Percent by Weight and Volume ```Name: D. D. Status: student Grade: 9-12 Country: USA Date: March 2009 ``` Question: What is v/v? How do you calculate it? How do you calculate a concentration base on percent by volume and percent by weight? On some msds' it is listed that a certain compound has a concentration of 35% by weight, and on others it is listed that a certain compound has a concentration of 35% by volume. Replies: Hello, This is a question that I remember being confusing myself because there are so many combinations, w/w, v/v, w/v etc. Take the amount of alcohol in beer, for example. In the US, it is generally indicated as abv (alcohol by volume), but in the UK it is generally indicated as abw (alcohol by weight). For ethanol, the abv is a larger number than abw and after I explain how each is calculated, you should understand why. I will use ethanol and water as the two parts of a mixture for the examples below. So before we begin, I expect you to be able to convert from grams to moles and vise verse. Let's list out some physical properties of ethanol and water. Ethanol: FW = 46, d = 0.798 (at 20C) Water: FW = 18, d = 0.998 (at 20C) d=0.914 v/v: To make a v/v solution, you simply take equal volumes of the two components and mix them together. One caveat however, mixing ethanol and water makes a funny thing occurs. The volume of the mixture will be less than the sum of the volume of water and ethanol. For example, if you have 1 liter of water and 1 liter of ethanol, then mix the two, what you end up with is only 1.92 liters of a 50/50 mixture (by volume). The reason for this is in the arrangement of the molecules. When water freezes, the arrangement of the water molecules take up more space than in liquid form, hence why ice becomes less dense than water and floats. When combined with ethanol, the mixture contracts and the molecules take up less space than the individual parts. So if you need to make a 30/70 v/v solution of ethanol in water, you can multiply 0.3 times the final volume you wish to achieve to figure out how much ethanol to add and do the same with 0.7 times the final volume to figure out how much water to add. Combine the two and viola, you have your mixture, which is almost the final volume you want. To compensate for this volume discrepancy, you can simply adjust your desired volume upward by 5-10%. The other situation is that you are given a unknown premixed solution and you wish to figure out the v/v ratio. I am not aware of a way that one can (simply) calculate this so the best option is then to look up empirical (experimentally derived) data on the Internet, the CRC Handbook or other reference texts. There are ways to calculate it, but this would require that you know the total number of moles of each component, or at least the mole ratio of the two. I will get more into this with w/w below. w/w: This is the other most commonly used ratio (w/v is not used much). To make a 35 w/w% solution, one simply weighs out the two components and mixes them together. This sounds like you are looking up concentrated hydrochloric acid, but I will stick to ethanol and water. If you want 1 kg of final solution, then 350 g of ethanol mixed with 650 g of water would give you a 35 wt/wt% solution of ethanol in water. Conversely, if you are given the w/w and a total mass, then calculating simply back calculate by multiplying the w/w% (35% in this example) by the total mass of the mixture and you will obtain the mass of ethanol (or component 1). But if you wish to calculate the wt/wt% of a unknown premixed solution, this gets a bit more tricky. The advantage to using w/w is that it is independent of the volume, and hence the contraction or expansion of the mixture due to molecular rearrangement, as noted above with v/v. Here you must know at a minimum the mole ratio between the two components, or the moles of one component and the total mass of the mixture Example 1: You have 1 mole of ethanol and the mass of the mixture is 100g. Since the FW of ethanol is 46, then 1 mole weighs 46 grams and you have a 46/100 = 46 w/w% mixture. Example 2: you have a mole ratio of 1:1 ethanol:water. In this case, even if you do not know the mass of the final mixture you can still figure out the w/w%. 1 mole of ethanol weighs 46 grams and 1 mole of water weighs 18 grams, so if you have equamolar amounts, then the total mass could be 64 grams, which would make this a 46/64 = 71.9 w/w% ethanol/water solution. I hope this helps! Matt Voss D. D., Percent by volume (v/v) is simply the volume of solute divided by the sum of the volume of solute and volume of solvent - prior to be mixed together - times 100. It is important to note that while the numbers may be close Vsolute/Vsolution is not equal to Vsolute/(Vsolute + Vsolvent) since Vsolution is not always equal to Vsolute + Vsolvent (since substances that mix well will interact and change the volumes upon mixing). Equivalently, percent by weight (w/w) is mass of solute/mass of solution * 100. In this case, since mass is not affected by mixing, mass of solution is always equal to mass of solute + mass of solvent. To convert between the two, you would need to go through a conversion factor of density. Remembering that density is mass/volume, you should be able to do the math. Greg (Roberto Gregorius) Hello D.D. %v/v = percent volume/volume; for example: 1 / 100 X 100% is 1%, therefore, 1mL / 100mL X 100% = 1% mL/mL, or, properly stated: 1% v/v %w/v = percent weight/volume; for example: 1g / 100mL is 1% w/v %w/w = percent weight/weight; for example: 1g / 100g is 1% w/w Hope this helps, Joel Jadus Click here to return to the Chemistry Archives

NEWTON is an electronic community for Science, Math, and Computer Science K-12 Educators, sponsored and operated by Argonne National Laboratory's Educational Programs, Andrew Skipor, Ph.D., Head of Educational Programs.

For assistance with NEWTON contact a System Operator (help@newton.dep.anl.gov), or at Argonne's Educational Programs