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Unformatted text preview: as those in chromatographic injection, may be compensated completely by normalizing the analyte peak areas to the IS peak area. Let’s do an experimental design exercise to prepare you for the quiz and Experiment II. Consider a piece of metallurgical grade silicon that is expected to have between 2 and 20 ppb iron (µg Fe per Kg of Si sample). You need to determine the amount of Fe by ICP‐MS, a sensitive method that requires dissolution of the sample. Let’s just assume for the moment that the ICP‐MS dynamic range is 0.10 ppb to 10 ppm (that is mg Fe per liter of solution) and that its sample volume requirement is 25 mL. The sample preparation is difficult and expensive because it requires dissolution of the Si in an oxidizing hydrofluoric acid bath (HF / HNO3 or some such abomination), so this should be considered and a large excess of Si should not be digested. You have at your disposal a 1000 ppm Fe solution and a range of common glassware including volumetric flasks of 10, 25, 50, 100, 250 and 500 mL, and volumetric pipets of 1, 2, 5, 10, 25, 50 and 100 mL. Use the following constraints: pipet no less than 1 mL, generate no more than 100 mL of any solution. Design a simple external calibration experiment. In the end you should be ableto report the following. 1.
4. Mass of Si sample to digest. Volume for digest. Possible dilutions for sample digest (vol flask and pipet volumes). Pipet and vol flask volumes and instructions for preparing five calibration standards. Note the following: unit ppth ppm ppb ppt Name in words Parts per thousand Parts per million Parts per billion Parts per trillion Definition 1 Grams per liter or Kg Milligrams per liter or Kg Micrograms per liter or Kg Nanograms per liter or Kg Definition 2 ‐ Micrograms per mL Nanograms per mL Picograms per mL Take 5 g Si dissolve and dilute to 25 mL. Conc is approx: 5g x 2ng/gSi/25 mL = 0.4 ppb on the lower end. This is acceptable. Conc is approx: 5g x 20ng/gSi/25 mL = 4 ppb on the upper end. This is acceptable. Cal stds should then be 0.1, 0.2, 0.5, 5, 10 ppb 1000 ppm x 1000 ppb / ppm = 1,000,000 ppb stock 1 mL dil to 100 mL = 10,000 ppb 1 mL dil to 100 mL = 100 ppb 100 ppb x 0.1 mL dil to 100 mL = 0.1 ppb 100 ppb x 0.2 mL dil to 100 mL = 0.2 ppb 100 ppb x 0.5 mL dil to 100 mL = 0.5 ppb 100 ppb x 5 mL dil to 100 mL = 5 ppb 100 ppb x 10 mL dil to 100 mL = 10 ppb Also note: Milli = 10‐3 Micro = 10‐6 Nano = 10‐9 Pico = 10‐12 pre m μ
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This homework help was uploaded on 04/04/2014 for the course CHEM 155 taught by Professor Terrill,r during the Spring '08 term at San Jose State.
- Spring '08