Identification of the unknown substance: C25723 For my research I found empirical formula and molecular weight of C25723 to find the molecular formula and I can use all the information I collected to determine my unknown volatile liquid C25723. Using molecular formula, I plan to look at “ChemExper Chemical Directory” website for more data. Once I have all of my data I will identify the volatile liquid C25723. To determine the name of my volatile liquid I had to find the empirical formula and to find the empirical formula I first had to do calculations to find molecular weight and use the equation “molecular weight= mass/moles.”. The mass of the sample was .386 g and the amount of moles is .0049 moles, to get moles I used the equation “n=pv/rt”. …show more content…
To get that formula you take the given value of 408832g CO2 and divide it by the total amount of CO2 and that is 44.011g CO2 then multiple that by 12.011g C and get 1.332669451g C. Then convert that to moles by dividing it by 12.011g C which will equal .11095408mol C. Then you must do the same with 1.9986g H2O, so you must divide that by the amount of H20, 18.016g H2O, then multiply it by 2.016 g H2 which is .223644405g H2. Then convert .223644405g H2 to moles by dividing it by 1.008g H which will give you .221869449mol H. Then to find the moles of oxygen I subtracted the total (2.0000g) and the grams of carbon (1.332669451g C) and the grams of hydrogen (.223644405g H) to get .443686144g O, then convert grams of oxygen to moles of oxygen and get .027730384mol O. Then to finally determine empirical formula you multiply moles of Hydrogen(.223644405mol H) by moles of Carbon(.11095408mol C) by moles of Oxygen(.027730384mol O) then divide that value by the smallest mole value which is Oxygen(.027730384mol O) and get the formula C4H8O. Then divide the mass of C4H8O1 by the molecular weight 83.85 and get .85996422 which rounded gives you 1 so the molecular formula is …show more content…
I first believed it to be “Cyclobutanol” because they have the same molecular formula but the densities are off because the unknown has a density of .7976 g/mol and “Cyclobutanol has a density of .92 g/mol.” Then there was the possibly of my unknown being “Ethyl vinyl ether” however the boiling points were different my unknown had a boiling point of 80.0 Celsius and “Ethyl vinyl ether had a boiling point of 36.0 Celsius.” Then there was the possibility of “2-Methoxypropene” being my unknown but the IR spectrum had a different peak then my unknown’s peaks. The closest substance with a molecular formula of C4H8O1 and a density of .7976g/mol and a boiling point of 80.0 degrees Celsius is “2-Butanone and that has a density of .806g/mol and a boiling point of 80.0 degrees Celsius” my density is slightly off from the found substance but there is experimental error. An experiment error could have been taking the flask off of the hot plate before all the substance was evaporated. Another error could have been when we touched the flask to see if it was cooled to room temperature before massing the flask and getting oil on the flask that can change the mass of the