Stoichiometry Sherlock Holmes, in Sir Arthur Conan Doyle’s A Study in Scarlet “In solving a problem of this sort, the grand thing is to be able to reason backward. This is a very useful accomplishment, and a very easy one, but people do not practice it much.”

The Mole 6.02 X 10 23

The Mole 1 dozen = 1 gross = 1 ream = 1 mole = x There are exactly 12 grams of carbon-12 in one mole of carbon-12.

= 6.02 x C atoms = 6.02 x H 2 O molecules = 6.02 x NaCl “molecules” (technically, ionics are compounds not molecules so they are called formula units) 6.02 x Na + ions and 6.02 x Cl – ions A Mole of Particles Contains 6.02 x particles 1 mole C 1 mole H 2 O 1 mole NaCl

Avogadro’s Number x is called “Avogadro’s Number” in honor of the Italian chemist Amadeo Avogadro ( ). Amadeo Avogadro I didn’t discover it. Its just named after me! 6.02 X 10 23

Just How Big is a Mole? Enough soft drink cans to cover the surface of the earth to a depth of over 200 miles. Enough soft drink cans to cover the surface of the earth to a depth of over 200 miles. If you had Avogadro's number of unpopped popcorn kernels, and spread them across the United States of America, the country would be covered in popcorn to a depth of over 9 miles. If you had Avogadro's number of unpopped popcorn kernels, and spread them across the United States of America, the country would be covered in popcorn to a depth of over 9 miles. If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years to count the atoms in one mole. If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years to count the atoms in one mole.

Everybody Has Avogadro’s Number! But Where Did it Come From? It was NOT just picked! It was MEASURED. It was NOT just picked! It was MEASURED. One of the better methods of measuring this number was the Millikan Oil Drop Experiment One of the better methods of measuring this number was the Millikan Oil Drop Experiment Since then we have found even better ways of measuring using x- ray technology Since then we have found even better ways of measuring using x- ray technology

The Mole 1 dozen cookies = 12 cookies 1 mole of cookies = 6.02 X 1023 cookies 1 dozen cars = 12 cars 1 mole of cars = 6.02 X 1023 cars 1 dozen Al atoms = 12 Al atoms 1 mole of Al atoms = 6.02 X 1023 atoms Note that the NUMBER is always the same, but the MASS is very different! Mole is abbreviated mol (gee, that’s a lot quicker to write, huh?)

Suppose we invented a new collection unit called a rapp. One rapp contains 8 objects. 1. How many paper clips in 1 rapp? 1. How many paper clips in 1 rapp? a) 1b) 4c) 8 2. How many oranges in 2.0 rapp? a) 4b) 8c) 16 a) 4b) 8c) How many rapps contain 40 gummy bears? 3. How many rapps contain 40 gummy bears? a) 5b) 10c) 20 Learning Check

6.02 x particles 6.02 x particles 1 mole 1 mole or or 1 mole 1 mole 6.02 x particles Note that a particle could be an atom OR a molecule! Avogadro’s Number as Conversion Factor

1. Number of atoms in mole of Al a) 500 Al atoms b) 6.02 x Al atoms b) 6.02 x Al atoms c) 3.01 x Al atoms 2.Number of moles of S in 1.8 x S atoms a) 1.0 mole S atoms b) 3.0 mole S atoms c) 1.1 x mole S atoms Learning Check

The Mass of 1 mole (in grams) The Mass of 1 mole (in grams) Equal to the numerical value of the average atomic mass (get from periodic table) Equal to the numerical value of the average atomic mass (get from periodic table) 1 mole of C atoms= 12.0 g 1 mole of C atoms= 12.0 g 1 mole of Mg atoms =24.3 g 1 mole of Cu atoms =63.5 g Molar Mass

Other Names Related to Molar Mass Molecular Mass/Molecular Weight: If you have a single molecule, mass is measured in amu’s instead of grams. But, the molecular mass/weight is the same numerical value as 1 mole of molecules. Only the units are different. (This is the beauty of Avogadro’s Number!) Molecular Mass/Molecular Weight: If you have a single molecule, mass is measured in amu’s instead of grams. But, the molecular mass/weight is the same numerical value as 1 mole of molecules. Only the units are different. (This is the beauty of Avogadro’s Number!) Formula Mass/Formula Weight: Same goes for compounds. But again, the numerical value is the same. Only the units are different. Formula Mass/Formula Weight: Same goes for compounds. But again, the numerical value is the same. Only the units are different. THE POINT: You may hear all of these terms which mean the SAME NUMBER… just different units THE POINT: You may hear all of these terms which mean the SAME NUMBER… just different units

Find the molar mass (usually we round to the tenths place) Learning Check! A.1 mole of Br atoms B.1 mole of Sn atoms =79.9 g/mole = g/mole

Mass in grams of 1 mole equal numerically to the sum of the atomic masses 1 mole of CaCl 2 = g/mol 1 mole Ca x 40.1 g/mol 1 mole Ca x 40.1 g/mol + 2 moles Cl x 35.5 g/mol = g/mol CaCl 2 1 mole of N 2 O 4 = 92.0 g/mol Molar Mass of Molecules and Compounds

Calculating Formula Mass Calculate the formula mass of magnesium carbonate, MgCO g g + 3(16.00 g) = g

Review: Chemical Equations Chemical change involves a reorganization of the atoms in one or more substances. C 2 H 5 OH + 3O 2  2CO 2 + 3H 2 O reactantsproducts 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water When the equation is balanced it has quantitative significance:

Solving a Stoichiometry Problem 1. Balance the equation. 2. Convert masses to moles. 3. Determine which reactant is limiting. 4. Use moles of limiting reactant and mole ratios to find moles of desired product. 5. Convert from moles to grams.

Working a Stoichiometry Problem 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed. 1. Identify reactants and products and write the balanced equation. Al+O2O2 Al 2 O 3 b. What are the reactants? a. Every reaction needs a yield sign! c. What are the products? d. What are the balanced coefficients? 43 2

Working a Stoichiometry Problem 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed? 4 Al + 3 O 2  2Al 2 O 3 = 6.50 g Al ? g Al 2 O 3 1 mol Al g Al 4 mol Al 2 mol Al 2 O 3 1 mol Al 2 O g Al 2 O x 2 x ÷ ÷ 4 =12.3 g Al 2 O 3 XXX

Limiting Reactant The limiting reactant is the reactant The limiting reactant is the reactant that is consumed first, limiting the amounts of products formed. that is consumed first, limiting the amounts of products formed.

molar mass Grams Moles Calculations with Molar Mass

Calculations with Moles: Converting moles to grams How many grams of lithium are in 3.50 moles of lithium? 3.50 mol Li = g Li 1 mol Li 6.94 g Li 45.1 X

Calculations with Moles: Converting grams to moles How many moles of lithium are in 18.2 grams of lithium? 18.2 g Li = mol Li 6.94 g Li 1 mol Li 2.62 X

Calculations with Moles: Using Avogadro’s Number How many atoms of lithium are in 3.50 moles of lithium? 3.50 mol Li = atoms Li 1 mol Li x atoms Li 2.11 x X

Calculations with Moles: Using Avogadro’s Number How many atoms of lithium are in 18.2 g of lithium? 18.2 g Li = atoms Li 1 mol Li6.022 x atoms Li 1.58 x g Li1 mol Li (18.2)(6.022 x )/6.94 XX

Aluminum is often used for the structure of light-weight bicycle frames. How many grams of Al are in 3.00 moles of Al? Aluminum is often used for the structure of light-weight bicycle frames. How many grams of Al are in 3.00 moles of Al? 3.00 moles Al ? g Al Converting Moles and Grams

1. Molar mass of Al1 mole Al = 27.0 g Al 2. Conversion factors for Al 27.0g Al or 1 mol Al 27.0g Al or 1 mol Al 1 mol Al 27.0 g Al 1 mol Al 27.0 g Al 3. Setup3.00 moles Al x 27.0 g Al 1 mole Al Answer = 81.0 g Al

Atoms/Molecules and Grams Since 6.02 X particles = 1 mole AND 1 mole = molar mass (grams) Since 6.02 X particles = 1 mole AND 1 mole = molar mass (grams) You can convert atoms/molecules to moles and then moles to grams! (Two step process) You can convert atoms/molecules to moles and then moles to grams! (Two step process) You can’t go directly from atoms to grams!!!! You MUST go thru MOLES. You can’t go directly from atoms to grams!!!! You MUST go thru MOLES. That’s like asking 2 dozen cookies weigh how many ounces if 1 cookie weighs 4 oz? You have to convert to dozen first! That’s like asking 2 dozen cookies weigh how many ounces if 1 cookie weighs 4 oz? You have to convert to dozen first!

molar mass Avogadro’s number Grams Moles particles molar mass Avogadro’s number Grams Moles particles Everything must go through Moles!!! Calculations

Atoms/Molecules and Grams How many atoms of Cu are present in 35.4 g of Cu? 35.4 g Cu 1 mol Cu 6.02 X atoms Cu 63.5 g Cu 1 mol Cu = 3.4 X atoms Cu

Learning Check! How many atoms of O are present in 78.1 g of oxygen? 78.1 g O 2 1 mol O X molecules O 2 2 atoms O 32.0 g O 2 1 mol O 2 1 molecule O 2

What is the percent carbon in C 5 H 8 NO 4 (the glutamic acid used to make MSG monosodium glutamate), a compound used to flavor foods and tenderize meats? a) 8.22 %C b) 24.3 %C c) 41.1 %C Percent Composition

Calculating Percentage Composition Calculate the percentage composition of magnesium carbonate, MgCO 3. From previous slide: g g + 3(16.00 g) = g

Formulas  molecular formula = (empirical formula) n [n = integer]  molecular formula = C 6 H 6 = (CH) 6  empirical formula = CH Empirical formula: the lowest whole number ratio of atoms in a compound. Molecular formula: the true number of atoms of each element in the formula of a compound.

Formulas (continued) Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio). Examples: NaClMgCl 2 Al 2 (SO 4 ) 3 K 2 CO 3

Formulas (continued) Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio). Molecular: H2OH2O C 6 H 12 O 6 C 12 H 22 O 11 Empirical: H2OH2O CH 2 O C 12 H 22 O 11

Empirical Formula Determination 1. Base calculation on 100 grams of compound. 2. Determine moles of each element in 100 grams of compound. 3. Divide each value of moles by the smallest of the values. 4. Multiply each number by an integer to obtain all whole numbers.

Empirical Formula Determination Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid?

Empirical Formula Determination (part 2) Divide each value of moles by the smallest of the values. Carbon: Hydrogen: Oxygen:

Empirical Formula Determination (part 3) Multiply each number by an integer to obtain all whole numbers. Carbon: 1.50 Hydrogen: 2.50 Oxygen: 1.00 x Empirical formula: C3H5O2C3H5O2

Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 1. Find the formula mass of C 3 H 5 O 2 3(12.01 g) + 5(1.01) + 2(16.00) = g

Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 3(12.01 g) + 5(1.01) + 2(16.00) = g 2. Divide the molecular mass by the mass given by the emipirical formula.

Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 3(12.01 g) + 5(1.01) + 2(16.00) = g 3. Multiply the empirical formula by this number to get the molecular formula. (C 3 H 5 O 2 ) x 2 = C 6 H 10 O 4

Chemical Formulas of Compounds (HONORS only) Formulas give the relative numbers of atoms or moles of each element in a formula unit - always a whole number ratio (the law of definite proportions). Formulas give the relative numbers of atoms or moles of each element in a formula unit - always a whole number ratio (the law of definite proportions). NO 2 2 atoms of O for every 1 atom of N NO 2 2 atoms of O for every 1 atom of N 1 mole of NO 2 : 2 moles of O atoms to every 1 mole of N atoms If we know or can determine the relative number of moles of each element in a compound, we can determine a formula for the compound. If we know or can determine the relative number of moles of each element in a compound, we can determine a formula for the compound.

Types of Formulas (HONORS only) Empirical Formula Empirical Formula The formula of a compound that expresses the smallest whole number ratio of the atoms present. Ionic formula are always empirical formula Molecular Formula Molecular Formula The formula that states the actual number of each kind of atom found in one molecule of the compound.

To obtain an Empirical Formula (HONORS only) 1.Determine the mass in grams of each element present, if necessary. 2.Calculate the number of moles of each element. 3.Divide each by the smallest number of moles to obtain the simplest whole number ratio. 4. If whole numbers are not obtained * in step 3), multiply through by the smallest number that will give all whole numbers * Be careful! Do not round off numbers prematurely

A sample of a brown gas, a major air pollutant, is found to contain 2.34 g N and 5.34g O. Determine a formula for this substance. require mole ratios so convert grams to moles moles of N = 2.34g of N = moles of N g/mole g/mole moles of O = 5.34 g = moles of O g/mole g/mole Formula: Formula: (HONORS only)

Calculation of the Molecular Formula (HONORS only) A compound has an empirical formula of NO 2. The colourless liquid, used in rocket engines has a molar mass of 92.0 g/mole. What is the molecular formula of this substance?

Empirical Formula from % Composition (HONORS only) A substance has the following composition by mass: % Na ; % B ; % H What is the empirical formula of the substance? Consider a sample size of 100 grams Consider a sample size of 100 grams This will contain grams of B and grams H Determine the number of moles of each Determine the simplest whole number ratio