Required Materials:
1.
Metric Ruler
2.
Plastic water cups
3.
50 ml graduated cylinders
4.
Mass balance beams
5.
Small glass beakers
6.
Thermometers
Objectives
1.
To understand how to convert numbers back and forth between normal format and scientific notation
2.
To understand how to convert values within the various units of the metric system
3.
To understand the concept of significant figures and how to express numbers with the correct number
Biology is a highly mathematical science. Recent advances in molecular biology, cell biology, physiology and genetics have happened over the last 50 years through the application of mathematics and its cousin, statistics. In biology, we quantify using the metric system and frequently in the form of exponential notation
EXPONENTIAL OR SCIENTIFIC NOTATION
When dealing with very large and very small numbers, it is usually easier to express them in scientific notation – a notation based on a multiple of 10. For example, the large number ten million or 10,000,000 can be written as 1 x 10 7 . In another example, the weight of a protein
(0.00000000000000000375 grams) can be written as 3.75 x 10 -18 grams.
To express a number in scientific notation, you do two things
1.
Move the decimal point so that there is only ONE non-zero digit to the left of the decimal point
2.
Determine the exponent by counting the number of places you moved the decimal point. If you moved the decimal point to the LEFT, the exponent is a POSITIVE number. If you moved the decimal point to the RIGHT, the exponent is a NEGATIVE number
Let’s look at some examples. a.
3,200 = 3.2 x 10 3  first you moved the decimal to the left so you created a single non-zero digit in front of a decimal point (e.g. 3.2), the you counted the number of places you moved the decimal point (three) and noted its direction (left = positive exponent). Moving it to the left two places would have created the number 32 x 10 2 . The number 32 is not a single, non-zero number – so this is incorrect. So be sure you move the decimal so that you create

a single number in front of the decimal point followed by the rest of the numbers after the decimal. b.
0.0032 = 3.2 x 10 -3  first you moved the decimal to the right so you created a single nonzero digit in front of a decimal point (e.g. 3.2), the you counted the number of places you moved the decimal point (three) and noted its direction (left = positive exponent) c.
1000 = 1 x 10 3 or just 10 3  for numbers where the single, non-zero digit is a 1, we simply ignore the 1 and just write the 10 and the exponent
To convert scientific notation back into a simple number, simply reverse the process. If the exponent is positive, move the decimal to the right and rewrite the number. If the exponent if negative, move the decimal to the left a.
6.3 x 10 3 = 6,300  this actually means take 6.3 and multiply it by one thousand b.
6.3 x 10 -2 = 0.063  this means to take 6.3 and divide it by one hundred
THE METRIC SYSTEM OF MEASUREMENT
The metric system was originally developed in France in the 18th century as a means of standardizing the various ways we had of measuring. It is based on units that are either multiples or divisions of 10. Because of that, it can also be called the ‘decimal system’ as ‘deca’ in latin means ‘ten’. In 1960, the International Conference on Weights & Measures (yes, there is actually a conference called that!) adopted the metric system as the System of International
Units (or SI Units). For the most part, it is used around the world as the standard means of measuring. While there are some notable ‘holdouts’ (the US and the UK come to mind), the metric system is used by all countries where science is concerned.
Length or distance: the basic unit for length is the meter (abbreviated ‘m’). It is a little longer than a yard. In the metric system, a thousand meters is known as a kilometer (abbreviated ‘km’).
This is frequently used to denote distances on the highway and is __________ of a mile.
Volume: the basic unit for volume is the liter (abbreviated ‘L’). It is slightly larger than a quart.
Many things are smaller than a liter. Something which is 1/1000 th of a liter is called a milliliter
(abbreviated ‘ml’ or ‘mL’). Got a bottle of water with you? Check out its volume.
Weight: the basic unit for weight is the gram (abbreviated ‘g’). It is actually 1/28 th of an ounce as is about the weight of a paperclip. Many weights are expressed as a 1000 grams or a kilogram
(abbreviated ‘kg’). A kg is equal to 2.2 pounds. What is your weight in kilograms?

Temperature: temperature is measured using the ‘Celsius’ scale (after Anders Celsius, the 18 th century Swedish physicist) or ‘centigrade’ scale. In this scale, water freezes at O°C and boils at
100°C. Compare this scale to our commonly used US Fahrenheit scale.
Converting from one metric unit to another
To change from one metric unit to another, you move the decimal point either to the left to make a number smaller or to the right to make a number larger. Metric units that are greater or smaller than the basic unit (by a factor of 10) are named by adding a prefix to the basic unit name. You should memorize these prefixes and their values as exponents. A prefix table is given below.

Here is another way to convert between metric units.
Express the metric unit you know as an exponent relative to the standard unit. Then express the metric unit you want to convert to as an exponent relative to the standard unit. For this, you will have to know your prefixes and what they mean. For example.
‘milli’ means 1/1000 th whereas ‘kilo’ means 1000x
1.
Subtract the exponents. This will tell you how many places you need to move the decimal point
2.
If the number you get is POSITIVE, it means the number is getting bigger and you move the decimal to the RIGHT. If the number you get is NEGATIVE, it means the number is getting smaller and you move the decimal to the RIGHT. e.g. 0.4 m = _______________ mm
In this example, we want to go from the standard unit – meter – to a smaller unit known as the millimeter
1.
Because the meter in this problem is the standard unit, we write 10 0 as the exponent for the 0.4m (10 0 means 1). Since a mm is 1/1000 th of the standard unit (meter), we write 10 -3 for the exponent for the unknown mm value. Remember 1/1000 th of a meter which is 10 -3
2.
0 minus -3 is +3. Which means we move the decimal to the RIGHT three places
So 0.4mm = 400mm e.g. 0.36 g = _____________ kg
1.
For this, we need to remember that the gram is the standard unit. So we write 10 0 as the exponent for the 0.36 g (10 0 means 1); and we write 10 3 for the exponent for the unknown kg value. Remember a kilogram is 1000 grams. 1000 is 10 3
2.
0 minus 3 is -3. Which means we move the decimal to the LEFT three places
So 0.36 g = 0.00036kg e.g. 0.13 L = _____________ mL
1.
We write 10 0 as the exponent for the 0.13 L (the liter is the standard unit); and we write
10 -3 for the exponent for the unknown mL value. Remember a mL is 1/1000 th of a liter which is 10 -3
2.
0 minus -3 is +3. Which means we move the decimal to the RIGHT three places
So 0.13 L = 130 mL

In all of these examples we were converting the standard unit to something that is non-standard.
But what if you want to convert a non-standard unit? e.g. 0.13 km = ______________mm
The kilometer is not a standard unit. It is 1000x bigger than the standard unit, the meter. Don’t worry though, we do this conversion the same way. Just write your exponents relative to the standard unit.
1.
We write 10 3 for the km; and we write 10 -3 for the exponent for the unknown mm value
2.
Subtract the exponents; 3 minus -3 is +6. Which means we move the decimal to the
RIGHT six placed
So 0.13 km = 130,000 mm.
Now what if you wanted to write this in scientific notation?
130,000 mm = 1.3 x 10 5 mm
Laboratory Exercise I: Measurements in the Metric System
Let’s try a few in-lab exercises that make you measure and then convert these measurements within the metric system. For this exercise you will need the following: a.
A metric ruler b.
A microscope slide c.
A plastic bottle d.
A 100 mL graduated cylinder
What is the width of your thumb in inches? In centimeters? In millimeters? In meters?

1.
___________________________ inches (in)
2.
___________________________ centimeters (cm)
3.
___________________________ millimeters (mm)
4.
___________________________ meters (m)
Take the plastic bottle and measure the following:
1.
The length of the bottle from base to neck _____________________
2.
The width of the bottle at the neck _______________________
3.
The width of the bottle at the base ________________________
4.
Using the beam balance, weigh the empty bottle and write this weight in grams
_________________
5.
Fill the plastic bottle with some water and re-weigh it in grams
__________________________
6.
How much does the water in the plastic bottle weigh? _________________________
7.
Using your 100 ml graduated cylinder, pour the water into it and measure the volume of water that was in the bottle ______________________________
With the values of weight and volume, you can calculate the density of the water
DENSITY = Weight (g) = Weight of the water (g)
Volume (mL) Volume of the water (mL)
So for example: if the weight of your empty bottle was 9.2 g and the weight of your bottle filled with water was 74 g

the weight of the water was 74 g – 9.2 g = 64.8 g if the volume of the water was 60.5 mL the density of water was 64.8 g divided by 60.5 mL = 1.07 g/mL or 1.1 g/mL
Go ahead and look up the density of water and see what you come up with
This example also illustrates an important point in scientific quantitation – the matter of decimal points and the numbers before and after them. When you perform mathematical operations using measurements (like what we just did above), there is the matter of how many figures before and after the decimal point do you use. Do you write the density of water as 1.07 or do you round up to 1.1?
To answer this, you have to know what a significant figure is. We call numbers before and after a decimal point significant figures (or digits)
The significant figures of a number are those digits that carry meaning by contributing to a number’s precision. Significant figures include all digits except:
 Leading zeroes – e.g. 007; in this number, the zeroes are leading and are insignificant. The same thing for a number like 0.003. The zeroes are insignificant but they do provide an indication of scale or size
 Trailing zeroes when they are merely placeholders to indicate the scale of the number – e.g. 12.3400; in this number the zeroes after the 4 do not affect the value of the number and may be omitted
When are Digits/Figures Significant?
Non-zero digits are always significant. Thus, 22 has two significant digits, and 22.3 has three significant digits. But with zeroes, the situation is more complicated. Here are some situations dealing with zeroes: a.
Zeroes placed before other digits are not significant; 0.046 has two significant digits. b.
Zeroes placed between other digits are always significant; 4009 kg has four significant digits. c.
Zeroes placed after other digits but behind a decimal point are significant; 7.90 has three significant digits. d.
Zeroes at the end of a number are significant only if they are behind a decimal point as in
(c). Otherwise, it is impossible to tell if they are significant. For example, in the number
8200, it is not clear if the zeroes are significant or not. The number of significant digits in
8200 is at least two, but could be three or four. To avoid uncertainty, use scientific notation to place significant zeroes behind a decimal point:
Numbers are often rounded to avoid reporting insignificant figures. For example, it would create false precision to express a weight as 12.34500 kg (with its seven significant figures) if the scales only measured to the nearest gram and gave a reading of 12.345 kg (which has five significant

1
2
3 figures). Numbers can also be rounded merely for simplicity rather than to indicate a given precision of measurement.
So let’s go back to our density problem. Should we round the 1.07 to 1.1 or 1 or just keep it as
1.07?
Well, here are the rules:
For adding and subtracting, you write the number so that it contains the fewest number of significant figures AFTER the decimal point e.g. 1.0056 + 7.113 + 0.81 = 8.9786. The number of significant digits in this operation are: 5, 4 and 2. So 8.9786 would round to 8.98  TWO significant digits past the decimal
For multiplying and dividing, you write the number so that it contains the fewest total number of significant digits (before AND after the decimal point) e.g. 9.3/0.62137 = 14.9669279. The number of significant digits in this operation are 2 and 5. So
14.9669279 would round to 15  TWO significant digits TOTAL.
Determining the Density of Water
Weight of empty bottle
(g)
Weight of bottle with water (g)
Weight of water (g)
Volume of water (ml)
Density of water (g/ml)
35 ml
70 ml

Review Questions
Answer the following in the spaces provided.
1. Convert the following into decimals and then into scientific notation. Use two significant digits after the decimal point
1.
1/100 = ______________________ = _______________________
2.
34/100 = ______________________ = _______________________
3.
479/100 = ______________________ = _______________________
4.
379/10,000 = ______________________ = _______________________
5.
267/100,000 = ______________________ = _______________________
2. Convert the following into scientific notation. Use two significant digits after the decimal point
1.
500 = ___________________
2.
379 =___________________
3.
14,890 = __________________
4.
0.03 = _____________________
5.
0.00345 = _________________
6.
0.0000000341 = ________________
7.
2 = __________________
8.
64 = __________________
3. What do the following prefixes mean in relationship to the standard unit? a.
Kilo- b.
Centi- c.
Deci- d.
Milli-
4. What do the following abbreviations mean? a.
lb = b.
g = c.
cc =

d.
mL = e.
oz = f.
sec = g.
cm = h.
L = i.
km =
5. Convert the following into decimal format numbers and then into scientific notation. Look at the units on the far right side to know what you are converting into! Use two significant digits after the decimal point
1.
42 cm = ______________________ = _______________________ m
2.
640 cm = ______________________ = _______________________ m
3.
6 mm= ______________________ = _______________________ cm
4.
598 mm = ______________________ = _______________________ cm
5.
68 m = ______________________ = _______________________ km
6.
54 m= ______________________ = _______________________ cm
7.
12 m = ______________________ = _______________________ mm
8.
26 cm= ______________________ = _______________________um
9.
26 cm = ______________________ = _______________________ m
10.
23 mL = ______________________ = _______________________ L
11.
345 L = ______________________ = _______________________ uL
12.
235 uL = ______________________ = _______________________ pL
13.
6 um = ______________________ = _______________________ m
14.
12 kg = ______________________ = _______________________ g
15.
6.2 ug = ______________________ = _______________________ kg
16.
2 ng = ______________________ = _______________________ g
17.
50°F = ______________________ = _______________________°C

18.
100°C = ______________________ = _______________________°F
19.
28 A° = ______________________ = _______________________m
20.
567 A° = ______________________ = _______________________nm
6. Calculate the following. Use one significant digit after the decimal point
1. (5.0 x 10 4 ) x (1.6 x 10 2 ) =
2. (4.0 x 10 0 )/(1.0 x 10 3 ) =
3. (9.0 x 10 9 ) x (2.0 x 10 5 ) =
4. (6.0 x 10 4 ) x (6.0 x 10 -4 ) =
5. (8.0 x 10 8 ) x (7.0 x 10 -6 ) =
6. (6.0 x 10 -2 )/(5.0 x 10 -8 ) =
7. (6.0 x 10 3 )/(3.0 x 10 -2 ) =
8. (5.0 x 10 4 ) x (1.6 x 10 2 ) =
9. (8.0 x 10 -3 )/(2.0 x 10 -6 ) =
10. (10.5 x 10 -2 ) x (5.0 x 10 -4 ) =
7. A box measures 1cmx6cmx5cm. What is its volume? ___________

8. Add 5.0 m + 6.25 m + 3.131 m = ________ (hint: use proper significant figures)
9. Add 1.2 g + 3 dg + 90 cg = ________ (hint: convert all numbers to the same unit)

Objectives
1.
An understanding of how to calculate the moles of a solute
2.
An understanding of how to calculate the molarity of a solution
3.
An understanding of molecular weight (molar mass) and how it relates to the chemical formula of the substance
4.
An understanding of molecular weight and how it relates to moles and molarity
5.
An understanding of what a ratio of a solution means
6.
An understanding of what the percent concentration of a solution means
7.
How to make a solution of a specific molarity if given information such as the molecular weight of a substance, the moles of the solute or the volume required
A solution consists of one or more substances (solutes) that are dissolved in a liquid. If the liquid is water, it is called an aqueous solution. The concentration of a solution tells you the amount of a solute dissolved in a specific volume of liquid. We can express the amount of a solute added to a liquid to make a solution in two ways:
1.
By the weight of the solute added to the solution
2.
By the number of particles in the solution
The amount of solute by weight can be expressed using ratios. Ratios indicate the proportion of solute (in grams) per volume liquid (in milliliters). For example, a vial labeled 1:1,000 Adrenaline would contain 1 g of epinephrine in 1,000 ml of solution. A 1:50 solution would contain 1 g in 50 mL of solution.
You can also express solute amounts using percentages. Percentage indicates the amount of solute (in grams) in 100 mL of liquid. For example, 3% saline means 3 g of NaCl in 100mL water.
7.5% sodium bicarbonate would contain 7.5 g of this chemical in 100 mL of solution.
The examples given about (ratios and %) are ways of expressing the concentration of a solution.
The properties and behavior of many solutions depend not only on the nature of the solute and solvent but also on the concentration of the solute in the solution. Chemists use many different units when expressing concentration; however, one of the most common units is molarity .
Molarity (abbreviated as M) is the concentration of a solution expressed as the number of moles of solute per liter of solution:
Molarity = moles of solute/ volume of solution
For example, a 0.25 M NaOH solution (this is read as 0.25 molar) contains 0.25 moles of sodium hydroxide in every liter of solution. Anytime you see the abbreviation M you should immediately think of it as mol/L.

In order to calculate the molarity of a solution, you need to know two things:
1.
the number of moles of solute in the solution
2.
the total volume of the solution
Once you know these two things – you can calculate Molarity. But first you need to know what a mole is. The amount of a solute is often expressed in moles. Just like a dozen of something is equal to 12, a mole is equal to 6 x 10 23 items. This is known as Avogadro’s number. So 2 moles is really 2 x (6 x 10 23 items).
To calculate molarity, you first calculate the number of moles of solute present in the solution.
Then you determine the number of liters of solution present. Divide the number of moles of solute by the number of liters of solution. This number is the molarity of the solution.
Example: What is the molarity of a solution if you dissolve 5.23 moles of a solute in 500 mL of solution?
First, you need to express the 500 mL of a solution in liters - 0.5 L. Once you convert your solution volume to liters – you are ready to plug it into the equation.
Molarity = moles/volume (L)
M = 5.23 moles/0.5 L
M = 10.46 M
This is pretty straightforward. But most of the time, you are not given the moles of the solute but have to calculate it first. To do this, you need to know the how much a mole weighs. The weight of a mole, for any element, is equal to the chemical’s weight in grams. We call this weight the molar mass. Molar mass is the grams of a chemical needed to make 1 mole. Molar mass is expressed as g/mole. We calculate molar mass using the periodic table of elements and the atomic mass for each element in the solute.
Example: What is the molar mass of water?
To do this, you need to know a few things:
1.
the chemical formula for water = its H
2
0
2.
the atomic mass of hydrogen = its 1.007 or 1.0
3.
the atomic mass of oxygen = its 15.999 or 16.0
So the atomic mass of H20 is about 18. Which means 1 mole of water would weight 18.0 g. So the molar mass of water is 18.0 g/mole.
Let’s look at another problem.
Example: What is the molarity of a solution prepared by dissolving 15.0 g of sodium hydroxide in
225 mL of solution?

Step #1: Calculate the number of moles in 15.0 g of sodium hydroxide.
The chemical formula for sodium hydroxide is NaOH.
The molar mass for O is 16.0 g/mol
The molar mass for H is 1.0 g/mol
The molar mass for Na is 23.0 g/mol
So the molar mass for NaOH is 40.0 g/mol. In other words, 1 mole of NaOH weighs 40 grams.
But you only have 15.0 g. This means that 15.0 g is really 15/40 or 0.375 moles.
Step #2: Convert your volume to liters
Easy. You have 225 mL. This is 0.225 L
Step #3: Plug your data into the equation for molarity
M = moles/volume (L)
M = 0.375/0.225
M = 1.6666666 or 1.7M
So dissolving 15.0 g of NaOH in 225 mL of water gives you a 1.7M solution.
Complete the following problems. Show your work and express your results using 2 significant digits.
1.
What is the chemical weight of lithium chloride (LiCl)? b. How much would 0.5 moles of LiCl weigh? c. How much would 2 moles of LiCl weigh?
2.
Calculate the chemical weight/molar mass of ammonium chloride (NH4Cl) b. How much would 0.15 moles of NH
4
Cl weigh? c. How much would 100 millimoles (i.e. 100mM) of NH
4
Cl weigh?

3. Calculate the number of moles required to make 100 mL of a 3.0 M solution of CuSO
4
.
4. A 4.0 g sugar cube (sucrose: C
12
H
22
O
11
) is dissolved in a 350 ml teacup filled with hot water.
What is the molarity of the sugar solution?
5. Calculate the molarity of the solution that results when 4.0 g of NaCl is dissolved in 30.0 ml of water.
6. How many grams of NaOH would you need to add to 0.35 liters of water to make a 1.50 M solution?
7. A chemist dissolves 98.4 g of FeSO
4
in enough water to make 2.0 L of solution. What is the molarity of the solution?

8. How many moles of KBr are in 25.0 mL of a 1.23 M KBr solution?
9. Battery acid is generally 3M H
2
SO
4
. How many grams of H
2
SO
4
are in 400 mL of this solution?
10. How would you make 1.50 L of a 1.62 M NaCl solution?
11. Calculate the molarity of the solution that results when 5.0 ml of CH
3
OH (methyl alcohol or methanol) is dissolved in 5.0 ml of water.
12. How would you make 2.0 Liters of 6.74 M CH
3
OH solution?
13. 45.0 g of Ca(NO
3
)
2
was used to create a 1.3 M solution. What is the volume of the solution?
14. How would you prepare 100mL of a 1:100 NaCl solution?

15. How would you prepare 10mL of a 1:20 glucose solution?
16. If a reagent bottle contains 10g of NaCl dissolved in 1L of water, what would its concentration be expressed as a ratio? What would it be as a percentage?
17. How would you prepare 100mL of a 2% solution of LiCl?
18. How would you prepare 500ml of a 0.5% saline solution? Remember saline is really NaCl.
19. How would you prepare 1L of a 10% calcium chloride solution?
20. If a reagent bottle contains 50g of glucose dissolved in 1L of water, what would its concentration be expressed as a percentage?