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• #14712
kitkat11
Participant

I’m confused with working out the final concentrations of these.

I added these to 6 separate flasks each containing 40 ml of ecoli in a glycerol medium:

a. 2.0 ml (10 mM) IPTG
b. 0.6 ml (1%) lactose
c. 0.6 ml (1%) lactose and 0.5 ml (8%) glucose
d. 0.6 ml (1%) lactose and 1.0 ml of 5-FU
e. 0.6 ml (1%) lactose and 1.0 ml of Chloramphenicol

ive already thought of doing these methods but they dont feel like they are right :

E.g If I am adding 0.6 mml of 1% lactose to a 40 ml culture of E.coli bacteria

Method 1: 1% of 0.6 ml is 0.06/40 = 0.0015 mM which would be 1.5 uM (i hate concentrations if you noticed? )

Method 2: c=n/v …. molecular weight of lactose is 342/(40 ml + 0.6 ml) = 8000 something which i would thing is wrong? unless it is 8423 uM and then it becomes 8.4 mM ?

Method 3: im running out of ides –> 0.6/0.01 = 60 mM?

• #104128
JackBean
Participant

1) you have 1% solution, so you canNOT get 0.0015 mM just from V1/V2!!!
2) use simply c1V1 = c2V2, you just need to keep the units of concentration (c1 and c2) the same, i.e. either % or M, but do not mix it!

• #104129
kitkat11
Participant

yes i noticed that mistake but i was so confused let people see where i was going wrong … umm can u explain that formula u gave though in a bit more detail (or e.g.) because i’ve never seen it? or heard of it? thanks!

• #104131
canalon
Participant

c1v1=c2v2
When calculating a concentration if you multiply the volume of original solution used (v1) by its concentration (c1) it is equal to the final concentartion (c2) time the final volume (v2)
i.e:
if you mix 1ml(v1) of a 1M solution (c1) in 100ml(v2) the final concentartion is c2=(c1xv1)/v2 or c2=0.01M

Beware your units if C1 is in %, c2 will be in % and if c1 is in Moles/L so will be c2

Knowing the Molecular weight of your compounds is necessary to convert between Moles and %

• #104135
kitkat11
Participant

ok so i think i get that formula … but how does moles then relate to percentage. so like for lactose it has a molecular weight of 132 (roughly) and the only formulas im thinking of include n=m/M and c=n/v …. but i don’t know how that relates to percentages?

• #104137
canalon
Participant

percentages are usually given v/v(volume to volume ratio) or w/v(or weight to volume ratio).

-The first is used for liquids (water, ethanol, detergent like triton…) and is the ratio of the substance to the solvent. i.e. a 13% v/v ethanol solution (aka wine) contains 13ml of ethanol for a final volume of 100ml

-The later is used for solids (salts, sugar,…) and represent the weight of solute(in grams) in a 100ml final solution. i.e. a 10% solution of arabinose contains 10g of arabinose in a 100ml volume. This is the kind of thing you are dealing with.

Now since you can calculate the number of moles in your solution since you know the amount of your solute and its molecular weight, you will have no problem calculating your concentrations (remember 100ml =/= 1L)

• #104139
kitkat11
Participant

ok so i think i understand … im just want to check.

so if i do the 0.6 ml of 1% lactose and put that in 40 ml of ecoli culture…. then:

i take 0.001 / 132 = 7.57 x 10^-6
and then i times that by 0.006 = 4.5 x 10^-8
and divide that by 0.0406 = 1.1 x 10^-6 which would be 1.1 uM?

sorry … im just having trouble now applying what you’ve said.

thanks so much though!

• #104143
canalon
Participant

c1 x v1=c2 x v2
with
c1=1%
v1=0.6ml

c2= you will calculate it
v2=40.6 ml

You have a concentration in % then you just convert that percentage in molarity by using the molecular weight of lactose (342.3g/mol)
1% = 1g/100ml= 10 g/L =10g/(MW of lactose in g/mol) mol/L

and your result is wrong by many order of magnitude.

• #104150
JackBean
Participant

I will explain the formula a little bit more.

as you wrote, you have n = c.V
so, if you dilute your solution, you add only pure solvent and thus the amount of your solute is not changing, thus if n1 is the amount before dilution and n2 after dilution, it is n1 = n2 and thus you can write
c1 . V1 = c2 . V2

this is for the moles and mol/dm3, but the same applies for the %, because if it is true, that n1 = n2, than must be true m1 = m2 (if we’re talking about the same substance and thus the same Mr;) and since
m = c . V
because in this case c is in [%] (w/v), that is, as canalon wrote, how many grams you have per 100 ml of solution
so you get again
c1 . V1 = c2 .V2
(but remember, that both c are in %!)

• #104246
kedw4123
Participant

so, can someone please give an example using B

Thanks! ๐

• #104247
kedw4123
Participant

so is this right for B:

c1 x v1 = c2 x v2

c1=1%
v1=0.6
c2=what we find
v2=40.6

MW of lactose =342.3g/mol
1%= 1g/100ml
= 10g/L
=10g/342.3g/mol
=0.029 mol/L

c2=c1 x v1/v2
= 0.029 M x 0.6 ml / 40.6 ml
= 0.0043 mol/L
=0.43 ยตM

Is that correct? Should the concentration be expressed like that?

Also, I’m having trouble on D

0.6 ml (1%) lactose and 1.0 ml 5-fluorouracil (FU)

• #104248
kedw4123
Participant

btw…following c2=c1 x v1/v2, my calculations are out,

I meant 0.029 M x 0.6 ml /40.6 ml
= 0.00043 mol/L
=0.43 mM

• #104252
JackBean
Participant

why are you calculating with the molecular weigth? Just use the percentage ๐

c2 [%] = (1%*0.6ml)/40.6ml

• #104257
kedw4123
Participant

If I do that, my answer is 0.015….but what are my units :S

• #104259
JackBean
Participant
quote JackBean:

c2 [%] = (1%*0.6ml)/40.6ml
• #104265
kedw4123
Participant

:/

• #104271
JackBean
Participant

let’s say, you will add 2 ml of 5% solution to 98 ml, OK?

So it will be
2 ml * 5% = (98+2) ml * c2 [%]
c2 = 0.1%