A 5.00 mL sample of ethanol in gasoline is mixed with 95.0 mL of chloroform. The

Question

A 5.00 mL sample of ethanol in gasoline is mixed with 95.0 mL of chloroform. The chloroform will function as your internal standard. The mixture is injected into a GC. The ethanol has a peak area of 14.33 and the chloroform has a peak area of 89.65. A standard is prepared with 1.00 ML of ethanol in 99.00 mL of chloroform. The standard is injected into the GC. The ethanol has a peak area of 8.73 and the chloroform has a peak area of 63.45. What is the volume percent of the ethanol in the gasoline sample.

Explanation / Answer

Ans. Given,

Peak area of ethanol in gasoline (5.0 mL sample) = 14.33

Peak area of ethanol standard (1.0 mL sample) = 8.73

It’s assume that the obtained peak area is directly proportional to H2O content.

Now,

            Peak area of 8.73 is equivalent to 1.0 mL pure ethanol

        Or, Peak area of 1.00 is equivalent to (1.0/ 8.73) mL pure ethanol

        Or, Peak area of 14.33 is equivalent to (1.0/ 8.73) x 14.33 mL pure ethanol

                                                            = 1.64 mL pure ethanol

Thus, 5.0 mL gasoline has 1.64 mL pure ethanol.

Now,

            % (V/V) ethanol = (volume of ethanol / total volume of gasoline sample) x 100

                                    = (1.64 mL/ 5.0 mL) x 100

                                    = 32.83%

Hence, volume % of ethanol in gasoline is 32.83%.

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