LAB MATH BASICS
There is very little that you have to “memorize” for Lab Math – it’s much more about figuring
out exactly what you need, and asking yourself if this makes sense. However, there are a few
things you DO need to just know – and from these you can figure out just about everything else.
Units. International units have special prefixes to indicate orders of magnitude, so that scientific
notation is not always necessary. A Kg is 1000 grams. We use a lot of small amounts in
biotech, so you should learn the following:
milli = 10-3, or 1/1000 of the unit in question (a mL is 1/1000 of a L or .001 L)
micro = 10-6 (abbr. with a Greek mu or and its usually OK to use the letter “u”)
nano = 10-9
pico = 10-12
To keep them straight, remember this:
-15
fempto = 10
My Uncle Norman Plays Fish Alot
atto
= 10-18
(compliments of Mary Hawley, BIT class of ’98)
Mole: 6.02 x 1023 atoms (or molecules)
The weight in grams of one mole of a substance is whatever 6.02 x 1023 molecules of it
weighs – the larger the molecule, the more one mole of it will weigh.
The formula weight gives you the number of grams that is in one mole of the substance.
(Generally formula weights are on the bottle, but you can calculate them from the
formula by summing the atomic weights of the components)
The abbreviation for mole is mol (not M, see below!)
Molarity – very important solution chemistry, it describes a concentration.
The molarity is the number of moles per liter. Its abbreviation is M
A 1 molar solution (1M) is 1 mol/L (note that 1 mmol/1mL is also 1 M!)
A 1 mM solution is 1 mmol/L (or 1 umol/mL)
Note: Do not confuse amount (mol) with concentration (M)
(Sometime you will see an N instead -- normality is used for acids and bases to express
the concentration of the ion that contributes to the pH. In a solution of H2SO4, two Hs
can dissociate from the SO4 and contribute H+ to the solution (raising the H+
concentration, thus dropping the pH), so a solution of 1 M H2SO4 (that many grams per
L) could also be expressed as 2N H2SO4
%
-- means parts of a hundred” (per cent !)
1% means one part out of one hundred: 1 g/100g (w/w)
Since, in solutions our solvent is mostly water (with a density of 1 g/mL), we often
express the solution as % w/v, or g/100 mL (i.e. 20% sucrose = 20 g sucrose/100
ml (final volume)
If the solute comes in liquid form, it may also be v/v (20 mL + 80 mL H2O, 100 mL final
vol)
Parts – sometimes solutions are given in parts. The example above 20% glycerol, could also be
expressed as 20 parts glycerol + 80 parts water, and would be written as 1:4
glycerol:water . You will sometimes see instructions for diluting something 1/5
(meaning total). That is 1:4.
HOW TO SOLVE LAB MATH PROBLEMS
Most of the problems that you have to solve in the lab relate to the preparation of specific
solutions. I solve maybe 95% of practical problems in the lab using either Dimensional Analysis
or the C1V1 equation. Learn how to use both of these, and apply them correctly.
Two things, very important, never to be forgotten:
1) when you have solved the problem, think about the answer .. DOES IT MAKE SENSE?
2) If it does, great – still check it – try to solve it another way, or backwards
Dimensional Analysis
This is a very useful, very easy technique. Look at what you have , and where you want to be,
and what you need to know to get there, and set up the necessary conversion factors to get you
there.
Consider: How many meters in a mile?
What you want to find out is the # meters = 1 mile.
The only conversion between metric and English distance that I have committed to
memory is 2.54 cm/in. I do also know that there are 5280 ft in a mile, and 12 inches in a
foot, and a cm is 1/100 of a meter.
Start with the original question , and then arrange the conversion factors so that the units cancel
approppriately:
# meters = 1 mile
x
5280 ft x 12 inch x
mile
1 ft
2.54 cm
inch
x
meter
100 cm
=
Keep your wits about you when you do this, especially on the “easy” metric conversions
(sometimes people slip and put 100 meter per cm, which makes the answer hugely wrong)
Once you cancel out the units, you will be left with meters, and then just crank the math.
It is just as easy with solution chemistry. Consider:
How many grams of NaCl do you need in order to make 200 mL a 0.5 M solution.? Or:
# grams NaCl = 200 mL of 0.5 M
First recognize that most of the time, when you set up a problem and say “of” it is a signal to
multiply. Recast the M as mol/L and you will see that if you do multiply, you will have almost
cancelable volume terms (just change the 200 mL to 0.2 L, or 1 L to 1000 mL.
# grams NaCl = 200 mL x 0.5 mol
1000 mL
Now you need to get find the number of grams in that many moles. Of course, the way to do
that is with ..... the formula weight. (For NaCl, the fw is 58.4, so there are 58.4 g in one mole)
Arrange this so that moles cancel and leave you with grams:
# grams NaCl = 200 mL x 0.5 mol x
1000 mL
58.4 g =
mole
g
That’s all, now just
do the arithmetic !
The C1V1 equation ... or actually
C1V1 = CfVf
(f = final)
This is used for making dilutions. We often make up concentrated versions of solutions and then
dilute them as needed because it is easier and more accurate than weighing out small amounts of
dry reagents all the time. This works becuase concentration x volume = amount. A solution
that is 4 M is 4 mol /L, if you multiply that by a volume in L, you will be left with an amount.
If you have 10 mL of 4 M NaCl, you have 4 mol/L x .01 L or .04 moles
To set up a C1V1 equation, consider the question: How much (volume) of a particular
concentration do I need to get the volume and concentration that I actually want (what is the
amount of the first solution that I need to get into my dilution?)
You want 300 mL of a 0.14 M solution of NaCl, and you have a stock solution of 4 M NaCl.
Say:
How many mL of 4 M NaCl do I need to give me 300 mL of 0.14 M?
And write :
(x mL) (4 M) = (300 mL) (0.14M)
(“of” tells you to multiply)
(Yes, I know, it’s really “V1C1” this way; that is my peculiar way of doing it because it makes sense to me. It doesn’t matter!)
Solve it:
Since all the units match, you can now just ignore them and solve, x will be in mL
4 x = 300 x 0.14; x = 42/4 = 10.5 mL. Add this to 289.5 mL and you have your
0.14 M solution.
Once you have solved the equation, you should do two things.
1. Ask yourself if it makes sense. If for example, you got a volume larger than the original
volume you had, perhaps you set things up wrong! Check!
2. Plug the solution back into the equation, and solve for one of the other terms.
(10.5 mL) (4 M) = 300 mL ( x M)
42 = 300x
x = 42/300 = .014
If your units do not match, make them: How many mL of 0.2 M solution do I need to make 5 L
of a 20 mM solution?
Set it up:
(x mL) (0.2 M) = (5 L )(20 mM)
But you cannot do the math yet, becuase the
units are not the same, so fix that:
(x L) (200 mM) = (5 L )(20 mM)
now they match, so do the math
Note: with this one, you probably can easily see that you did not really have to do a
C1V1 on paper – 20 mM is ten-times less concentrated than 0.2 M, so you can just do a 1/10 dilution – for 5 L that
would mean 0.5 L and then 4.5 L of water to bring it to 5 L. You should get accustomed to doing that sort of thing
in your head but until you are good at it, use the C 1V1 equation.
The biggest mistake people make with the C1V1 equation is when they try to use an amount
instead of a concentration. THIS DOES NOT WORK!
A couple of special applications for this useful C1V1 tool:
One thing the C1V1 is very helpful for is when you have to adjust the concentration of a solute
that is contributed from more than one source. Here is an example:
Lets say you are doing a restriction enzyme digest in two steps.
In your first step, you have a 20 uL reaction that has (among other things) 50 mM NaCl. After
an hour, you want to add a different enzyme and increase the volume to 40 uL, and also increase
the salt to 100 mM (because the 2nd enzyme works better at the higher salt). Clearly, if you just
add 20 uL of enzyme and water to the first reaction, you will dilute the salt to 25 mM, so you
will have to add some concentrated salt as well. But how much? Lets say the stock is 1 M.
The question: How much 1 M NaCl do I add to 20 uL of 50 mM NaCl so that when I make the
final 40 uL it will be 100 mM in salt.
The equation: (x) (1 M) + (20 uL) (50 mM) = 40 uL (100 mM)
you have a term for each salt source
Fix the units: (x) (1000 mM) + (20 uL) (50 mM) = 40 uL (100 mM)
OK, now they are all the same, so that means that x will be in uL, and you can now ignore the
other units and crank the numbers:
1000 x + 20 x 50 = 40 x 100
1000 x + 1000 = 4000
1000 x = 3000,
so x = 3 uL
*************
Another variation: Most of the time when you make a dilution, you are able to figure the stock
amount to add, and then adjust the water to give you the correct final volume. Sometimes you
will be adjusting a solution and wish to get the correct final concentration, but you cannot add
water.
Example: How much 8M urea will you add to 20 mL of a solution such that the final
concentration is 1 M? .
You cannot set it up as usual (x mL) (8 M ) = (20 mL ) (2M)
Why is this wrong? Because if you add some of the urea to 20 mL it won’t be 20 mL anymore!
It will be a new volume, greater than 20 mL by exactly the amount you added!
So.. take the amount to be added into consideration for the final volume:
(x mL) (8 M) = (20 + x)ml (2 M)
I am sure you all remember the distributive property from algebra:
So the above becomes
a ( b + c ) = ab + ac
8 x = 40 + 2 x
6 x = 40, so x = 40/6 = 6.7 mL
to see if that’s correct, set it up and solve for something else:
(6.7 mL)(8 M) = (26.7 mL)( x M) ; 53.6 = 26.7 x; x = 2 M as we had hoped