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tiL12214_appe_643-698.indd Page 677 09/10/10 8:37 AM user-f463
tiL12214_appe_643-698.indd Page 677 09/10/10 8:37 AM user-f463 Volume/207/MHDQ243/tiL12214_disk1of1/0073512214/tiL12214_pagefile

−3
_ m _ 5.7 × 10 g
ρ mineral

SG = and ρ mineral = and V = L × W × H __

ρ water V 4 _ _
3
0.16 cm

π ( )

__ = 3 __
2
m

g
(L × W × H)
__ _

∴ SG = 1
3
ρ water cm
25.09 g
−3
___ _
5.7 × 10

g
(2.89 cm)(1.86 cm)(1.79 cm)
___ π ( ) 3 _
4 _
_
0.16

= 3 _ cm 3
2
_
g
_

1 = 1

g
cm 3 _

cm 3
g
25.09
__ __ −3

5.7 × 10

__ __ = _
(2.89)(1.86)(1.79) (cm)(cm)(cm)

= 4 _ 3
g

1 _ π0.08

3
cm 3 _
−3
5.7 × 10

25.09
_ = −3
= 2.1 × 10
9.62
= 2.61 = 2.7
The feldspar grains will float.
Examination of Figure 17.11 shows sodium plagioclase
crystallizes at low temperature on Bowen’s reaction series. 17.5. The bond type is determined from the electronegativity
difference between the elements in the mineral. Figure 9.10
17.3. In this case, determine volume by using the formula for the provides electronegativity values, and Table 9.5 explains the
volume of a sphere. difference in absolute electronegativity in terms of the bond type.
D = 3.2 cm Moissanite
m = 72.04 g electronegativity of Si = 1.8
g
_ electronegativity of C = 2.5
ρ water = 1
cm 3 bond type = ?
SG = ? difference = ∣2.5−1.8∣
_ m _ 4 _ _ ) 3 = 0.7
D
ρ mineral

SG = and ρ mineral = and V = π (

ρ water V 3 2 Moissanite has polar covalent bonds.
_
m
4 _ D _ Fluorite

3

π ( ) electronegativity of Ca = 1.0

3 _
2

∴ SG =
ρ water electronegativity of F = 4.0
72.04 g
_ bond type = ?
3.2 cm difference = ∣4.0−1.0∣

3
4 _ _

π ( 2 )

3 _
= g = 3.0

_
1 Fluorite has ionic bonds.

cm 3
_ From Table 17.2, fluorite has a hardness of 4, so it is softer.
72.04
4 _ 3.2 _ g

3 _
π ( ) 17.6. Use the silicon to oxygen ratio to determine the percent

3
_
2

= 3 _ cm
g
1 _ composition. Determining percent composition is described in

cm 3 example 10.3.
_
72.04

= Olivine Step 1: Write the formula for a single tetrahedron based
4 _

π1.6 3

3 on ratio: SiO 4
72.04
_ Step 2: Determine the formula weight of a single tetrahedron.
=
17
Atoms Atomic Weight Total
= 4.2
1 of Si 1 × 28.09 u = 28.09 u

17.4. Determine the specific gravity of the mineral and compare it to 4 of O 4 × 16.00 u = 64.00 u

the specific gravity of liquid bromoform. Formula weight = 92.09 u
D = 0.16 cm Step 3: Determine the percentage of silicon.
−3
m = 5.7 × 10 g __ ×
(28.09 u Si)(1)

g
_ 92.09 u SiO 4 100%SiO 4 = 30.5% Si
ρ water = 1
cm 3 Hornblende
SG = ?
Step 1: Write the formula for a double chain based on
_ m _ 4 _ _ ) 3
D
ρ mineral
SG = and ρ mineral = and V = π ( ratio: Si 4 O 11

ρ water V 3 2
Step 2: Determine the formula weight of a double chain.
_
m

4 _ D _

3

π ( ) Atoms Atomic Weight Total

3 _
2

∴ SG = 4 of Si 4 × 28.09 u = 112.36 u
ρ water
11 of O 11 × 16.00 u = 176.00 u
Formula weight = 288.36 u
E-35 APPENDIX E Solutions for Group A Parallel Exercises 677