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Two chemistry questions?

1. A certain organic compound in a vapor state has a density of 3.27 g/L at 95 degrees C and 758 torr. What is the molecular weight? If the compound contains 24.2% C, 4.1% H, and 71.1% C, What is the molecular formula?

For the first one, use the ideal gas law (PV=nRT), and rearrange it to get n/V = P/(RT). This will give you moles per volume. If you plug in the given values for P, R, and T and then cancel out all the units, you get 0.0330 moles per Liter. This is equal to the 3.27 grams per Liter you were given originally. In other words, 3.27 grams of your compound equals 0.0330 moles. Divide 3.27 by 0.0330 and you get 99.04 grams per mole.

Did you make a typo on the percentage compositions? If you add up 24.2, 4.1, and 71.1, you only get 99.4%. Also, you listed out carbon as being both 24.2 and 71.1%. I’m not sure if that was genuine or if you meant 24.2 % of some other element. Anyway, assuming that these were not typos, the easiest way to do this is trial and error (in my opinion). You know its an organic compound which means a backbone of carbon. The general rule is that an organic compound will have 2n+2 hydrogens where n is the number of carbon atoms. I tried n=7, which gives 7 carbons and 16 hydrogens. That’s 100 g/mol, but does not match the percentages you gave.

Lastly, the balanced equation is 2 C2H6 + 7O2 = 4 CO2 + 6 H2O. The initial conditions (200 mL, 20 C, 750 torr, etc) means you can use the ideal gas law to back compute the number of moles of CO2 produced (0.008205). Using the stoichiometric ratio of 4 moles CO2 to 2 moles ethane, you get that 0.00410 moles ethane were burnt originally.

As a general rule, combustion reactions are complete which means that all the ethane was consumed. Again using the ratio of 6 moles water to 2 moles ethane means 0.0123 moles water vapor was produced. Rearrange the ideal gas law to the form of V = nRT/P, plug in the given values (24 C, 790 torr, etc), and cancel out units to get 0.288 L.

As an aside, I’ve assumed that the 4 C higher temperature and the 20 torr greater pressure will not affect the degree of combustion. Strictly speaking, this isn’t true since a higher temperature disfavors the products in an exothermic reaction (which combustion is) and a higher pressure disfavors the products in a reaction which increases the number of gas molecules present (9 on the reactant side vs. 10 on the product side). However, I doubt this will be a big change since combustion reactions are normally total. Additionally, unless you’ve been given an equilibrium constant or measurements of partial pressure, I doubt you could compute the change in moles on the product side resulting from the higher temperature and pressure.

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