| 1 grams (g) | 1 M | 1 L |
| 1.000 milligrams (mg) | 1.000 mM | 1.000 mL |
| 1.000.000 micrograms (µg) | 1.000.000 µM | 1.000.000 µL |
| 1.000.000.000 nanograms (ng) | 1.000.000.000 nM | 1.000.000.000 nL |
https://pubs.acs.org/doi/10.1021/ed5007376
Molarity (M) = moles of solute / volume of solution (in liters)
1 M of NaCl in your solutions is the same as saying 1 mol of NaCl per liter, e.g. 6 x 10²³ NaCl molecules in your solution (this comes from Avogadros constant).
Normality and Equivalent concentration
In chemistry, normality (N) is sometimes used to specify the equivalent concentration of the compound when considering a specific process, which might be:
- in acid-base chemistry the number of deprotonatable site, 1M of the monoprotic hydrochloric acid (HCl) is identical to 1 N, but 1 M of the diprotic sulfuric acid (H2SO4) is identical to 2 N, because (in theory) the acid could lose two protons. The more unstable carbonic acid (H2CO3) in fact easily gives off two protons, where this nomenclature is handy.
- similarily in redox reaction where the number of electrons are identical with the normality, e.g. 1 M of benzylviologen is 2 N, because of the two redox transitions at -350 mV and -540 mV, respectively
- but also in precipitation reaction, with the number of ions to precitate
Converting mass concentrations into molar concentration
0.5 mg/L of NaCl (M=58.443 g/mol) –> 0.0005 g/L
1.) Produce molar solutions from solid chemicals
Question: You want to produce 1 liter of a 0.5 M (or 500 mM) NaCl solution. So you have NaCl powder and want to know how much NaCl do I have to add in order to get the molarity in my solution?
What you need?:
- final volume V ( here 1 L)
- final concentration c (here 0.5 M)
- molecular weight M of the solute (here for NaCl, its 58.44 g/mol)
Equation:
m = M ⋅ c ⋅ V
mol and liters cancel and the result will be the amount of solute in gramm
For the question above that would be:
m = M ⋅ c ⋅ V m = 58.44 g/mol ⋅ 0.5 mol/L ⋅ 1 L m = 29.22 g
That means we need 29.22 g of NaCl in order to produce one liter of a 0.5 M
General advice: First try to dissolve the whole amount of NaCl in 800 mL of water and add the rest in order to get to 1 L of the final volume.
2.) Produce procentual solutions from solid chemicals
Question: You want to produce 1 liter of a 5% Agarose solution. So you have Agarose powder and want to know how much Agarose do I have to add in order to get the molarity in my solution?
What you need?:
- final volume V ( here 1 L)
- final concentration in % (here 5% that is 0.05, if 1 is 100%)
Equation:
m = % ⋅ V
For the example this means, 0.05 times 1000 mL equals to 50 mL (= 50 g). So you have to weight in 50 g and dissolve them in 1000 mL of water.
3.) Produce molar solutions from fluid chemicals
Producing molar solutions from fluid chemicals generally just involved dilution.
Here a simple equation applies:
c1 ⋅ V1 = c2 ⋅ V2
Example: You have a 3 M NaOH and need 500 mL of a 0.5 M NaOH solution. How much mL of 3 M NaOH you need to get this solution?
What you need?
- initial concentration of the reagent (c1 = 3 M)
- final concentration of the reagent (c2 = 0.5 M)
- final volume of the reagent (V2 = 500 mL)
By applying the formular above, you get:
3 M ⋅ x = 0.5 M ⋅ 500 mL x = 83 mL
Same can be applied to percentual solutions. Here a small example:
You have a 70% ethanol stock and need 300 mL of a 20% ethanol solution.
70% ⋅ x = 20 % ⋅ 300 mL x = 85.7 mL
4.) Produce molar solution from fluid percentual chemicals
Example: You should produce 500 mL of 3 M acetic acid from 100% acetic acid stock solution.
What you need?
- molecular weight M of the chemical (here, 60.05 g/mol)
- density p of the chemical (here, 1.05 kg/L or g/mL)
- percentage of the initial solution (here, 100% i.e. 1)
- final concentration of the reagent (c2 = 3 M)
- final volume of the reagent (V2 = 500 mL)
Short notice about the density:
g/cm³ = g/mL = 0.001/0.001 kg/L = kg/L
Equation:
c1 = (p / M) ⋅ % AND c1 ⋅ V1 = c2 ⋅ V2
First you need to calculate the molarity of the stock solution (c1), by using the molecular weight (M), the density (p) and the percentage (%).
c1 = (1.05 kg/L / 60.05 g/mol) ⋅ 1 c1 = (1050 g/L / 60.05 g/mol) ⋅ 1 c1 = 17.5 M
So, a 100% acetic acid as a molarity of 17.5 M.
To calculate the amount needed use the same example as above.
17.5 M ⋅ x = 3 M ⋅ 500 mL x = 85.7 mL
Example 2: You have a 37% hydrochloric acid stock solution and would like to produce 100 mL of a 3 M HCl-solution.
p (@c=37%) ~ 1.184 kg/L = 1184 g/L
M = 36.46 g/mol
c1 = (1184 g/L / 36.46 g/mol) ⋅ 0.37 c1 = 12.015 M 12.015 M ⋅ x = 3 M ⋅ 100 mL x = 24.97 mL
About 25% HCl to 75% water.
Example 3: What is the molar concentration of glycerol, if dilute 5 mL of a 100% glycerol solution in one liter of media.
Molar mass: M = 92.09 g/mol
Density p = 1.261 g/mL = 1261 g/L
c = p/M * % = 1261 g/L / 92.09 g/mol * 1 (1= 100%) = 13.69 M
A 100% Glycerol stock solution is 13.69 M.
v1 * c1 = v2 * c2
5 mL * 13.69 M = 1000 mL * c2
c2 = 0.06846 M = 68.46 mM
Your final concentration of glycerol is 68.46 mM.
Example 4: Production of 5 mL of a 10% APS solution from powder:
Ammonium persulfate (APS) is an important compounds in SDS gels.
M = 228.18 g/mol
d = 1.98 g/cm-3 = 1.98 g/mL = 1980 g/L
10 % solution: 1980 g/L / 228.18 g/mol * 0.1 = 0.867 M
We need 0.99 g to produce a 5 mL of a 10% APS solution.
5.) Produce percentual solution from fluid molar chemicals
You have a 5 M NaCl stock solution and want to have 800 mL of a 20 % NaCl solution.
What you need:
- molar weight of the stock solution chemical (here, 58.44 g/mol)
- initial concentration (5 M)
- final volume (800 mL)
- final concentration (20%)
m = M ⋅ c ⋅ V V = m / (M ⋅ c)
20% of 800 mL are 160 mL (~160 g), This is m.
m = M ⋅ c ⋅ V V = 160 g / (58.44 g/mol ⋅ 5 mol /L) V = 0.548 mL
6.) Calculating the molarity of percentual solution
Normal saline is a commonly used compound in medicine to treat wounds, dry eyes and even intraveneously to avoid dehydration. It is a 0.9% NaCl solution, which is important because the exact same concentration can be found in the human body. Therefore, the treatment doesn’t lead to any osmotic effects.
But what does 0.9% NaCl meaning in molar concentration?
0.9% NaCl actually denotes a mass concentration, meaning “0.9% m/v” (mass/volume) NaCl. This is going to say, that a saline solution is produced by dissolving 0.9 g of solute (the NaCl salt) in a final volume of 100 mL. Scaling this up, you can use 9.0 g per liter to produce a saline solution at home (this won’t be sterile though).
Since we know, that the molar mass of NaCl is 58.44 g/mol, we can easily calculate the molarity:
m = M ⋅ c ⋅ V 9 g = 58.44 g/mol ⋅ c ⋅ 1 L c = 9 g / (58.44 g/mol ⋅ 1 L) c = 0,154 M
Normal saline therefore has a molarity of 0.154 M (= mol / L).
The thing about the use of percent(%) in biology
In contrast to the “0.9% m/v” (mass/volume) equating to % w/v (weight/volume) of grams/milliliters (g/mL) as seen above for the saline, we can also have “mass percent solutions” (m/m = m% = mass solute/mass total solution after mixing) ), or “volume percent solutions” (v/v = v% = volume solute per volume of total solution after mixing).
Since in biology, we mostly deal with the most abundant solvent, i.e. water with an approximate density of 1 g/cm3 (1 g/mL), the expression 0.9% m/v is effectively identical to 0.9% m/m, however not scientifically accurate. Therefore always describe what you mean.
To make matters worse, water solutions containing relatively small quantities of dissolved solute, this ratio of grams solute per mL solution and be found multiplied by 100, resulting a “mass/volume percentage” or “1 m/v %.” (= 1 g per 100mL).
Calculating with proteins
Since proteins are such gigantic beasts masses, these are often given in Dalton (Da), kiloDalton (kDa) or MetaDalton (MDa). A dalton (abbreviated with Da or u) is the unified atomic mass unit and defined by 1/12 of the mass of a carbon-12 atom.
This maybe sounds familiar, as Avogadro’s constant, the number of atoms in one mol of any compound was derived from that.
See Wikipedia
The value of the Avogadro constant was chosen so that the mass of one mole of a chemical compound, in grams, is numerically equal (for all practical purposes) to the average mass of one molecule of the compound, in daltons (universal atomic mass units); one dalton being 1/12 of the mass of one carbon-12 atom, which is approximately the mass of one nucleon (proton or neutron).
1 Da = 1 g/mol
1 mole of a 50,000 Da protein is 50,000 g.
Generally, as a rule of thumb one aminoacids weights approximate 100 Da. Therefore a protein with approximately 500 aminoacids weights approximately 50,000 Da or 50 kDa.
Let’s say, you have a 3.94 µg/µL protein concentration of a 149 kDa protein and could like to know the molar concentration, use:
Suppose, that 35ng of 47kDa protein were diluted in 86µl of H2O and you need to know the concentration in mol/L (M).
First, use the following equation:
m[g] = Q[mol] x Mw[Da] Q[mol] = m[g] / Mw[Da] Q[mol] = 3.5 x 10-8 g / 47000 Da Q[mol] = 7.44 x 10-13 mol = 0.00074468 nmol (or 7.44 x 10-4 nmol)
m [g] – weight of protein [gram] : here 35 ng
Mw [Da] – molecular weight of protein [Dalton] : here 47 kDa
Q [mol] – quantity of protein [mole]; unknown
Now we only need to set this quantity in relation to the volumn
c[M] x V[L] = Q[mol] c[M] = Q[mol] / V[L] c[M] = 7.44 x 10-13 mol / 0.00086 L c[M] = 8.66e-9 M
Cheat Sheets for Computational Biochemistry 