This document provides information and examples about making dilutions in a laboratory setting. It begins with definitions of key terminology used in dilutions. It then discusses how to make stock solutions at specific molar concentrations or weight/volume percentages. Several methods for making dilutions are covered, including diluting to an unspecified or specified final volume in one or multiple steps. Serial dilutions are also explained. Sample problems demonstrate how to calculate amounts needed to make solutions of a given concentration from more concentrated stock solutions.

1. Lab Math
Why do I care?
• Between 44,000 and 98,000 people die every year as a result in
medical errors.
• Medical errors are more deadly than breast cancer, motor
vehicle accidents, or AIDS.
• Sebastian Ferrero died at Shands Hospital in Gainesville, Florida
in October 2007, at age 3, as a result of a medication overdose
followed by a series of preventable medical errors. Check out his
story:
• http://www.sebastianferrero.org/Sebastian-story.php
• http://floridaphysician.med.ufl.edu/2008/12/features/how-one-boy-changed-
the-college-of-medicine/
Terminology
Aliquot
C1 x V1 = C2 x V2
Cross multiplication
Diluent
Dilution
Dilution factor
Metric system
Molar
Ratio
Stock solution
v/v
w/v
Background
Dilutions with Sample Problems
1. Making stock solutions
a. Molar concentrations (w/v)
b. Concentrations weight/volume (w/v %)
2. To dilute a solution to an unspecified final volume in just one step
3. To dilute a solution to a specified final volume in just one step
4. To dilute a solution to an unspecified final volume in several steps
5. To dilute a solution to a specified final volume in several steps
6. Serial dilutions

2. Making stock solutions
a. Molar concentrations (w/v)
b. Concentrations weight/volume (w/v)
Molar concentrations (w/v)
• A 1 molar solution is a solution in which 1 mole of a compound is dissolved in a total
volume of 1 liter.
• First look up the M. W. (molecular weight) of the compound of interest – it’s always
written on the original container. Sometimes it is expressed as F.W. (formula weight). is
now more correctly called the relative molar mass (Mr)
For example sodium chloride has a M. W. of 58.44
Therefore you add 58.44 g of NaCl to water, fill up to 1L – which equals a 1M solution.
Concentrations weight/volume (w/v %)
Another, somewhat less often used, concentration unit is weight/volume percent (w/v %). It
is defined as:
Concentration solute (w/v %) = [(mass of solute (g)/ volume of solution (ml)] x 100
You need to know that 1 g = 1 ml (which is correct for water)
For example, suppose we dissolve 1.2 g of NaCl in enough water to make 160 mL of
(saline) solution, what is the w/v % of NaCl?
[(1.2 g / 160 ml] x 100 = 0.75 w/v % NaCl
We interpret this number as 0.75 g of NaCl in 100 mL of solution. 500 mL of 0.75 w/v %
NaCl solution would contain 0.75 x 5 g = 3.75 g of NaCl.
Diluting a Solution to an Unspecified Volume in Just One Step
 One must first calculate how many times to dilute (this is called the dilution factor) the
initial material (stock solution) to obtain the final concentration. To accomplish this
type of dilution, use the following formula:
Dilution Factor (DF) = Initial concentration (IC)/Final Concentration (FC)
 For example, if you want to dilute a solution with an initial concentration of solute of
5% down to 1%, using the above formula gives:
IC/FC = DF
5%/1% = 5
 Thus, in order to obtain a 1% solution from a 5% solution, the latter must be diluted five
times. This can be accomplished by taking one volume (e.g. ml) of the initial
concentration (5%) and adding four volumes (e.g. ml) of solvent for a total of five
volumes. Stated another way, 1 ml of a 5% solution + 4 ml of diluent will give a total of 5
ml, and each ml contains 1% instead of 5%.

3. Diluting a Solution to a Specified Volume in Just One Step
 First, calculate the number of times the initial concentration must be diluted by dividing
the final concentration (FC) into the initial concentration (IC), i.e. the DF.
 Second, divide the number of times the initial concentration must be diluted into the
final volume specified to determine the aliquot (or portion) of the initial concentration
to be diluted.
 Third, dilute the aliquot of the initial concentration calculated in step 2 by the volume
specified.
 For example, you have a 10% solution and want a 2% solution. However, you need 100
ml of this 2% solution.
IC/FC = DF = 10%/2% = 5
 Divide the number of times the 10% solution must be diluted (DF) in to the final volume
specified:
100 ml/5 = 20ml
Dilute the portion of the 10% solution to the volume specified:
20ml of the 10% solution + 80ml of diluent = 100ml (each milliliter = 2%)
 Another method for performing this type of dilution is to use the following formula.
 This is the most important formula you will encounter for all your lab calculations
C1V1 = C2V2
or
C1/C2 = V2/V1
C1 = standard concentration available
C2 = standard concentration desired
V2 = final volume of new concentration
V1 = volume of C1 required to make the new concentration
 For example, if you want to prepare 100ml of 10% ethyl alcohol from 95% ethyl alcohol,
then
Diluting a Solution to an Unspecified Volume in Several Steps
 Frequently in the microbiology laboratory, large dilutions must be employed. They
cannot be done in one step because they are too large. As a result, they must be done in
several steps to conserve not only amounts of diluent to be used but also space. For
example, a 0.5 g/ml solution diluted to 1 μg/ml is a 500,000-fold dilution.
0.5g = 0.5 x 106
mg = 500,000 μg
(0.5 g/ml) / (1 μg/ml ) = 500,000 fold dilution

4.  To obtain a solution containing 500,000 μg/ml in one step would require taking 1ml of
0.5g/ml stock and adding 499,999ml of diluent. As you can see, it would be almost
impossible to work with such a large fluid volume.
 A 500,000 times dilution can be easily performed in two steps by first taking 1ml of the
initial concentration and diluting it to 500ml and second by diluting 1ml of the first
dilution to 1,000ml.
 Even these volumes are impractical in the real lab setting. In practical terms a lab
scientist would make the first 1:500 dilution, by adding 10 μl to make 5 ml diluent and
then take 10 μl of the first dilution and add it to 9.99 ml of diluent to obtain the 1:1000
dilution.
To check 500 x 1000 = 500,000
0.01ml of 500,000 μg/ml + 4.99ml of diluent = 1,000 μg/ml
0.01ml of 1,000 μg/ml + 9.99ml of diluent = 1 μg/ml
Using this two step procedure, we have cut down the volume of diluent used from
499,999ml to ~15 ml (4.99ml + 9.99ml)
Dilutions Ratios Used in This Module
 According to the American Society for Microbiology Style Manual, dilution ratios may
be reported with either colons (:) or slash (/), they are basically interchangeable in our
examples.
o A slash indicates the ratio of a part to a whole, same as a fraction: for
example ½ mean one of two parts, with a total of two parts.
o A colon, similarly, indicates the ratio of one part to two parts. Thus, ½
equals 1:2
Note 1:1 is NOT a dilution – it refers to the original (1 part of the whole)
1:2 is a 50% dilution, where 1 part of the stock is mixed with 1 part of the diluent
Diluting a Solution to a Specified Volume in Several Steps
 This type of dilution is identical to all previous dilutions with the exception that the
specified final volume must be one factor of the total dilution ratio.
 For example, you want a 1/10,000 dilution of whole serum (undiluted) and you need 50
ml.
 Divide dilution needed by the volume
10,000/50 = 200
 200 (1/200 dilution) = the first step in the dilution factor; the second is 1/50, obtained as
follows:
1ml of serum + 199 ml of diluent = 1/200 dilution
1 ml of 1/200 dilution + 49 ml of diluent = 1/50
To check: 50 x 200 = 10,000
Serial Dilutions
 The usefulness of dilutions becomes most apparent when small volumes of a material
are required in serological procedures. For example, if 0.01 ml of serum in a certain

5. test, instead of measuring out this small volume with a consequent sacrifice of
accuracy, I t would be more advantageous to dilute the serum 100 times. One ml of this
1/100 dilution would then contain 0.01 ml of the serum. Each ml of this dilution would
be equivalent to the required 0.01 ml of undiluted serum.
 Dilutions represent fractional amounts of a material and are generally expressed as the
ratio of one volume of material to the final volume of the dilution. Thus, a 1/10 dilution
of serum represents 1 volume of serum in 10 volumes of dilution (1 volume of serum + 9
volumes of diluent). Undiluted serum may be expressed as 1/1.
1 ml of serum + 1 ml of saline may be expressed as 1/2. Each milliliter of this dilution is
equivalent to 0.5 ml of undiluted serum
1 ml of serum + 2 ml of saline may be expressed as 1/3. Each milliliter of this dilution is
equivalent to 0.33 ml of undiluted serum
1 ml of serum + 99 ml of saline may be expressed as 1/100. Each milliliter of this dilution
is equivalent to 0.01 ml of undiluted serum
 From the above, one can see that dilution expressions are fractions written as ratios
where the numerator is the unit and the denominator is the dilution value.
 Division of the numerator by the denominator will give the amount of material per
milliliter of the dilution. For example, in a 1/25 dilution, 1/25 = 0.04. Therefore, each
milliliter of this dilution contains 1/25 or 0.04 ml of the original material.
 To convert a ratio into a dilution expression, divide both the numerator and
denominator by the value of the numerator. For example, in a mixture consisting of 4
ml of serum and 6 ml of saline,
4 ml of serum + 6 ml of saline = 10 ml of serum dilution
Ratio of serum dilution = 4/10
Dividing both the numerator and denominator by the numerator value (4),
(4/4) / (10/4) = 1/2.5 = serum dilution
 In the preparation of dilutions, any multiple of the constituent volumes may be used.
For example,
1/30 dilution = 1 ml serum + 29 ml saline
= 0.5 ml serum + 14.5 ml saline
= 2 ml serum + 58 ml saline
 Serial dilutions indicated that an identical volume of material is being transferred from
one vessel to another. The purpose of this procedure is to increase the dilutions of a
substance by certain increments. For example, in a twofold dilution, the dilution is
doubled each time (1/2, 1/4, 1/8, etc)

6. Sample Problems
Molar concentrations
• Problem A1: How you would make 500mL of a 0.05M NaCl solution
o You need the formula weight of NaCl, which is 58.453
o Next, you need to take 500mL and go to L, which is then .5 L
o Multiply .5L with .05M, to get the amount of moles, which is then .025 moles
o Then, take .025 and multiply by the FW (58.453), you get: 1.461325 g
o Therefore, to make 500mL of a 0.05M NaCl solution, you need to weigh out 1.461325g of NaCl
• Problem A2: How much arginine do you need to make a 0.01M solution of arg (MW = 174.20)?
o As stated earlier, that the molecular weight of NaCl is equal to 1L and 1M of solution, we can
deduce that the molecular weight is equal to 1M.
Therefore, 174.20=1M, we want .01 M
174.20 (.01) = 1.742g
• Problem A3: How much arginine do you need to make 100 ml of a 0.01M solution of arg (MW = 174.20)?
o The MW of arginine is 174.2
o 100ml is .1L
o .1 x .01 = .001 moles
o .001 moles x 174.20 = 0.1742g
Concentrations weight/volume (w/v)
• Problem B1: How much NaCl do you need to weigh out to make1L of a 0.9 % NaCl
solution?
 .009M NaCl solution
 58.453
 1L x .009 = .009 moles
 .009 x 58.453 = .526077
 You need to weigh out .526077g of NaCl to make a 0.9% solution of NaCl
Diluting a Solution to an Unspecified Volume in Just One Step
 Problem 0 : If you have a 1M arg stock and you want on inject a 100 ml of a 1mM
solution– What do you use to dilute the stock?
o M1V1=M2V2
o 1M (x) = (.1L)(.001M)
o X= .0001 mL of NaOH
 Problem 1. Dilute a solution that has an initial concentration of solute of 10% down to
2%.
o You need to dilute it five times because 10%/2%= 5
 Problem 2. Dilute a solution that has an initial concentration of 0.01% to 0.0001%.
o You need to dilute it 100 times because 0.01%/0.0001% = 100

7. Diluting a Solution to a Specified Volume in Just One Step
 Problem 3. You have a 0.01% solution. You want a 0.001% solution and you need 25 ml.
o You need to dilute it 100 times and divide that into the target volume
o 25mL/100 = .25 mL of solution with 99.75mL of diluent
 Problem 4. Prepare 50 ml of a 3% solution from a 4% solution.
o ¾ = .75
o 50mL/.75 = 66.67mL with 33.33mL of diluent
 Problem 5. Prepare 100 ml of 10% alcohol from 95% alcohol.
o C1 = 95%, C2 = 10%
V2 = 100ml
V1 = X
95% x X = (10%) x (100ml)
X = [(10%) x (100ml)]/95%
X = 10.5ml
o 10.5ml of 95% alcohol + 89.5ml (100 ml – 10.5 ml) of H2O = 100ml of 10%
alcohol.
 Problem 6. You need 45 ml of 50% alcohol. How can you prepare this from 70%
alcohol?
o C1= 70%, C2= 50%
V2=45 ml
V1=X
70% x X = (50%) x (45ml)
X= [(50%)x(45ml)]/(70%)
X=32.1ml
32.1ml of 70% alcohol with 12.9mL of H2O= 45ml of 50% alcohol
 Problem 7. What volume of a 20% dextrose broth solution should be used to prepare
250 ml of 1% dextrose broth?
o C1V1=C2V2
o 20% x = 250ml (1%)
o X= 250/20
o X=12.5ml of 20% dextrose broth solution should be used with 237.5 ml of H20
 Problem 8. Prepare a 45 ml of a 2% suspension of RBCs from a 5% suspension.
o C1= 5%, C2= 2%
o 5% x = 2% (45)
o X= 2% (45) / 5%
o X= 18 mL of 5% RBCs with 27mL of H2O
 Problem 9. What volume of a 0.02% solution can be prepared from 25 ml of a 0.1 %

8. solution?
o C1V1=C2V2
o 25ml (.1%)= .02 (x)
o 125mL can be prepared
 Problem 10. What percent concentration of alcohol is prepared when 10 ml of 95%
alcohol are diluted with H20 to make a final volume of 38 ml?
o C1V1=C2V2
o 10ml (95%)= 38ml (x)
o X= 25%
 Problem 11. What percent of dextrose is prepared when 3 ml of a 10% dextrose solution
are mixed with 12 ml of broth?
o C1V1=C2V2
o 3ml (10%) = C2(12ml)
o C2= 2.5%
Diluting a Solution to an Unspecified Volume in Several Steps
 Problem 12. You have a stock solution of protein containing 10 grams per ml. You want
a concentration of 2 mg/ml. How would you perform this dilution in just three steps?
o 10g/ml divided by .002g/ml is equal to 5,000
o This can be done in three steps by first taking 1ml of initial and diluting it to
50ml, the second by taking 1ml of that and diluting it to 100ml, and then in the
third step, taking 1ml of that and diluting it to 1,000
 Problem 13. You have a stock solution of 10 mg/ml of vitamins and want to obtain a
solution of 0.5 mg/ml. How would you perform this dilution in just three steps?
o 10mg/ml divided by 0.5 mg/ml provides a 20 fold dilution
o In just three steps, you can take 1ml of initial and diluting it to 2ml, take 1ml of
that and diluting it in 5ml, and taking 1ml of that and diluting it in 4ml
Diluting a Solution to a Specified Volume in Several Steps
 Problem 14. You want a 1:128 dilution of serum and you need 4 ml. How would you
perform this dilution in several steps?
o 128/4 =32
o 1ml of serum + 31 ml of diluent = 1/32 dilution
o 1 ml of 1/32 dilution + 3ml of diluent = 1/4
 Problem 15. You want a 1:3,000 dilution of serum and you need 2 ml. How would you
perform this dilution in several steps?
o 3,000/2= 1,500
o 1 ml of serum + 1,299ml of diluent = 1/1,500 dilution
o 1 ml of 1/1,500 dilution + 1 ml of diluent = 1/2
 Problem 16. How would you prepare 1 ml of a 1:5 dilution of sera?
o 5/1 = 5
o 1 ml of serum + 4ml of diluent = 1/5 dilution
 Problem 17. How would you prepare 8 m1 of a 1:20 dilution of sera?
o 20/8= 2.5
o 1 ml of serum + 1.5 ml of diluent = ½.5 dilution
o 1 ml of ½.5 dilution + 7ml of dilution = 1/8 dilution

9. Serial dilutions
 Problem 18. Based on the following dilutions, how many bacteria were present in the
original sample?
Going from the original sample to the final plate containing 201 colonies, each dilution is a
1/10 dilution, so going back per one would make it increase tenfold, therefore in the original
sample, there was 20, 100, 000
 Problem 19a. How many bacteria were present in the original sample (CFU/ml)?
Dilution Volume Plated Number of Colonies Counted
10-3
1.0 ml TNTC
10-4
1.0 ml 256
10-5
1.0 ml 21
10-6
1.0 ml 7
10-7
1.0 ml 0
10-8
1.0 ml 1
2, 560, 000 were in the original sample
 Problem 19b. How many bacteria were present in the original sample (CFU/ml)?
Dilution Volume Plated Number of Colonies Counted
10-2
0.1 ml 241
10-3
0.1 ml 35
10-4
0.1 ml 7
10-5
0.1 ml 0
241, 000 were in the original sample
Sample 9 ml 9 ml 9 ml9 ml 201 colonies counted
on plate
1 ml 1 ml 1 ml 1 ml
1 ml

10.  Problem 20. If 0.1 ml of a urine culture from a 10-6
dilution yielded 38 colonies, how
many bacteria were there per ml in the original sample?
3.8 x 10 8
were in the original sample
 Problem 21. How many bacteria were present in the following sample?
Dilution Volume Plated Number of Colonies Counted
10-7
0.1 ml 26
2.6x109
 Problem 22. How many bacteria were present in the following sample?
4.6 x 1013
 Problem 23. How many bacteria were present in the original sample?
I decided to attach my work on paper that I did for this question as it was really extensive:
Undiluted
sample
10
-2
dilution
46 colonies counted
on plate
0.1 ml 0.1 ml 0.1 ml
0.1 ml
10
-2
dilution
10
-2
dilution
Sample
1.0 ml
9.0 ml 4.5 ml
84 colonies counted
on plate
0.5 ml 3.0 ml 1.0 ml
0.01 ml
27 ml 99 ml 9 ml 18 ml
1.0 ml 2.0 ml

12. Supplemental Material
Smartphone/iPad Apps useful for lab math
• Biomath calculator (Promega)
• Daily Calcs (Life Technologies)
• Lab Tools (Biolegend)
• Converter
• Dilution
Volumes May Change, Dilution Factor Stays the Same
1:100 dilution
1 + 99
10 + 990
0.1 + 9.9
1:5 dilution
1 + 4
10 + 40
0.1 + 0.4
1:20 dilution
1 + 19
10 + 190
0.1 + 1.9
1:10 dilution
1 + 9
10 + 90
0.5 + 4.5
3 + 27
1:8 dilution
1 + 7
5 + 35
0.5 + 3.5
1:15 dilution
1 + 14
5 + 70
0.5 + 7.0
Metric and English Measurement Equivalents
The Metric System
The metric system comprises three basic units of measurement: distance measured in
meters, volume measured in liters, and mass measured in grams. In order to designate larger

13. and smaller measures, a system of prefixes based on multiples of ten is used in conjunction with
the basic unit of measurement. The most common prefixes are:
kilo = 103
= 1,000
centi = 10-2
= 0.01
milli = 10-3
= 0.001
micro = 10-6
= 0.000001
nano = 10-9
= 0.000000001
Scientific Notation
Microbiologists often have to deal with either very large or very small numbers, such as
5,550,000,000 or 0.00000082. The mere manipulation of these numbers is cumbersome. As a
result, it is more convenient to express such numbers in scientific notation (standard
exponential notation). Scientific notation is a set of rules involving a shorthand method for
writing these numbers and performing simple manipulations with them. Scientific notation
uses the fact that every number can be expressed as the product of two numbers - one of which
is a power of the number 10. Numbers greater than 1 can be expressed as follows:
1 = 100
100,000 = 105
10 = 101
1,000,000 = 106
100 = 102
10,000,000 = 107
1,000 = 103
100,000,000 = 108
10,000 = 104
1,000,000,000 = 109
In the above notations, the exponent to which the l0 is raised is equal to the number of
zeros following the 1. Numbers less than 1 can be expressed as follows:
0.1 = 10-1
0.000001 = 10-6
0.01 = 10-2
0.0000001 = 10-7
0.001 = 10-3
0.00000001 = 10-8
0.0001 = 10-4
0.000000001 = 10-9
0.00001 = 10-5
In the above notations, the number of the negative exponent to which 10 is raised is
equal to the number of digits to the right of the decimal point.
Numbers that are not an exact power of 10 can also be dealt with in scientific notation.
For example, a number such as 1234, which is greater than 1, can be expressed in the following
ways:
123.4 x 10 123.4 x 101
12.34 x 100 12.34 x 102
1.234 x 1,000 1.234 x 103
0.1234 x 10,000 0.1234 x 104
0.01234 x 100,000 0.01234 x 105
pH and pH Indicators
pH is a measure of hydrogen ion (H+
) activity. In dilute solutions, the H+
activity is
essentially equal to the concentration. In such instances,
pH = -log [H+
]

14. The pH scale ranges from 0 ([H+
] = 1.00
M) to 14 ([H+
] = 10-14
M).
A pH meter should be used for accurate pH determinations, observing the following
precautions:
1. Adjust the temperature of the buffer used for pH meter standardization to the same
temperature as the sample. Buffer pH changes with temperature; for example, the pH of
standard phosphate buffer is 6.98 at 0o
C, 6.88 at 20o
C, and 6.85 at 37C.
2. It is important to stir solutions while measuring their pH. If the sample is to be stirred
with a magnetic mixer, stir the calibrating buffer in the same way.
3. Be sure that the electrodes used with tris buffers are recommended for such use by the
manufacturer. This is necessary because some pH electrodes do not give accurate readings with
tris (hydroxymethyl) aminomethane buffers.
In instances where precision is not required, such as in the preparation of routine media,
the pH may be checked by the use of pH indicator solutions. By the proper selection, the pH can
be estimated within ± 0.2 pH units. Some common pH indicators and their useful pH ranges are
listed in the following table.