Chapter 24: QE-E (page 662)

When 1.06 mmol of 1-pentanol and 1.53 mmol of 1-hexanol were separated by gas chromatography, they gave peak areas of 922 and 1570 units, respectively. When 0.57 mmol of pentanol was added to an unknown containing hexanol, the peak areas were 843:816 (pentanol:hexanol). How much hexanol did the unknown contain?

Short Answer

Expert verified

The unknown contains 0.47 mmol of hexanol.

Step by step solution

Formula

Therefore, for a standard mixture the formula needs to be used for quantitative analysis with internal standard

$\begin{array}{rcl} \frac{A_{X}}{X} & = & F [\frac{A_{S}}{S}] \\ \frac{V C_{X}}{X} & = & F [\frac{V C_{S}}{S}] \\ \frac{C_{X}}{X} & = & F [\frac{C_{S}}{S}] \\ A_{X} & = & A r e a o f a n a l y t e s i g n a l \\ A_{s} & = & A r e a o f i n t e r n a l s \tan d a r d \\ X & = & C o n c e n t r a t i o n o f a n a l y t e s i g n a l \\ S & = & C o n c e n t r a t i o n o f i n t e r n a l s \tan d a r d \end{array}$

Calculation

Let, P stands for pentanol and H stands for hexanol.

As the volume is unknown therefore concentration must be substituted as concentration is proportional to mmol.

$\begin{array}{rcl} \frac{C_{H}}{H} & = & F [\frac{C_{P}}{P}] \\ \frac{1570}{1.53} & = & F [\frac{922}{1.06}] \\ F & = & 1.18 \end{array}$

Now it is stated that when 0.57 mmol of pentanol was added to an unknown containing hexanol, the peak areas were 843:816 (pentanol: hexanol). Therefore, for an unknown mixture we can write

$\begin{array}{rcl} \frac{C_{H}}{H} & = & F [\frac{C_{P}}{P}] \\ \frac{816}{H} & = & 1.18 [\frac{843}{0.57}] \\ H & = & 0.47 \end{array}$