[Solidus]

Hey, I am new to this site and was wondering if you guys could help me with this task.

I conducted an experiment where i measured the time it took to evaporate 1 drop of gasoline and 1 drop of 2-propanol.

I was given a list of some other organic compounds with their enthalpy of vaporization values and asked to measure the time it took to evaporate a drop of each:

Compund Enthalpy of vaporization (kJ/mol)  time in seconds

Water               43.98                            155
Methanol           37.43                             5
Ethanol             42.43                             27
1-propanol         47.45                             25
1-butanol           52.35                            41
1-pentanol         57.02                            42
Heptane            36.57                            15
Octane              41.47                            22
Nonane             46.55                            59
Decane              51.42                           65


The time it took for gasoline to evaporate = 8 seconds
and 2-propanol = 22 seconds

I am now supposed to calculate the enthalpy of vaporization of gasoline and 2-propanol by comparing the times to the given compounds.

Now I was wondering if you guys can give me some tips on how to do this. What should i consider when trying to calculate the enthalpy of vaporization? It is strange that water has such a lower enthalpy of vaporization value than for example 1-propanol but it still took a lot more time to vaporize a drop of water than a drop of 1-propanol. Also, what compounds does gasoline consist of?

I would appreciate any help i can get

Thanks.

[Yggdrasil]

Try making a scatter plot of the enthalpies of vaporization v. the times of vaporization then constructing a linear regression line.  This may give you a rough estimate of the relationship between the two variables.

I think gasoline is a mixture of hydrocarbons around 7 or 8 carbons in lenght.  So I think its mostly isomers of octane and heptane.  Wikipedia may have a better answer.

[Solidus]

Yes that is helpful. I was thinking of drawing one scatter plot with a line of best fit for the alkanes and one for the alcohols for more accurate results. Maybe i should exclude water? Once again, does anyone know why water has such a high evaporation time value but a low enthalpy of vaporization value?

[Yggdrasil]

Grouping compounds by structural similarity is a great idea especially if the data indicate that the two sets of data bhave differently.

Excluding water is probably a good idea.  My guess as to why its evaporation time is unusually high is due to its surface area.  Water is very cohesive and it forms drops instead of spreading out over a surface.  This limits the surface available for evaporation and slows evaporation.

[Solidus]

yes and also since we evaporated one drop of each substance, more water molecules form in a drop on water than a drop of 2-propanol for example. So the volume of water that was evaporated was considerably more than the other substances and thus the longer evaporation time. Correct?

[Yggdrasil]

That's another good explanation.

[Solidus]

One last problem.

Each substance was evaporated 3 times and the times were recorded.

So for water for example the times were

Time 1 = 168.12 ± 0.5 seconds
Time 2 = 145.52 ± 0.5 seconds
Time 3 = 171.93 ± 0.5 seconds

so for the average, (168.12+145.52+171.93) ÷ 3 = 161.86 ± ?

Does anyone know what method can be used to calculate the error calculation on the mean? I was thinking of using largest deviation but then we would get an error calculation of 26 seconds or so hehe. Is there a better method?

Thanks

[Dan]

One last problem.

Each substance was evaporated 3 times and the times were recorded.

So for water for example the times were

Time 1 = 168.12 ± 0.5 seconds
Time 2 = 145.52 ± 0.5 seconds
Time 3 = 171.93 ± 0.5 seconds

so for the average, (168.12+145.52+171.93) ÷ 3 = 161.86 ± ?

Does anyone know what method can be used to calculate the error calculation on the mean? I was thinking of using largest deviation but then we would get an error calculation of 26 seconds or so hehe. Is there a better method?

Thanks

I would have thought that...

mean = (168.12 ± 0.5 + 145.52 ± 0.5 + 171.93 ± 0.5)/3
        = (168.12 + 145.52 + 171.93 ± 1.5)/3
        = [(168.12 + 145.52 + 171.93)/3] ± 0.5

but I'm no statistician.

[Solidus]

That is what I thought as well, but I think you have to use standard deviation or largest deviation or something like that.

Does anyone know a program where you can easily write mathematical formulas? It kinda sucks to write them in word as it is limited.

[mir]

If your sample distrubution is like a gauss curve, you could use the ordinary formula for standard deviation (s).

And then from the number of freedom (n-1), find t and the confidence limit (95%).

For your three sample large population, t = 4.3

In this interval, you find with 95% certainty a random sample in your population.

But of course, before you do this, you have to check for outliers.

[Yggdrasil]

I agree with mir.  The standard deviation of your measurements is larger than the inherent error of your instruments, so you would report the standard deviation as the error in your measurements.

[Solidus]

The problem is that, lol, I did not understand much of what mir said. Could you explain it a bit more with maybe an example?

[Yggdrasil]

Standard deviation is explained here http://www.robertniles.com/stats/stdev.shtml

If you have more questions about standard deviation, I can post an example if you need it.  Excel can calculate standard deviation using stdev(x1, x2, x3, ...), where the x1, x2, x3,... are your measurements.

You can ignore what mir said about confidence intervals, degrees of freedom, and outliers.  That's overkill for a data set consisting of only three data points.

[mir]

You can ignore what mir said about confidence intervals, degrees of freedom, and outliers.  That's overkill for a data set consisting of only three data points.

Hehe :-)

I say: You should use every occasion to use what you have learned. Three data points, should not scare you - You should laugh, and just do it 

[Solidus]

One thing that is puzzling me is why methanol, even though it has hydrogen bonds, is such a liquid with such low viscosity and enthalpy of vaporization value.


Also one thing that is yet unclear to me:

                        kJ/mol                          seconds

Water               43.98                            155
1-propanol         47.45                             25

Why does water have a lower enthalpy value than one 1-propanol but it takes over 6 times as long to evaporate water than 1-propanol?

I thought the definiton of enthalpy of vaporization was the energy required to transform a given quantity of a substance into a gas that is usually measured in kJ/mol.

Alright maybe in this experiment, there is no given quantity as it was a drop of a substance. However, as i see it, even if it would be a GIVEN quantity of water and
1-propanol, water still has far stronger intermolecular forces than 1-propanol and yet it has a lower enthalpy value? I simply dont understand.