Sec. 10.4: Empirical & Molecular Formulas Chapter 10 : The Mole Sec. 10.4: Empirical & Molecular Formulas

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

Synthetic chemists are often involved in developing new compounds for industrial, pharmaceutical and home uses.

It is the analytical chemist’s job to identify the elements that a compound contains and determine its chemical formula.

Percent Composition The first step in this chemist’s job is to determine the percent by mass of the elements in the compound. Recall that the percent by mass of any element in a compound is found by: Mass of Element x 100 Mass of Compound

The percent by mass of each element in a compound is the percent composition of a compound. Percent composition can be determined from experimental data the chemical formula

From Experimental Data A chemist has a 100.0 g sample of XY. The sample contains 55.0 g X and 45.0 g Y. Percent of X = 55.0 g/100.0 g x 100 = 55% Percent of Y = 100 – 55 = 45%

From the Formula Assume you have 1 mol of the compound. 1 mol of H2O = 18.0 g In 1 mol of H2O, there are 2 moles of H atoms and 1 mol O atoms 2 mole H atoms x 1.0 g = 2.0 g H 1 mol H atoms 1 mol O atoms = 16.0 g % H = 2.0g/18.0g = 11 % % O = 100 – 11 = 89%

Practice Problems Calculate the percent composition of sodium sulfate (Na2SO4). Which has the larger percent by mass of sulfur, H2SO3 or H2S2O8? What is the percent composition of a compound that contains 2.644 g of gold and 0.476 g of chlorine? 36.11% Ca 63.89% Cl

Empirical Formula The empirical formula is the simplest formula for a compound. A molecular formula is the same as the empirical formula or it is a multiple of the empirical formula; it is the actual number of atoms of each element in one molecule or formula unit of the compound. For example, if the empirical formula of a compound is C3H8 , its molecular formula may be C3H8 , C6H16 , etc.

An empirical formula expresses the ratio of atoms in a compound An empirical formula expresses the ratio of atoms in a compound. The molecular formula will be a whole-number multiple of the empirical formula Empirical formulas can be calculated from experimentally measured mass percent data. An additional experiment (for example, a gas density measurement) would be required to get a molecular formula.

Empirical Formula Benzene Molecular formula C6H6 Empirical Formula CH Spacefilling model of benzene Benzene is a common and important compound with formula C6H6. The empirical formula is CH. Example of a compound whose molecular formula differs from its empirical formula.

Empirical Formula Acetylene Molecular Formula C2H2 Empirical Formula ? Spacefilling model of acetylene Acetylene is a common and important compound with formula C2H2. The empirical formula is CH. Example of a compound whose molecular formula differs from its empirical formula. Acetylene Molecular Formula C2H2 Empirical Formula ?

Empirical Formula Glucose Molecular Formula C6H12O6 Spacefilling model of glucose Glucose is a carbohydrate important to sustaining life. It is a compound with formula C6H12O6. The empirical formula is CH2O. Example of a compound whose molecular formula differs from its empirical formula.

Empirical Formula Carbon Dioxide Molecular Formula CO2 Spacefilling model of carbon dioxide Carbon dioxide is a common and important compound with formula CO2. The empirical formula and the molecular formula are the same, CO2. CO2 is one of many examples of compounds whose empirical formulas and molecular formulas match. Carbon Dioxide Molecular Formula CO2 Empirical Formula ?

Empirical Formula The empirical formula may or may not be the same as the molecular formula. If they are different, the molecular formula will be a multiple of the empirical formula.

To Determine Empirical Formula Molar Mass

Empirical Formula The percent composition of a compound was found to be 40.05% S and 59.95% O. What is the empirical formula for the compound? Al2S3

Empirical Formula Step One: Assume you have 100 g of the compound**; Step Two: use molar mass to determine moles of each element. 40.05 g S x 1 mol S = 1.248 mol S 32. 1 g 59.95 g O x 1 mol O = 3.747 mol O 16 g ** If you are given mass data instead of % composition, determine moles directly from the grams.

Empirical Formula Step 3: Divide both mole values by the smaller of the two to get the mole ratio of the elements. 1.248 mol S = 1 mol S 1.248 3.747 mol O = 3 mol O

Empirical Formula Now use the simplest, whole number mole ratio as subscripts in the empirical formula: 1 mol S: 3 mol O SO3

Practice Problems What is the empirical formula for a compound that contains 10.89% Mg, 31.77% Cl, and 57.34% O? Determine the empirical formula for a compound that contains 74.19% Na and 25.81% O. When an oxide of potassium is decomposed, 19.55 g of K and 4.00 g of O are obtained. What is the empirical formula of the compound? N2O3

Honors Practice Problem Determine the empirical formula for a compound that contains 35.98% aluminum and 64.02% sulfur. The chemical analysis of aspirin indicates that the molecule is 60.00% C, 4.44% H, and 35.56% O. Determine the empirical formulas for aspirin.

Molecular Formula Compounds with the same empirical formula can have very different properties. Remember, the empirical formula does not always indicate the actual number of moles in the compound. So, different compounds can have the same empirical formula. Acetylene (C2H2) and benzene (C6H6) are different compounds with the same empirical formula, CH.

Molecular Formula In order to distinguish between different compounds with the same empirical formula, a chemist must go one step further and determine the compound’s molecular formula. The molar mass of the compound is determined through experimentation and compared with the molar mass represented by the empirical formula.

Molecular Formula Suppose a compound has an empirical formula of ClCH2 and a molar mass of 98.96 g/mol. How can its molecular formula be determined? Step One: Find the molar mass of the empirical formula. ClCH2 = 35.5 + 12.0 + 2(1.0) = 49.5 g/mol

Molecular Formula Step 2: Divide the molar mass of the compound by the molar mass of the empirical formula. 98.96 g/mol = 1.999 = 2 49.5 g/mol The molecular formula is this multiple of the empirical formula.

Molecular Formula Step 3: Multiply the subscripts in the empirical formula by this multiple. ClCH2 becomes Cl2C2H4. This is the molecular formula of the compound.

Practice Problems The empirical formula of a compound is found to be C2H3O2. It has a molar mass of 118.1 g/mol. Determine the molecular formula for the compound. The molar mass for a compound having the empirical formula of CH is found to be 78.1 g/mol. What is the molecular formula for the compound?

Honors Practice Problem A colorless liquid composed of 46.68% nitrogen and 53.32% oxygen has a molar mass of 60.01 g/mol. What is the molecular formula? A compound was found to contain 49.98 g C and 10.47 g H. The molar mass of the compound is 58.12 g/mol. Determine the molecular formula. N2O2