Lecture 2: Units and Implications of Measurement

Course

Unit 1

Chapter 1

General Chemistry

Molecular Structure and Properties

Introduction

This lecture is part of the MEEP curriculum on General Chemistry. Information about Project MEEP and other General Chemistry lectures are available below.

Project MEEP

General Chemistry

Recommended Time: 1 Hour

Lecture Schedule

Lecture Preview

How do we describe the amount of something and how is it standardized?
How are accuracy and uncertainty implied in measurements?
How can we convert between different units of measurements?
What is the concept of a mole?

Lecture Content

SI Units

Suppose we are doing an experiment, and from that experiment, we generated approximately a tablespoon of solid products. As researchers and followers of the scientific method, we need to accurately record how heavy that tablespoon of solid is, because a simple “a tablespoon worth of product” is not academically rigorous enough and is prone to misinterpretations.

We need a standardized system to describe “how much there is of something” so that it leaves no room for misinterpretations. Introducing the International System of Units, or more commonly known as the SI Units. In this system, there are 7 base units that are defined from fundamental features and increasingly abstract and idealized formulation. The rest of the SI units can be derived from those 7 base units. Let’s look at the base units first.

Symbol

Name

Described Quantity

second

time

meter

length

kilogram

mass

ampere

electric current

kelvin

temperature

mol

mole

amount of substance

candela

luminous intensity

Visualization of the Base Units

We can go through the definitions of each base unit, but as mentioned above, the definition for each base unit is now increasing abstract and idealized. For example, let’s quickly compare the 1889 definition of the kilogram and the 2019 definition of the kilogram.

1889

2019

1 kilogram is defined to be exactly the weight of the International Prototype of the Kilogram (IPK) – a platinum-iridium alloy that is stored in a vault in Paris.

1 kilogram is defined by taking the fixed numerical value of the Planck constant $h$ to be $6.62607015 \times 10^{−34}$ when expressed in the unit $J\cdot s$ , which is equal to $kg\cdot m^{2}\cdot s^{−1}$ , where the metre and the second are defined in terms of $c$ and $\Delta v_{Cs}$ .

In fact, in 2019, a revision was undergone to change 4 out of 7 of the definitions of the SI base units to be based on natural physical constants rather than human-made artifacts. One of the reasons that artifacts vary over time – the IPK was fluctuating throughout its years of service, with an increase of tens of micrograms.

However, practically speaking, there is no drastic consequence of the revision of the SI base units – 1 kilogram after 2019 is practically the same as 1 kilogram before 2019, and same for the other revised units.

Let’s visualize the SI base units to make it more approachable.

1 second is approximately the duration of 3 rapid blinks of our eyes.
1 kilogram is approximately 4 medium-sized potatoes.
1 meter is approximately the length between our fingers to our opposite shoulder when we t-pose.
1 ampere is approximately the amount of electricity that passes through a 100-watt lightbulb in a US household.
Room temperature (25 Celsius or 77 Fahrenheit) is 298.15 kelvins.
Approximately 18 grams (just above 3 teaspoons) of water corresponds to 1 mole of substance.
1 candela is approximately the brightness emitted by a common candle.

Visualization of 1 second, 1 kilogram, and 1 meter.

Deriving Other Units

We can use the SI base units to define other units. Common derivations are:

The unit for speed – defined by $m/s$ .
The unit for area – defined by $m^{2}$ .
The unit for volume – defined by $m^{3}$ .
The unit for density – defined by $kg/m^{3}$ .
The unit for force – defined by $kg\cdot m / s^{2}$ .
The unit for pressure – defined by $kg / (m\cdot s^{2})$ .

The list goes on, and we will see more of those derived units in physics. In chemistry, we will stick to a few common units.

Unit Prefixes

However, simply having these defined SI units (both the base ones and their derivatives) are not enough. Measurements can span orders of magnitudes. For bigger magnitudes, we usually use scientific notation, such as the mass of the sun, which is around $2\times 10^{30} kg$ . For smaller magnitudes, on the other hand, we like to use unit prefixes to imply the exponent (the $\times 10^{n}$ part of the expression).

Below is a list of prefixes alongside their implications.

Prefix Name

Symbol

Factor

pico

$10^{-12}$

nano

$10^{-9}$

micro

$\mu$

$10^{-6}$

milli

$10^{-3}$

centi

$10^{-2}$

deci

$10^{-1}$

kilo

$10^{3}$

mega

$10^{6}$

giga

$10^{9}$

tera

$10^{12}$

Here are some examples:

1 km = 1000 m; 1 m = 0.001 km
1 mg = 0.001 g; 1 g = 1000 mg
1 $\mu$ L = 0.000001 L; 1 L = 1000000 $\mu$ L

Simply speaking, it’s a matter of moving the decimal point to the left or right. However, beware that when converting units with exponents using prefixes, the exponents need to be applied to the factoring value as well. For example, if we want to convert $m^{3}$ to $cm^{3}$ :

$2.4 m^{3} = 2.4 \times (1 m)^{3} = 2.4 \times (100 cm)^{3} = 2.4 \times 100^{3} cm^{3} = 2.4 \times 10^{6} cm^{3}$

This could be a bit tricky, but this situation isn’t as common in chemistry as in physics. However, it is always good to be aware of unit prefix conversions.

Also, here is a unit that we need to remember – the liter.

$1L = 1000mL = 1000cm^3 = 1dm^3 = 0.001 m^3$

Significant Figures

When measuring things, there is always a certain extent of uncertainty – we can only accurately measure things to an extent. To reflect the level of accuracy we have in our measurements, we use the power of significant figures. Let’s call them sig figs for convenience. Sig figs allow us to see how accurate and uncertain our measurements are.

Counting Sig Figs

For the most part, counting sig figs is simple – we start from the most left non-zero number and see how many digits are on its right.

$58271958 \rightarrow 8$ sig figs
$370.4999 \rightarrow 7$ sig figs
$15.8000000 \rightarrow 9$ sig figs
$0.000000000000001 \rightarrow 1$ sig fig
$0.0000000000000010 \rightarrow 2$ sig figs

The only ambiguity present is when we have a number that ends in 0 but doesn’t have a decimal point, like 135000. We do not know which 0 digit is part of the measurement – it could be there to simply imply magnitude. It’s easy when our measurement yields 135000.0, which implies that all the 0’s are part of the measurement, leading to a conclusion that 135000.0 has 7 sig figs. As a result, to minimize ambiguity for ambiguous numbers, it is recommended to use scientific notation or appropriate unit prefixes to describe them. For this course, we will view those 0’s (like the ones in 135000) as non-sig figs.

$1.35\times 10^{6}\rightarrow 3$ sig figs
$1.350\times 10^{6}\rightarrow 4$ sig figs
$1.3500\times 10^{6}\rightarrow 5$ sig figs
$1.35000\times 10^{6}\rightarrow 6$ sig figs
$1.350000\times 10^{6} = 135000.0\rightarrow 7$ sig figs

Calculating with Sig Figs

There are 3 rules when calculating with sig figs.

Rule 1: Rounding

If the digit dropped is $> 5$ , we increase the retained digit by 1.
If the digit dropped is $< 5$ , we keep the retained digit as the same.
If the digit dropped is $= 5$ , we either increase or maintain the retained digit such that it’s even. But if there are non-0 trailing digits, round up. (It’s based on reliable statistics aimed at avoiding bias)

Rule 2: Adding and Subtracting

Observe the number of decimal places. Round the result to the same number of decimal places as the number with the fewest decimal places in the calculation.

Rule 3: Multiplying and Dividing

Observing the number of sig figs. Round the result to the same number of sig figs as the number with the fewest sig figs in the calculation.

Let’s work with some examples.

Example 1: Rounding numbers to 3 sig figs.

$0.032971\rightarrow 0.330$
$15824\rightarrow 15800$
$6.8550001\rightarrow 6.86$ (non-0 trailing digits)
$6.8550000\rightarrow 6.86$ (retained digit is even)
$6.8650000\rightarrow 6.86$ (retained digit is even)

Examples 2: Basic algebra with sig figs.

$1.0023g+4.383g=5.3853g=5.385g$ (3 decimal places)
$486mL-421.23mL=64.77mL=65mL$ (0 decimal places)
$0.6238cm\times 6.6cm = 4.11708cm^{2}=4.1cm^{2}$ (2 sig figs)
$421.23g/486mL=0.866728...g/mL=0.867g/mL$ (3 sig figs)

Remember: We can only add or subtract values with the same units. We cannot add or subtract values with different units.

Note that those calculation rules only apply to measurement values. Numbers used for scaling purposes, numbers that are defined constants, and conversion values are not measurements and thus will not follow the sig fig rules. For example, if there are 2 apples that weigh exactly 40.08 grams each, they have a total of $2 \times 40.08g = 80.16g$ because 2 is not a measurement – it’s a scaling factor. Similarly, converting 25.000 Celsius to kelvins will be $25.000^{\circ}C+273.15=298.150K$ because 273.15 is a conversion factor.

Unit Conversion

There are multiple units that describe mass, volume, length, etc., and there are numerous ways to convert between those units. We will encounter these in chemistry, especially in lab. Let’s look at some examples to warm ourselves up. Remember, sig fig rules do not apply to unit conversion factors.

Question: An anti-freeze has a volume of 4.00 quarts (qt) and a mass of 9.26 pounds (lb). What is the density of this anti-freeze in g/mL?

Hint:

$1L=1.0567qt$
$1kg=2.2046lb$

Solution:

Step 1: Converting the mass

$9.26lb\times \frac{1kg}{2.2046lb}\times \frac{1000g}{1kg} = 4.20\times 10^{3}g$

Step 2: Converting the volume

$4.00qt\times \frac{1L}{1.0567qt}\times \frac{1000mL}{1L} = 3.78\times 10^{3}mL$

Step 3: Calculating the Density

$\frac{4.20\times 10^{3}g}{3.78\times 10^{3}mL} = 1.11 g/mL$

Answer: The anti-freeze has a density of 1.11 g/mL.

Notice that while converting units, we put the units we want to convert to at the numerator.

Question: Convert 310.2 kelvins into Fahrenheit.

Hint:

$T_K = T_{^{\circ}C} + 273.15$
$T_{^{\circ}F} = T_{^{\circ}C} \times 1.8 + 32$

Solution:

Step 1: Converting kelvins to Celsius

$T_{^{\circ}C} = 310.2-273.15=37.0^{\circ}C$

Step 2: Converting Celsius to Fahrenheit

$T_{^{\circ}F} = 37.0^{\circ}C \times 1.8 + 32 = 98.6^{\circ}F$

Answer: 310.2 kelvins will be 98.6 Fahrenheit.

Here is a refresher of converting between Celsius, Fahrenheit, and kelvin. We will talk more about kelvin in another chapter, but it’s always to be aware of it earlier so it will be easier as we go on.

where

$T_{^{\circ}C}$ denotes temperature in Celsius

$T_{^{\circ}F}$ denotes temperature in Fahrenheit

$T_K$ denotes temperature in kelvin.

The Mole

Recall that previous in this lecture, we talked about the mole being one of the SI base units. A quick refresh: The mole describes the amount of a substance. So, what is the definition of the mole?

1 mole is defined as the amount of substance of $6.02214076\times 10^{23}$ elementary entities. This number is the fixed numerical value of the Avogadro constant, $N_A$ , when expressed in the unit $mol^{-1}$ .

Let’s simplify this definition. Recall that chemistry is the study of matter. Matter (objects, solutions, gases, basically everything) is composed of very small particles like atoms, ions, and molecules (don’t worry, we’ll talk about them next lecture). For example:

A standard bottle of water (the 16.9oz or 0.500L ones) contains around $1.67\times 10^{25}$ water particles.
A 1.00cm x 1.00cm x 1.00cm block of gold contains around $5.91 \times 10^{22}$ gold particles.

However, those numbers are too big for us to comprehend or have a clean calculation. So, we scale those numbers down using a defined constant to ensure consistency. It’s like buying groceries: we buy “a pair of” shoes (2 shoes), a “dozen” eggs (12 eggs), or a “baker’s dozen” of donuts (13 donuts). If we say that we will buy a “mole” of basketballs, we will be buying around $6.02 \times 10^{23}$ basketballs. This defined “scaling constant” we use is called the Avogadro constant (or Avogadro’s number) $N_A$ , which has a value of $6.02214076 \times 10^{23}$ . For our convenience, we’ll take the Avogadro constant as:

$N_A = 6.02 \times 10^{23} \ mol^{-1}$

The unit is included by definition. Using this newfound tool, we discovered our first formula.

where

$n$ denotes the amount of substance in moles

$N$ denotes the number of particles of said substance

$N_A$ denotes the Avogadro constant in $mol^{-1}$ .

Using this constant, we can rephrase our examples above.

$1.67\times 10^{25}$ water particles become $27.8\ mol$ of water.
$5.91 \times 10^{22}$ gold particles become $0.0981\ mol$ or $98.1\ mmol$ of gold. (Recall unit prefixes)

Converting number of particles to moles.

To summarize, 1 mole of some matter contains $6.02 \times 10^{23}$ particles of that matter. Here is a Ted-Ed video explaining what a mole is.

Once we learn more about the particles that make up matter, we will introduce more chemistry-related units and calculations that we will use to describe the properties of a matter or a reaction with the mole being a central unit. For now, start getting used to the scope of the mole.

Assignments

1. Preview Questions

How do we describe the amount of something and how is it standardized?
How are accuracy and uncertainty implied in measurements?
How can we convert between different units of measurements?
What is the concept of a mole?

2. Lecture Worksheet

The lecture worksheet is available as a pdf file below. Remember, practice makes perfect!

3. Further Reading

SI Units: https://en.wikipedia.org/wiki/International_System_of_Units

Video – “Seven Dimensions” by Kieran Borovac (a linear algebra approach of units)
Video – “The kg is dead, long live the kg” by Veritasium (an explanation on the 2019 revision of SI base units)