A certain compound was found to have a molar mass of   74.122 g.?
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
961946
one mole of a substance is 6.02x10^23 (avogadro's number) atoms. so... you have one mole.\nso from this, there would be 22.99 grams.\n\nexplained:\nn(moles) = m(mass) / M(molecular wt.)\n1 = m / 22.99\nm = 1 * 22.99\nm = 22.99 g \n\n:-)
196290
Forumula- F2Se \nCAS Rxn #- 14017-34-6 \nMolecular Weight- 116.9568\n\nNot too much info about this compound on the internet. Sorry.
169996
Calcium(II) Telluride\n\nCalcium ion Ca2+\nTelluride ion 2-\n\nFormula Weight 167.678 g/mol
924159
It depends on what you have, if you know the mole of the molecule, you can determine the mass by converting molar mass (g/mol) into mass (g). The mole of the molecule can be found on the periodic table, but first you need to know the molecule formula of the compound. Then add it up.
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
524657
molarity = moles/volume\n\n125 g C2H5OH / 46 g/mol = 2.717 moles\n\nmolarity = 2.717/0.250 = 10.87 M
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.
768474
3.24 kmol= 3.24*10^3mol=3.24*10^3*261.992 [molecular wt of the compd.] g=848854.08 g
464799
I assume that you have an equation such as A and B react together to give C. Work out the molar ratios of each, i.e. is it 1 equivalent of A reacted with 1 equivalent of B to give 1 equivalent of C?\nAssuming it is a 1:1:1 ratio, then look at the amounts you actually reacted - chances are that you did not use 1 mole of each of the reactants. Work out the molar quantities of each reactant. (mass used/molecular weight of compound)\n\nIf you used 0.5 moles of each, then you would get 0.5moles of product and just convert this to a mass.\n\nIf one of the reactants is in excess, e.g. 0.5 moles of A and 0.8moles of B was used, ignore B because as there is only 0.5 moles of A you would only make a maximum of 0.5 moles of C.\n\nHope this helps
872523
You need to specify a compound before we can tell you what its molecular geometry is!
290781
MgSO4(s) + 2HCL(aq) ---> H2SO4(aq) + MgCl2 (aq)
65886
Synthesis - a compound forms from simpler reactants
324518
molecular weight's 305.4188 g/mol
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
882624
Update from my other answer:\n\nThe reaction is  ethanol + K2Cr2O7 -> products. This means that 1 mol ethanol will react with 1 mol K2Cr2O7. 1 mol K2Cr2O7 weighs 2*39.1 + 2*52 + 7*16 = 294.2 grams. 1.1 mg was consumed which is 3.74 * 10^-6 mol. Therefore the testsample had also 3.74 * 10^-6 mol of ethanol. Ethanol is C2H5OH, and therefore 1 mol weighs 2*12 +6 + 16 = 46 gram. \nTherefore there was 46 * 3.74 * 10^-6 = 1.72 x 10^-4 gram = 0.172 mg of ethanol.
433211
This is simple-\n3Mg(S) + N2(G) (2 is subscript) Yields (arrow) Mg3N2(S)