Empirical+Formulae

Empirical Formula [91--93] --From percentage to formula

__Ex__: The following compounds, CH2 ,C2H8, C4H16 all have the same percentage composition. Therefore, the emperical formula (simplest formula) is CH2.
 * __Definition__:** Empirical formula is a formula that gives the smallest ratios of atoms in a compound.
 * The lowest whole number ratio of elements in a compound.
 * The molecular formula the actual ratio of elements in a compound.
 * The two can be the same.
 * CH2 empirical formula
 * C2H4 molecular formula
 * C3H6 molecular formula
 * H2O both.

The steps to determine the empirical formula: __ Example#1: __ Calculate the empirical formula of a compound composed of 38.67% C, 16.22% H, and 45.11% N. __ Solution: __ Assume 100g so that 36.67g C x 1 mol C/ 12.01g C = 3.220 mole C 16.22g H x 1 mol H/1.01g H =16.09 mole H 45.11g N x 1 mol N/14.01g N =3.219 mole N If we divide all of these by the smallest one, it will give us the empirical formula. __ Example#2: __ A compound is 43.64% P and 56.36% O. What is the empirical formula? • Divide both by the lowest one The ratio is 3.5mol O/1.4mol P = 2.5mol O/1mol P P 1 O 2.5 Multiply the result to get rid of any fractions 2 x P 1 O 2.5 = P 2 O 5
 * 1) Pretend that you have a 100 gram sample of the compound.
 * 2) Then, change the % to grams.
 * 3) Convert the grams to mols for each element.
 * 4) Write the number of mols as a subscript (like the 2 in H2O) in a chemical formula.
 * 5) Divide each number by the least number.
 * 6) Multiply the result to get rid of any fractions.
 * 3.220 mole C
 * 16.09 mole H
 * 3.219 mole N
 * C3.22H16.09N3.219
 * The ratio is 3.220mol C/3.219mol N = 1mol C/1mol N
 * The ratio is 16.09mol H/3.219mol N = 5mol H/1mol N
 * C1H5N1 is the empirical formula
 * 43.6g P x 1mol P/30.97g P = 1.4mole P
 * 56.36g O x 1mol O/16.0g O = 3.5mole O
 * P 1.4 O 3.5