Question

Preparing a Gross Sample for Laboratory Analysis

A carload of Hg-ore containing grains of cinnabar (86% Hg by mass; density= 8.19 g/cm3) and grains of basalt (containing no Hg; density=3.05 g/cm3) is to be sampled and analyzed for mercury. The average density of the bulk material is found to be 3.22 g/cm3. To keep the sampling error below 0.5% relative it is determined that 1.15×106 particles are needed in the gross sample. Supposing the grains in the sample had an approximate spherical radius of 3.0 mm, then the gross sample should weigh about 4.19×105 g (roughly half a ton). This is an impracticably large amount for lab analysis. Therefore, the carload of ore is repeatedly crushed or ground, mixed and divided until the grains are of a sufficiently smaller radius such that only a 150-gram gross sample will be required in the lab to obtain the same level of sampling uncertainty.

In the final ground and mixed sample, what should be the average particle mass?

/particle

In the final ground and mixed sample, what should be the approximate particle radius?

Answer 1

Total mass of laboratory sample = 150g.

Average mass of grains of the sample = 6 x 10^-5g/grain
Hence number of grains = 150/6 x 10^-5 = 2.5*10^6 grains.
Now average density = 3.22g/cm3,

volume of the 150g-sample = mass/density = 150g/3.22g/cm3 = 46.58cm3.

This is the volume of 2.5*10^6particles.

Hence average volume = 46.58/2.5*10^6 = 1.86 x 10^-5cm3
Volume of a sphere = 4/3 pi (radius)^3 = 1.86 x10^-5cm3
(radius)^3 = (1.86 x 10^-5 x 3)/4 x 3.142 = 4.38 x 10^-5

Hence r = 0.0352 cm