How can I find the molecular formula of a gas from experimental data?
609831
pv=nrt\nmeasure pressure, volume, and temperature to find the number of moles, n\n\nfind the mass of the gas\n\nmass of gas/ moles of gas= molecular mass (g/mol)
841680
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 5.752 g x (12 / 44) = 1.569 g\nhydrogen: 2.354 g x (2 / 18) = 0.262 g\noxygen: 2.356 - 1.569 - .262 = 0.525 g\n\n2) convert to moles.\ncarbon: 1.569  / 12  = 0.131 mol\nhydrogen: 0.262  / 1  = 0.262 mol\noxygen: 0.525 / 16 = 0.0328 mol\n\n3) divide each molar amount by the lowest value\ncarbon: 0.131  / 0.0328  = 4\nhydrogen: 0.262 l / 0.0328  = 8\noxygen: 0.0328 / 0.0328 = 1\n\nWe have now arrived at the answer: the empirical formula of the substance is C4H8O
905163
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 65.45% x 4.13 g = 2.703 g\nhydrogen: 5.49% x 4.13 g = 0.227 g\noxygen: 29.06% x 4.13 g = 1.200 g\n\n2) convert to moles.\ncarbon: 2.703 / 12 = 0.225 mol\nhydrogen: 0.27 / 1 = 0.227 mol\noxygen: 1.200 / 16 = 0.0750 mol\n\n3) divide each molar amount by the lowest value to obtain small integers\ncarbon: 0.225 / 0.075 = 3\nhydrogen: 0.227 / 0.075 = 3\noxygen: 0.075 / 0.075 = 1\n\nThe empirical formula of the substance is (C3H3O)k\nThe atomic masses add up to 55. To find k we need the molar mass of the compound. Use PV=nRT and plug in the data given *in the correct SI units*. Solve for n the number of moles in the sample. You problably find 5.51/55 or 5.51/110 so that k=1 or 2
205383
Divide each mass by their molar weight to get the number of moles of each gas. 0.856g He/(4.003 g He/mole) = 0.214 mole of He. Do that for all the masses and add them up. I got a total of 1.034 moles. \nFrom that you could find the total pressure using PV=nRT.\nNext, find the mole fraction (I think this is the same as molar concentration) which is the # of moles of each gas/total number of moles.\nThen multiply each mole fraction by the total pressure to get each partial pressure. \nHope this helped. Let me know if you want me to go through each step.
568814
The general formula for alkanes is CnH2n+2
712915
i have not heard of spectroscopy being used to determine the amounts of gases, but i guess you could by looking at how deep the colours are. If the colours are paleish = not alot,etc
700450
Letters A, B, D.\nIn not sure about C, its wording is tricky.\nIdeal gas law : PV = nRT\nReal gas law : (P + (a(n / V)^2)) * (v - nb) = nRT\nThe corrected pressure times the corrected volume.
1153522
The quantities you have listed are in fact related via the ideal gas law.  So, assuming you aren't in huge temperature or pressure extremes, you can use the ideal gas law with decent results.\n\nIdeal Gas Law\nPV = nRT\n\nwhere P = pressure (N/m^2)\nV = volume (m^3)\nn = number of moles of gas\nR = universal gas constant = 8.3145 J/(mol*K)\nT = temp (in Kelvins)\n\nSo, by looking at the equation you can not convert air pressure directly to volume without also knowing the temperature and number of moles of gas.
453509
i'm not familiar with the unit barg, but just use the ideal gas equation PV=nRT solved for n so n=(PV)/(RT) with an apropriate value for the universal gas constant (R).  that will give you the number of moles of N2 which you can then determine the volume under standard conditions if you wish using the same equation.
878166
To answer this, you can use the ideal gas law, V = nRT/P\n\nYou know R , T  and P, andso you need to find n.\n\nTo find n , note that 2 moles of KClO3 produces 3 moles of O2. Given the mass of KClO3, you kind find the moles of this compound by dividing the mass by the molar mass. Assuming that there is 100% conversion, you can use the ration above to find the moles of O2.\n\nLastly, I'm not sure, but I think you need to use the gage pressure, which is the barometric pressure - the vapor pressure.
317042
www.chemfinder.com might help
777427
C:H = 85.63:14.37 (mass ratio)\n= 85.63/12.01:14.37/1.008 (molar ratio)\n= 7.13:14.26\n= 1:2\nSo the empirical formula is CH2 and its empirical molecular weight ≈ 14.03\n\nP1V1/T1 = P2V2/T2\n0.980 * 1/(273.2 + 51.7) = 1.V2/273.2 (at STP of 1 atm and 0°C)\nSo V2 = 0.980 * 273.2/324.9\n= 0.824 L\nThus 3.09 g occupies 0.824 L\nSo 3.09 * 22.4/0.824 g occupies 22.4 L at STP\nie 1 g mole = 84\nThus MW of this gas is 84\nSo number of empirical units = 84/14.03 ≈ 6\nSo the molecular formula is C6H12
860730
Multiply moles by the molecular weight (grams per mol) to get mass (grams).
386525
Convert metric tons to grams.  Divide by the molecular weight of CO2.  Multiply by Avagadro's number.  Use scientific notation.  It's a really big number.
53064
Assuming your gas is an ideal gas, you can use the ideal gas law.\n\nPV=nRT where \nP=Pressure in atm (1 atm= 760 mm Hg so 730 mm Hg= 0.96atm\nV=volume in L\nn= moles\nT=Temperature in K (C=K+273 so 30 C = 303 K)\nR= the universal gas constant= 0.08206 L-atm/mol-K\n\nSolve the ideal gas equation for n=PV/RT\n\nn=[(.96atm)(10L)]/[(.08206 L-atm/mol-K)(303K)]\n\nn=0.386 moles
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.
756159
n   = pV/RT = 1.03*1.02/284*0.0821 = 0.045moles\n\nmoles of Nitrogen = 0.045*0.78=0.0351\n\nmoles of oxygen = 0.022*0.045 = 0.0098mole\n\nweight of the gases = 0.0098*32 + 0.0351*28.02 =0.316+ 0.986=1.3g\n\ndensity 1.3 g/l