1. The problem statement, all variables and given/known data Known: 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, lets say 15mg. And I measure a mass of compound B, lets 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. Relevant equations C=nv etc simple chem equations 3. 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.