If molar extinction coefficient is unknown, can concentration still be found in grams rather than moles?
952788
A=E*d*C\n\n(d=length of the path of the light through the solution)\n\nIf you have only one value of absorbance then obviously there is no way of determining the concentration without the molar extiction coefficient (E).\n\nIn practice even if you know E, you don't use it to find the unknown concentration. Instead you measure the A of standard solutions and do a standard curve. From the equation of that curve (which you find using linear regression) and the value of A of your sample you calculate the unknown amount. \nThe standard curve doesn't need to be A vs mole/lt. It can be A vs anything e.g. g/ml, mg/ml, ppm or even just g, mole, etc. the only thing that changes is what the slope represents.\nFor example for A vs mole/lt the slope corresponds to E*d, for A vs g/lt the slope corresponds to E*d/MW, etc.\n\nIf you  always have the same volume of sample you can do the curve A vs grams directly. Then if V is the volume of the sample, the slope corresponds to E*d*V/MW\n\nEven if your standard curve is expressed in moles/lt you can convert it to grams. You just need the MW of the compound and the volume, not E...
334265
I think the concentrations would be too small to be visible.  you would only be able to react with the available OH-'s, which is 10^-7 mol/L, in theory the maximum quantity of Pb(OH)2 you could make then would be 5x10^-8 mol/L.  If Le Chatelier's principle came into play maybe a little more, but that's assuming everything reacted, which probably wont happen.
32482
To find how many grams are in a solution, you need to know how many moles there are. \n\n1 M= 1 mole/L\nor Molarity = moles/volume\n\nmoles= Molarity*volume= (0.033 L)(0.150 M)=.00495 moles.\nYou have to convert mL to L because molarity is in L.\n\nGrams is moles multiplied by molar mass.\nMolar mass is calculated by adding the masses of the individual elements.\nmolar mass= 2N + 8H + 1S + 4O\n=2(14.0) + 8(1.00) + 1(32.1) + 4(16.0)\n=164.2 g/mol\n\ngrams= (164.2 g/mol)(.00495 mol)= .81279 g (NH4)2SO4\n\ngrams= 0.813 g (NH4)2SO4.  (sig figs)
810972
Cool stuff, I used to get confused by ugly chemistry calculations, too :D\n\nYou have 3.88 g acid in 250cm3, so in the 25 cm3 you titrated there are 0.388 g of the acid. You need to know how many moles that is, so that you can calculate molar mass.\n\nThe amount of NaOH used up is 46.5 cm3 × 0.095 mol/dm3, that is 0.0465 dm3 × 0.095 M, which equals 0.00442 mol (It's not exactly that much, but I'll stick with three significant figures).\n\nSince NaOH reacts with monoprotic acids in a 1:1 ratio, that means that there was exactly the same amount of acid in the solution.\n\nNow we know that 0.388 g of the acid is 0.00442 mol, so we can divide the mass by the number of moles to get molar mass. That is 0.388g/0.00442 mol = 87.8 g/mol.\n\nRelative molecular mass is simply molar mass in grams, but without the unit, so Mr of the acid is 87.8.
646163
Presuming that the base is monobasic.\n\na) with 25 ml added, you are at the equivalence point.  The pH is 7.\nb) after adding 25.01 ml, you will have 0.01 ml of base that did not get reacted.  The molarity of the base in solution will then be 0.1 * (0.01/50.01) = 2.0 x 10^-5.  Your pOH is 4.70, pH would be 9.30\nc) Now, it's 0.1 ml of unreacted base.  Molarity is 0.1 * (0.1/50.1) = 2.0 x 10^-4.  Your pOH is 3.70, pH is 10.30.\nd) Now 1 ml of unreacted base.  Molarity is 0.1 * (1/51) = 2.0 x 10^-3.  pOH is 2.70, pH is 11.30.\ne) Now 25 ml of unreacted base.  Molarity is 0.1 * (25/75) = 3.33 x 10^-2.  pOH is 1.48, pH is 12.52.\n\nIn each case, you have to figure out the amount of unreacted titrant and recalculate the molarity as if you were simply diluting it, using McVc = MdVd, where the volume is now the total volume of the solution (the unreacted amount plus the 25 ml of acid and the 25 ml of base that have neutralized each other).
961122
You have to be more specific. As one knows, in titration, we determine the Normality or Molarity of solution which is titrated. Normality or Molarity of any solution depends on molecular weight. Mol Wt/1000 results in 1 N solution.... I think this is what you asked.
968320
here, you just use the ideal gas law, PV=nRT.  since molarity is equal to moles of solute per liters of solution, M=n/V, so P=MRT.\ntherefore,\nKp=Pco*Ph2o/(Ph2*Pco2)\n  =[CO]RT*[H2O]RT/ ([H2]RT*[CO2]RT)\nall the RT coefficients end up canceling out, so:\nKp=[CO][H2O]/[H2][CO2]\n  =Kc\nsince there were equal numbers of gas molecules on each side of the equation, Kp was equal to Kc.  as a general rule, Kp=Kc*(RT)^N, where N is the change in the number of gas molecules.
715558
OK, but you need to look up atomic weights or calculate molar mass.\n\n   Grams = Moles x molar mass  OR  moles x atomic mass\n\n    Moles = Grams/(molar mass) OR Grams/(atomic mass)
232172
ΔT = 80.6 °C - 75.4 °C = 5.2 °C; m =  ΔT/ Kfp = 5.2 °C/(6.9 °C/m) = 0.75 m; m = moles solute / 1000 g solvent, so that 1.00 g/ 12.5 g = x g/1000 g, x = 80 g, and 80 g = 0.75 mol, or molar mass = 80 g/0.75 mol = 1.1×10^2 g/mol.\n\nHope this helps,\nUltimeaciax
543827
If you know the grams, which you do on one...its easy.  To find the grams from ml, you need to know the density.  Density is in grams/ml, so if the density was 2 g/ml and you had 4.5 ml, then you would have 9 grams.  So you take the grams and you need to find moles to calculate percent yield.  To do this, find the total molecular weight (for instance, water is 18 b/c of two H's which are 1 and one O which is 16).  When you have the grams and molecular weight, divide the grams by the molecular weight.  This will give you the moles, since the molecular weight is grams per mol.  Do this for every reagent or thing you use.  Whicever one has the lowest amount of moles is your limiting reagent because once that is gone, the reaction stops!  Now the theoretical yield is found by doing stoiciometry.  So whatever your limiting reagent is...that number is your theoretical yield for this problem.  Then weight the product when you are done, and find the amount of moles you have by oncve again dividing how many grams you have by the molecular weight.  Then divide that by your Theoretical yield and that is your percent yield!  Looks harder than it is.
905387
Molarity is simple.  The formula is Moles of Solute divided by Liters of solution. \n\nThe solute is the smaller amount, if you get confused with the solute and solvent.  Many times the solvent is water.
1177722
I know my chem, but have never heard of it nor could I find it.\n\nPerhaps you mispelled it?  It should be described in your book if they're going to use it as a problem.
304287
This is called Base line correction.\nYour solvent may be having some absorption, or the impurities in it may be having some absorption..so to measure your compunds absorption exactly, you need to take the solvent (n whcih you are going to dissolve your compound later) and do this base line correction.\nIn a way you are telling instruments "NOT TO WORRY ABOUT THESE ABSORPTIONS"\n\nThats it\nYou got it?
920352
Did you measure the absorbance of a standard in addition to the unknown sample?\n\nIf so, you could plug the values into (As / Au) = (Cs / Cu)\nWhere\nAs = Absorbance of the Standard\nAu = Absorbance of the Unknown\nCs = Concentration of the Standard\nCu = Concentration of the Unknown?\n\nAlternatively, did you get a value for the molar absorptivity of PNP at the wavelength you used? \nIf so, you could plug it into\n\nA = (epsilon)(b)(c)\n\nWhere\nepsilon is the molar absorptivity,\nb is the length of the path through which the light passed,\nc is the concentration of the chromogenic substance.\n\nAnd you could rearrange that to solve for c.\n\nHope it helps
144551
since the density is 1 it doesn't matter how the % was calculated. usually you need to be specific.\n3% is 3mls peroxide/100 mls or 30 mls(30gms)/1 L\n30/molar mass = 0.88M for concentration.
498666
Aaaaaaaaalright... Formulas depend on what you're solving for but since they all involve using the ratio of moles it shouldn't be too different from equation to equation.\n\nLet's go over the mole first to help you better understand what's going on in stoichiometry.  The mole is the amount of a substance that contains as many particles as there are atoms in exactly 12g of carbon-12.\n\nWHOA there... hold on for a second.  The mole is simply a counting unit, just like a dozen is.  We don't usually buy 12 or 24 doughnuts; we order half dozen, one dozen, or two dozen.  Similarly, a chemist may want 1 mol of carbon, or 2 mol of iron, or 2.567 mol of calcium.  \n\nThe number of particles in a mole is known as Avogadro's number, named after an Italian chemist Avogadro. (Go figure)\n\nAvogadro's number is 6.022 x 10^23... which is the number of particles in exactly one mole of a pure substance.  One mol of a substance is going to equal its AMU on the periodic table, or, the mass. \n\nFor example, one mol of Helium is 4g.  If you wanted to find out how many grams are in 3 mols of Helium, you would do a conversion like this:\n\n3 mols He x 4g He/1 mol He\n\nIn words that would be... 3 mols of Helium multiplied by however many grams per mol Helium is, which is 4 grams per mole.  The mols would cancel out leaving us with the units in grams.  Do 3x4 and you get 12 grams of Helium in 3 mols of it.\n\nIf you want to find out how many mols of something you have you would take the amount of grams you have of it and set up another conversion like this:\n\nAmount of g x 1 mol/gram in that mol\n\nFor example, if you have 20 grams of Helium and you want to find out how many mols of it you have, the equation would look like this:\n\n20g Helium x 1 mol/4g He per mol\n\nThe grams would cancel out, leaving the answer in mols.  You would then take 20 divided by 4 and get 5 mols of Helium.\n\n(Hoping this is making sense)\n\nNow stoichiometry is just pretty much comparisons of stuff. (Mols)\n\nAlright, when you're asked to calculate the amount in moles of one substance that will react with or be produced from the given amont in moles of another substance, you must use this simple mole conversion problem.\n\namount of given substance (mols) ----> amount of unknown substance (mols)\n\nIn order to solve this type of equation, use this conversion factor.\n\ngiven quantity x mole ratio = unknown quantity\n\n\nLet's try an example now.\n\nCOsub2 + 2LiOH ---> Lisub2COsub3\nsub=subscript\n\nLet's calculate how many moles of Lithium Hydroxide are required to react with 20 mol of COsub2\n\nGiven information = amount of COsub2 = 20 mols\nUnknown = amount of LiOH in moles\n\nSince the equation is balanced, we need to use the ratio that would give us LiOH in the final answer, therefore, meaning LiOH must be on the top of the ratio.\n\nThe correct ratio would be (mols LiOH/mols COsub2)\n\nIn order to calculate this, set up the equation now. (given q x ratio = unknown)\n\nmols COsub2 x (mols LiOH/mols COsub2) = mols LiOH \n\nsince the mols of COsub2 are on the bottom, it cancels out with the mols of COsub2 on the other side of the multiplication side because that's the rule in math, therefore leaving you with the units of LiOH.\n\nsubstitute the values into the equation and calculate the answer.\n\n20 mols COsub2 x (2mols LiOH/1 mol COsub2) = 40 mol LiOH\n\nsince the mols of COsub2 cancel out, you're left with 20 x 2 mols LIOH which equals 40 mols of LiOH.\n\nThe ratio of LiOH/COsub2, 2:1, was made from the balanced equation because for every mol of COsub2 you have you need 2 mols of LiOH to react with it to produce the right amount of products.\n\nHope this helped, good luck.
470567
To do this problem we need to do some convertions.  Convertion factors are simply ratios that allow us to change the units of a certial quantity.  For example: the molar concentration (molarity) is a convertion factor that is a ratio of the moles of dissolved substance to the volume of solution (moles/L). Using the concentration  we can convbert moles to liters or liters to moles depending on what we have to start with.  Another convertion factor that we will need is the molar mass of KOH.  This is the ratio of the mass of KOH to the moles of KOH (this is usually measeured in grams per mole (g/mol)).  We can find this number by looking up the molar masses of K, O, and H on a periodic table and adding them together. K =39.0983g/mol, O = 15.9994g/mol, and H = 1.00794g/mol. \nAdding these together we have a molar mass of 56.10564g/mol for KOH.\nSchematicaly this is the road map we will follow to find our answer:\n\n-------(moles/L)-------(g/mol)\nvolume <------> mole <-------> mass\n\nWhere volume, moles, and mass are actuall ammounts and the convertion factors between them are on the arrows.\n\nNow for your specific problem we will start with the given volume 750mL.  First we will convert it from mL to L then convert it to moles of KOH then finally to grams of KOH.\n\n750ml(L/1000mL)(15.0mol KOH/L)(56.1056g KOH/mol KOH)= 631.188g KOH\n\nTo three significant figures our answer is 631g KOH.  This is a big number but a 15.0 M solution is very concentrated so it is not surprizing.