Practice worksheet on accuracy, precision, and percent error in scientific measurements.

Let's solve each question step by step and explain the reasoning.

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1) You are given a block of metal that has a 'known' mass of 110.2 grams. You measure its mass three times and get the following mass readings: 100.2 g, 100.1 g, 100.3 g.

#### a. What was your average mass?

To find the average:

$$
\text{Average} = \frac{100.2 + 100.1 + 100.3}{3} = \frac{300.6}{3} = 100.2 \text{ g}
$$

✔ Answer: 100.2 g

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#### b. What was the range of your data?

Range = Maximum value – Minimum value
= 100.3 g – 100.1 g = 0.2 g

✔ Answer: 0.2 g

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#### c. Which is better, the accuracy or the precision of your measurements?

- Precision refers to how close the measurements are to each other (reproducibility).
Your measurements are very close: 100.1, 100.2, 100.3 → high precision.

- Accuracy refers to how close the measurements are to the true value.
The known mass is 110.2 g, but your average is 100.2 g → large difference → low accuracy.

So, precision is better than accuracy.

✔ Answer: Precision is better.

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#### d. What is your percent error?

Percent Error formula:
$$
\text{Percent Error} = \left| \frac{\text{Experimental} - \text{Accepted}}{\text{Accepted}} \right| \times 100\%
$$

$$
= \left| \frac{100.2 - 110.2}{110.2} \right| \times 100\% = \left| \frac{-10.0}{110.2} \right| \times 100\% \approx 9.07\%
$$

✔ Answer: ~9.07%

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2) Consider the following data for the length of an object as measured by three students. The length is known to be 14.54 cm. Circle the letter of the conclusion which summarizes the data shown in the table below.

| Student | Trial 1 | Trial 2 | Trial 3 | Trial 4 | Trial 5 |
|--------|--------|--------|--------|--------|--------|
| A | 14.8 | 14.1 | 14.5 | 14.6 | 14.2 |
| B | 14.8 | 14.2 | 14.6 | 14.5 | 14.8 |
| C | 14.6 | 14.5 | 14.5 | 14.4 | 14.6 |

True value = 14.54 cm

We need to evaluate accuracy (how close to 14.54) and precision (how close the trials are to each other).

#### Student A:
- Values: 14.8, 14.1, 14.5, 14.6, 14.2 → spread from 14.1 to 14.8 → low precision
- Average = (14.8+14.1+14.5+14.6+14.2)/5 = 72.2/5 = 14.44 → off by ~0.1 → moderate accuracy

#### Student B:
- Values: 14.8, 14.2, 14.6, 14.5, 14.8 → spread from 14.2 to 14.8 → moderate precision
- Average = (14.8+14.2+14.6+14.5+14.8)/5 = 72.9/5 = 14.58 → very close to 14.54 → high accuracy

#### Student C:
- Values: 14.6, 14.5, 14.5, 14.4, 14.6 → all values between 14.4 and 14.6 → very consistent → high precision
- Average = (14.6+14.5+14.5+14.4+14.6)/5 = 72.6/5 = 14.52 → very close to 14.54 → high accuracy

So:
- Most precise: Student C (tight cluster)
- Most accurate: Both B and C are close, but C’s average (14.52) is closer to 14.54 than B’s (14.58)

Wait — let’s check:

- Student C: 14.52 → |14.52 - 14.54| = 0.02
- Student B: 14.58 → |14.58 - 14.54| = 0.04
- So C is more accurate than B.

Thus:
- Most precise: C
- Most accurate: C

✔ Answer: a. Student C has done the most precise and accurate work.

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3) While working in the laboratory, a student finds the density of a piece of aluminum to be 2.85 g/cm³. The accepted value for the density of aluminum is 2.6999 g/cm³. What is the student's percent error?

Use the formula:

$$
\text{Percent Error} = \left| \frac{2.85 - 2.6999}{2.6999} \right| \times 100\%
= \left| \frac{0.1501}{2.6999} \right| \times 100\% \approx 5.56\%
$$

✔ Answer: ~5.56%

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4) A student experimentally determines the specific heat of water to be 4.29 J/g°C. She then looks up the specific heat of water in a reference book and finds that it is 4.18 J/g°C. What is her percent error?

$$
\text{Percent Error} = \left| \frac{4.29 - 4.18}{4.18} \right| \times 100\%
= \left| \frac{0.11}{4.18} \right| \times 100\% \approx 2.63\%
$$

✔ Answer: ~2.63%

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✔ Final Answers Summary:

#### 1)
a. Average mass = 100.2 g
b. Range = 0.2 g
c. Better: Precision
d. Percent error = ~9.07%

#### 2)
Correct choice: a. Student C has done the most precise and accurate work.

#### 3)
Percent error = ~5.56%

#### 4)
Percent error = ~2.63%

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