 g At 44.01 g. And in one month of carbon dioxide at one moment of carbon I get 1.2 falls a carbon 7.657 g of H 20. And in one mall Of H 2 0. of the Molar Mass. And it's too 1.008g Plus 15.999 g. Get 18.02 g and one more of H malls of hydrogen. 0.08 0. 98 balls of hydrogen. And now for the massive oxygen we're gonna need to use mass equals the number of most times similar mass. The mass of oxygen get the mask of the unknown. Find us a mask of carbon and it's a mass of hydrogen for the mass of oxygen. Get 22.98 g -1.275 balls of carbon, 12.011 g. Remote 20 0.84 and eight malls of hydrogen 1.008g ramon for the mass of oxygen. We get 6.809 g. And I wanted to find the most of oxygen. It's 6.809 grams. About 15 .999 g per mole. We get 0.4256 walls of oxygen. And then we can divide by the smallest number of moles of carbon At 1.275 five x 0.4256. I get three hydrogen. I have 0.84 98. Down by 0.42. Our sex two. Our oxygen 0.42 By 0.42. R. Six. Okay one. These are going to be a subscript in the empirical formula. The empirical formula NBC three shoot out. So now to find them like a formula, I need to find what number two. Most private subscript by to get them. Like in formula we're gonna use eggs was the molar mass. What about the empirical formula mask, We get 54 08. And for the mass of c three H two L three 12.011 g plus two Have 1.008g plus 15 999 g. Get 54 805 g. And this is approximately one For the most part are subject by one. Our molecular formula is the same as an empirical formula, And it's AC three H. Two. Huh? Thanks for watching my video and I hope it was helpful.

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Key Concepts

Osmotic Pressure

Combustion Analysis

Empirical and Molecular Formulas