Chemical Mixture Problems

Recently, I tutored a few students in math, and they were having the same problem: How to solve a chemical mixture equation?

Chemical mixture problems can pretty much always be solved with the equation: p₁(v₁) + p₂(v₂) = p₃(v₁ + v₂), where "p" refers to the various percentages of the chemical involved and "v" refers to the various volumes of the chemical involved. The question should give either all three percentages as well as one volume, or both volumes as well as two percentages.

When solving a problem such as this, it is unimportant whether the percentage is listed as a number such as 75, or a decimal, such as 0.75, as long as the same method is used throughout the problem.

Here is a sample question below:

A chemist needs 150 milliliters of a 50% saline solution but has only 18% and 78% solutions available. Find how many milliliters of each that should be mixed to get the desired solution.

So, we can start equating variables with values: p₁ = 18, p₂ is 78, p₃ = 50, and v₁ + v₂ = 150. Solving for v₁, we have v₁ = 150 - v₂. Thus, 18(150 - v₂) + 78(v₂) = 50(150). After simplifying, we have the following: 2,700 - 18v₂ + 78v₂ = 7,500. Simplifying further: 60v₂ = 4,800. Our answer? v₂ = 80. However, we still need to find v₁, so we substitute v₂ back into our equation for v₁, so v₁ = 150 - 80. Our final solution leaves us with v₁ = 70 mL (of the 18% solution), v₂ = 80 mL (of the 78% solution).

Now say that we have the same question, given different values:

A chemist needs a 50% saline solution but has only 18% and 78% solutions available. If he has 70 mL of the 18% solution, find how many milliliters of the 78% that that he needs to get the desired solution.

So, again, we have very similar variables: p₁ = 18, p₂ is 78, p₃ = 50, and v₁ = 70. Thus, our equation looks like this: 18(70) + 78(v₂) = 50(70 + v₂). After simplifying, we have the following: 1,260 + 78v₂ = 3,500 + 50v₂. Simplifying further: 28v₂ = 2,240. Our answer? v₂ = 80 mL, and v₁ + v₂ = 150 mL total.

The one last type of this problem that one may see is that of the missing percentage:

A chemist can make a 50% saline solution with 70 mL of an 18% solution and 80 mL of a saline solution with an unknown percentage. What percent salinity does the second solution have?

Here we have the following variables: p₁ = 18, p₃ = 50, v₁ = 70, and v₂ = 80. Our equation is 18(70) + p₂(80) = 50(70 + 80). Simplifying: 1,260 + 80p₂ = 50(150). After another round of simplification, we have: 80p₂ = 7,500 - 1,260. And again: 80p₂ = 6,240; leaving us with the result that the percent salinity of the second solution is as follows: p₂ = 78%.

For those who may need more math help than I can provide, Khan Academy has tutorials on many subjects, math among them, from Kindergarten all the way up to higher-level math such as Calculus!

I hope that this post is helpful to both aspiring engineers and struggling mathematicians alike. If any clarifications would aid understanding, please let me know via the comments section below or my contact page. Thanks for reading!