In gas chromatography, the Kovats retention index (shorter Kovats index, retention index; plural retention indices) is used to convert retention times into system-independent constants. The index is named after the Hungarian-born Swiss chemist Ervin Kováts (1927–2012), who outlined the concept in the 1950s while performing research into the composition of the essential oils. The retention index of a chemical compound is retention time interpolated between adjacent n-alkanes. While retention times vary with the individual chromatographic system (e.g. with regard to column length, film thickness, diameter and inlet pressure), the derived retention indices are quite independent of these parameters and allow comparing values measured by different analytical laboratories under varying conditions and analysis times from seconds to hours. Tables of retention indices are used to identify peaks by comparing measured retention indices with the tabulated values.
Isothermal Kovats retention index The Kovats index applies to organic compounds. The method interpolates peaks between bracketing n-alkanes. The Kovats index of n-alkanes is 100 times their carbon number, e.g. the Kovats index of n-butane is 400. The Kovats index is dimensionless, unlike retention time or retention volume. For isothermal gas chromatography, the Kovats index is given by the equation:
I i = 100 [ n + l o g ( t i − t 0 ) − l o g ( t n − t 0 ) l o g ( t n + 1 − t 0 ) − l o g ( t n − t 0 ) ] , {\displaystyle I_{i}=100\left[n+{\frac {log(t_{i}-t_{0})-log(t_{n}-t_{0})}{log(t_{n+1}-t_{0})-log(t_{n}-t_{0})}}\right],}
where the variables used are:
I i {\displaystyle I_{i}} , the Kováts retention index of peak i
n {\displaystyle n} , the carbon number of n-alkane peak heading peak i
t i {\displaystyle t_{i}} , the retention time of compound i, minutes
t 0 {\displaystyle t_{0}} , the air peak, void time in average velocity u = L / t 0 {\displaystyle u=L/t_{0}} , minutes The Kovats index also applies to packed columns with an equivalent equation:
I i = 100 [ n + l o g ( V i 0 ) − l o g ( V n 0 ) l o g ( V n + 1 0 ) − l o g ( V n 0 ) ] {\displaystyle I_{i}=100\left[n+{\frac {log(V_{i}^{0})-log(V_{n}^{0})}{log(V_{n+1}^{0})-log(V_{n}^{0})}}\right]}
… excerpt ends here. Continue reading the full article.
