SME
Physical Geology · Problem type
A mineral’s density and specific gravity
Use this for “A 50 g brassy mineral raises the water from 40 mL to 50 mL. Find its density and specific gravity.”
Diagram
The water rose from 40 mL to 50 mL: the object takes up 10 cm³.
Mass m = 50 g
Density ρ = 5 g/cm³
This is an example. Type your own values to start a new problem.
V (cm³)
=
V₂ (mL)
−
V₁ (mL)
ρ (g/cm³)
=
m (g)
÷
V (cm³)
G
Specific gravity
calculated
Formulas
V
=
V₂
−
V₁
10 = 50 − 40
ρ
=
m
V
5
=
50
10
G
=
ρ
1.00
5
=
5
1.00
Assumptions
•
1 mL of water displaced = 1 cm³ of mineral; the sample sinks and has no air pockets.
•
Specific gravity is the density over water’s 1.00 g/cm³, so it has no unit.
•
Most silicates are 2.6–3.4; metal ores are heavier (pyrite about 5, galena 7.5).
Step-by-step
WE KNOW
m = 50 g, V₁ = 40 mL, V₂ = 50 mL
FIND
volume (V), density (ρ), specific gravity (G)
Step 1 · Find volume (V)
Use V = V₂ − V₁
The water rises by the mineral’s volume: 1 mL is 1 cm³.
V
=
50
−
40
V
=
10
cm³
Step 2 · Find density (ρ)
Use ρ = m ÷ V
Density is the mass in each cubic centimeter.
ρ
=
50
10
ρ
=
5
g/cm³
Step 3 · Find specific gravity (G)
Use G = ρ ÷ 1.00
Compare with water: divide by 1.00 g/cm³, and the units cancel.
G
=
5
1.00
G
=
5
Check
Put the numbers back in. Both sides should match.
10 = 50 − 40 ✓
5
=
50
10
✓
5
=
5
1.00
✓
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A silicate structure’s formula and charge
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A rock’s origin from its texture
Topic check
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