what is the importance of the mole ratio in stiochiometry calculations?
145088
its nice, because the molecular weight of a compound in grams = 1 mol of that compound. It makes it easy this way to figure out equations involving mol gram ratios. \n\nAlso, remember 6.022 x 10 to the 23rd power. \n\n1 mol X atoms - 6.022 x 10^23 X atoms\n6.022 x 10^23 X atoms = blank grams X\n1 mol X atoms - blank grams X
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.
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?
202815
100g O2 x (1 mole O2)/(32g O2) x (22.4 L O2 @ STP)/(1 mole O2 @ STP) x (1/1.5) x (298/273)\n\n= 50.9 L\n\nThe 22.4 L = 1 mole is only true @ STP.  The higher pressure reduces that volume and the higher temperature increases it.
422875
U have not mentioned the gas gravity to determone deviation factor to have an exact answer...generally..taking z=1\npsia ( absolute value) will also be used \nPsc*Vsc/Tsc=PV/T\n\nso: Vsc/V=PTsc/PscT=53.1/T \n\nThis is the simplest correct answer.... make is easier but worse that the process is isothermal so\n\nVsc/V=102 --> 1cu ft at TVD= 102 SCF\n\nThe most exact solution can be provided if u want..with taking temperatire and deviation,composition in account
823430
the buoyancy of the object and its volume.
20952
The mass _ratio_ of elements in a defined compound is a constant. Therefore for a given mass of that compound, the mass of the element in that sample depends on the total mass (i.e. amount) of the compound.
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.
24946
For gases, we speak of "standard gas density." This is the density of the gas (expressed in grams per liter) at STP. If you discuss gas density at any other set of conditions, you drop the word standard and specify the pressure and temperature. Also, when you say "standard gas density," you do not need to add "at STP." STP is part of the definition of the term. It does no harm to say "standard gas density at STP," it just looks a bit silly. \n\nYou can calculate the ideal standard gas density fairly easily. Just take the mass of one mole of the gas and divide by the molar volume. \n\nMore explanation at the site linked below.
313383
Molarity= moles/liters
294763
Well, stoichiometry is not a method or a device. It is simply calculations of reactions in chemistry. It doesn't really have a role but in smelting, since smelting is not a chemical change. It is actually a physical change of state.
960715
stoichiometry is \n\nCalculation of the quantities of reactants and products in a chemical reaction. \n\nThe quantitative relationship between reactants and products in a chemical reaction.
524657
molarity = moles/volume\n\n125 g C2H5OH / 46 g/mol = 2.717 moles\n\nmolarity = 2.717/0.250 = 10.87 M
922214
If you write the formula wrong, then you will calculate the formula weight wrong (too many or too few of one atom in a formula will translate to an incorrect formula weight). Then, when you try to figure out how many moles of the compound you have, your moles will be wrong too. And THEN, if you try to work out the stoichometry of a reaction using the incorrect formula, your mole RATIOS will be wrong, and thus you will get a wrong number when you try to determine the expected yield of a product in a reaction. (Not to mention it probably won't balance with an incorrect formula). If you're making up a solution using a wrong number of moles (see above), your molarity (moles per liter solution) will be wrong too. Everything will be wrong if you don't get your chemical formula right!
1062497
What everyone has said is correct, but look at it this way: Consider a square. We know from the theorem of pythagoras that the ratio of the hypotenuese to the length of one side = sqrt(2) or, if you like, the ratio of one side to the hypotenuese = 1/sqrt(2). These constant ratios are independent of the size of the square. We have the ratio of the perimeter of the square to the hypotenuese = 4/sqrt(2) = 2*sqrt(2), which is independent of the size of the square.  Similarly the ratio of the circumference of a given circle to the longest chord(the diameter) has an exact value, and similarly this value is independent of the size of the circle. As has been mentioned, for any regular n-gon(square, pentagon, hexagon etc) the ratio of the perimeter to the length of the longest line segment that can be drawn within the n-gon will have a specific value depending on n, and this value can be determined. As we let n approach infinity, the n-gon approaches a circle and the ratio of the perimeter of the n-gon to the longest "chord" approaches a specific limit and this limit is pi, the ratio of the circumference to the diameter.  And, as mentioned above, analytic techniques can (and have) determined the value of the constant pi to a trillion or so digits.\n\nFootnote: SK, determining the limit of a series or even that the limit exists does not necessarily require calculus.\nFor example, the sum of a geometric series such as Sum[1/(2^n)] = 1/2 + 1/4 + 1/8+....+1/2^n is proved without calculus to be 1- 1/2^n and the limit of it (without calculus) as n---> inf. = 1. Check out the below link regarding pi.\n\n http://mathworld.wolfram.com/ArchimedesAlgorithm.html
125296
use spectrophotometric method(uv-vis spectrometry).Few things u shud remember.....\n1) Use Labert-Beer's law\n2) u shud know the molar extingtion coefficient value of that particular species\n3) concentration of the species\n4) Calculate lambdamax. using LB law\nIt is an indirect method.\n\nUV-VIS. Spectroscopy will give u the lambdamax value directly...alongwith absorbance
316922
Ok.... \nfirst: 1atm = 760 torr\n Now, as per the gaseous  STP conditions., 1 mole of oxygen will occupy 22.4 lit volume at STP conditions.\nSTP: T=273K\n        P= 760 torr\n       V =22.4 lit\n       Mass of O2 at STP = 32g\n\nNow. second set of conditons:\n P2= 1400 torr\nT2 =41 K\nMass= ?\n\nPV = nRT\nPV = (w/Mwt) XRT\n\nP1/w1T1 = (P2 /W2 T2)\n\nSubstitute these values;;\nW2= 392g of oxygen.\n\nIs the solution is clear enough?