Some jargon
• sample – the material you are testing
• analyte – the chemical substance you are determining the concentration of
• matrix – everything in the sample that isn’t the analyte
• technique – the basic equipment or chemical process used in the analysis, e.g. titration, UV/VIS
spectrometer, pH electrode
• method – the set of steps used in the analysis
• procedure – the same as method
Suggested method of determining sample amount and necessary dilutions
1.
Start with 1 g or 1 mL of sample. Calculate how much analyte is in this amount. Work this out
in the unit specified in the standard range, eg if the range is in mg/L, use mg.
2.
Determine the concentration (in mg/L or g/L) if the 1 g/mL of sample is dissolved in a 100 mL
volumetric flask. This is very easy – just multiply by 10! (per 100 mL up to per L).
3.
Identify how much dilution or concentration is required to get the analyte level to the required
range. Use simple proportions (eg 2x, ÷5).
4.
Work out how this can be achieved in practice by changing sample amounts or dilution (via
pipette/vol. flask).
Points to consider
a) Sample masses should not be less than 100 mg or >25 g.
b) Pipette volumes should be between 5 and 50 mL.
c) Volumetric flasks should be between 50 and 500 mL.
d) Small decreases in concentration are done by reducing the sample amount or increasing the
volumetric flask volume, larger decreases by dilution. Avoid serial dilutions where possible – try
a less sensitive technique.
e) Increases in concentration are best done by increasing the sample amount. If this is not possible,
a more sensitive technique is probably the better option.
EXERCISE 1
Explain the reason for Points (a) to (e).
EXAMPLE 1
How much analyte is in (a) 1 g of 123 mg/kg and (b) 1 mL of 4.56 %w/v?
What will the concentration be when this is dissolved in 100 mL of solution?
(a)
123 mg/kg means 123 mg of analyte in 1 kg of sample, or 1000 g of sample.
In 1 g of sample, there will 123 ÷ 1000 mg of analyte, i.e. 0.123 mg.
When dissolved in 100 mL, we have 0.123 mg/100 mL. To get to a more normal unit, e.g.
mg/L, multiply by 10 => 1.23 mg/L.
EXAMPLE 1 (CONT’D)
(b) 4.56 %w/v means 4.56 g in 100 mL.
In 1 mL, there will be 4.56 ÷ 100 = 0.0456 g, or 45.6 mg.
When dissolved in 100 mL, we have 45.6 mg/100 mL, and multiply by 10 to get to 456 mg/L.
EXERCISE 2
1.
How much analyte in the specified unit in 1 g or mL of the sample? What is the concentration
when “dissolved” in 100 mL?
(a)
11.1 mg/kg (in mg)
(b)
222 g/L (g)
(c)
3.33 %w/w (mg)
(d)
0.444 g/100 mL (mg)
(e)
5.55 mg/kg (ug)
(f)
0.666 mg/L (ug)
2.
What dilution (or concentration) would be appropriate for the samples in 1(a-f) given the
following standard ranges?
(a)
5-20 mg/L
(b)
0.5-2 g/L
(c)
10-100 mg/L
(d)
1-5 mg/L
(e)
10-50 ug/L
(f)
10-50 ug/L
EXAMPLE 2
The fluoride in toothpaste is to be analysed. The range of standards for the technique is 5-100 mg/L.
The sample is expected to be 0.1%w/w fluoride. What is a suitable sample amount and dilution?
0.1%w/w means 0.1 g or 100 mg fluoride per 100 g of toothpaste.
Therefore 1 g of toothpaste contains 1 mg of fluoride.
1 g of toothpaste dissolved in 100 mL means 1 mg F/100 mL = 10 mg/L.
This is at the bottom end of the standard range, so it would be OK, but really a little low, so we need
more sample. 2 g will give 20 mg/L, which is better. 5 g will give 50 mg/L, which would be as high as
you would go.
EXERCISE 3
Determine a suitable sample amount and/or dilution for the following:
(a) The calcium concentration in milk is typically 0.15 g/100 mL. The standards are 1-100 mg/L.
(b)
A soil sample has an expected lead level of 10 mg/kg. The analytical procedure requires a
solution with a concentration between 25-100 ug/L.
(c)
The expected acid concentration of vinegar is 40 g/L. The standards are 1-10 g/L.
(d)
A purity check on AR grade sodium nitrate requires the preparation of a solution of
approximately 50 mg/L nitrate. FWs: Na 23, O 16, N 14