1. Chemistry 1210: General Chemistry Dr. Gina M. Florio 13 September 2012 Jespersen, Brady, Hyslop, Chapter 4 The Mole and Stoichiometry

2. Conversion Factors Conversion Factor – relates one quantity to another – used to convert between two units in chemistry What is my height in centimeters (cm) if I am 5 feet 4 inches tall? 1. How many inches are in a foot? 2. How many inches are in a centimeter? 12 inches = 1 foot 1 inch = 2.54 cm

3. Factor Label Method The factor-label method, or dimensional analysis lets us treat a numerical problem as one involving a conversion from one kind of units to another using conversion factors. A recipe calls for 32 grams of cheese. My kitchen scale only displays weight in ounces. How many ounces of cheese do I need to use? Example: Conversion Factor Factor Label Method

4. Atomic Mass and Molecular Mass Atomic Mass– the mass of an individual atom measured in atomic mass units, u – listed in the periodic table beneath the element symbol Molecular Mass – the mass of an individual molecule measured in u – the sum of the atomic masses of the atoms in the molecular formula Example: The molecular mass of water, H 2 O, is twice the mass of hydrogen (2 x 1.008 u) plus the mass of oxygen (15.999 u) = 18.015 u Ch 4.1

5. Formula Mass – the mass of an individual formula unit measured in atomic mass units, u – used for ionic compounds – calculated the same as a molecular mass Example: The formula mass of calcium oxide, CaO, is the mass of calcium (40.08 u) plus the mass of oxygen (15.999 u) = 56.08 u Ch 4.1

6. The Mole: Connecting the macroscopic & molecular The mole is a conversion factor – relates mass to the number of atoms or molecules for a chemical substance (element, molecular compound, ionic compound) – one mole of a substance has a mass (g) equal to its formula (or molecular or atomic) mass (u) Example H2OH2O Molecular Mass = 2(H) + (O) = 2(1.0 g) + 16.0 g = 18.0 g 1 mole H 2 O = 18.0 g H 2 O Ch 4.1

7. How many water molecules are in a mole of water? Need the conversion factor between mass (g) and u: 1 u = 1.66 x 10 -27 kg 1 kg = 1000 g Answer: 1 mole H 2 O = 18.0 g H 2 O Recall: 1 molecule H 2 O = 18.0 u (molecular mass) Need the conversion factor between g and kg: Ch 4.1

8. Avogadro, how many ______ are in a mole? 6.023 x 10 23 ______ are in a mole. The “blank” can be anything: Always. Cats Flowers Molecules This conversion factor (6.023 x 10 23 objects/mole) is known as Avogadro's number. Grains of sand Ch 4.1

9. 1. Notice the magnitude of Avogadro's number (6.023 x 10 23 ): ~10 23 Avogadro’s Number Why is this number so HUGE ? Because atoms and molecules are so tiny. A huge number of atoms or molecules are needed to make a lab-sized sample. Two important points: Ch 4.1 2. Avogadro’s number links moles and atoms, or moles and molecules, and provides an easy way to link mass and atoms or molecules.

10. Example Problem 1. Convert mass (g) to moles using molar mass 2. Convert moles to number of atoms (or molecules) using Avogadro’s number Assuming that pennies are 100% Cu, how many Cu atoms are found in a penny weighing 3.00 g? Ch 4.1

11. Mole-to-Mole Ratios: Chemical Formulas Water (molar mass 18.015 g/mol) as an example: 1 mole H 2 O  6.023 x 10 23 molecules H 2 O 1 mole H 2 O  18.015 g H 2 O 18.015 g H 2 O  6.023 x 10 23 molecules H 2 O Within chemical compounds, moles of atoms always combine in the same ratio as the individual atoms themselves. 1 mole H 2 O  2 moles H 1 mole H 2 O  1 mole O Mole-to-Mole Ratio: Recall: Ch 4.2

12. Stoichiometry Ch 4.2 – relates the masses of reactants needed to make a compound – the study of the mass relationships in chemical compounds and reactions – MOLE to MOLE ratios

13. Example Problem ANALYSIS: 15.0 g Fe 2 O 3  ? g Fe KNOWN: 1 mol Fe 2 O 3  2 mol Fe 1 mol Fe 2 O 3  159.7 g Fe2O3 1 mol Fe  55.85 g Fe How many grams of iron are in a 15.0 g sample of iron(III) oxide? Ch 4.2 SOLUTION:

14. Percent Composition Ch 4.2 Percentage composition (or percentage composition by mass) The percentage by mass is the number of grams of the element in 100 g of the compound and can be calculated using: From our previous example:

15. Percent Composition, Why Should I Care? Hmmm…what in the world is this stuff? Detective, send that sample back to the lab for analysis!

16. Sample Problem Ch 4.2 A sample was analyzed and found to contain 0.1417 g nitrogen and 0.4045 g oxygen. What is the percentage composition of this compound? ANALYSIS:Find sample mass and calculate % SOLUTION: KNOWN: Mass of the whole sample

17. Molecular Formula & Empirical Formula Ch 4.3 Hydrogen peroxide consists of molecules with the formula H 2 O 2. This is called the molecular formula. However, the simplest formula for hydrogen peroxide is HO and is called the empirical formula. We can calculate the empirical formula for a compound from mass data. The empirical formula thus contains the simplest mole-to-mole ratio of the atoms in the compound.

18. Example Problem Ch 4.3 A 2.012 g sample of a compound contains 0.522 g of nitrogen and 1.490 g of oxygen. Calculate its empirical formula. ANALYSIS: We need the simplest whole number mole ratio between nitrogen and oxygen SOLUTION: Convert to moles Find the simplest whole number mole-to-mole ratio

19. Empirical Formula (a few final notes) The formula for an ionic compound is the same as the empirical formula. Ch 4.3 Example: The empirical formula of hydrazine is NH 2, and its molecular mass is 32.0 g/mol. What is its molecular formula? For molecules, the molecular formula and empirical formula are usually different. The molecular formula will be a common multiplier times all the coefficients in the empirical formula.

20. Indirect Analysis of Empirical Formula Combustion Analysis: an indirect method used to determine the empirical formula of a unknown compound containing only C, H, and O. Ch 4.3 When a compound made only from C, H, and O burns completely in pure oxygen (O 2 ) only carbon dioxide (CO 2 ) and water (H 2 O) are produced: We can collect and quantify all of the carbon dioxide (CO 2 ) and water (H 2 O) produced to find the empirical formula of the original compound.

21. Example Problem The combustion of a 5.217 g sample of a compound of C, H, and O gave 7.406 g CO 2 and 4.512 g of H 2 O. Calculate the empirical formula of the compound. ANALYSIS: This is a multi-step problem. Ch 4.3 First determine the mass of each element. Then use the masses of the elements to calculate the empirical formula of the compound.

22. Example Problem SOLUTION: Ch 4.3 The combustion of a 5.217 g sample of a compound of C, H, and O gave 7.406 g CO 2 and 4.512 g of H 2 O. Calculate the empirical formula of the compound.

23. The coefficients of a balanced chemical equation provide the mole-to-mole ratios for the substances involved in the reaction. Stoichiometry & Chemical Equations Whenever a problem asks you to convert between different substances, the calculation usually involves a mole-to-mole relationship. Example: If 0.575 mole of CO 2 is produced by the combustion of propane, C 3 H 8, how many moles of oxygen are consumed? The balanced equation is: C 3 H 8 + 5 O 2  3 CO 2 + 4 H 2 O Ch 4.4

24. Balancing Chemical Equations Chemical equations provide quantitative descriptions of chemical reactions. Conservation of mass is the basis for balancing equations. To balance an equation: 1. Write the unbalanced equation. 2. Adjust the coefficients to get equal numbers of each kind of atom on both sides of the arrow Ch 4.4

25. Balancing Chemical Equations: some tips Guidelines for Balancing Equations: 1. Balance elements other than H and O first 2. Balance as a group any polyatomic ions that appears unchanged on both sides of the arrow 3. Balance separately those elements that appear somewhere by themselves 4. As a general rule, use the smallest whole-number coefficients when writing balanced chemical equations _ NH 3 + _ O 2  _ NO + _ H 2 O Example: ElementReactantsProducts N O H ElementReactantsProducts N O H Ch 4.4 _ NH 3 + _ O 2  _ NO + _ H 2 O

26. Limiting Reactants (aka Limiting Reagent) All reactions eventually use up a reactant and stop. Any reagent that is not completely consumed during the reactions is said to be in excess. The reactant that is consumed first is called the limiting reagent (reactant) because it limits the amount of product that can form. The computed amount of product is always based on the limiting reagent. Ch 4.5

27. Example: How many grams of NO can form when 30.0 g NH 3 and 40.0 g O 2 react according to: Example Problem 4 NH 3 + 5 O 2  4 NO + 6 H 2 O Ch 4.5

28. Percentage Yield The amount of product isolated from a chemical reactions is almost always less than the calculated, or maximum, amount. The actual yield is the amount of the desired product isolated. The theoretical yield is the amount that would be recovered if no loss occurred (the calculated, maximum amount). Ch 4.6 The percentage yield is the actual yield as a percentage of the theoretical yield: