Lecture 9

Percent Composition and Empirical Formulas 

Readings: Sections 3.7-3.8 in Kotz and Treichel 4th ed.

A. Expressing the Constant Composition of a Compound. Recall Proust's Law of Constant Composition: the ratio of masses of the elements pure samples of a given compound does not vary. We can express this in three ways:

• By atom (or mole) ratios of the elements, e.g. P₄S₃ has an atom ratio of 4 atoms of phosphorus to 3 atoms of sulfur in each molecule, or 4 moles of phosphorus to 3 moles of sulfur in each mole of P₄S₃ molecules.
  
• By mass ratios, e.g. 56.29 g of P react with 43.71 g of S to form 100.00 g of P₄S₃.

  Thus the mass ratio of P:S for P₄S₃ is 56.29 g P to 43.71 g S, or more simply,
  1.288 g P to 1.000 g S.

• By percent composition of the compound. Again, taking P₄S₃ as the example:

  If 56.29 g of P react with 43.71 g of S to form 100.00 g of P₄S₃, then the percentages of P and S are:

  % P in P₄S₃ is 56.29% P
  % S in P₄S₃ is 43.71% S

   

B. Deducing the Chemical Formula from % Composition

We have learned to calculate the molar mass of a compound according to this schematic process:

We also know how to calculate the % of a given element in the compound when the formula is known:

However, chemists often make new compounds and need to determine their chemical formulas from experimental data. Now surely if we can get % composition from the chemical formula, we should be able to reverse the process and get the formula from the composition. In other words, we will go from % composition to the formula by following this schematic process:

Example. Suppose you have a compound of phosphorus and oxygen. You analyze it and find that it contains 43.64 % P and 56.36 % O. What is its empirical formula? (The empirical formula is the formula that has the lowest whole-number ratio of the atoms.)

Solution. Step 1. Mass % to mass of each element. Let's assume that we have 100 g of the compound. Then we know from the percentages that 100 g of the compound will contain 43.64 g P and 56.36 g O.

Step 2. Mass to moles of each element.

(43.64 g P) x (1 mol P/30.97 g P) = 1.409 mol P
(56.36 g O) x (1 mol O/16.00 g O) = 3.523 mol O

Step 3. Moles to mole ratio of the elements. Now we determine the mole ratio: divide the moles of oxygen by the moles of phosphorus.

3.523 mol O/1.409 mol P = 2.500 mol O/mol P

That is, for every mole of phosphorus, we have 2.500 moles of oxygen.

Step 4. Mole ratio to empirical formula of the compound. The empirical formula must contain whole numbers of atoms. The ratio 2.5:1 is the same as the ratio 5:2. Thus, the empirical formula of our unknown compound is P₂O₅.

C. Empirical Formula vs. Molecular Formula

In Part B above, we determined that the simplest formula for our compound is P₂O₅. Is this the actual formula of an individual molecule of P₂O₅? One way to find out is to experimentally determine the molar mass of our compound by techniques that will be discussed later in the course.

Suppose that we conduct an experiment and learn that the molar mass of our compound is 283.9 g/mol. We calculate the molar mass of P₂O₅:

molar mass P₂O₅ = (2 mol P) x (30.97 g P/mol) + (5 mol O) x (16.00 g O/mol)

= 141.94 g/mol if the formula is P₂O₅

However, the experimental molar mass is 283.9 g/mol, exactly twice the calculated molar mass:

283.9 g/mol (exptl.) / 141.94 g/mol (calcd.) = 2.000

This means that we have to have twice as many atoms of P and O to make a molecule. Thus the molecular formula of our compound is P₄O₁₀.

D. Hydrates

Many compounds are known that can exist as an anhydrous form (anhydrous means "without water") and as one or more hydrated forms. One such compound is cobalt(II) chloride. In this particular case, the colours of the hydrated and anhydrous forms are dramatically different:

CoCl₂ (anhydrous cobalt(II) chloride) is deep blue
CoCl₂.6H₂O (cobalt(II) chloride hexahydrate) is pink

Note the way the formula is written for the hexahydrate. "CoCl₂.6H₂O" means that each mole of CoCl₂ combines with 6 moles of H₂O to form one mole of the new compound, CoCl₂.6H₂O. (See p. 99 in Kotz and Treichel, or look on Chapter 3 of your CD-ROM to see a movie of the color changes.)

The empirical formulas of many hydrates can be determined by heating the hydrate until all the water is driven off. If the sample is weighed before and after heating, the weight loss is due to water, and the sample weight after heating is the weight of the anhydrous compound. The empirical formula can then be determined:

mass of water lost ---> moles of water
mass of anhydrous compound ---> moles of compound

mole ratio water/compound ---> formula of hydrate

(See Example 3.13, p. 137 in Kotz and Treichel for a worked example.)

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Last modified September 29, 2000 by M. F. Richardson
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