Question

The components of a 10.2 mg sample are sufficiently separated by a chromatographic column that is 2.5 cm in diameter and 50 cm long using a flow rate of 0.56 mL/min. To separate 205 mg of the sample with a similar separation of the components, what size column diameter and length should be used

Answer 1

The question involves increasing the size of a chromatographic column to separate a larger sample while maintaining resolution, which requires consideration of various chromatographic parameters and is often optimized experimentally.

Step-by-step explanation:

The question relates to scaling up chromatographic separations, specifically how to modify column dimensions to separate a larger sample size while achieving similar separation of components. The initial separation used a chromatographic column 2.5 cm in diameter and 50 cm long for a 10.2 mg sample.

To separate 205 mg of the sample, one would need to increase the column size. However, the exact calculation for this scaling is not straightforward and depends on factors such as the chromatographic technique, type of separation (isocratic or gradient elution), and equipment limitations.

This would typically require a higher volume of mobile phase and possibly a longer column length or larger diameter to accommodate the increased sample load while maintaining resolution.

Precise calculations would be made based on column volume, flow rate, sample capacity, and the properties of the compounds being separated. For practical purposes, this is often done experimentally to optimize separation without loss of resolution.

Answer 2

To separate 205 mg of the sample with a similar separation of the components, you would need a column with a diameter of approximately 4.19 cm and a length of 1000 cm.

Step-by-step explanation:

To separate 205 mg of the sample with a similar separation of the components, you would need to determine the size of the column diameter and length that should be used.

First, we can calculate the flow rate of the new sample using the given flow rate of 0.56 mL/min and the mass ratio between the new sample and the initial sample. Since the initial sample is 10.2 mg and the new sample is 205 mg, the flow rate for the new sample would be:

(205 mg / 10.2 mg) × 0.56 mL/min = 11.06 mL/min

Using the equation for flow rate, Q = A × v, where Q is the flow rate, A is the cross-sectional area of the column, and v is the linear velocity, we can rearrange to solve for the cross-sectional area:

A = Q / v = 11.06 mL/min / (0.2 cm/min) = 55.3 cm²

Now, we can calculate the diameter of the new column using the formula for the area of a circle:

Area = π × r², where r is the radius and π is a constant. Rearranging to solve for the radius, we get:

r = √(Area / π) = √(55.3 / π) ≈ 4.19 cm

Finally, we can double the length of the original column since we want to separate a sample that is approximately 20 times larger, so the new column length would be:

50 cm × 20 = 1000 cm

Therefore, to separate 205 mg of the sample with a similar separation of the components, you would need a column with a diameter of approximately 4.19 cm and a length of 1000 cm.