An analysis of ore assays about 1.5% iron. What minimum sample mass should be taken if the relative error resulting from a 0.5 Ng loss is not to exceed -2%?

Steps to Solve

1. Define Relative Error Relative error is calculated as the difference between the measured value and the true value, divided by the true value. In this case, the formula can be expressed as follows:

$$ \text{Relative Error} = \frac{\text{Measured Value} - \text{True Value}}{\text{True Value}} $$

2. Set Up the Equation From the problem, we know that the relative error must not exceed -2% for a loss of 0.5 Ng. We convert -2% to its decimal form:

$$ -2% = -0.02 $$

3. Express the Measured Value Let ( m ) be the true sample mass. Given a loss of 0.5 Ng, the measured value after loss can be expressed as:

$$ m - 0.5 , \text{Ng} $$

Now we can plug this into the relative error formula:

$$ \frac{(m - 0.5) - m}{m} = -0.02 $$

4. Simplify the Equation We simplify the equation from step 3:

$$ \frac{-0.5}{m} = -0.02 $$

5. Solve for Sample Mass $m$ Now we solve for ( m ):

Multiply both sides by ( m ) and then by -1:

$$ 0.5 = 0.02m $$

Now divide by 0.02:

$$ m = \frac{0.5}{0.02} $$

Calculate the value:

$$ m = 25 , \text{Ng} $$