Calculate and give the empirical formula of hydrocarbon?
961780
Both ethene and etyne have two carbon atoms\n\nThus x*12.8/100 = 24\n\nSo, x = 187.5\n\nNumber of Bromine atoms = 187.5*100/(85*80) is approximately 2\n\nSo, it is ethene having the empirical formula CH2.
363124
detlaH=H(products) - H(Reactants)\n\n200KJ/mol
1068373
If you know you're dealing with an organic compound, you take the formula weight and divide by 12.  This gives the number of carbon atoms.  You look at the remainder and determine if it could be the number of hydrogen atoms.  Alkanes have 2n + 2 hydrogen atoms (where n = number of carbons).  Alkenes have 2n and alkynes have 2n - 1.
876756
Depends on what base was used.  Was it NaOH?\n\nOne mole of anything is 6.02 x 1023 (Avogadro's number) of that thing. One mole of an element is 6.02 x 1023 atoms of that element and is a mass of that element equal to its atomic weight in grams. One mole of a compound is 6.02 x 1023 molecules of that compound and equals its formula weight (the sum of the atomic weights of all the atoms in the formula) in grams. For example, 1 mole of C = 12 g C = 6.02 x 1023 atoms; 1 mole CO2 = 44 g CO2 = 6.02 x 1023 molecules of CO2. And (grams of compound) × (formula weight) = number of moles of compound (number of moles of compound) × (formula weight) = grams of compound.\n\n\nNa = 23\nO = 16\nH = 1\n1 mole of NaOH = 40 grams
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
961793
Call it (CxHyOz)\nMass of C = 12x\nMass of H = 2y\nMass of O = 16z\n-----------------------\nIts molecular weight M = 12x + 2y + 16z = 90\n\nPercentage of C: (12x / M) *100% = 26.7% <-> 12x = 90 * 26.7 / 100 <-> 12x = 24 <-> x = 2\n\nYou can do the same with H and O to find y = 2 and z = 4.\n\nThe formula is, hence, C2H2O4.
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
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)
222097
OH-(aq) + HC2H3O2(aq)---> H2O(l) + C2H3O2-(aq)
801550
You have\nmole CH3COOH= Ca*Va= 0.2*0.2=0.04\nmole KOH= Cb*Vb= 0.2*0.2= 0.04\n\nCH3COOH + KOH -> CH3COOK + H2O\n\nThe stoichiometry is 1:1 and you have equal mole of acid and base therefore they will react completely and you will have only CH3COOK in the final solution\nHowever this is the salt of a weak acid with a strong base so it will hydrolyze according to\n\n.. .. .. ..  .. CH3COO- + H2O <=> CH3COOH + OH-\nInitial .. .. .. .. C\nReact .. .. .. ..x\nProduce .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. x .. .. .. .. .. . x\nAt Equil. .. .. C-x .. .. .. .. .. .. .. .. .. .. .. x .. .. .. . .. .. x\n\nKb=[CH3COOH][OH-] / [CH3COO-] = x^2/(C-x)\n\nBut Kb=Kw/Ka= 10^-14/(1.82*10^-5)\nC=moleCH3COOK/Vfinal = Ca*Va/(Va+Vb)=0.04/0.4\nsubstitute Kb,C and find x (you can assume that C>>x and C-x=C. this simplifies the equation to x^2/C=Kb. If you calculate x and it is truly much smaller than C the assumption is valid. Otherwise you have to solve the quadratic equation)\n\npOH=-logx and pH=14-pOH
834362
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 4.203 g x (12 / 44) = 1.146 g\nhydrogen: 3.441 g x (2 / 18) = 0.382 g\noxygen: 3.060 - 1.569 - .262 = 1.532 g\n\n2) convert to moles.\ncarbon: 1.146 / 12 = 0.0955 mol\nhydrogen: 0.382 / 1 = 0.382 mol\noxygen: 1.532 / 16 = 0.0958 mol\n\n3) divide each molar amount by the lowest value to obtain small integers\ncarbon: 0.0955 / 0.0955 = 1\nhydrogen: 0.382 / 0.0955  = 4\noxygen: 0.0957 / 0.0955 = 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.
954475
An empirical formule simply states the ratio of the atoms making up the compound, but not the total number or their arrangement. This is what is determined by elemental analysis.\n\nA molecular formula shows the total number and their relation to each other in structure...much more important for organic chem than inorganic.\n\neg(1)...C3H6O is the empirical formula for acetone, CH3COCH3, but also for methyl vinyl ether,  CH3OCH=CH2...very different compounds\n\neg(2)...CH2 is empirical for many hydrocarbons, including 2-butene, CH3-CH=CH-CH3 and cyclohexane, ...CH2CH2CH2CH2CH2CH2...(this is a cyclic compound, imagine the 'tails' connected).
195778
1.  C(s)+O2--> CO2(g) - 393.5 kj\n2.  H2(g)+1/2 O2(g)--> H20 -285.8 kj\n3.  2C2H2(g)+5O(g)-->4CO2(g)+2H2O(... -2598.8\nCALCULATE the value for the reaction 2C+H2-->C2H2\n\nFirst balance eqn 3 with 5O2 (g) then reverse it so that you have 2C2H2(g) on the RHS. Of course, you will change the sign of the eqn as well. Now halve it so you have one C2H2 instead of 2. Let's call the new eqn number 4\n\nSecond, double the 1st eqn. Let's call it eqn 5\n\nLastly, add eqns 3, 4 and 5:\n\n4 2CO2(g)+H2O --> C2H2(g)+5O2(g) +1299.4 kJmol-1\n5 2C(s)+ 2O2--> 2CO2(g) -787.0 kJmol-1\n3 H2(g)+1/2 O2(g)--> H20 -285.8 kj\n----------------------------------------------\n  2C(s)+ H2(g) --> C2H2(g) + 226.6 kJmol-1
954811
C3H8 + 5O2 --> 3CO2 + 4H2O
872375
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 7.793 g x (12 / 44) = 2.125 g\nhydrogen: 3.190 g x (2 / 18) = 0.354 g\noxygen: 3.192 - 2.125 - .354 = 0.713 g\n\n2) convert to moles.\ncarbon: 2.125 / 12 = 0.177 mol\nhydrogen: 0.354 / 1 = 0.354 mol\noxygen: 0.713 / 16 = 0.0446 mol\n\n3) divide each molar amount by the lowest value to obtain small integers\ncarbon: 0.177 / 0.0446 = 4\nhydrogen: 0.354  / 0.0446 = 8\noxygen: 0.0446 / 0.0446 = 1\n\nThe empirical formula of the substance is (C4H8O)n\nThe atomic masses add up to 72.1 g, which is also the molar mass of the compound. Therefore n=1 and we find C4H8O
895405
use dalton's law to get the pressure of CH4(total pressure - H2 pressure)\nthen use ideal gas law to get the moles of CH4(PV=nRT)\n*watch your units*\nchange moles of CH4 to grams of CH4(moles x molar mass)
903485
in C6H6 = 6 6 0\n    C14H10 = 14 10 0\n    C8H10 = 8 10 0\n    C6H4(CH3)2 = 8 10 0\n \nand in total C=36\n                   H=36\n                   O=0