> #chemistry , #mining , #analysis > The methods described to be used for analysis of ore that assay about 1.2% of gold. What minimum sample mass should be taken if relative error resulting from 0.4mg loss is not exced a) -0.1% b)-1.2% ?
Jan 13

# The methods described to be used for analysis of ore that assay about 1.2% of gold. What minimum sample mass should be taken if relative error resulting from 0.4mg loss is not exced a) -0.1% b)-1.2% ?

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Jan 13
To determine the minimum sample mass that should be taken, we need to calculate the maximum allowable relative error resulting from a 0.4mg loss.

First, we need to find the mass of gold in the ore sample. Let's assume the mass of the ore sample is M grams.

Mass of gold = 1.2% of M = (1.2/100) * M = 0.012M grams

Now, we need to find the maximum allowable relative error resulting from a 0.4mg loss. Let's assume the maximum allowable relative error is E%.

E% = (0.4mg / 0.012M grams) * 100

To find the minimum sample mass, we need to solve for M in the equation above.

a) For a maximum allowable relative error of -0.1%:
-0.1 = (0.4mg / 0.012M) * 100
-0.1 * 0.012M = 0.4mg
-0.0012M = 0.4mg
M = 0.4mg / -0.0012
M ≈ -333.33 grams

Since mass cannot be negative, this solution is not valid.

b) For a maximum allowable relative error of -1.2%:
-1.2 = (0.4mg / 0.012M) * 100
-1.2 * 0.012M = 0.4mg
-0.0144M = 0.4mg
M = 0.4mg / -0.0144
M ≈ -27.78 grams

Again, mass cannot be negative, so this solution is also not valid.

Therefore, there is no valid minimum sample mass that can be taken to ensure that the relative error resulting from a 0.4mg loss does not exceed -0.1% or -1.2%.
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