# Mass percent of the elements is ethylene glycol

Submitted by kellyann16AP on Mon, 10/17/2011 - 18:42
ethylene glycol is used as an automoblie antifreeze and in the manufacture of polyester fibers. the name of glycol stems from the sweet taste of this poisonous compound. combustion of 6.38mg of ethylene glycol gives 9.06mg CO2 and 5.58mg H2O. the compound contains only C H and O. what are the mass percentage of the elements in the ethylene glycol? I know that the formula is C2H6O2 (only because I looked it up). The anwser is in the back of my book, but I want to know how to solve it myself.

All the hydrogen in the sample turned into water, and all the carbon in the sample turned into carbon dioxide. From the formulas of water and carbon dioxide you know what fraction of each combustion product is the element of interest

(2.016g H/18.017g H2O) and (12.011g C/44.009g CO2)

Multiplying by those ratios, you will find the amount of H and C in the sample

(5.58mg H2O)(2.016g H/18.017g H2O) = 0.624mh hydrogen, and

(12.011g C/44.009g CO2) = 2.473mg carbon.

As those two elements add up to 3.10mg, you can calculate the amount of oxygen as the rest of the sample

6.38mg sample - 3.10mg C & H = 3.28mg O

From those amounts you could calculate mass percentages, and/or the empirical formula.

For the empirical formula, you would figure that you had

(0.623mg H)/(1.008g H/1mol H atoms) = 0.619mmol H atoms

(2.473mg C)/(12.011mg C/1mole of C atoms) = 0.206mmol C atoms

(3.28mg O)/(15.999mg O/1mol O atoms) = 0.206mmol O atoms

So you see that for each C atom you have 1 O atom  and 3 H atoms for the empirical formula CH3O.

0.206mmol O atoms/0.206mmol C atoms = 1 O atom/1 C atom

0.619mmol H atoms/0.206mmol C atoms = 3 H atoms/1 C atom

If you could determine the molecular weight, you would realize that there are double that number of atoms in a molecule of ethylene glycol, and would find the formula C2H6O2.