Semi OT: Calculating dilution

[KJ1I]

I used to know how to do this, but having not needed to do so for a long time, I have forgotten the process.

I have 1355 grams of a solution containing 20% 2-Butoxyethanol. I need to raise the percentage of 2-Butoxyethanol to 25%. How much 100% 2-Butoxyethanol do I need to add? Total weight after addition is irrelevant.

[ed_h]

Assuming all percentages are by weight, you should add 90.33g of the 2-BE.

Ed

[old mart]

As near as dammit 90 grams.

[MattiJ]

You need to add 5% of the stuff. 1355/100*5=67.75(g)

Hopefully you are not a pharmacist
(271+61.75)/(1355+67.75)*100% = 23.3%

[tomato coupe]

Hopefully you are not a pharmacist
(271+61.75)/(1355+67.75)*100% = 23.3%

I think you have a typo in your numerator. (=> 23.8%)

[KJ1I]

Assuming all percentages are by weight, you should add 90.33g of the 2-BE.

Ed

Thanks,

Your assumption is correct, everything is by weight. Adding the 90 g will be easy (shop scale reads to grams). So what formula did you use to get there, just in case I ever have to do it again? The solution is for an airbrush cleaner for acrylic paint. I used all the 2-BE I had on hand, but now that I have resupplied, I want to bring the solution to the "proper" percentage.

[3 Phase Lightbulb]

I think you have a typo in your numerator. (=> 23.8%)

It's 23.387% unless you suck at math.

[KJ1I]

Hopefully you are not a pharmacist
(271+61.75)/(1355+67.75)*100% = 23.3%

You lost me there: where did the 61.75 come from? I get the 271 (20% of 1355) and the 67.75 (suggested addition).

Thinking further: should the 61.75 actually be 67.75 (the addition)?

[MattiJ]

You lost me there: where did the 61.75 come from? I get the 271 (20% of 1355) and the 67.75 (suggested addition).

Typo.
And luckily I'm not pharmacist either

[tomato coupe]

It's 23.387% unless you suck at math.

See posts #9 and #10. MattiJ just had a typo, that's all, and it gave him the wrong number.

[ed_h]

Thanks,

Your assumption is correct, everything is by weight. Adding the 90 g will be easy (shop scale reads to grams). So what formula did you use to get there, just in case I ever have to do it again? The solution is for an airbrush cleaner for acrylic paint. I used all the 2-BE I had on hand, but now that I have resupplied, I want to bring the solution to the "proper" percentage.

Your 20% solution has 0.2 x 1355 g = 271 g of BE

You want to add an unknown weight, X, of BE to bring the solution to 25%.

The new weight of BE will be 271 + X.

The new weight of the entire solution will be 1355 + X.

So

271 +X = 0.25(1355 + X)

Solve for X = 90.33 g

Ed

[KJ1I]

Once more, thank you.

Kj

[MichaelP]

Kevin,

Sorry, I deleted my post by mistake. To recoup, I stated that if, by your request, we ignore the final solution weight, you'd need to add 67.75g of BE. And I noted that, in reality, the final concentration will become 23.8% instead of the needed 25%.

This difference is because we ignored the new total solution weight.

To reach 25% solution in real life, you need to follow Ed's calculations. That's how it's done properly. As you can see, the difference is stunning.

[dian]

complicated thinking here. 5% of 1355 = 67.75. thats it.

[cameron]

complicated thinking here. 5% of 1355 = 67.75. thats it.

And what is "it", other than the incorrect answer to the question?