An unknown substance has the composition of 77.87% c, 2.63% h and 55.62% o. the compound has a molar mass of 345.2 g/mole. what is the molecular formula?

If we use 100 grams of compound:
m(C) = 0,7787 · 100 g = 77,87 g.
n(C) = 77,87 g ÷ 12 g/mol = 6,49 mol.
m(H) = 0,0263 · 100 g = 2,63 g.
n(H) = 2,63 g ÷ 1 g/mol = 2,63 mol.
m(O) = 0,5562 · 100 g = 55,62 g.
n(O) = 55,62 g ÷ 16 g/mol = 3,48 mol.
n(C) : n(H) : n(O) = 6,49 mol : 2,63 mol : 3,48 mol.
n(C) : n(H) : n(O) = 2,5 : 1 : 1,33.
From molecular mass only possible is C₁₂H₉O₁₂.

; lc

  remember that hf is only the energy of one photon. not two. not three. one.  1) 5.12 * 10^3 photons  find the energy of one photon, then determine how many photons are needed to reach the sensitivity.  e = hf  = (6.626 * 10^-34 js)(5.95 * 10^19 hz)  = 3.94247 * 10^-14 j per photon  (2.02 * 10^-10 j) / (3.94247 * 10^-14 j/photon) = 5123.69149 photons  round to three sig figs.  2) 5.77 * 10^-17 j  energy of pulse = energy per photon * number of photons  = hfn  = (6.626 * 10^-34 js)(1.66 * 10^13 hz)(5250)  = 5.774559 * 10^-17 j  round to three sig figs.  3) 4.07 * 10^8 kj/mol  10^12 pm = 1 m  c = fλ  f = c/λ  e = hf  = hc/λ  = [(6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.94 * 10^-12 m)](1 kj/1000 j)(6.02 * 10^23 photons/mol)  = 40702571.4 kj/mol  round to three sig figs.  4) 9.94 * 10^-1 m  e = hc/λ  λ = hc / e  = (6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.00 * 10^-25 j)  = 0.9939 m  round to three sig figs.  5) 9.06 * 10^3 photons  energy of pulse = energy per photon * number of photons  number of photons = energy of pulse / energy per photon  = energy of pulse / (hf)  = (3.87 * 10^-13 j) / [(6.626 * 10^-34 js)(6.45 * 10^16 hz)]  = 9055.23695 photons  round to three sig figs.  6) 2.78 * 10^-15 j  this is problem 1, except in reverse.  sensitivity = minumum number of photons * energy per photon  = (5930)(6.626 * 10^-34 js)(7.077 * 10^14 hz)  = 2.78070758 × 10^-15 j  round to three sig figs.