# Calculation of Silia*MetS* to Use

> back to Silia*MetS* metal scavengers

To get an effective metal removal, the amount of Silia*MetS* Metal Scavenger used is very important. You can determine by two different ways how much scavenger will be needed;

- from the residual concentration (
*more accurate method*) - from the amount of metal catalyst used (
*when the residual metal concentration is unknown*)

## From residual metal concentration (*ppm*)

Knowing the palladium (Pd) level in 800 g of material is 500 ppm, (the oxidation state does not affect the calculation).

**Data needed:**

- Loading of the scavenger (SiliaMetS Thiol): 1.2 mmol/g
- Metal molecular weight: Ex. Pd = 106.42 g/mol
- Amount of product to be treated containing the Pd: Ex. 800 g
- Residual concentration of metal: Ex. 500 ppm of Pd

### 1. Determine the amount of palladium to be scavenged

### 2. Calculate the amount of scavenger (SiliaMetS Thiol) to use (1 equivalent)

To scavenge 400 mg of palladium, 3.13 g of Silia*MetS* Thiol is needed if using only one equivalent. However, it is highly recommend trying a minimum of 4 equivalents at first. In this case, the amount of Silia*MetS* Thiol will be 4 times higher (4 x 3.13 g = 12.40 g).

Sometimes, the metal residual concentration is unknown. In such a case, the amount (g) of palladium to be scavenged can be replaced by the amount of metal catalyst used for the reaction:

## From amount of metal catalyst used

- Amount of metal catalyst used: Ex. 10 g of Pd(PPh
_{3})_{4} - Metal catalyst molecular weight: Pd(PPh
_{3})_{4}= 1155.56 g/mol

**Data needed:**

### Determine the amount of palladium to be scavenged

The amount of Silia*MetS* Thiol to be used can then be determined as stated above (see point 2. above). In this particular case, one equivalent of Silia*MetS* Thiol corresponds to 7.20 g.