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### Topic: What is meant by 'Mol%'  (Read 2015 times)

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#### foxthreefour

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« on: October 24, 2018, 08:45:09 PM »
Hi, I'm reading an article on the reduction of ketones using H2, Pd/C. In the work up it states the ratios as:

1 mmol substrate
10ml ETOH
10 mol% Pd/C

Now the loading of the Pd/C is 5%. Is the 'mol%' part a typo? Does it simply mean that for every 1 mol of substrate 0.1mol of Pd/C is added, or 0.1mol of Pd (relative to its loading of 5%) is added?

#### chenbeier

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« Reply #1 on: October 25, 2018, 06:23:07 AM »
Another word is mole fraction for it:

https://en.wikipedia.org/wiki/Mole_fraction.

10 mol% means you you from the sum of mole of palladium and carbon 10% Pd.

In case you have 1 mol Pd/C then you have 0,1 mol Pd and 0,9 mol C.

For 5% it is 0,05 mol Pd and 0,95 mol C.

#### foxthreefour

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« Reply #2 on: October 25, 2018, 03:27:15 PM »
Another word is mole fraction for it:

https://en.wikipedia.org/wiki/Mole_fraction.

10 mol% means you you from the sum of mole of palladium and carbon 10% Pd.

In case you have 1 mol Pd/C then you have 0,1 mol Pd and 0,9 mol C.

For 5% it is 0,05 mol Pd and 0,95 mol C.

So would I be right in saying if the reaction was carried out on a 1 mol scale, the amounts would be:

1 mol substrate
10,000ml EtOH
212g Pd/C

For Pd/c method was:

Pd mol (106.42) / 10% = 10.642 x 20 (for 5% loading on C) = 212.84g Pd/C

Is this right?

#### chenbeier

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« Reply #3 on: October 25, 2018, 03:35:20 PM »
If you have 1  mol Pd/C and the percentage is 5% then you have 0,05 mol Pd and 0,95 mol C.

0,05 mol Pd = 5,321 g
0,95 mol C = 11,4 g

#### foxthreefour

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« Reply #4 on: October 25, 2018, 03:43:43 PM »
Thanks Chenbeier. Here is an extract of the article, I'm not sure I've communicated it effectively.

'In a typical procedure all reactions were performed on a 1
mmol scale. The substrate was generally dissolved in 10 ml of
EtOH, 10 mol% of the catalyst was added and the mixture
degassed twice under vacuum (using a water pump) and
refilling with hydrogen each time.'

The catalyst in this reaction is 5% Pd/C. In other reactions the catalyst varied between 5% Pd/CaCO3, 5% Pd/Al2O3, 10% Pd/C etc. Can you see where the confusion comes in? I feel like I'm dealing with a percentage of a percentage, just want to make sure I get it right.

#### kriggy ##### Re: What is meant by 'Mol%'
« Reply #5 on: October 26, 2018, 02:17:03 AM »
You have to calculate, because teh actualy catalyst is the Palladium itself so if you have to use 5mol% of catalyst on 1mmol scale its about 5mg of pure palladium which then means 50mg of 10% Pd/C or 100 mg of 5% Pd/C (you weight the amount of catalyst depending on the one you have available actualy, higher loading means less amoung weighted but there is a same amount of active catalytic centres)