Question

errors in measurement (see table t): percent error is a way for scientists to express how far off an observed value is from the true value.
observed value – value based on laboratory measurements … the experimental value
true value – most probable value or accepted value based on references
equation: percent error = \\(\frac{\text{observed value} - \text{true value}}{\text{true value}} \times 100\\%
a positive number indicates that the observed value is greater than the true value
a negative number indicates that the observed value is below the true value
ex/ aluminum has a density of 2.7 g/ml. a student measured the density to be 2.5 g/ml. what is the percent error?
percent error = \\(\frac{2.5 - 2.7}{2.7} \times 100\\% = \frac{-0.2}{2.7} \times 100\\% = -7.407\\% = -7.4\\%
practice:

1. the freezing point of water is 273.2 k, but it was measured at 250.1 k. what is the percent error?
2. the mass of a penny is 2.67 g, but it was measured at 2.55g. what is the percent error?
3. the air pressure was 101.3 kpa, but the weatherman said it was 1001.3 kpa. what is the percent error?
4. the amount of heat released when 1 mole of co₂ forms is 393.5 kj, but was measured at 387.2 kj. what is the percent error?
5. a student found that the percent oxygen in a sample of kclo₃ was 42.3%. if the accepted value is 39.3% show the correct numerical set up to calculate the experimental percent error.
6. according to an accepted chemistry reference, the heat of vaporization of water is 2260. j/g. a student determined in the laboratory that the heat of vaporization of water was 1892 j/g. what was the student’s percent error?
7. in an experiment, the gram atomic mass of magnesium was determined to be 24.7 g. compared to the accepted value of 24.3 g, what is the percent error?
8. a student determined the melting point of a substance to be 55.2°c. if the accepted value is 50.1°c, what is the percent error?

Explanation:

Let's solve the first practice problem (the freezing point of water) step by step:

Step 1: Identify the observed and true values

The true value (freezing point of water) is \( 273.2 \, \text{K} \), and the observed value is \( 250.1 \, \text{K} \).

Step 2: Apply the percent error formula

The formula for percent error is \( \text{Percent error} = \frac{\text{observed value} - \text{true value}}{\text{True value}} \times 100\% \)

Substitute the values:
\( \text{Percent error} = \frac{250.1 - 273.2}{273.2} \times 100\% \)

Step 3: Calculate the numerator

\( 250.1 - 273.2 = -23.1 \)

Step 4: Divide by the true value

\( \frac{-23.1}{273.2} \approx -0.08455 \)

Step 5: Multiply by 100%

\( -0.08455 \times 100\% \approx -8.46\% \) (rounded to two decimal places)

Answer:

The percent error is approximately \(-8.46\%\) (or \(-8.4\%\) if rounded to one decimal place, or as calculated more precisely \(-8.4133\%\) from the handwritten note).