Question

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2) Consider anarrow bore (0.25 mm ID) thin-film (0.10 um) column with 5000 plates/meter. Consider also,la wide-bore (0.53 mm diameter, thick-film (5.0 um) column with 1500 plates per meter. The density of the stationary phase is 1.0 g/mL. a) What mass of stationary phase is in each column in a length equivalent to one theoretical plate? How many nanograms of analyte can be injected into each column if the mass of analyte is not to exceed 1% of the mass of stationary phase in one theoretical plate? b)

Answer 1

2 (a) Narrow-bore column –

Method 1:

V(stationary phase) = V(outer cylinder) – V(inner cylinder)

V(stationary phase) in a meter tube = πd_outer²L/4 - πd_inner²L/4

V(stationary phase) in a meter tube = π(1 m)/4{[(0.25 mm)²-[0.25 mm – 2(0.0001 mm)]²}

V(stationary phase) in a meter tube = π(1000 mm)/4[0.0625 mm²-(0.2498 mm)²]

V(stationary phase) in a meter tube = π/4(9.996 x 10^-2 mm³) = 7.85 x 10^-2 mm³

Method 2: Using a thin cylinder approximation, we can also find the stationary phase volume. See below:

V(stationary phase) in a meter tube = πdtL (where t = thickness)

V(stationary phase) in a meter tube = π(0.25 mm)(0.0001 mm)(1000 mm) = 7.85 x 10^-2 mm³ Mass(stationary phase) in a meter tube=(7.85 x 10^-2 mm³)(1.0 g/mL)(0.001 cm³/mm³)(1 mL/cm³)(10⁹ ng/g)= 7.85 x 10⁴ ng

Mass(stationary phase) in a meter tube = 7.85 x 10⁴ ng

Mass in one theoretical plate = (7.85 x 10⁴ ng/m)(1 m/5000 theoretical plate) = 16 ng/plate

Wide-bore column –

V(stationary phase) in a meter tube = πdtL

V(stationary phase) in a meter tube = π(0.53 mm)(0.005 mm)(1000 mm) = 8.325 mm³

Mass(stationary phase) in a meter tube=(8.325 mm³)(1.0 g/mL)(0.001 cm³/mm³)(1 mL/cm³)(10⁹ ng/g)

Mass(stationary phase) in a meter tube = 8.325 x 10⁶ ng

Mass in one theoretical plate = (8.325 x 10⁶ ng/m)(1 m/1500 theoretical plate) = 5.55 x 10³ ng/plate

2. (b) Narrow-bore column –

Injection limit = mass fraction of stationary phase x mass of stationary phase

Injection limit = (0.01)(15.7 ng) = 0.16 ng

Wide-bore column –

Injection limit = (0.01)(5.55 x 10³ ng) = 56 ng