The percentage composition of any element is 100% (of the element).

        i)  calculate the molecular weight (or formula
             weight) by summing up the masses of all
             atoms present.
        ii) calculate the mass of the of any one
             particular element present.
        iii) compute: element mass/MW x 100
                 - this is the percentage composition
                   of that element in the compound!
        iv) repeat for all elements present
        v)  check your work: the sum of all %'s
              should be 100.

An example: (if you can do this one yourself, you'll have no problems!)

#40a) Calculate the mass percent of potassium in "penicillin G potassium," C₁₆H₁₇KN₂O₄S.

i) Calculate MW...

MW(C₁₆H₁₇KN₂O₄S) =

 16 x C = 16 x 12.011 = 192.18   u
 17 x H = 17 x 1.0078 =   17.133 u
   1 x K =  1 x 39.098   =   39.098 u
   2 x N = 2 x 14.0067 =   28.013 u
   4 x O = 4 x 15.9994 =   63.998 u
   1 x S  = 1 x 32.066   =   32.066 u
                                          372.49   u

ii) Calculate mass of potassium: only 1 atom
    therefore = 39.098 u!

iii) Compute the percentage:

    (39.098 u/372.49 u) x 100 = 10.50 %

iv) Repeat for all elements present.
(necessary to check our work,
but unnecessary otherwise for this problem...)

carbon = (192.18 u/372.49 u) x 100 = 51.59 %
hydrogen = (17.133 u/372.49 u) x 100 = 4.60 %
nitrogen = (28.013 u/372.49 u) x 100 = 7.52 %
oxygen = (63.998 u/372.49 u) x 100 = 17.18 %
sulfur = (32.066 u/372.49 u) x 100 = 8.61 %

v) These add up:

(10.50 + 51.59 + 4.60 + 7.52 + 17.18 + 8.61) % = 100 %

        i)   Compute the amount of each element (in grams)
              present in a 100 g sample of the substance
        ii)  convert each amount to moles
        iii) convert the numbers obtained into the
              smallest whole numbers possible
                    - divide each number by the smallest
                      one computed first
                    - choose a multiplier which will convert
                       all of the results into whole numbers
                            * the multiplier may be obtained
                               by inspection

                               X.25  ---> multiplier = 4
                               X.33  ---> multiplier = 3
                               X.50  ---> multiplier = 2

Example:

#14) A substance is 68.42 % chromium and
31.58 % oxygen...  What is it's empirical formula?

i) 100 g contains 68.42 g of Cr and 31.58 g of O

ii) 68.42 g Cr x 1 mole Cr      =  1.316 moles Cr
                           51.996 g Cr

    31.58 g O x   1 mole O      =  1.974 moles O
                         15.9994 g O

iii) Our formula currently is: Cr_1.316O_1.974

divide both numbers by 1.316 ---> Cr₁O_1.5
multiply by 2 according to the chart and we get:
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To get the molecular formula we need the molecular weight (by mass-spec or some other means).   We then multiply the empirical formula by some integer which gives us a molecular mass consistent with our experimental results
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An example:

A pure liquid is 92.26 % carbon and the remainder is hydrogen.  In the mass spectrum of this substance there is a tall peak at 78.11 u (m/e).  What is the molecular formula for this substance?

i) The percentage of hydrogen is 7.74%.  100 g of this substance contains 92.26 g of carbon and 7.74 g of hydrogen.

ii) 92.26 g of carbon x 1 mole carbon
                                         12 g carbon
    = 7.69 moles of carbon
 
    7.74 g hydrogen x 1 mole hydrogen
                                       1 g hydrogen
    = 7.74 moles of hydrogen

iii) divide both numbers by 7.69: CH_1.006

It is quite clear that the empirical formula is CH.

iv) Since the molecular weight is 78.11, and our empirical formula weight is 13 we need to multiply the empirical formula by 78.11/13 = 6.

Our molecular formula is therefore C₆H₆ (benzene).

Three things are necessary to experimentally measure the percentage of carbon and hydrogen in a sample.

        1. The initial mass of the substance burned
        2. The mass of the CO₂ produced
        3. The mass of the water produced

After that, all that's left is the calculation!

#52) A 0.3654 g sample of p-cresol is burned and yields 1.0420 g of CO₂ and 0.2437 g of water.  It's molecular mass (from mass-spectroscopy) is 108.1 u.

a) The percentage composition is given by:

1.0420 g CO₂ x 1 mole CO₂ x 1 mole carbon
                           44.01 g CO₂      1 mole CO₂

= 0.02368 moles C  x 12.011 g C
                                         mole C

= 0.2844 g C

= (0.2844 g/0.3654 g) x 100 = 77.83% C

= 0.02706 moles H  x 1.0078 g H
                                         mole H

= 0.02727 g H

= (0.02727 g/0.3654 g) x 100 = 7.463 % H

100 - 77.83 - 7.463 = 14.71 % O

77.83 % C, 7.46 % H, 14.71 % O

i) 100 g contains 77.83 g C, 7.46 g H, 14.71 g O
ii) Dividing each by their molar atomic masses we get:

6.480 moles C, 7.40 moles H, 0.9194 moles O

for a formula of:  C_6.48H_7.40O_0.9194

iii) dividing each by the smallest subscript (0.9194)
we get:
                        C₇H₈O

    carbon         hydrogen     oxygen
(7 x 12.01) + (8 x 1.008) + 16.00 = 108.1 u

The empirical formula IS the molecular formula.
para-cresol is C₇H₈O

The "formal" charge associated with an atom in a molecule is termed it's oxidation state.

This is a formalism, a constructive/instructive tool, but one which may not reflect reality.

How do we assign oxidation states in molecules or ionic substances?
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