C₁V₁=C₂V₂ (aka M₁V₁ = M₂V₂) is probably the most useful thing I can ever share with you! It’s my FAVORITE equation! I (legit) LOVE this simple equation that helps you figure out how much of something to add to get to a specific concentration, or what the concentration of that thing is if you’ve added a certain amount of it. It can come in handy all the time – even when you’re baking!⠀

text from 11/29/20, video’s new, formula’s timeless!⠀

Say you want to make almond chocolate chip cookies with 25 chocolate chips per cup of dough and 10 almonds per cup. And you want to make 10 cups of dough. How much to add, what should you do? Look to C₁V₁=C₂V₂!⠀

I learned it M₁V₁ = M₂V₂, where M refers to Molarity which is a measure of concentration (more on this below). But the concentrations don’t have to be molar, so you’ll often see/learn it as C₁V₁=C₂V₂ (with C for concentration) because the concentrations don’t have to be molar, they can be things like chocolate chips per cup – they just have to be in the same type of units on both sides. But I learned it as M and it’s stuck so that’s how I always think of it, but I’m gonna go with C₁V₁=C₂V₂ in this post cuz it’s hopefully less confusing. In either case, V stands for volume. ⠀

C₁V₁=C₂V₂ tells you that the initial concentration times the initial volume equals the final concentration times the final volume. So if you don’t know one of those values, you can stick an “x” there and then solve for x. ⠀

And it works for anything even if you have multiple things in the same total volume (chocolate chips AND almonds!). So If you want a mixture of things in the same total volume you can C₁ (thing1)(V₁ )(thing1)= C₂(thing1)V₂ AND C₁(thing2)(V₁)(thing2)= C₂(thing2)V₂. You can have chocolate chips and almonds in your cookies! Go nuts! And you don’t have to just follow the recipe, you can use C₁V₁ = C₂V₂ to modify it to fit your tastes…⠀

Say the recipe calls for 10 chocolate chips per cup of dough (for those not from the US a cup isn’t just any old cup – it’s an actual unit of measurement (equivalent to about 240mL)). So if you wanted to make 1 cup of batter, you’d need to add 10 chocolate chips. And if you wanted to scale up – say double the recipe and make 2 cups, you’d need to double the amount of chocolate chips (so add 20).⠀

But that doesn’t sound like very chocolate-chippy cookies… So you can add more chocolate chips to the same amount of dough to get a higher concentration of chocolate chips. Concentration refers to how much of something there is in a certain amount of 3D space (volume). The higher the concentration of a thing, the more “crowded” the area is in that thing. But concentrations of different things are independent – so, for example, you can have a high chocolate chip concentration but a low almond concentration in almond chocolate chip cookies. If you were super tiny & were to “walk through” the dough of such cookies you’d frequently encounter chocolate chips, but less frequently run into an almond.⠀

So if we wanted to make 1 cup of almond chocolate chip cookie dough (“fancy dough”) with concentrations of 25 chocolate chips per cup and 10 almonds per cup, how we would do this?⠀

Instead of counting out individual chocolate chips and almonds, baking recipes usually go by volumes (like cups & tablespoons) – and how many chocolate chips or almonds are in that volume depends on the concentration – which, for one-component things (like pure chocolate chips) depends on how many can fit. I don’t know how many chocolate chips would actually fit in a cup, but let’s say 100 chocolate chips per cup is our starting concentration (C₁) of our chocolate chip “stock” and the final concentration we want is 25 chocolate chips per cup.⠀

And we’ll go with 50 almonds per cup as our almond “stock concentration” And technically almonds are “drupes” not nuts, but I’m gonna call them nuts because it’s shorter, but use almonds cuz they’re easy to draw… (so plant people don’t get mad) And I’m gonna call plain dough “batter” to distinguish this “filler” from the final “fancy dough” product which is a mix of all the ingredients (so foodie folks don’t get mad!)⠀

Then we take the components one at a time. We’ll start with the chocolate chips and write out a C₁cV₁c = C₂cV₂, filling in what we know. (the little c’s are for chocolate chips) We could write out:⠀

(100 chips/cup)(V1c) = (50 chips/cup)(1 cup)⠀

The thing we need to find is V₁c (how many cups of chocolate chips to add), so we can rearrange to get V₁c alone…⠀

(100 chips/cup)(V₁c) = (50 chips/cup)(1 cup)⠀

V₁c = [(50 chips/cup)(1 cup)]/(100 chips/cup)]⠀

V₁c = 0.5 cups ⠀

So you’ll need 1/2 a cup of chocolate chips – but you don’t add this to 1/2 a cup of plain batter! “V₂” is your FINAL volume. The chips will take up space – 1/2 of the space (unless it melts or dissolves) which is why when making water-based (aqueous) solutions you want to mix the stuff & then use water to bring the solution to the exact volume you want). So the amount of plain batter is only 1 cup total – 0.5 cups chips = 0.5 cup plain batter – but wait! – you still need to account for the almonds! ⠀

We said our starting almond concentration is 50 nuts/cup and we want 10 nuts/cup final. So we can write out another C₁V₁ = C₂V₂. The V₂ is the same V₂ as we used for the chocolate chips, because we want the same final volume of “fancy dough” – 1 cup⠀

We could write out:⠀

(50 nuts/cup)(V₁n) = (10 nuts/cup)(1 cup)⠀

V₁n = [(10 nuts/cup)(1 cup)]/(50 nuts/cup)]⠀

V₁n = 0.2 cups ⠀

So we need to use 0.2 cups of nuts. So the amount of plain batter we’ll need to get 1 cup of our fancy dough is only 1 cup (final volume) – 0.5 cup chocolate chips – 0.2 cup nuts = 0.3 cups plain batter. ⠀

Oh no! You realize you’ve made a mistake with your dough! – you went nutty and added 0.25 cups of nuts instead of 0.2! What are you to do? You *do* have to make more, but you DO NOT have to start all over! Instead you can add more plain batter & chocolate chips to get to that nut concentration you wanted. And you can use C₁V₁ = C₂V₂ again to help. Except that now your initial V₂ is your starting V₁. So to avoid confusion (hopefully), we’ll write⠀

C₂nV₂n = C₃nV₃⠀

We’ll start by figuring out what new final volume we need to get the nut concentration right. And then we’ll figure out how many chocolate chips we need to add to get our chocolate chips back up. So, the search for V₃ – what could it be?⠀

Firstly, what is the C₂ of nuts we’re currently at after the boo-boo? (C₂n)⠀

(50 nuts/cup)(0.25 cups) = (C₂n)(1 cup)⠀

C₂n = [(50 nuts/cup)(0.25 cup)]/(1 cup)]⠀

C₂n = 12.5 nuts/cup⠀

So our C₂ of nuts is 12.5 nuts/cup (instead of the desired 10 nuts/cup which we’ll use as our C3), our V₂ is 1 cup, & V₃ is the new final volume of “fancy dough” we’ll need…⠀

C₂nV₂ = C₃nV₃⠀

(12.5 nuts/cup)(1 cups) = (10 nuts/cup)(V₃)⠀

V₃ = [(12.5 nuts/cup)(1 cup)]/(10 nuts/cup)]⠀

V₃ = 1.25 cups⠀

So instead of making 1 cup of dough, we’re gonna make 1 and a quarter. Once again, we’ll use the plain batter as “filler” to get us to the volume we want, but we also have to add more chocolate chips because, even though we did things right with them the first time, we now have a different final volume we need to adjust for. ⠀

Usually how I do this is to pretend like I haven’t made any, then figure out the difference in volume and find how much I need for the extra volume. So I need to make 1.25-1 = 0.25 more cups-worth. So I go back to C₁cV₁c = C₂cV₂ like we did first except the V₂ is now 0.25 cups.⠀

So⠀

(100 chips/cup)(V₁c) = (50 chips/cup)(0.25 cup)⠀

V₁c = [(50 chips/cup)(0.25 cup)]/(100 chips/cup)]⠀

V₁c = 0.125 cups ⠀

So we can add 0.125 cups of chocolate chips to our botched dough and then add plain batter to get to the final volume we want (1.25 cups). .25-.125 = 0.125cups⠀

Phew! We saved it! Cookie catastrophe averted!⠀

Now let’s look at a more lab-by use – mixing different stock solutions (of things like salts and pH stabilizers) to make buffers (pH-stabilized salt waters). Instead of chocolate chips per cup we’re gonna need to talk in things like salt ions per liter, and salt ions (things like Na⁺ & Cl⁻ of table salt) are really tiny. Like really really tiny so you’re usually dealing with tons and tons of them. And if you thought counting individual chocolate chips would be a hassle…. ⠀

This is where that whole “molarity” thing comes in – it’s a measurement of how many of a thing there are in one liter. And when the “thing” you’re dealing with is a teeny tiny molecule instead of a chocolate chip, there are gonna be a LOT of them in a liter. So instead of talking about individual molecules, for accounting purposes we can talk in terms of groups of them – kinda like when you’re keeping a tally and you make marks in groups of 5) – or how you might talk about 2 dozen bagels instead of “24 bagels” – but we need a much bigger grouping than 5 or 24…⠀

A mole is just like the biochemist’s version of a dozen. Actually it’s Avogadro’s version – A mole is Avogadro’s number (6.02×10²³) of something. Which is way more convenient than dealing with individual salt molecules. But it’s a bit of overkill when dealing with chocolate chips….⠀

There’s no way you’re gonna be able to get 1M cookie dough – you’d have to stuff 602 sextillion (602 billion trillion) of them into a cup of dough which is not physically possible. In fact, I couldn’t find the stat for chocolate chips specifically (though if you want to measure the volume of a chocolate chip & do the math go ahead!) but if you assume the volume of a chocolate chip is similar to that of an M&M, if you were to coat the continental US (sorry Alaska and Hawaii) with 1 mole of chocolate chips, that coating would be 84km (52 miles) deep! http://bit.ly/33ha24m⠀

Yet, 1 mole of water molecules only takes up 18mL (at normal temp) & weighs just 18g. As I like to say, “Dream big, think really really really small!” (and use Avogadro’s help to keep track of it all!) ⠀

But often, even for really tiny things, you have less than a mole of them you need to do calculations with. Just like you can use centimeters when a meter’s too big and millimeters when a cm is too big, you can use metric prefixes to talk about smaller numbers while still using mole as your base unit. So, for example, a millimole is 1 thousandth of a mole and 1 millmole per liter gives you a concentration of 1 millimolar (1mM).⠀

So, if a cup’s 240mL, this is the same as 0.240 L. And in that cup we have 50 chocolate chips, we can do dimensional analysis to find out the molarity of our dough in terms of chocolate chips and walnuts. Dimensional analysis basically means, line terms up and multiply in a way that you get to cancel out all the units except the one you want. it’s another of those super handy dandy tools in the biochemist’s toolbox. more here: http://bit.ly/dimensionalanalysising

starting with the chocolate chips:⠀

(50 chips/0.24L)(1 mol chips/6.02×10²³ chips) = ~3.5 x 10⁻²², which needs a matrix prefix that barely makes it only the NIST website’s chart. Apparently 10⁻²⁴ (a septillionth) gets the prefix “yocto” (abbreviated y), so our cookie concentration is about 350 yoctomolar (350 yM)⠀

and the almonds?⠀

(10 nuts/0.24L)(1 mol nuts/6.02×10²³ nuts) = ~6.9 x 10⁻²³, so our nut concentration is about 69 yoctomolar (12 yM)⠀

Before we get into solution making, a cool side note: the “Avogadro” of “Avogadro’s number” is the same Avogadro (Amadeo Avogadro) as we talked about with the ideal gas law – and the concepts are related – he came up with the theory that the same number of molecules of any gas will take up the same volume (because the gas molecules are spaced so far apart) even though they might weigh different amounts (have different masses) because different gas molecules have different numbers of atoms and some of those are heavier than others – this is why helium balloons float better than your hand (or mouth) blown ones – helium’s lighter than the gas molecules in normal air). http://bit.ly/buoyancyballoons ⠀

Anyways – back to the buffers – I spend a lot of time making these pH-stabilized salt waters to keep my proteins happy. But how much salt? And how much buffering agent (pH-stabilizer)? Depends. At different times I’ll want different things. Not only do I sometimes want different concentrations of them (how much of the dissolve thing (solute) compared to the overall volume) – but I also want different volumes at different times.⠀

I can calculate how much to weigh out based on the formula weight, which tells you how much the thing weighs per mole. But if I had to do this each time, I’d spend all my time weighing stuff out and wouldn’t have time to purify the proteins I want to bathe in the solutions I’m making. If I make stock solutions at a higher concentration than I will ever actually want to use it at I can then just add different amounts of that liquid version depending on what I want the final concentration at. And this is a HUGE time saver, especially when it comes to things that aren’t as straight forward as just sticking the solid in water and it instantly dissolving. Some solids take a while to dissolve or have to be pH-adjusted, etc.⠀

So I keep stock solutions of the things I use a lot. A few convenient ones I use the most are 1M Tris (a buffering agent), 5M NaCl, and 3M KCl.

You want to make sure that your stock solution is high enough that you can add it and whatever else you want in the solution without going over the volume you want. And the higher the concentration of your stock, the less often you’ll have to make more. But there’s only so much of something you can get to dissolve, so you have to make sure not to go over the solubility limit.⠀more on making stock solutions here: http://bit.ly/sciencestocksolutions

In the pics you’ll see a real-life example – even for simple ones I often write it out because it’s easy to make stupid mistakes – especially when you’re trying to multitask.⠀

If you’re doing a serial dilution, where you do a series of dilutions where you dilute something by the same ratio each time (such as 1/2-ing something, then 1/2-ing that 1/2-ed something to get a 1/4 of the original concentration and then 1/2 that to get 1/8 of the original concentration…) you can use C₁V₁=C₂V₂ where one dilution’s C₂ becomes C₁ for the subsequent dilution. You can use some shortcuts to get the calculate the concentration of any dilution without having to go through the whole series. You just need to know the first concentration, the number of dilutions since then, and the Dilution factor (DF), which is the amount by which you’re diluting by each step. So in this example, the dilution factor was 2/1 = 2. To get the concentration at a certain dilution, multiply the original concentration by )1/DF raised to the power of the # of dilutions). For the third dilution, for example, we’d multiply the original concentration times (1/2)³ = original concentration times 1/8, which is what we figured out above! More on this here: http://bit.ly/serialdilutions

You can learn a lot more about concentrations here: http://bit.ly/solutionconcentrations ⠀

more math-y stuff on a new blog page I added: http://bit.ly/mathyposts

more on topics mentioned (& others) #365DaysOfScience All (with topics listed) 👉 http://bit.ly/2OllAB0⠀

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