Lab Fundamentals: Equipment, Error, and Data Analysis实验基本功:仪器、误差与数据处理

AP Science Practice 5 · Mathematical Routines~35 min

Learning goals

  • Choose equipment for the precision a task needs, and record readings to that equipment’s precision
  • Describe the precision and accuracy of data with the mean, range and percent error
  • Read meaningful quantities from the slope and intercept of a linear graph

Before you start

  • Reliable measurements (1.1.5)
  • Error and percent error (2.1.3)
  • Experimental design and error analysis (4.3.5)

What every lab question builds on

Each lesson in 4.4 covers one kind of AP lab. Whatever the lab, FRQs test three things: choosing the right equipment, recording data to the equipment’s precision, and calculating a result from the data and judging it. This lesson practices all three.

Equipment and its precision

When reading a scale, read to one digit past the smallest marking (the last digit is estimated). For equipment with a single mark (a volumetric pipet or flask), the equipment itself sets the volume.

EquipmentRecord toUsed for
Analytical balance0.0001 gWeighing samples
Buret (50 mL)0.01 mLTitrations
Pipet (25 mL)25.00 mLOne exact portion
Volumetric flask (250 mL)250.0 mLMaking solutions
Graduated cylinder (100 mL)0.1 mLRough volumes
BeakerEstimate onlyHolding, dissolving

Use precise equipment for quantities that go into a calculation (masses and volumes used to find the result); a quantity that isn’t used in the calculation, such as water added to dissolve a solid, only needs a rough measurement.

Choose the equipment: pick the best tool for each task (some tasks have two good choices)
  1. Measure exactly 25.00 mL of an acid into a flask for titration
  2. Prepare 250.0 mL of a 0.1000 M standard solution
  3. Measure the volume of NaOH added during a titration (the amount needed isn’t known in advance)
  4. Add about 30 mL of distilled water to dissolve a weighed solid acid in a flask
  5. Weigh out about 0.2 g of a solid sample and record its exact mass

Describing a set of data

  • Mean: the average of several trials, reported as the final result.
  • Range极差: largest value − smallest value. A small range means high precision (the data agree with each other).
  • Percent error百分误差: |experimental − accepted| ÷ accepted × 100%. A small percent error means high accuracy (close to the true value).
  • If every result is off in the same direction, there is probably a systematic error; data scattered both ways show random error.

Using straight-line graphs

When two quantities plot as a straight line, the slope and intercept are often the quantities you want: mass vs volume gives density as the slope; absorbance vs concentration gives εb (4.4.3); ln[A] vs time gives −k (4.4.6).

  • Draw a best-fit line最佳拟合直线: as close as possible to all the points, with about as many on each side; it doesn’t have to pass through every point.
  • To find the slope, pick two points far apart on the line, not two data points.
  • The slope has units: y-axis units ÷ x-axis units.

Example 1: concentration from buret readings

25.00 mL of hydrochloric acid is titrated with 0.1000 M NaOH. The buret reads 1.25 mL at the start and 23.70 mL at the end. Find the acid concentration and say how many significant figures it has.

Step-by-step solution
  1. Volume of NaOH added. Final reading minus initial reading:

    V=23.70−1.25=22.45 mL=0.02245 L\begin{aligned} V &= 23.70 - 1.25 = 22.45\ \text{mL} \\ &= 0.02245\ \text{L} \end{aligned}
  2. Moles of NaOH.

    n(NaOH)=0.02245×0.1000=2.245×10−3 mol\begin{aligned} n(\ce{NaOH}) &= 0.02245 \times 0.1000 \\ &= 2.245 \times 10^{-3}\ \text{mol} \end{aligned}
  3. Acid concentration. HCl and NaOH react 1 : 1:

    [HCl]=2.245×10−30.02500=0.08980 M\begin{aligned} [\ce{HCl}] &= \frac{2.245 \times 10^{-3}}{0.02500} \\ &= \boxed{0.08980\ \text{M}} \end{aligned}
  4. Significant figures. 22.45 mL, 0.1000 M and 25.00 mL all have four significant figures, so the result has four.

Example 2: judging a set of results

Three titrations give the concentration of an NaOH solution as 0.1012, 0.1008 and 0.1027 M; the accepted value is 0.1000 M. Find the mean, range and percent error, and comment on the data.

Step-by-step solution
  1. Mean.

    cˉ=0.1012+0.1008+0.10273=0.1016 M\begin{aligned} \bar{c} &= \frac{0.1012 + 0.1008 + 0.1027}{3} \\ &= 0.1016\ \text{M} \end{aligned}
  2. Range. 0.1027 − 0.1008 = 0.0019 M.

  3. Percent error.

    error=∣0.1016−0.1000∣0.1000×100%=1.6%\begin{aligned} \text{error} &= \frac{|0.1016 - 0.1000|}{0.1000} \times 100\% \\ &= \boxed{1.6\%} \end{aligned}
  4. Comment. All three results are above the accepted value, which points to a systematic error (for example, a problem standardizing the NaOH). The third trial is well above the other two, so the precision is only fair; another trial would show whether it’s an outlier that can be discarded.

Practice: compared with the accepted value, which describes this data?
  1. Accepted 0.1000 M; measured 0.1001, 0.0999, 0.1000
  2. Accepted 1.000 g/mL; measured 1.052, 1.051, 1.053
  3. Accepted 36.0%; measured 33.1%, 38.9%, 36.4%
  4. Accepted 8.31; measured 8.30, 8.32, 8.31
  5. Accepted −55.8 kJ/mol; measured −49.6, −49.9, −49.7
  6. Accepted 0.250 M; measured 0.231, 0.262, 0.247
  7. Accepted 63.55 g/mol; measured 63.54, 63.57, 63.55
  8. Accepted pKa 4.76; measured 4.95, 4.96, 4.94

Common mistakes

Mistake 1: “Writing more digits is always safer; more digits means more precise.” A reading can only be estimated to one digit past the smallest marking, so extra digits mean nothing. Writing 52.400 mL from a 100 mL graduated cylinder only shows a misunderstanding of significant figures.

Mistake 2: “If the accuracy is high, the precision must be high too.” They’re different. Widely spread data whose mean happens to land near the true value are accurate but not precise; tightly grouped data that are all off in one direction are precise but not accurate.

Mistake 3: “A best-fit line must pass through every point and through the origin.” A best-fit line shows the overall trend, with points on both sides. The intercept can mean something too: if a container wasn’t subtracted when weighing, the intercept is the container’s mass.

Quick check

A 50 mL buret has markings every 0.1 mL. A reading should be recorded to:

Which is the correct way to calculate percent error?

A graph of a liquid’s mass (y-axis) against its volume (x-axis) is a straight line. What does the slope represent?

A liquid is measured in a 100 mL graduated cylinder marked every 1 mL. Which recorded value is appropriate?

Which reading comes from an analytical balance?

When drawing a best-fit line, you should:

AP-style practice

AP-style free response

To find the density of ethanol, a student places a dry beaker on an electronic balance (without taring it), adds different volumes of ethanol from a buret, and records the total mass of beaker and ethanol each time:

V (mL)Total mass (g)
5.0055.98
10.0059.83
15.0063.72
20.0067.58

(a) A graph of total mass against volume is a straight line. What does its slope represent?

(b) Estimate the slope from the first and last data points, with units.

(c) The line’s y-intercept is about 52.1 g. What does it represent?

(d) The accepted density of ethanol at this temperature is 0.789 g/mL. Calculate the percent error of the result in (b).

(e) The measured density is lower than the accepted value. The beaker sat open on the balance throughout. Suggest a reason that explains this result.

(f) If a 100 mL graduated cylinder (marked every 1 mL) were used to add the ethanol instead, how would the precision of the measured density change? Explain.

Model answer and scoring

(a) Slope = Δm ÷ ΔV, the mass per milliliter of ethanol, which is its density.

(b) Slope = (67.58 − 55.98) g ÷ (20.00 − 5.00) mL = 11.60 g ÷ 15.00 mL = 0.773 g/mL.

(c) The total mass when V = 0, which is the mass of the empty beaker (about 52.1 g). Not taring doesn’t change the slope, so the density isn’t affected.

(d) Percent error = |0.773 − 0.789| ÷ 0.789 × 100% = 2.0%.

(e) Ethanol is volatile. With the beaker open, some ethanol evaporated between adding it and weighing, so the measured masses are too small while the volumes come from the buret readings; m/V is too low. (The larger the volume and the longer the wait, the more evaporates, so the slope is too low as well.)

(f) The precision decreases. A graduated cylinder can only be estimated to 0.1 mL, while a buret reads to 0.01 mL; the volume uncertainty is about 10 times larger, so the points scatter more and the slope is less certain.

Scoring: (a) the slope is the density; (b) a correct two-point slope with units of g/mL; (c) the intercept is the beaker’s mass; (d) the percent error setup and value; (e) the chain from evaporation to masses that are too small; (f) lower precision, explained with the reading precision of the two pieces of equipment.

Summary

  • Use precise equipment (analytical balance, buret, pipet, volumetric flask) for quantities used in calculations, and read to one digit past the smallest marking; other quantities only need rough measurement.
  • The range describes precision and percent error describes accuracy; results that all lean one way usually mean a systematic error.
  • The slope and intercept of a straight-line graph are often the quantities you want. Give the slope units, and calculate it from two points far apart on the best-fit line.

AP tip: When an FRQ asks why a buret is used rather than a graduated cylinder, answer in terms of precision and give numbers: “a buret reads to 0.01 mL, while a graduated cylinder reads only to 0.1 mL.”

Sources: density of ethanol (0.789 g/mL at 20 °C) from the Wikipedia article “Ethanol”; other data are illustrative.

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