Evaluate density methods and systematic errors
Track how a mistaken mass or volume changes the inferred ratio and improve the relevant measurement.
Cross-referenced to the 2028–2029 0653 section using original teaching from the related single-science courses. A 0653-specific objective-by-objective and Core/Supplement review is in progress. Written practice is self-checked and does not demonstrate supervised laboratory competence or award an official Cambridge grade.
Before you begin
- Use density as a ratio.
- Recognise separate mass and volume measurements.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
With unchanged mass, what happens to m/V if the volume denominator is too large?
What you will learn
- Predict density overestimates and underestimates from method errors.
- Distinguish repeatability from a corrected systematic method.
Track the numerator and denominator
If an irregular solid's true mass is 48 g and volume 16 cm³, its density is 3.0 g/cm³. Trapped air increasing measured displacement to 20 cm³ gives 2.4 g/cm³, an underestimate. Losing liquid so that the recorded rise is only 14 cm³ instead gives about 3.43 g/cm³, an overestimate.
For liquid density, forgetting to subtract an empty container adds unwanted mass to the numerator and makes density too high. A regular-solid dimension error can make calculated volume too high or low. State which measured quantity is affected, then explain the ratio change; saying only that the experiment is inaccurate does not identify the mechanism.
Choose an improvement that addresses the cause
Remove attached bubbles before displacement, lower a solid without splashing, read the meniscus at eye level and use a zeroed balance. Each action addresses a particular source of error. Repeating and averaging careful trials can reduce the influence of random scatter, but repeating the same container-mass omission or unit error cannot remove that bias.
Illustrative repeated liquid-density results 0.79,0.81,0.80 g/cm³ have mean 0.80 g/cm³. Their closeness supports repeatability, not proof of accuracy or completion of a practical assessment. Compare methods under matched material and temperature conditions, retain sensible precision, and seek supervised practical experience and teacher feedback for the real laboratory skills.
Sources for this lesson
Pause and explain
A liquid calculation repeatedly includes the empty container's mass. Does averaging ten such trials fix it?
Worked example
A 48 g solid truly occupies 16 cm³ but a lost-liquid error gives a measured displacement of 14 cm³. Find the inferred density and error direction.
Show the worked solution
- True density = 48/16 = 3.0 g/cm³.
- Inferred density = 48/14 ≈ 3.43 g/cm³.
- The underestimated volume causes density to be overestimated.
Answer Approximately 3.43 g/cm³, an overestimate.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A 48 g sample gives a mistaken 14 cm³ displacement rather than its true 16 cm³. Compare densities.
Give me a hint
Calculate both ratios with the same mass.
Compare my reasoning
- True density is 3.0 g/cm³.
- Mistaken density is 48/14 ≈ 3.43 g/cm³.
- A too-small measured volume produces a too-large density.
3.0 versus approximately 3.43 g/cm³; an overestimate.
Look for these in your work
- Calculate both values.
- Explain the denominator effect.
Repeated results are 0.79,0.81,0.80 g/cm³. Find their mean and say whether agreement proves accuracy.
Give me a hint
Average the values, then distinguish scatter from bias.
Compare my reasoning
- Sum = 2.40 g/cm³.
- Mean = 2.40/3 = 0.80 g/cm³.
- Agreement supports repeatability but does not exclude a shared systematic error.
0.80 g/cm³; agreement does not prove accuracy.
Look for these in your work
- Compute the mean.
- State the evidence limit.
An irregular solid's density is too low because bubbles cling to it, while a liquid's density is too high because its container mass was included. Suggest a separate correction for each.
Give me a hint
Identify the wrong denominator in one method and numerator in the other.
Compare my reasoning
- Remove trapped air and remeasure displacement for the solid.
- Subtract or tare the empty container for the liquid.
- Repeat the corrected methods, pairing each sample's mass and volume.
Remove bubbles; subtract or tare the container; repeat the corrected measurements.
Look for these in your work
- Match each correction to its cause.
- Avoid claiming repetition alone fixes bias.
Common mistakes
- Explain which part of m/V changed.
- Repeatability does not establish accuracy.
- Averaging does not correct a persistent systematic omission.
What can you explain now?
What is the mean of 0.79,0.81,0.80 g/cm³?
Compare with the explanation
0.80 g/cm³.
Sum the three results and divide by three, retaining the density unit.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Explain one high-density and one low-density method error using the numerator and denominator.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.