Calculate mass of compound given the mole % of total solution

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SUMMARY

The discussion focuses on calculating the mass of compound A required to achieve a 10% molar fraction in a solution containing compounds A and B. Given the mass of compound A (15 mg, MM=200 g/mol) and compound B (6 mg, MM=700 g/mol), the total concentration of the solution is maintained at 20 mg/mL. Participants emphasize the importance of converting mass concentrations to molar concentrations to simplify calculations, allowing for a straightforward determination of the necessary amount of compound A to meet the specified molar fraction.

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Homework Statement


Known: [/B]
Mass of compound A: 15mg (weighed in excess), MM=200g/mol
Mass of compound B: 6mg, MM=700g/mol
Concentration of each compound listed above(Concentration of solution of compound A or solution of Compound B): 20mg/L
Total concentration of the solution: 20mg/mL
Molar percent(percent of total number of moles in solution) of compound A: 20%
Problem:
I will list everything that is given to me by the problem as (Given:)

So I measured a mass of compound A, let's say 15mg. And I measure a mass of compound B, let's say 6mg. (Given: concentration of each compound is 20mg/mL). Compound A is measured in excess. So using the mass, and the given concentration, I calculate the volume required for each compound(using the given concentration of 20mg/mL) and mix each compound into their required volumes to create 2 solutions. However, (Given: 10% molar percent[that is to say 10% of the total number of moles in the final solution]) of Compound A, what is the mass and volume of Compound A that I need to add to Compound B in order to get that mole percent? However, I cannot change the total concentration of the solution(Given: 20mg/mL)

2. Homework Equations

C=nv etc simple chem equations

The Attempt at a Solution


Moles A: 15mg/1000/200g/mol=7.5*10^(-5)
Moles B: 6mg/1000/700g/mol=8.57*10^(-7)

Total Moles in sol'n*(1-0.2)=Moles Compound B
Total Moles in sol'n=1.07*10^(-6) moles
Moles of Compound A needed=1.07*10^(-6) moles*0.2=2.14*10^(-7)
Mass of Compound A=2.14*10^(-7)*200g/mol*1000=0.043mg
Volume of Compound A to add=0.043mg/(20mg/mL)=0.00214mL
^This answer is wrong however.
 
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It is still not clear what you are trying to do and what you are calculating.
 
Borek said:
It is still not clear what you are trying to do and what you are calculating.
Hi, I have two powders, compound A and compound B. I need to make 2 solutions each containing the two powders, so I use the concentration of each solution(20mg/mL) to determine how much of the solvent I need to add. Now I have 2 solutions with powder A and powder B dissolved in them. I need to add a certain amount of the compound A solution(since it was made in excess) to compound B solution so that the entire final solution has 10% molar mass of compound A. I am trying to calculate the mass Compound A and consequently the volume (as made in my solution containing compound A) that I have to add to the solution containing compound B . However, I must keep the final concentration of the solution at 20mg/mL. I hope this clarifies my question. Thank you!
 
Last edited:
Lucy345 said:
Hi, I have two powders, compound A and compound B. I need to make 2 solutions each containing the two powders, so I use the concentration of each solution(20mg/mL) to determine how much of the solvent I need to add. Now I have 2 solutions with powder A and powder B dissolved in them.
OK, at this stage you have one solution containing 20 mg/mL of A, and a second solution containing 20 mg/mL of B.

I need to add a certain amount of the compound A solution(since it was made in excess) to compound B solution so that the entire final solution has 10% molar mass of compound A.

10% molar mass or 10% molar fraction?

This is getting a bit convoluted, as molar fraction should take the solvent into account. Somehow I feel that's not what you are interested in.

If you are going to work with molar fractions it will be better to convert mg/mL concentrations into molar concentrations. Then calculating how much solution of A should be added to a given amount of B is trivial, as all you need is to find how many moles of B are already present, and calculate how many moles of A are needed so that the molar ratio translates into 10% molar fraction of A.

If you mix solutions that are 20 mg/mL the final solution will be 20 mg/mL as well (just it will be a concentration of mixture, not of a pure substance).
 

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