Question

How to calculate amount of substance in phycocyanin tablet? please help

a. You want to determine the amount of Phycocyanin in a 500 mg herbal tablet. Before analysis by UV-Vis spectroscopy, five Phycocyanin tablets (each 500 mg) were crushed in a pestle and mortar. 1.2 g of the crushed tablets were then dissolved in 100 ml of ethanol. 50 ml of the solution was filtered to remove undissolved particles. 2 ml of the filtered solution was then diluted by a factor of 10 (i.e., dilution factor is 10). The diluted sample was analysed using UV-Vis spectroscopy and had an absorbance of 0.324. Show each calculation step clearly.

What was the amount in mg of Phycocyanin in a 500 mg tablet?

What is the percent (%w/w) of Phycocyanin in a tablet?

Molar mass of phycocyanin=586.69

Answer 1

I got `"346 mg"` per tablet, and `69% "w/w"`. To two sigs, `"350 mg"`.

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Since the first goal is to get the `"mg"` of phycocyanin in the tablet, we ought to be able to get the concentration at some point. And since absorbance is related to concentration, that's where we should start.

Recall Beer's law:

`A = epsilonbc`

where:

• `A` is the absorbance.
• `epsilon = "229 L/mol"cdot"cm"` is the molar absorptivity (extinction coefficient) of phycocyanin typically in `"L/mol"cdot"cm"`.
• `b = "1 cm"` is the path length of a typical cuvette.
• `c` is the concentration of the sample in typically `"mol/L"`.

Let's track what has been going on.

1. Five tablets were prepared, `"1.2 g"` in `"100 mL"` of ethanol. The initial concentration is `"12 g tablet/L"` (and some particles were filtered out later that did not dissolve past `"12 g/L"`).
2. When `"50 mL"` of the solution was used, the concentration was unaltered from the `"100 mL"` solution.
3. `"2 mL"` of the solution was diluted 10-fold, so you now have `"1.2 g/L"` of the tablet in solution. The tablet has a certain mass percentage of phycocyanin.
4. This particular sample of unknown concentration has an absorbance of `0.324`, of only the phycocyanin.

Note that `A` is of the phycocyanin itself, and the concentration of `"1.2 g/L"` is of the tablet.

So, we will need to find the concentration using the absorbance, not through the dilution logic.

`c = A/(epsilonb) = 0.324/("229 L/mol"cdotcancel"cm" cdot cancel"1 cm")`

`=` `"0.00141 mol/L phycocyanin"`

Or, using its molar mass would give it in `"g/L"`:

`(0.00141 cancel"mol")/"L" xx "586.69 g phycocyanin"/cancel"mol"`

`=` `"0.830 g/L"`

Now, the key point is that it was of the `bb"20 mL"` of solution put into the UV-Vis spectrophotometer, since a 10-fold dilution was done on a `"2 mL"` sample, so there are

`"0.830 g phycocyanin"/(1000 cancel"mL") xx 20 cancel"mL" = "0.0166 g phycocyanin"`

in `bb"2 mL"` of solution (since adding solvent does not change the mass of the solute).

[At this point we COULD determine the concentration of the phycocyanin in the starting solution already, but I think the following argument is more useful.]

Since mass and volume are extensive, and concentration is intensive we can say that

`"0.0166 g phycocyanin"/"2 mL solution" = "0.415 g phycocyanin"/"50 mL solution"`

after filtering undissolved particles out. Before filtering undissolved particles out,

`"0.415 g phycocyanin"/"50 mL solution" = color(green)("0.830 g phycocyanin"/"100 mL solution")`

which, by the way, is also `"8.30 g phycocyanin"/"L solution"`.

That's what we expected from back-calculating from a 10-fold dilution of the `"0.830 g phycocyanin /L"` measured sample: the concentration should be `bb10` times of what it was compared to when it was measured.

Now that we know how much of the `"1.2 g"` was phycocyanin that was dissolved in `"100 mL"` of ethanol, we now consider it before it was dissolved.

We can say that before grinding the tablets,

`"0.830 g phycocyanin"/"1.2 g crushed tablets" = "1.73 g phycocyanin"/"2.5 g whole tablets"`

or

`"1730 mg phycocyanin"/"2500 mg whole tablets"`

or

`color(blue)("346 mg phycocyanin"/"500 mg tablet")`

Therefore, there are `ulbb"346 mg"` of phycocyanin per tablet, and the mass percent is:

`= color(blue)ul(69%"w/w")`

to two sig figs.