11.4 – Empirical and Molecular Formulas

11.4 Objectives: Explain what is meant by the percent composition of a compound. Determine the empirical and molecular formulas for a compound from mass percent and actual mass data.

Percent Composition Every chemical compound has a definite composition. What law is this referring to? The composition of a compound is usually stated as the percent by mass of each element in the compound. The type of chemist whose job it is to identify the elements & their percent by mass in a compound is an analytical chemist.

The equation used to determine the percent composition of an element in a compound is:

Summary of Determining Percent Composition: Find the mass of each component element in the compound Divide by the total mass of the compound Multiply that answer by 100 to get a percent

Percent Composition Example Problems: 1) Determine the percent by mass of each element in calcium chloride (CaCl 2 ). K:UK: 36.11% Ca and 63.89% Cl

2) What is the percent of oxygen in H 3 PO 4 ? K:UK: 65.31% oxygen

3) Calculate the percent composition of a compound that is 29.0 g of Ag with 4.30 g of S. K:UK: 87.1% Ag and 12.9% S

4) Which has the larger percent by mass of sulfur, H 2 SO 3 or H 2 S 2 O 8 ? K:UK: H 2 SO 3  39.06% S

Empirical Formula The information from the percent composition can be used to determine the formula for a compound. The empirical formula is the simplest whole-number ratio of atoms of elements in the compound. In many cases, the empirical formula is the actual formula for the compound. EX: H 2 O or H 2 SO 4

Calculating Empirical Formulas The data used to determine the formula for a compound may be in the form of percent composition or it may be the actual masses of the elements in a given compound. If percent composition is given: 1. Assume that the total mass of the compound is g 2. Percentages of each element equals the mass in grams. 3. Turn grams to moles by using the molar mass of each element 4. Find the mole ratio by dividing everything by the smallest # of moles 5. If those numbers are whole numbers you have just found the subscripts for the formula If actual masses are given you can skip to step 3.

Example Problem: What is the empirical formula for a compound that contains 10.89% magnesium, 31.77% chlorine and the rest oxygen? K:UK: MgCl 2 O 8

Determine the empirical formula of a compound containing 2.644g of gold and 0.476g of chlorine. K:UK: AuCl

BUT…often in determining empirical formulas, the calculated mole ratios are still not whole numbers. In such cases all the mole ratio values must be multiplied by the smallest factor that will make them whole numbers.

More Examples: 1. A blue solid is found to contain 36.84% nitrogen and 63.16% oxygen. What is the empirical formula for the solid? K:UK: N2O3N2O3

2. Propane is a hydrocarbon. It is composed of 81.82% carbon and 18.18% hydrogen. What is the empirical formula? K:UK: C3H8C3H8

Molecular Formula For many compounds, the empirical formula is not the true formula. Examples: CH 2 (empirical formula) vs. C 2 H 4 (molecular formula) The molecular formula identifies the actual number of elements in a molecule. Sometimes the empirical and molecular formulas are the same. Ex: Water H 2 O To determine the molecular formula the molar mass of the compound must be determined through experimentation and compared with the mass represented by the empirical formula.

Notice that the molecular formula for acetic acid (C 2 H 4 O 2 ) has exactly twice as many atoms of each element as the empirical formula (CH 2 O). The molecular formula for a compound is always a whole-number multiple of the empirical formula.

In order to determine the molecular formula for an unknown compound, you must know the molar mass of the compound in addition to its empirical formula. Then you can compare the molar mass of the compound with the molar mass represented by the empirical formula.

This is done using the following equation: You get a number to multiply the subscripts of the empirical formula by to get the molecular formula. Let’s do some practice problems.

Practice Problems: 1) Maleic acid is a compound that is used in the plastics and textiles industries. The composition of maleic acid is 41.39% carbon, 3.47% hydrogen, and 55.14% oxygen. Its molar mass is g/mol. Calculate the molecular formula for maleic acid. K:UK:

Start by determining the empirical formula: What is the mole ratio of the elements? 1C:1H:1O

So the empirical formula is: CHO Next, calculate the molar mass represented by the empirical formula g/mol As stated in the problem, the molar mass of maleic acid is known to be g/mol. To determine the molecular formula for maleic acid, calculate the whole number multiple to apply to its empirical formula.

This calculation shows that the molar mass of maleic acid is four times the molar mass of its empirical formula CHO. Therefore, the molecular formula must have four times as many atoms of each element as the empirical formula. Thus, the molecular formula is C 4 H 4 O 4

To Review:

= n

More Practice Problems: 2) Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. It has a molar mass of 194 g. What is its molecular formula? K:UK:

3) A compound was found to contain g carbon and g hydrogen. The molar mass of the compound is g/mol. Determine the molecular formula. K:UK: C 4 H 10