Use density, mass and volume
Find density, mass and volume with compatible compound units, interpret liquid displacement and compare materials before combining explicitly additive volumes.
Section 4.4 · Measures
Exam scope and assessment notes
Foundation and Higher share the seven published F:4.4A–G Measures rows. The Higher table says See Foundation Tier; there are no additional H4.4 rows. The chosen eight-lesson split covers instrument scales, time intervals, estimates, nearest-degree angle measurement, three-figure bearings, average speed, density and pressure. Eight lessons and thirty-two original worked/staged cases are StudyForA’s design, not a syllabus-prescribed count. Each structural source link requires the corresponding complete teaching record carrying that source ID, a worked explanation and all Guided, Independent and Transfer stages. F4.4F requires the complete speed lesson; broad F4.4G requires all three speed, density and pressure lessons, not a speed-only partial link. Source binding and substantive structure must belong to the same record; a separate untagged duplicate cannot supply missing evidence. These are structural links, not exhaustive subskill or specialist approval. Preserve the actual calibrated instrument ticks, pointer/meniscus positions, ruler endpoints, units and protractor alignment action. Read labelled scales rather than assuming CSS pixels are millimetres. Use only diagrams explicitly supplied as accurate for measurement; ordinary examination diagrams need not be to scale and must not be measured without instructions. Numerical reference readings, protractor openings and the measured bearing remain hidden before a calibrated learner attempt; revealing a reference is not proof of instrument operation. Keep ideal model values separate from approximate human readings and nearest-degree rounding; no universal automatic tolerance or measurement competence is established. Bearings start at the named origin and run clockwise from declared north, with parallel north assumptions stated and leading zeros retained. Distinguish reverse rays, interior route angles and forward heading changes; no magnetic or globe-navigation correction is inferred. Clock intervals retain am/pm, noon/midnight and day order. Average speed uses total route distance over the full named elapsed interval, including stated stops; displacement and instantaneous speed are different. Calculating time by division requires positive speed: at constant zero speed a positive distance is impossible, while zero distance alone does not determine elapsed time. Keep density mass/volume units compatible and volume positive; liquid displacement is final minus initial level. Combined volumes are added only under the stated no-void/no-volume-change assumption. The pressure formula is supplied in every pressure stage; do not infer that the density or average-speed formula is also supplied. Use a provided force in newtons, positive contact area, correct squared-area conversion and an explicitly limited average-pressure model. No real-world safety or route-feasibility certification follows. The source assessment context uses SI units and may require mathematical instruments. These lessons do not complete4.5 nearest-mm line drawing, formal shape/compass constructions or scale-drawing problems,5.2 transformations,4.8 bearing trigonometry,4.9/4.10 full mensuration or4.11 map/similarity work. No blanket formal-proof waiver is imported. Written practice is self-checked, not automatically marked; acknowledgements and attempts do not award marks or establish mastery. Specialist approval and formal assessment validation remain unestablished. Publisher navigation is separate evidence: literal F4.4D appears in Book2 Shape and Space6, map physical10/pages128–160, and Hub36 Bearings and Scale Drawings, co-tagged F4.5C/F4.11B; those co-tags confer no construction/map credit here. Literal F4.4G appears in Book1 Number5, map physical5/pages255–268, and Hub13 Compound Measures. A/B/C/E/F have no literal4.4 chapter or Hub tag in that map; do not invent one from nearby time or scale-drawing topics. Book1 public uncorrected-proof contents start Number5/Compound measures at256 and end its summary at269, conflicting with map255–268; preserve both references. Book2 chapter start128 and next start161 agree with its map range, but no explicit bearings contents heading locates every skill. Neither relevant chapter body is present in the public samples. Full final-book/Teaching Hub content and pagination alignment remain unverified; chapter links are not full-book or modular teaching credit. The current2026/2027 manual lists linear4MA1 F/H in November2026 and June2027; FR/HR are June2027 only, based on centre eligibility. Vietnam is on the regional-centre list; learner residence alone does not choose an option. The table points to page2 although the list is on physical/printed3. Modular4XMAF/4XMAH cash-ins and4WM units remain separate entries for international centres. This entry context does not create a2027 revision of the Issue2 specification.
Before you begin
- Multiply and divide decimals and rearrange a quotient.
- Read supplied liquid-volume levels and subtract them.
- Use the supplied equivalences 1 mL=1 cm³ and 1 g/cm³=1000 kg/m³ without claiming a complete volume-conversion programme.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For a sample of mass m and volume V, which relationship gives density?
What you will learn
- Use density=mass/volume and rearrange to find either mass or volume.
- Convert between g/cm³ and kg/m³ and interpret a displaced-volume diagram.
- Compare materials in common units and state the assumptions needed for a combined-volume calculation.
A uniform solid sample has mass 540 g and volume 200 cm³. Find its density in g/cm³ and kg/m³. Then find the mass of 350 cm³ of the same uniform material. Use 1 g/cm³=1000 kg/m³ and explain the unit conversion.
Read, draw and calculate before comparing. Every requested part and explanation needs your own attempt. The teal protractor and orange rays show your actual placements, including wrong ones. Comparisons do not award marks or establish practical instrument competence.
Coordinates use the original worksheet grid. The entire drawing and your tools rotate together; clockwise bearings retain their direction. Case or orientation changes clear all responses, placements and references. Named given points can be placed exactly; other drawing clicks use the nearest half-unit grid point and never snap to a solution.
Given panel 1
- Mass (g)
- 540
- Volume (cm³)
- 200
- New sample volume (cm³)
- 350
The board retains the last submitted valid placement. Editing a draft invalidates its submission without moving the last drawing; submit again before comparison.
Density compares mass with volume
Density is mass divided by volume. If a uniform sample has mass m and volume V, write density=m/V. A larger sample of the same uniform material has a larger mass in the same ratio, while its density stays the same. Attach the compound unit: grams divided by cubic centimetres gives g/cm³, not simply grams.
In these ordinary material models, mass, volume and density are positive. The density calculated from an object’s total mass and external volume is its average or bulk density; it need not identify the density of a constituent material if the object contains cavities. Do not infer a substance’s identity from a single quotient. A zero volume gives an undefined division and is not a valid input to this model.
Find the missing mass or volume
Rearrange the relationship before substituting: mass=density×volume and volume=mass/density. Choose the form that has the requested quantity on its own. If a sample displaces liquid, find the displaced volume as final level minus initial level, using the supplied readings. The final liquid reading alone includes the original liquid and therefore is not the object’s volume.
For a displacement task, the assumptions matter: the object is fully submerged, liquid is not lost, no air is trapped and the object does not dissolve. In these tasks use 1 mL=1 cm³. The before-and-after cylinder levels are given measurements to interpret, not extra unknown quantities. Keep the initial and final levels visible in your working so the subtraction can be checked.
Convert before comparing or combining
To convert a density, change both mass and volume units. One gram is 0.001 kg, while 1 cm³ is 0.000001 m³, so 1 g/cm³=1000 kg/m³. Consequently a number in kg/m³ is divided by 1000 to express it in g/cm³. Comparing the numerical parts before conversion can reverse the intended conclusion.
For combined samples, add their masses and their stated volumes only when the question explicitly says that the volumes combine without change or voids. Then divide total mass by total volume. The simple average of two densities works for equal volumes under that additive-volume assumption, but not as a universal mixing rule. The relationship is taught here; no promise that a density formula will be supplied in an assessment is inferred from the separate pressure-formula note.
Sources for this lesson
Pause and explain
Which is equivalent to 2.3 g/cm³?
Worked example
A uniform solid sample has mass 540 g and volume 200 cm³. Find its density in g/cm³ and kg/m³. Then find the mass of 350 cm³ of the same uniform material. Use 1 g/cm³=1000 kg/m³ and explain the unit conversion.
Show the worked solution
- Use density=mass/volume: 540÷200=2.7 g/cm³. This is a mass-per-volume value, not a new mass for the sample.
- Convert both parts of the compound unit. Since 1 g/cm³=1000 kg/m³, 2.7 g/cm³=2700 kg/m³.
- For the new volume use mass=density×volume with compatible g and cm³ units: 2.7×350=945 g.
- The same uniform-material assumption keeps density fixed as volume changes. The larger 350 cm³ sample has a larger mass than the original 200 cm³ sample, which checks the direction.
Answer Density: 2.7 g/cm³=2700 kg/m³. Mass of 350 cm³: 945 g.
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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 uniform sample has mass 448 g and volume 160 cm³. Find its density. For the same material, find the mass of 75 cm³ and the volume occupied by 140 g. Show a suitable form of the relationship for each requested quantity.
Give me a hint
First calculate density. Then multiply for a mass and divide for a volume.
Compare my reasoning
- Calculate density=448÷160=2.8 g/cm³. Keep this common density for the same uniform material.
- For a 75 cm³ sample, use mass=density×volume: 2.8×75=210 g. The cm³ units cancel, leaving the requested mass in grams.
- For a 140 g sample, use volume=mass/density: 140÷2.8=50 cm³. Dividing by grams per cm³ recovers a volume, not another mass.
- Check the last result by multiplying 2.8 g/cm³×50 cm³=140 g. The units and the inverse operations agree.
Density: 2.8 g/cm³. Mass of 75 cm³: 210 g. Volume of 140 g: 50 cm³.
Look for these in your work
- Find and label the 2.8 g/cm³ density first.
- Use multiplication for the new mass and division for the new volume.
- Check the 50 cm³ result by recovering 140 g with compatible units.
An object of mass 561.6 g is fully submerged without dissolving, trapped air or lost liquid. The labelled measuring-cylinder levels rise from 350 mL to 422 mL. Using 1 mL=1 cm³, independently find the object’s volume and density. Then find the volume occupied by 156 g of the same uniform material. Interpret both levels in the supplied before-and-after diagram.
Give me a hint
Subtract the two level readings before dividing mass by volume.
Compare my reasoning
- The displaced liquid volume is 422−350=72 mL. Using the supplied equivalence, the object volume is 72 cm³.
- Calculate density=561.6÷72=7.8 g/cm³. Do not divide by 422 cm³, because that includes the original liquid.
- For 156 g of the same uniform material use volume=156÷7.8=20 cm³. Keep the density at 7.8 g/cm³ because the material model is unchanged.
- Check by multiplying 7.8×20=156 g and retain the stated submersion assumptions. They justify treating the liquid increase as the object volume.
Object volume: 72 cm³. Density: 7.8 g/cm³. Volume for 156 g: 20 cm³.
Look for these in your work
- Use the actual before-and-after level difference from the diagram.
- Find density with the 72 cm³ volume and the g/cm³ unit.
- Rearrange to obtain 20 cm³ and retain the physical assumptions.
Material A has density 2.5 g/cm³ and material B has density 2700 kg/m³. Compare the masses of 100 cm³ samples of each. If these two stated volumes then combine without voids or any volume change, find the combined mass, volume and average density in g/cm³. Explain the condition that makes the volume addition legitimate.
Give me a hint
Convert B to g/cm³ before comparing; add masses and explicitly additive volumes before finding the combined density.
Compare my reasoning
- Convert 2700 kg/m³ to 2700÷1000=2.7 g/cm³. Thus B has the larger density in the same unit, rather than merely the larger original numerical label.
- For 100 cm³ samples, A has mass 2.5×100=250 g and B has mass 2.7×100=270 g.
- The combined mass is 250+270=520 g. The task expressly allows additive volumes, so the combined volume is 100+100=200 cm³.
- Average density is 520÷200=2.6 g/cm³. It happens to be the mean of 2.5 and 2.7 here because the volumes are equal and additive; do not generalise that as a rule for arbitrary mixtures.
A: 250 g; B: 270 g. Combined mass: 520 g; volume: 200 cm³; average density: 2.6 g/cm³ under the stated additive-volume assumption.
Look for these in your work
- Convert B to 2.7 g/cm³ and compare like units.
- Find both sample masses before adding them.
- Use the explicit no-void/no-volume-change condition and explain why this is not a universal averaging rule.
Common mistakes
- Dividing by the final cylinder level instead of the change in liquid volume.
- Changing only the numerator or only the denominator when converting a compound density unit.
- Averaging two densities without the necessary volume assumptions, or claiming that a quotient identifies a material.
What can you explain now?
A solid is fully submerged and the liquid level changes from 80 mL to 105 mL. What volume enters its density calculation, using 1 mL=1 cm³?
Compare with the explanation
25 cm³.
Use the displaced volume, 105−80=25 mL=25 cm³. The final 105 mL includes the original liquid, so it is not the object’s volume.
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Make it stick
Recall density=mass/volume, mass=density×volume and volume=mass/density. State the displaced-volume subtraction and the factor between g/cm³ and kg/m³, including why both unit parts matter.
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.