Question

1. describe what percent error is measuring: measuring the accuracy measured value - accepted value/accepted value x 100
2. what would a negative percent error indicate? the measured value is smaller than the accepted value
3. if a student’s percent error for the mass of a substance is 100% and the actual mass of the substance is 50 grams, what was the student’s measured mass?

\frac{100 - 50}{50}

1. a student in a laboratory determined the boiling point of a substance to be 71.8°c. the accepted value for the boiling point of this substance is 70.2°c. what is the percent error of the student’s measurement?

Explanation:

Step1: Recall percent - error formula

The percent - error formula is $\text{Percent Error}=\frac{\vert\text{Measured Value}-\text{Accepted Value}\vert}{\text{Accepted Value}}\times100\%$.

Step2: Solve problem 5

Given $\text{Percent Error} = 100\%$ and $\text{Accepted Value}=50$ grams. Let the measured value be $x$.
Substitute into the formula: $100=\frac{\vert x - 50\vert}{50}\times100$.
First, divide both sides by 100: $1=\frac{\vert x - 50\vert}{50}$.
Then, multiply both sides by 50: $\vert x - 50\vert=50$.
Case 1: $x−50 = 50$, then $x=100$.
Case 2: $x - 50=-50$, then $x = 0$.

Step3: Solve problem 6

The measured value $x = 71.8^{\circ}C$ and the accepted value $y = 70.2^{\circ}C$.
Substitute into the percent - error formula: $\text{Percent Error}=\frac{\vert71.8 - 70.2\vert}{70.2}\times100\%$.
First, calculate the numerator: $\vert71.8 - 70.2\vert=1.6$.
Then, $\text{Percent Error}=\frac{1.6}{70.2}\times100\%\approx2.28\%$.

Answer:

1. The measured mass could be 0 grams or 100 grams.
2. The percent error is approximately $2.28\%$.