Page 304 - Theory and Problems of BEGINNING CHEMISTRY

APP.] SCIENTIFIC CALCULATIONS 293

if y = 14 and z = 5.0, then we simply substitute the given values and solve:
14
x = = 2.8
5.0
In chemistry and other sciences, such equations are used continually. Since density is defined as mass divided
by volume, we could use the equation
y
x =
z
with y representing the mass and z representing the volume, to solve for x, the density. However, we find it easier
to use letters (or combinations of letters) that remind us of the quantities they represent. Thus, we write
m
d =
V
with m representing the mass, V representing the volume, and d representing the density. In this way, we do not
have to keep looking at the statement of the problem to see what each variable represents. However, we solve
this equation in exactly the same way as the equation in x, y, and z.
Chemists need to represent so many different kinds of quantities that the same letter may have to represent
more than one quantity. The necessity for duplication is lessened in the following ways:
Method Example
Using italic letters for variables and roman (regular) m for mass and m for meter
letters for units
Using capital and lowercase (small) letters to mean V for volume and v for velocity
different things
Using subscripts to differentiate values of the same V 1 for the first volume, V 2 for the second, and so on
type
Using Greek letters π (pi) for osmotic pressure
Using combinations of letters MM for molar mass

Each such symbol is handled just as an ordinary algebraic variable, such as x or y.

EXAMPLE A.1. Solve each of the following equations for the first variable, assuming that the second and third variables
are equal to 15 and 3.0, respectively. For example, in part a, M is the first variable, n is the second, and V is the third. So n
is set equal to 15, and V is set equal to 3.0, allowing you to solve for M.
(a) M = n/V
(b) n = m/MM (where MM is a single variable)
(c) v = λν (λ and ν are the Greek letters lambda and nu.)
(d ) P 1 V 1 = P 2 (2.0) (P 1 and P 2 represent different pressures.)

Ans. (a) M = 15/3.0 = 5.0 (c) v = (15)(3.0) = 45
(b) n = 5.0 (d) P 1 = (3.0)(2.0)/15 = 0.40

EXAMPLE A.2. Solve the following equation, in which each letter stands for a different quantity, for R:
PV = nRT
Ans. Dividing both sides of the equation by n and T yields
PV
R =
nT
EXAMPLE A.3. Solve the following equation for F in terms of t:
t 5
=
F − 32.0 9