Density Determination Using Various Methods to Measure Volume

Experiment 1

Experiment 1

Goal:To accurately determine densities of objects

Method:Determine volume of objects (3 methods)Calculate densities from mass and volume

DensitiesSubstance Density

(g/mL)wood 0.35water 1.00quartz 2.65

diamond 3.51Al 2.7

brass 8.4Au 19.3Os 22.4

Increasing density

=Increasing

“heaviness”

Density from lab data

Vm

==volumemassρ ← Analytical balance

← Three methods

Volume measurements:

–by geometry

–by water displacement

–by pycnometry

Uncertainties/Error

Mass: balance

±0.0001 g

Geometry: calipers

±1 of last digit cm

Displacement: grad. cylinder ±0.1 mL probably

Pycnometry: pynometer

from ⟨average⟩

Propagate: all but mass fractions

Procedure

1. Four cylindersbrass

aluminum plugged hollow

Record:2. Number

#

3. Type

type

4. Mass

(analytical balance)

m#

1) Volume by geometry

1. Lengths

and diameters

l, d2. Calculate volume

and error

V, σV

length, l

ldπldπV 22

42=⎟

⎠⎞

⎜⎝⎛=

diameter, d

1) Al example –

density using volume by geometry

Mass

17.4640 gLength

5.08 cm

Diameter 1.29 cm

32 632085

2291

464017cm

g

cm.cm.π

g.Vm .d ===

⎟⎠⎞

⎜⎝⎛

( )( )

3

3

050

2632 291010

085010

46401700010

cmg

cm.cm.

cm.cm.

g.g.

cmg

ddlm

.

.

ddlm

±=

⋅++±=

+++±= σσσσρ ρσ

Using vernier

calipers

Object falls between 3.3 and 3.4 cm

Line on auxiliary scale matches at 8

Section 3B in manual

Length:3.38 ±

0.01 cm

2) Volume by water displacement

1.

Record initial volume

of water

Vi

2.

Add metal cylinder3.

Record final volume

Vf

4.

Calculate volume

and error

V, σV

initialfinalcylinder VVV −=

2) Al example using volume by water displacement

mass

17.4640 gVwater+cylinder

66.55 mL ±0.05mL

−

Vwater

60.00 mL ±0.05mL

Vcylinder

6.55 mL ±0.1mL

mLg

mL.g.

Vm .672556

464017 ===ρ

( ) ( ) mLg

mL.mL.

g.g.

mLg

Vm ..Vm 040672 6610

46401700010 ±=+±=+±= σσ

ρ ρσ

Water displacement

46.5 mL

Meniscus:

liquid’s curved surface

66.0 mL

66.0 mL–

45.5 mL20.5 mL

(object’s volume)

3) Volume by pycnometry

Pycnometry: pertaining to specific gravityV = mH2O

/ρH2O

1.

Make pycnometer2.

Calibrate

Calibration exampleCalibration –

multiple trials (how well you fill to the mark)

trial

masswater+pycnometer

1

92.7829 g2

92.7825 g

3

92.7826 g4

92.7831 g

Average:

92.7828 gσmass

:

±0.0003g (using big equation)

Pycnometry continued

3.

Add H2

O to mark record initial mass

mi

4.

Add metal cylinder 5.

Remove H2

O until at mark6.

Record final mass

mf

7.

Calculate volume

and error

V, σV

waterwaterdisplaced waterdisplaced cyl ρ

m VV 1×==

Pycnometry continued

llArea 22

d42

d ππ =⎟⎠⎞

⎜⎝⎛=llArea 2

2

d42

d ππ =⎟⎠⎞

⎜⎝⎛=

waterdisplacedBobjectA massmassmassmass +=+

A B

ρmV =

3) Al example using volume by pycnometrymassA

92.7828 g (calibration average) ±0.0003g

massB

103.7227 g

±0.0003gmasscylinder

17.4640 g

±0.0001g

g.)g.g.g.(g.g.g.g.massmassmassmass

water displacedmass

BobjectAwater displaced

0007000030000100003052416782892464017782892

±=++±=

=++=−+=

σ

( ) ( ) ( ) mL.mL.VV

mL.g.

mLg . VV

g.g.

mcylmcylV

waterdisplaced cyl

mm

cyl000705386

538699780

52416

5241600070 ±=±=±≈+±=

=×==

σρσσ ρσ

( ) ( ) mLg

mL.mL.

g.g.

mLg

Vm

mLg

mL.g.

Vm

..

. Vm 000306712

6712

538600070

46401700010

5386464017

±=+±=+±=

===σσ

ρ ρσ

ρ

Record temperature, T

Density of Water vs. Temperature

0.9950

0.9960

0.9970

0.9980

0.9990

1.0000

15 20 25 30

Temperature (oC)

Den

sity

(g/m

L)

4) Volume of void in hollow cylinder

voidmetalcyl VVV +=

pycnometrym ρmetal

V =

Example void calculation –

brass cylinderIf Vcylinder

= 4.970 mL (by pycnometry)& mcylinder

=34.5964 g (analytical balance)& ρbrass

= 8.387g/mL

(pycnometry of solid cylinder)

( )mL.

g.mL.

VVV

g.mL

metalcylvoid

8450

5964349704 38781

=

⋅−=

−=

No error propagation required for this

5) Mass Fractions in Mixed Cylinder

brassAlcyl mmm +=

Let:

mAl

= X.mcyl

mbrass

= (1-X)mcyl

( ) cylcylcyl mXmXm −+= 1

Mixed cylinder

brassAlcylinder VVV +=

brassAl ρρbrassAl

cylmmV += mAl

= X.mcyl

mbrass

= (1-X)mcyl

m ρV =

( )brassAl

XXρρ

cylcylcyl

m1mV

−+= ÷

by mcyl

( )brassAl

XXρρρ−

+=11

cyl

Collect terms with X

brassAlbrass

XXρρρρ

−=−11

cylSolve for X

⎟⎟⎠

⎞⎜⎜⎝

⎛−=−

brassAlbrass

Xρρρρ

1111

cyl÷

to isolate X

Mixed cylinder

X = Al fraction

1-X

= brass fractionbrassAl

brassX

ρρ

ρρ11

11cyl

−

−=

No error propagation required for this

Pycnometry:

volume Vmixed

Balance:

mass mmixed

Mixed cylinder –

example

X

= Al fraction (here: 13.97%)

1-X

= brass fraction (here: 86.03%)

860301

13970

38781

67121

38781

45761

11

11

.X

.

..

..X

mLg

mLg

mLg

mLg

brassAl

brasscyl

=−

=−

−=

−

−

=

ρρ

ρρ

No error propagation required for this

Use pycnometry data to find:

density of pure brass 8.387 g/mL

density of pure Al (solid cylinders)

2.671 g/mL

density of mixed cylinder (example)

6.457 g/mL

ReportAbstractResults/ Samples calculations including:

Mass and volume by each methodVolume of void Mass fractionError analysis (parts 1 – 4)Volume of objects (3 methods)

yes

Volume of inner void noMass fractions for mixed cylinder

no

Discussion/review questions

Equipmentcylinders: brass, aluminum, mixed, hollowVernier calipers50 mL Erlenmeyer flasklab markerPasteur pipet100 mL graduated cylinder400 mL beakerthermometer(analytical balance)

Caution

NO

YES