Questions for “Making solutions…”
1. HEPES (a commonly used buffer) has a molecular weight of 238.3. How many grams
of HEPES do you need to weigh out to have one mole?
2. I have weighed out 327.4 g of HEPES, how many moles is this?
3. A bottle is labeled 1 M MgCl2 (MW 95.21): how many moles of the salt is there in one
liter of the solution?
4. The same bottle in question three: How many millimoles of salt is there in one
milliliter of the solution?
5. The same bottle in question three; how many grams of the salt is there in one liter of
solution?
6. We can use this shortcut to finding out how many grams of a chemical (with
molecular weight = MW) is needed for a solution with a certain molarity M:
Grams = (volume in liters) x (molecular wt, g/mole) x (concentration in molarity)
Remember: you add the grams of chemical to about ¾ s of the final volume (the chemical
takes up volume), then top it off to final volume after all chemicals are added.
Show how the units in the equation above cancel (to the right of the equal sign, put each
term as a fraction and show how the same units in the numerator will cancel a
denominator).
7. Use the formula in question 6: I want to make up a solution of glycine at 15 mM, with
a final volume of 150 ml (glycine MW is 75).
8. How many grams are needed: 100 ml final volume, 5 M stock solution of CaCl2 (MW
of 111).
9. A very large protein has a molecular weight of 400,000. You want to make a solution
of the protein at 1% (most proteins will start precipitating out of solution at about 1-2%;
so this protein solution is very very concentrated). How many grams of protein is needed
for 10 ml?
10. Remember the 2 equations for dilutions:
DILUTION FACTOR = (STOCK CONC.)/(FINAL DILUTE CONCEN.)
VOL. OF STOCK TO BE ADDED = FINAL VOLUME OF THE DILUTE SOLN
DILUTION FACTOR
Remember to keep units constant; don’t use liters in one place, and ml in another.
You want 25 ml of 0.5 M CaCl2, 1 M MgS04, and have the following two stock
solutions: 5M CaCl2, and 2.5 M MgS04. What is the volume of each stock solution to
add together to form the desired solution? After adding the appropriate volume of the
two stock solutions, you would top off to 25 ml.
11. I want to find out the concentration of DNA: so I set the spectrophotometer to 260
nm (not 280, as with proteins) and obtain an optical density (OD, or absorbance) of 0.5.
the extinction coefficient for double stranded DNA is 0.02. This extinction coefficient
can be used to find the concentration of DNA not in % (as we did with insulin), but gives
the concentration in mM. What is the concentration of DNA in mM? Write out the
equation: OD = e (concentration in mM) and substitute in the numbers.
12. I want a solution of 1 mM Valine (10 ml; MW 117). Show me the equation and
calculations – how many grams of valine do I add? Can I measure this amount of valine
accurately? Most scales go only to 0.1 mg and this means your error will be great when
you have to measure something out to 0.1 mg. It can be done but it is typically not
accurate.
Selected Answers: do not look at these until you have tried to answer the questions….
1. HEPES (a commonly used buffer) has a molecular weight of 238.3. How many grams
of HEPES do you need to weigh out to have one mole?
238.2 g
2. I have weighed out 327.4 g of HEPES, how many moles is this?
327.4/238.3 = 1.37 moles of HEPES.
3. A bottle is labeled 1 M MgCl2 (MW 95.21): how many moles of the salt is there in
one liter of the solution? 1 mole
4. The same bottle in question three: How many millimoles of salt is there in one
milliliter of the solution?
1 mole per liter = 1 millimole per milliliter
5. The same bottle in question three; how many grams of the salt is there in one liter of
solution?
95.21 g
6. We can use this shortcut to finding out how many grams of a chemical (with
molecular weight = MW) is needed for a solution with a certain molarity M:
Grams = (volume in liters) x (molecular wt, g/mole) x (concentration in molarity)
Remember: you add the grams of chemical to about ¾ s of the final volume (the chemical
takes up volume), then top it off to final volume after all chemicals are added.
Show how the units in the equation above cancel (to the right of the equal sign, put each
term as a fraction and show how the same units in the numerator will cancel a
denominator).
Liters x (g/mole) x (moles/liter) =
Liters x g x moles
Moles x liters
so moles and liters cancel—leaving only grams
7. Use the formula in question 6: I want to make up a solution of glycine at 15 mM, with
a final volume of 150 ml (glycine MW is 75). How many g do I add to 150 ml?
(no answer; use equation above)
8. How many grams are needed: 100 ml final volume, 5 M stock solution of CaCl2 (MW
of 111).
5 x 111 x 0.1 L = 55.5 g
9. A very large protein has a molecular weight of 400,000. You want to make a solution
of the protein at 1% (most proteins will start precipitating out of solution at about 1-2%;
so this protein solution is very very concentrated). How many grams of protein is needed
for 10 ml?
(no answer)
10. Remember the 2 equations for dilutions:
DILUTION FACTOR = (STOCK CONC.)/(FINAL DILUTE CONCEN.)
VOL. OF STOCK TO BE ADDED = FINAL VOLUME OF THE DILUTE SOLN
DILUTION FACTOR
Remember to keep units constant; don’t use liters in one place, and ml in another.
You want 25 ml of 0.5 M CaCl2, 1 M MgS04, and have the following two stock
solutions: 5M CaCl2, and 2.5 M MgS04. What is the volume of each stock solution to
add together to form the desired solution? After adding the appropriate volume of the
two stock solutions, you would top off to 25 ml.
0.5/5 x 25 ml = 2.5 ml of CaCl2 stock
½.5 x 25 ml = 10 ml of MgS04 stock
11. I want to find out the concentration of DNA: so I set the spectrophotometer to 260
nm (not 280, as with proteins) and obtain an optical density (OD, or absorbance) of 0.5.
the extinction coefficient for double stranded DNA is 0.02. This extinction coefficient
can be used to find the concentration of DNA not in % (as we did with insulin), but gives
the concentration in mM. What is the concentration of DNA in mM? Write out the
equation: OD = e (concentration in mM) and substitute in the numbers.
OD= 0.5 = (0.02) (x mM)
x= 25 mM
12. I want a solution of 1 mM Valine (10 ml; MW 117). Show me the equation and
calculations – how many grams of valine do I add? Can I measure this amount of valine
accurately? Most scales go only to 0.1 mg and this means your error will be great when
you have to measure something out to 0.1 mg. It can be done but it is typically not
accurate.
1.17 mg; your weighing out would probably not be accurate as you might weigh out 1.27,
or 1.07 as your scales might not read accurately ± 0.1 mg.
For a more accurate solution, you would want to make up a larger volume- say 100 ml.
Then you would have to weigh out 23.40 mg and any error ( plus or minus 0.1 mg) would
be reduced. Check this: so you could have 23.5 mg instead of correct 23.4 mg; this is
only 23.5/23.4 = 0.4% error (make the division, remember to multiply by 100 to get a
percentage). If you weigh out 1.27 instead of 1.17, this is 1.27/1.17 = 8.5% error an
increase from 0.4 to 8.5. This is a 21 fold increase in error: 8.5/0.4= 21 fold increase in
error.