The Chemistry Maths Book, Second Edition

26 Chapter 1Numbers, variables, and units

This has dimensions of inverse length and is normally reported in units of the

reciprocal centimetre, cm

− 1

. For the sodium example,λ 1 = 1 5.8976 1 × 110

− 5

cm and

The second line of the sodium doublet lies at 16973 cm

− 1

, and the fine structure

splitting due to spin–orbit coupling in the atom is17 cm

− 1

.

In summary, the characteristics of the radiation observed in spectroscopy can

be reported in terms of frequencyνin Hz (s

− 1

), wavelengthλin (multiples of ) m,

energy ∆Ein units of eV, molar energy in units of kJ mol

− 1

, and wavenumber 9 in

units of cm

− 1

. These quantities are related by equations (1.14) to (1.17). Conversion

factors for energy are

1 eV 1 = 1 1.60218 1 × 110

− 19

J 1 ≡ 1 96.486 kJ mol

− 1

1 ≡ 1 8065.5 cm

− 1

0 Exercises 106

Approximate calculations

Powers of 10 are often used as a description of order of magnitude; for example, if a

length Ais two orders of magnitude larger than length Bthen it is about 10

2

1 = 1100

times larger. In some calculations that involve a wide range of orders of magnitude

it can be helpful, as an aid to avoiding errors, to calculate the order of magnitude of

the answer before embarking on the full detailed calculation. The simplest way of

performing such an ‘order of magnitude calculation’ is to convert all physical quantities

to base SI units and to approximate the magnitude of each by an appropriate power of

ten, possibly multiplied by an integer. Such calculations are often surprisingly accurate.

EXAMPLE 1.20Order of magnitude calculations

(i) For the calculation of volume in Example 1.1 (ignoring units),

Two estimates of the answer are

(a) (b)

(ii) For the calculation of the moment of inertia of CO in Example 1.18 (ignoring

units),μ= 1 1.1385 1 × 110

− 26

1 ≈ 110

− 26

andR 1 = 1 1.1281 1 × 110

− 10

1 ≈ 110

− 10

, and an order of

magnitude estimate of the moment of inertia isI 1 = 1 μR

2

1 ≈ 1 (10

− 26

) 1 × 1 (10

− 10

)

2

1 = 110

− 46

(accurate value1.4489 1 × 110

− 46

).

0 Exercise 107

V≈

××

=.×

−

−

10 8 300

10

24 10

1

5

3

V≈

××

=

−

−

10 10 10

10

10

12

5

3

V

nRT

p

==

.×. ×

=. ×

−

0 1 8 31447 298

10

2 478 10

5

3



ν=

×

=

−

−

1

5 8976 10

16956

5

1

. cm

cm