SME
Grade 12 Science · Problem type
A mineral’s density by water displacement
Use this for “A 26.5 g mineral raises the water from 50 mL to 60 mL. What is its density, and which mineral could it be?”
Diagram
The water rose from 50 mL to 60 mL: the object takes up 10 cm³.
Mass: 26.5 g
Density: 2.65 g/cm³
This is an example. Type your own values to start a new problem.
ρ (g/cm³)
=
m (g)
÷
(
V₂ (mL)
−
V₁ (mL)
)
V
Volume
calculated
Formulas
V
=
V₂
−
V₁
10 = 60 − 50
ρ
=
m
V
2.65
=
26.5
10
Assumptions
•
1 mL of water displaced = 1 cm³ of mineral.
•
The sample sinks and has no air pockets.
•
Density is a clue, not proof: compare it with hardness and streak too.
•
Common minerals in g/cm³: feldspar 2.6, quartz 2.65, calcite 2.7, pyrite 5, hematite 5.3, galena 7.5, gold 19.3.
Step-by-step
WE KNOW
m = 26.5 g, V₁ = 50 mL, V₂ = 60 mL
FIND
volume (V), density (ρ)
Step 1 · Find volume (V)
Use V = V₂ − V₁
The water rises by the mineral’s volume: 1 mL is 1 cm³.
V
=
60
−
50
V
=
10
cm³
Step 2 · Find density (ρ)
Use ρ = m ÷ V
Share the mass over the cubic centimeters of the mineral.
ρ
=
26.5
10
ρ
=
2.65
g/cm³
(close
to
quartz,
2.65
g/cm³)
Check
Put the numbers back in. Both sides should match.
10 = 60 − 50 ✓
2.65
=
26.5
10
✓
Topic check
See what you know: 5 questions on this topic
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