In a gas chromatography experiment, a 2.4 x 10-6 L sample of wine is found to contain 3.1 x 10-7 L of ethanol. a. What is the percentage of ethanol (by volume) in the wine?

Answer 1

Hi, Chelsea,

Both values for the total sample and the ethanol are in liters. Agreed that they are inconveniently small numbers, but as long as the units are the same, they will cancel.

The % by volume is the volume of (ethanol/total volume)*100%

Ethanol = 3.1 x 10^-7 L

Wine sample = 2.4 x 10^-6 L

Percent ethanol in the wine = 3.1 x 10^-7 L/(2.4 x 10^-6 L) x 100%

I'm going to eliminate a lot of the x 10^xx by adding/subtracting their exponents

This can be written as (3.1/2.4)*(10^-7 * 10²)/10^-6 (the Liters cancel out, leaving the %)

3.1/2.4 = 1.3 (2 sig figs)

The exponents on top are added (-7 + 2 = - 5) and the ones on bottom are subtracted from the ones on top (- 5 - (-6) = 1).

So the result is 1.3 x 10¹, or 13%

I hope this helps,

Bob

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