Compound C (MW = 550) is provided at a concentration of 25g/100mL. 1mL is taken and compound D is to be added (MW = 1025, concentration = 75mM) to give a final ratio of compound C:D of 4:1. What volume of compound D is to be added?
Jenny has posted this problem before and I responded by asking if the 4:1 ratio is by grams or by molarity. It makes a difference. Don't guess at the answer; find out which ration is needed and repost.
the ratio is moles
To solve this problem, we need to calculate the volume of compound D that needs to be added to achieve a final ratio of 4:1 between compound C and compound D.
First, let's determine the number of moles of compound C in 1 mL (the volume taken):
1 mL of compound C is equivalent to 1/100 mL, or 0.01 mL.
To convert mL to grams, we can use the concentration given:
25 g/100 mL = 0.25 g/mL
0.01 mL * 0.25 g/mL = 0.0025 g of compound C
Next, let's calculate the number of moles of compound C:
moles of C = mass / molecular weight
moles of C = 0.0025 g / 550 g/mol = 4.5 x 10^-6 mol
Now, let's calculate the number of moles of compound D needed to achieve the desired ratio:
moles of D:moles of C = 1:4 (final ratio)
moles of D = 4 * moles of C
moles of D = 4 * 4.5 x 10^-6 mol = 1.8 x 10^-5 mol
Finally, we can calculate the volume of compound D using its concentration:
moles of D = concentration * volume
volume = moles of D / concentration
volume = (1.8 x 10^-5 mol) / (75 mmol/mL) = 2.4 x 10^-7 mL
Therefore, approximately 2.4 x 10^-7 mL of compound D needs to be added to achieve the desired ratio of 4:1 between compound C and compound D.