If 0.5 mg of precipitate loss causes a relative error of -1.0% at 50 mg, what will be the relative error at 500 mg?
Understand the Problem
The question is asking how the relative error changes when the mass of the sample increases from 50 mg to 500 mg while considering the loss of a precipitate. We need to determine the new relative error at the increased mass based on the initial conditions.
Answer
The new relative error is \( \frac{1}{10} e_1 \).
Answer for screen readers
The new relative error is ( \frac{1}{10} e_1 ).
Steps to Solve
- Identify Initial Conditions
Let the initial mass of the sample be ( m_1 = 50 , \text{mg} ) and let the loss of precipitate be ( x , \text{mg} ).
- Calculate Initial Relative Error
The relative error can be calculated using the formula:
$$ \text{Relative Error} = \frac{\text{Loss of Precipitate}}{\text{Initial Mass}} $$
So, the initial relative error ( e_1 ) is given by:
$$ e_1 = \frac{x}{m_1} = \frac{x}{50} $$
- Identify New Conditions
For the increased mass, let ( m_2 = 500 , \text{mg} ). The loss of precipitate remains the same.
- Calculate New Relative Error
The relative error at the new mass can be recalculated as follows:
$$ e_2 = \frac{x}{m_2} = \frac{x}{500} $$
- Express in Terms of Initial Relative Error
We can relate the new relative error ( e_2 ) to the initial one ( e_1 ):
From the earlier equation:
$$ e_1 = \frac{x}{50} $$
Thus,
$$ e_2 = \frac{50}{500} \times e_1 = \frac{1}{10} e_1 $$
This shows that the new relative error is one-tenth of the initial relative error.
The new relative error is ( \frac{1}{10} e_1 ).
More Information
This situation illustrates how increasing the mass reduces the relative error in measurements, making results more reliable. In practical terms, a larger sample helps account for errors, especially when the absolute loss is small compared to the total mass.
Tips
- Failing to recognize that the loss of precipitate remains constant while the mass changes, which can lead to incorrect calculations.
- Misapplying the relative error formula, especially confusing absolute error with relative error.
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