Ch. 6 Chemical Composition and Stoichiometry The Mole Concept [6.2, 6.3] Conversions between g mol atoms [6.3, 6.4, 6.5] Mass Percent [6.6, 6.7] Empirical and Molecular Formula [6.8, 6.9] Bring your calculators! This chapter covers the meat of chemical calculations! CHEM 100 F07 1 The Mole Concept In the lab, balances measure amounts of chemicals in grams, but reactions happen on the atomic scale. How do we reconcile between these units? The MOLE!!!! Definition: the number of atoms in exactly 12 g of 12 C. Abbreviation: mol Avogadro s Number 6.022142 x 10 23 things = 1 mole What are the things? Atoms, ions, molecules, electrons, pennies, chickens. CHEM 100 F07 2 1
The Mole Concept In your lifetime, you may own 400 shoes. How many moles of shoes is that? 1 mole = 6.022142 x 10 23 shoes 400 shoes x 1 mole 6.022142 x 10 23 shoes = 7 x10-22 moles of shoes During a reaction, 0.563 mol of oxygen gas react. How many molecules react? Atoms? 0.563 mol O 2 x 6.0221 x 1023 molecules = 3.39 x 10 23 molecules 1 mol 2 atoms 3.39 x 10 23 molecules x 1 molecule = 6.78 x 1023 atoms CHEM 100 F07 3 Problems How atoms of carbon are in 0.750 moles MSG: NaC 5 H 8 NO 4? Step 1: Do you have the formula? The formula provides a ratio of atoms in a molecule Step 2: Start with the number of moles, use N A to convert to molecules Step 3: Use the formula to convert molecules to atoms 0.750 moles MSG x 6.022 x 1023 molecules 1 mole x 5 atoms C 1 molecule MSG = 2.26 x 1024 atoms C Step 2 Step 3 CHEM 100 F07 4 2
Problems How many atoms of oxygen are in 1.500 moles of aluminum sulfate? CHEM 100 F07 5 Molar Masses Conveniently, 1 amu = 1 g/mol. The atomic masses found on the periodic table are now the number of grams of the element in one mole. The same relation can be made for formula weights as well. Problem: Determine the molar masses for the following pure substances. Mo N 2 Cu 2 SO 4 C 6 H 12 O 6 CHEM 100 F07 6 3
Using Molar Masses The mole is the SI unit for amount, and allows us to do comparisons. But there are no balances that weigh in moles! We will use the molar masses to convert between moles and grams. A reaction requires 0.475 mol of Magnesium turnings. How much magnesium, in g, is that? Step 1: Do you have a formula? Step 2: Determine the molar mass. Step 3: Use the molar mass to convert g mol 0.475 mol x 24.305 g 1 mol = 11.5 g CHEM 100 F07 7 Using Molar Masses How many atoms of sodium are in 13.0 g of sodium metal? For an experiment that I am performing, I need to add 12.5 mmol of iron (II) sulfate. How many grams must I weigh out? CHEM 100 F07 8 4
Practice Problems What is the mass of 4.91 x 10 21 platinum atoms? How many atoms of copper are in a 133 kg pure copper statue? CHEM 100 F07 9 Practice Problems Determine the number of sulfur atoms in 1.59 mmol of carbon disulfide. Determine the number of H atoms are in 9.88 mol of NH 3. CHEM 100 F07 10 5
Mole-to-Mole Comparisons All chemical comparisons must be made in moles! Example: You begin a reaction with 12.3 g of Na 3. How many grams of sodium ions are in the reaction? Path: g Na 3 mol Na 3 mol Na atoms Na Don t forget to make sure you have a formula, and determine molar masses! 12.3 g Na 3 x 1 mol Na 3 163.94 g x 3 mol Na x 6.022 x 1023 atoms Na = 1.36 x 10 23 atoms Na 1 mol Na 3 1 mol CHEM 100 F07 11 Mole-to-Mole Comparisons A reaction requires 10.0 g of iron atoms. How many grams of iron (II) hydroxide are required to complete the reaction? CHEM 100 F07 12 6
Mass Percent Another way to represent amounts of atoms in a compound is through the mass percent. Mass % X = mass of X in the whole mass of the whole x 100 % Problem: Determine the mass percent of oxygen in FeSO 4. Use that value to determine the grams of oxygen in 98.75 g of FeSO 4. CHEM 100 F07 13 Determining Empirical Formula The primary use of mass percents is in the experimental determination of emprical formula. Problem: The pre-hormone androstenedione (commonly called andro) has been in the sports news in recent years owing to its arguable contribution to the breaking of the homerun record. The product has a composition that is 79.68% carbon (C), 9.15% hydrogen (H), and 11.17% oxygen by weight. What is the empirical formula for this compound? CHEM 100 F07 14 7
Practice Problems Determine the empirical formula for acetominophen, the active ingredient in Tylenol, from the elemental analysis. C 63.56%, H 6.00%, N 9.27%, O 21.17% CHEM 100 F07 15 Practice Problems Determine the empirical formula for glucose. C 40.00%, H 6.72%, O 53.29% CHEM 100 F07 16 8
Empirical vs Molecular Formula When we determined the formula in the previous problem, it was not the actual formula for glucose, but rather the empirical formula. Empirical Formula: The lowest, whole-number ratio of atoms in a molecule. Molecular Formula: The actual ratio of atoms in a molecule. Example: Glucose has an empirical formula of CH 2 O. It has a molar mass of 180.2 amu. What is the molecular formula? CHEM 100 F07 17 Practice Problem Adipic acid is used in the commercial manufacture of Nylon. The composition of the acid is 49.3% C, 6.9% H, and 43.8% O by mass. The molecular weight is 146 amu. What is the molecular formula for Adipic Acid? CHEM 100 F07 18 9