Calculate the density, g/cm3, of propane   C3H8   at   95.0oC   and   874 torr.?
918968
(See answer to very similar question at the second source listed below)\n\nAssuming ideal gas behavior, we have that the molar density (moles per unit volume) of a gas is given by:\n\nmolar density = n/V = P/(R*T)\n\nwhere R is the gas constant = 62.363 (liter*torr)/(mol*K) -see source\nP is the pressure\nT is the absolute temperature\n\nTo convert from molar density to mass density (mass per unit volume) we simply multiply by the molecular mass of the material in question. In this case, the molecular mass of propane is 44.1. \n\nPutting all this together, and plugging in values:\n\nmass density = (44.1 gm/mol)*(874 torr)/((62.363 liter*torr/mol*K)* (273.15+95)K)\n\nmass density = 1.679 gm/liter = 1.679e-3 gm/cm^3
189614
Gas density d is formulated as d=MP/RT where M is molar mass, P is pressure, R is universal gas constant (R=0.0821) and T is Kelvin temperature. \n\nSTP is standart pressure and temperature and after 1997 IUPAC (International Union of Pure and Applied Chemistry) started to use 0°C (273K) and 1 bar (0.987 atm).\n\nSo for Argon:\n\nd = [(39.95 g/mol) x (0.987 atm)] / [(0.0821 atm-L/mol-K) x (273K)] \n\nd = 1.759 g/L
851916
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 + 51.5) = 1.V2/273.2 (at STP of 1 atm and 0°C)\nSo V2 = 0.980 * 273.2/324.7\n= 0.825 L\nThus 3.09 g occupies 0.824 L\nSo 3.09 * 22.4/0.825 g occupies 22.4 L at STP\nie 1 g mole = 83.9\nThus MW of this gas is 83.9\nSo number of empirical units = 83.9/14.03 ≈ 6\nSo the molecular formula is C6H12
655398
Here's how to solve it (I am convinced that you should do your own homework, but hints are OK as long as you learn something):\n\nStart solving by finding the molecular mass of the compound.\nA. Volume occupied by one mol of gas is 22.4 L at standard temperature and pressure (STP: 760 mmHg and 273 K (0 C))\nTo find the volume of gas that weighs 2.57 g at STP, use formula: P1V1/T1=P2V2/T2, where R is 0.08206, P1=319 mmHg, V1=2 L, T1=298 K; P2=760 mmHg, T2=273 K and V2 is unknown (it will be somewhat less than 1 L).\nAfter you get this volume, see how many mols do you have in it: divide 22.4 by V2 you just got. Then multiply this number by 2.57 and get the molecular mass of your compound (M(X)).\n\n2. Multipy M(X) separately by 0.6481, 0.136 and 0.2159 to get number of atomic weight units that are allocated to C, H, and O respectively. \nThen divide first number by 12, 2nd by 1 and 3rd by 16 to see how many atoms it amounts to. The resulting integer numbers are X, Y, Z.\n\n3. CxHyOz is the empirical formula of your compound. Google it to get the possible molecular ones. This is a trick question, so there will be more than one but less than 8. All of them are alcohols or ethers.
842808
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 9.847 g x (12 / 44) = 2.686 g\nhydrogen: 5.038 g x (2 / 18) = 0.560 g\noxygen: 4.146 - 2.686 - .560 = 0.900 g\n\n2) convert to moles.\ncarbon: 2.686 / 12 = 0.224 mol\nhydrogen: 0.560 / 1 = 0.560 mol\noxygen: 0.900 / 16 = 0.056 mol\n\n3) divide each molar amount by the lowest value\ncarbon: 0.224 / 0.056 = 4\nhydrogen: 0.560 / 0.056 = 10\noxygen: 0.056 / 0.056 = 1\n\nThe empirical formula of the substance is (C4H10O)n\nThe atomic masses add up to 74, which is also the molar mass of the compound. Therefore n=1 and we find C4H10O
763458
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 5.601 g x (12 / 44) = 1.528 g\nhydrogen: 4.586 g x (2 / 18) = 0.510 g\noxygen: 4.078 - 1.528 - .510 = 2.040 g\n\n2) convert to moles.\ncarbon: 1.528 / 12 = 0.127 mol\nhydrogen: 0.510 / 1 = 0.510 mol\noxygen: 2.040 / 16 = 0.128 mol\n\n3) divide each molar amount by the lowest value to obtain small integers\ncarbon: 0.127 / 0.127 = 1\nhydrogen: 0.510  / 0.127 = 4\noxygen: 0.128 / 0.127 = 1\n\nThe empirical formula of the substance is (CH4O)n\nThe atomic masses add up to 32, which is also the molar mass of the compound. Therefore n=1 and we find CH4O
905654
Propane is higher.  It's because C3H8 has a higher density than CH4.  There's a formula in "Perry's Chemical Engineer's Handbook" that lets you estimate the heat value of a hydrocarbon fuel, based on its density, which I don't have available right now...however, the following should be useful:\n\nhttp://cta.ornl.gov/bedb/appendix_a/Heat_Content_for_Various_Fuels.xls
899695
The molecular weight of C2H6 is 30 g/mol, so 138 g = 138 g / 30 g/mol = 4.60 mol\n\nThe balanced combustion reaction of ethane is this:\nC2H6 + 3.5 O2 --> 2 CO2 + 3 H2O\n\nSo, for each mole of C2H6, you need 3.5 moles of O2, so you need 3.5 * 4.60 mol = 16.1 mol\n\nUse the ideal gas law to find the volume:\nPV = nRT\nP = 100 kPa\nn = 16.1 mol\nR = 0.008314 m3 kPa / (mol °K) : the ideal gas constant\nT = 27°C = 300°K\n\nV = nRT/P = 16.1 * 0.008314 * 300/100 m3\nV =  0.402 m3
912712
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 4.793 g x (12 / 44) = 1.307 g\nhydrogen: 2.452 g x (2 / 18) = 0.272 g\noxygen: 2.018 - 1.307 - .272 = 0.439 g\n\n2) convert to moles.\ncarbon: 1.569 / 12 = 0.109 mol\nhydrogen: 0.272 / 1 = 0.272 mol\noxygen: 0.439 / 16 = 0.0274 mol\n\n3) divide each molar amount by the lowest value to find small integers:\ncarbon: 0.109 / 0.0274 = 4\nhydrogen: 0.272  / 0.0274 = 10\noxygen: 0.0274 / 0.0274 = 1\n\nThe empirical formula of the substance is (C4H10O)n\nThe atomic masses add up to 74, which is also the molar mass of the compound. Therefore n=1 and we find C4H10O
759497
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 7.432 g x (12 / 44) = 2.027 g\nhydrogen: 4.057 g x (2 / 18) = 0.451 g\noxygen: 3.383 - 2.027 - .451 = 0.905 g\n\n2) convert to moles.\ncarbon: 2.027 / 12 = 0.169 mol\nhydrogen: 0.451 / 1 = 0.451 mol\noxygen: 0.905 / 16 = 0.0566 mol\n\n3) divide each molar amount by the lowest value to obtain small integers\ncarbon: 0.169 / 0.0566 = 3\nhydrogen: 0.451  / 0.0566 = 8\noxygen: 0.0566 / 0.0566 = 1\n\nThe empirical formula of the substance is (C3H8O)n\nThe atomic masses add up to 60, which is also the molar mass of the compound. Therefore n=1 and we find C3H8O
552184
C3H8 + 5O2 = 3CO2 + 4H2O\n\n1 mole of propane is required.\n\nCheck:\nC: 3 on both sides\nH: 8 on both sides\nO: 10 on both sides
885936
we have PV = m/M * R * T\nT = t(celsius) + 273\nR = 0.082 = const\nM is the molecular mass\nP is the pressure (1atm = 760 mmHg)\n\nso M = (m*R*T)/(P*V) = 3.6 * 0.082 * (23.4+273)/(4042.4/760 * 0.183 )= 90 g/mol\n\nwith the percentages we can find the compound \n\nthe formula is CxHyOz (x,y,z is the number of atoms)\n\nx : y : z = %C/M(C) : %H/M(H) : %O/M(O) = 40/12 : 6.71/1 : 53.28/16 = 3.33 : 6.71 : 3.33 = x : 2x : x\n\nwe can use this table to find the answer\nx=1 ; M = 30 : CH2O\nx=2 ; M = 60 : C2H4O2\nx=3 ; M = 90 : C3H6O3\nso the formula is C3H6O3
889941
Use the state equation of the ideal gases and solve for volume V:\n\np*V = n*R*T, V = n*R*T/p\n\nNow p = 657 Torr = 657/760 atm = 0.86 atm and R = 0.082 L*atm/mol*K, T = 283.8 oC = (283.8 + 273) K = 556.8 K so:\n\nV = 2.82*0.082*556.8/0.86 = 148.940 L
96968
mass of C4H10 = 48.0amu + 10.1amu = 58.1 amu\n(100.0g C4H10/1)(1 mole C4H10/58.1g)= 1.72 moles C4H10\nSo 100 grams of C4H10 is 1.72 moles.\n\n(1.72 moles C4H10/1)(13 moles O2/2 moles C4H10)= 11.2 moles 02\nSo you need 11.2 moles of oxygen to burn 100 grams of C4H10.\n\n(11.2 moles O2/1)(32.0g/1 mole O2)= 358g O2\n\nYou need 358 grams of oxygen.
847153
PV = nRT  .... n/R = P/RT\n\n800 torr x (1.3158×10^−3 atm/1 torr) = 1.053 atm\n\nn/R = (1.053 atm)/(8.206 x 10^-2 L atm/Kmol  x 298 K)\nn/R = 0.043 mol/L\n\nNH3 mass is (14 + 1x3)=17 g/mol\n\ndensity = n/R but we need it in g/mL\nn/R = 0.043 mol/L x (17 g/mol NH3) x (1L/1000mL)\nn/R = 7.31 x 10^-4 g/mL
412124
Do you know the Ideal Gas Law:\n\nPV = nRT?\n\nP = pressure in kPa\nV = volume the gas occupies\nn = the number of MOLES of gass\nR = 8.381\nT = Temperature in Kelvins\n\nYou might need to check for the units for volume and pressure, but that's pretty much it.  Solve for n, and use the density given to find the mass (assuming you know the MW for methane [16.04 g/mol])\n\nI'm not gonna do it for ya :-p, but it should be self explanatory
954811
C3H8 + 5O2 --> 3CO2 + 4H2O