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CHAPTER 1. UNITS OF CHEMISTRY AND SIGNIFICANT DIGITS

I. | SI UNITS |
In chemistry, the characterization protocol of a substance often requires physical measurements. These measurements are reported in either base or derived SI units. There are seven base units; they are called base units because none of them can be expressed as combinations of the other six.

QUANTITY Length Mass Time Temperature Amount of Substance Electric Current Luminous Intensity
NAME/DEFINITION Meter Kilogram Second Degree (kelvin or Celcius) mole ampere candela
SI UNIT m kg s K or 0C mo A cd

Other units can be built from the seven base units. Some widely used combinations of the base units are given special names to make working with them easier.

QUANTITY Area Volume Density Speed Acceleration Force Pressure Energy
NAME/DEFINITION Meter Squared Meter Cubed Mass per Unit Volume Distance traveled per unit time Speed change per unit time Mass times acceleration Force per unit Area Force times Distance
SI UNIT m2 m3 kg/m3 m/s m/s2 kg/(m/s2) kg/(m2/s2)

Note that in order for the SI units to follow a certain pattern, the kilogram and the cubic meter are the SI units of mass and volume respectively. However, these quantities are impractical for use in chemistry. Instead, the gram and the liter are most widely used. One gram is equal to 1 X 10-3 kg and one liter is equal to 1 X 10-3 cubic meters. These quantities have been written with powers of ten. This is very commom in chemistry because the values to be reported are often larger or smaller than the base SI units. Such quantities are written with prefixes that represent multiples of ten.

Prefix Symbol Exponent Prefix Symbol Exponent
yotta Y 1024 yocto y 10-24
zetta Z 1021 zepto z 10-21
exa E 1018 atto a 10-18
peta P 1015 femto f 10-15
tera T 1012 pico p 10-12
giga G 109 nano n 10-9
mega M 106 micro 10-6
kilo k 103 milli m 10-3
hecto h 102 centi c 10-2
deca da 101 deci d 10-1

When solving problems with derived units it is recommended to write such quantities in terms of the the base units and carry along the units. This is known as the factor label method or dimensional analysis.

EXAMPLE 1.1
A 180.0 kg sample of chemical Δ has a volume of 140.0 L. Compute the density of chemical Δ in g/cm3.

SOLUTION
First let's get the number of grams:
180.0 kg X [103 g/kg] = 180.0 X 103 g of chemical Δ
Note that the units are treated as algebraic quantities and since [kg/kg] = 1, the kg units are said to cancel.
For the volume,
140.0 L X [ 103 cm3/L] = 140.0 X 103 cm3 of chemical Δ
And the density,
ρ of chemical Δ = [180.0 X 103 g] / [140.0 X 103 cm3] = 1.286 g/cm3

EXAMPLE 1.2
Compare the volumes of three spheres of equal mass but made of different materials: lead ρ = 11.3 g/cm3, silver ρ = 10.5 g/cm3, and aluminum ρ = 2.7 g/cm3.

SOLUTION
Let's take a look at the density formula,
ρ = Mass / Volume
*With the volume constant, mass is directly proportional to the density.
*With the mass constant, volume is inversely proportional to the density.
And volume of a sphere is given by,
Vsphere = [4/3] π r3 = [1/16] π d3
The spheres (drawn to scale) would look like this,


The formulas are not actually required, just know that volumes of high density materials are heavier than equivalent volumes of low density materials. Because the density of silver (10.5 g/cm3) is comparable to the density of lead (11.3 g/cm3), the two spheres have nearly the same volume. Since the density of aluminum is a lot smaller than both silver and lead, the aluminum sphere is a lot larger.

EXAMPLE 1.3
The "G temperature scale" has been calibrated so that,
0 0G = -11.5 0C
100 0G = 197.60C
Draw a thermometer that uses the G scale. Compute the freezing point of water in 0G (G degrees) and convert 150.0 0G to its equivalent in 0C.

SOLUTION
The G scale:

There are [197.6 -(-11.5)] = 209.1 0C between 0 and 100 0G, so the conversion factor in terms of magnitude will be:
1 0G = [100 0C] / [209.1 0G]
The 0C scale is -11.5 at 0 0G,
0C to 0G <-- add 11.5
0G to 0C <-- subtract 11.5
Note that the order in which the operations are performed will change the final answer!
For the freezing point of water,
{[100 0C] / [209.1 0G]} X [00C + 11.5] = 5.50 0G
For the conversion of 150 0G --> 0C,
{[150.0 0G X 209.1 0C] / [100.0 0G]} - 11.5 = 302.2 0C

EXAMPLE 1.4
The Fahrenheit temperature scale has been calibrated so that,
32 0F = 0 0C
212 0F = 1000C
Convert the normal body temperature (98.6 0F) to its equivalent in 0C and the melting point of Nitrogen (-345.8 0F) to their equivalent in 0F.

SOLUTION
There are [212 - 32] = 180 0F between 0 and 100 0G, so the conversion factor in terms of magnitude is:
1 0F = [100 0C] / [180 0F]
The 0C scale is 0 at 32 0F,
0F to 0C <-- add 32
0C to 0F <-- subtract 32
Note that the order in which the operations are performed will change the final answer!
For the normal body temperature,
{[100 0C] / [180 0F]} X [98.6 0F - 32] = 37.0 0C
For the boiling point of Nitrogen,
{[180 0F X -209.9 0C] / [100 0C]} + 32 = 345.8 0F

EXAMPLE 1.5
Due to unsettled differences with France, the United States of America decides to make its own length/mass standard in the form of a Titanium-Nickel alloy bar. The longest edge of the bar will be the meter standard and has been measured to be 1.000000 m. The width and height of the bar are 0.015000 m and 0.010000 m respectively. The mass of the bar is 1.00000 kg and will be the kg standard. Calculate the percent Titanium by mass in the bar. The density of Titanium is 4.51 g/cm3 and the density of nickel is 8.90 g/cm3. Assume volume additivity.

SOLUTION
Volume of alloy bar = 100 cm X 1.50 cm X 1.00 cm = 150 cm3 = 150 mL

Note that 1 mL = 1 cm3 so we can use either unit of volume.

Density of alloy bar = ρ = 1000 g / 150 mL = 6.67 g/mL
ρ = M / V --> V = M / ρ --> Total M / Average ρ = [M1 / ρ1] + [M2 / ρ2]

This means that the mass of Titanium plus the mass of Nickel has to equal the mass of the alloy, here 1000 g. It is simpler to set this value equal to 1 g, we could use 1000 g and still get the same answer because we are working with relative values.

1000 g --> 1 g
M = 1 g = Total mass --> M1 = &Psi = mass of Titanium and M2 = (1Ψ) g = mass of Nickel
1 g / 6.67 g/mL = [Ψ g / ρ1] + [(1-Ψ) g / ρ2]
1 g / 6.67 g/mL = [Ψ g / 4.51 g/mL] + [(1-Ψ) g / 8.90 g/mL]

The grams cancel and the Lowest Common Denominator is (4.51 X 8.90)

0.150 mL X (4.51 X 8.90) = [{Ψ X (4.51 X 8.90)} / (4.51) mL-1] + [{(1-Ψ) X (4.51 X 8.90)} / (8.90) mL-1]
6.02 mL = [(Ψ X 8.90) mL] + 4.51 mL [(Ψ X 4.51) mL]
6.02 mL - 4.51 mL = [(Ψ X 8.90) mL] [(Ψ X 4.51) mL]
1.51 mL = Ψ X 4.39 mL --> 1.51 mL / 4.39 mL = Ψ = 0.344
% Titanium by mass = 34.4 %


EXAMPLE 1.6
One liter (1.00 L) of an ethanol-water solution (commonly known as Vodka) weights 900.0 grams. Ethanol has a density of 0.789 g/cm3 and water has a density 1.00 g/cm3. Compute the percent ethanol by mass in the solution. Assume volume additivity.

SOLUTION
Can be solved in a similar way as problem 1.5. Volume additivity means that the volume of Vodka equals the volume of ethanol plus the volume of water. This is not always true and some substances have either positive or negative partial volumes. Ethanol is an example, but drunks and bartenders don't know that or choose to ignore the fact so we will do the same.

Mass of water = Ψ
Mass of ethanol = 1 - Ψ
Volume of water = 1.00 L X [1.00 X 103 cm3/1.00 L] = 1.00 X 103 cm3
Average density = 900 g / [1.00 X 103 cm3] = 0.900 g/cm3
0.900 g/cm3 = 1.00 g/cm3 (Ψ) + {0.789 g/cm3 X (1 - Ψ)}
0.900 g/cm3 = 1.00 g/cm3 (Ψ) + 0.789 g/cm3 - 0.789 g/cm3 (Ψ)
0.900 g/cm3 = 0.789 g/cm3 - 0.211 g/cm3 (Ψ)

Put Ψ's on one side, plain numbers on the other,

0.111 g/cm3 = 0.211 g/cm3 (Ψ)
[0.111 g/cm3]/[0.211 g/cm3] = Ψ
0.526 = Ψ <-- mass of water (relative amount)
0.526 - Ψ = 0.474 <-- mass of ethanol (relative amount)
water = 52.6 %, ethanol = 47.4 %

EXAMPLE 1.7
In light of the oil crisis, the Saudis are using some type of beer-powered dromedary transport (single hump camel) as an alternative the standard fossil fuel powered transport (car). Certain camel has a beer capacity of 50.0 L and burns 2.75 kg of beer (ρ = 0.795 g/cm3) per kilometer of travel. How far can this camel go in a full hump?

SOLUTION
Camel's hump capacity in cm3 = 50.0 X [103 cm3/L] = 50.0 X 103 cm3
Mass of beer = 50.0 X 103 cm3 X 0.795 g/cm3 = 39.8 X 103 g = 39.8 kg
Distance the camel can travel in a full hump of beer = [39.8 kg] / [2.75 kg/km] = 14.5 km
Note that camels do not store beer in their humps as this problem suggests. Their humps are a reservoir of fatty tissue, while beer is stored in their blood. This allows them to survive days on end without food and water. Our apologies to the dromedary folks out there.

II. | SIGNIFICANT DIGITS |
Note that the answers have been rounded off to a certain number of significant digits because experimental measurements are inaccurate to some extent. In Example 1.5 the number of significant digits for the mass and dimensions of the alloy bar are indicative of the high degree of certainty required to make an object that is to be used as a standard. Such an object would have to be ground down and measured with high precision calipers to within a certain tolerance. As a rule, there is an uncertainty of one unit in the last significant digit unless otherwise specified. So the number 1.000000 m is automatically assumed to mean 0.999999 or 1.000000 or 1.000001 m.

Thus when rounding off numbers:

-If the digit to the immediate right is 5,6,7,8, or 9, add 1 to the place we are rounding to.
-If the digit to the immediate right is 1,2,3, or 4, simply eliminate that digit.

And for the number significant digits:

-All non-zero digits are significant.
[911 = 3 significant digits]

-Zeros between non-zero digits are significant.
[22.409 = 5 significant digits]

-Zeros to the left of the first non-zero digit are not significant.
[2.02 = 3, 0.25 = 2, 0.025 = 2 significant digits]

-If a number ends in zeros to the right of the decimal point, those zeros are significant.
[2.0 = 2, 2.00 = 3 significant digits]

-If a number ends in zeros to the left of the decimal point, those zeros may or may not be significant.
[110, if written as 11.0 X 101 = 2, if written as 1.10 X 102 = 3 significant digits]

The last two rules seem complicated but they are not. It is only a question of accuracy, which is part of day to day activities in the chemistry laboratory.

When multiplying or dividing numbers, the result of your calculation has the same number of significant digits as the operand with the fewest number of significant digits.
For example:
1.87 X 3.0 = 5.6
1.87 (3 significant digits) X 4.0 (2 significant digits) = 5.6 (2 significant digits)

When adding or subtracting numbers, keep the fewest number of decimal places that are in all of the numbers. Keep in mind that the total number of significant digits is irrelevant.
For example:
5.67 + 6.5432 = 12.21
5.67 (2 decimal places) + 6.5432 (3 decimal places) = 12.21 (2 decimal places)

17.089 - 10.2200 = 6.869
17.089 (3 decimal places) + 10.2200 (4 decimal places) = 6.869 (3 decimal places)

When taking the logarithm of a number, the number of decimal places in the result must be the same as the number of significant digits in the number you started with.
For example:
log10(24) = 1.38

Because 24 has two significant digits, 1.38 has two decimal places. The digits to the left of the decimal point in a logarithm reflect the value of the exponent of the argument of the logarithm expressed in scientific notation. These digits do not contribute to the significant digits of either the argument or the logarithm. The digits to the right of the decimal point in a logarithm reflect the significant digits of the mantissa of the argument. So if you write 24 as 2.4 X 101, the logarithm is:

log10(2.4 X 101) = (0.38) + (1) = 1.38.

If you write 24000 as 2.40 X 104, the logarithm is:

log10(2.40 X 104) = (0.380) + (4) = 4.380.
Notice that 2.40 X 104 has three significant digits, hence the term 0.380, and the 4 comes from the exponent and is thus irrelevant to the number of significant digits retained

Keep in mind that those rules apply to measured quantities which are non-exact. If a number is exact then there is no uncertainty and the number of significant digits is infinite. For example people, with the exception of half-wits, come in integer numbers so there can be either 10 or 11 people but not 10.5.


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