Cyclopropane, a gas used with oxygen as a general anesthetic,
is ...

Question

Cyclopropane, a gas used with oxygen as a general anesthetic,
is composed of 85.7% C and 14.3% H by mass. 
(a) If 1.56 g of cyclopropane has a volume of 1.00 L at 99.7
kPa and 50.0 °C, what is the molecular formula of cyclopropane?

Accepted Answer

Hi everyone. Let's take a look at this problem. Xylene is a hydrocarbon used as a gasoline additive. It is made up of 95% carbon and 9.5% hydrogen by weight at 101.325 killer pascal's and 100 degrees Celsius 2.34 g of Xilin occupies a volume of 729.1 Middle leaders. We need to determine the molecular formula of xylene. All right. So we need to determine the molecular formula of xylene and we're told it's a hydrocarbon. So that means it only has carbon and hydrogen. So, we want to figure out what is our X. And our Y. Okay, we have C. X H. Y. And we need to determine that. So let's go ahead. And because we're dealing with a gas, we need the ideal gas law and the ideal gas law is PV equals N. R. T. What we can do is we can calculate the molar mass of xylene and we can calculate the molar mass of xylene by rearranging this ideal gas law to solve for moles which is R. N. So if we divide both sides by R. T. We get and is going to equal P. V over R. T. And remember since we're looking for molar mass molar mass is equal to grams over mole. We already know what our grams is. They tell us we have 2.34 g. Okay, they tell us that right here. So once we saw for moles we can figure out the molar mass. So let's go ahead and do that first p represents pressure. Okay? So in the problem they tell us that RP Is equal to 101. Kill a pascal's. We need to convert this to A. T. M. Okay? And the reason is because we need all of our units to match and our constant has the unit of A. T. M. So to convert this to A. T. M. The conversion is 1 80 M. Is equal to 101.3 - five killer Pascal's. Okay. So here we can see we have one A. T. M. That is our pressure. We move this over so it doesn't cut off the screen. So we have one a. t. m. Okay. Our killer pascal's canceled. R. V. Represents volume and we're told we have 721.9 729. ml. We need to convert this to leaders. Okay. And one millimeter There is 10 to the - leaders. So our middle leaders canceled. And our volume is going to be 0.72 leaders. Okay. Our temperature or are Is a gas a constant. We should know 0.0 eight 206 Leaders time A. T. M. Over more times kelvin and T is temperature. Our temperature we're told is 100 and 30 degrees Celsius. We need to convert this to kelvin because our unit for our gas continent is in kelvin. So we go from degrees Celsius to kelvin by adding 273.15. So our temperature is 403. Kelvin. So now we have all of our variables to solve four moles. Okay, So let's go ahead and solve four moles of Xilin. So we get N. Is equal to P Is one a. t. m. Our volume is 0.7291 leaders over are 0. heaters time A. T. M over more times Calvin. And our temperature is 0.15 kelvin. Okay, so because we're looking for moles here, all of our units should cancel except moles. So our A. T. M cancel our leaders cancel and our kelvin cancels. So we're left with moles which is what we're looking for. Okay, so let's go ahead and solve. And when we do we get N is equal to zero point 0- moles. And this is of Shailene. Okay, so now we can plug this into our molar mass so we can figure out what is our molar mass of xylene. So we're going to say our grams of Xilin which is 2.34 g. That was given in the problem divided by the moles. We just solved for gives us a molar mass of 106.17 g Permal. Okay, so now that we have our molar mass, we can go ahead and solve for moles of both carbon and hydrogen. We're going to assume we have 100 g of our compound because we're given percentage 90.5% carbon and 9.5% hydrogen. If we assume that we have 100 g, we can go ahead and convert those percentages and two whole numbers. So let's go ahead and do that. So we have 0.5 g of carbon and we want to go from grams of carbon two moles of carbon. So we could do that using molar mass. So we can do that using the molar mass of carbon. So in one mall of carbon We have 12.01 g of carbon. Our grams of carbon cancel. And we're left with moles of carbon. So we have 7.557. 354 moles of carbon. And we'll do the same thing. Figure out how many moles of hydrogen we have. So we're going to same thing, assume we have 100 g of the compound. And so that gives us 9.5 g of hydrogen. And we need to convert this two moles of hydrogen in one mole of hydrogen. We have 1.008 g of hydrogen. Okay, so are grams of hydrogen cancel and we have 9. moles of hydrogen. Okay, so now that we know how many moles of each we have, we can divide both by the smallest number of moles. So our smallest number of moles here is our moles of carbon by dividing by the smallest number of moles. It's going to give us a whole number or it's going to give us a number that we're going to want to be a whole number. Okay, so our 7.535 moles of carbon divided by the smallest number of moles of each, which is the moles of carbon. So I write that here, this is our smallest. We're going to divide both by the smallest. This is going to give us one carbon. Okay And then for our moles of hydrogen we have 9.4246 malls of hydrogen divided by the smallest number of moles which is our carbon. So 7.5354 moles of carbon gives us 1.25 hydrogen. So we need this to be a whole number. So we need to multiply By the smallest number that will make this a whole number. And we're going to multiply both by that. So if we multiply this 1.25 x four, we get five hydrogen. So that is the whole number that we need. So we're going to also multiply this by five. So we're going to have five hydrogen and five carbons. So this is our empirical formula. What we just calculated is our empirical formula. And so our empirical formula. So we have our empirical formula is what we just calculate. So we have five carbons. Well let me just make sure here We have four carbons and five hydrogen. Okay, yes, I made a mistake here. So this should be multiplied by four. Okay, so we have four carbons and five hydrogen. This is our empirical formula. So this is not what they asked us for. They asked us for the molecular formula. So from our empirical formula we can derive a ratio and that ratio is going to give us a whole number that will multiply our empirical formula by to get the molecular formula. So our ratio is going to be our molar mass of xylene over our empirical formula. Our empirical molar mass. I'm sorry. So the molar mass of xylene over our empirical our empirical formula molar mass is going to give us our ratio that we're going to multiply our empirical formula by the molar mass of xylene Is a 106.17 grams per mole. And our empirical moller mass, the c four H five. That molar mass is 53. grams per mole. Okay, so when we do that, we get a ratio of two. So now we're going to multiply This empirical formula by 2 to get the molecular formula. So our last thing are molecular formula. When we multiply our empirical formula by two, we get C8 age 10. Okay, and this is our final answer. This is the molecular formula of xylene. That's the end of this problem. I hope this was helpful.

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