The Mole (mol) The mole is a very large number used for counting particles. 6.022 x 1023 of anything is a mole Defined as the number of atoms in exactly 12 grams of carbon-12. 6.022 x 1023 is called Avogardro’s Number

Representative Particles ATOMS are the smallest particle of an ELEMENT MOLECULES are the smallest particle of MOLECULAR COMPOUNDS (Covalent)   FORMULA UNITS are the smallest particle of IONIC COMPOUNDS 1 mol = 6.022 x 1023 atoms, molecules, formula units

Mole Conversions Example 1: Convert 1.35 moles of carbon disulfide(CS2) into molecules?

Mole Conversions Example 2: How many moles are in 3.59 x 1021 formula units of sodium nitrate, NaNO3

Molar Mass (molecular weight) Mass of 1 mole of a substance. Represented by M & has the units of grams/mol To determine the molar mass of an element, look at the element’s atomic mass from the periodic table. To determine the molar mass of a compound, add up the atomic masses of the elements that make it up.

Find the Molar Mass of A single carbon atom. 1 mole of carbon atoms 1 mole of chlorine 12.01 amu 12.01 grams 35.45 x 2 = 70.90 grams

Examples Example 6: What is the molar mass of sulfur dioxide? Example 7: Calculate the molar mass of Al2O3 2 (26.98) + 3(16.00) =101.96 g/mol

Mole Conversions 1 mol = molar mass (g) Example 8:   Example 8: How many moles of calcium carbonate are in a stick of chalk with a mass of 14.8 g.

Mole Conversions 1 mol = molar mass (g) Example 9:   Example 9: How many atoms are in 0.551 g of potassium ?

Mole Conversions 1 mol = molar mass (g) Example 10:   Example 10: How many hydrogen atoms are in 72.5 g of C3H8O ?

Gases & The Mole Many of the chemicals we deal with are gases. They are difficult to weigh. Need to know how many moles of gas we have. Two things effect the volume of a gas Temperature and pressure Compare at the same temp. and pressure.

Standard Temperature and Pressure 0 ºC and 1 atm pressure abbreviated STP At STP 1 mole of gas occupies 22.4 L Called the molar volume Avogadro’s Hypothesis - at the same temperature and pressure equal volumes of gas have the same number of particles.

Mole Conversions 1 mol = 22.4 L Example 11   Example 11 What is the volume of 4.59 mole of CO2 gas at STP?

Percent Composition Percent of each element a compound is composed of. Find the mass of each element, divide by the total mass, multiply by a 100. Easiest if you use a mole of the compound.

Example 12: Calculate the mass percent of carbon in C2H6O So the percentage of carbon in ethane is… %C = (2)(12.01 amu) (46.08 amu) 24.02 amu 46.08 amu = x 100 = 52.1 % © 2009, Prentice-Hall, Inc.

Calculate the % composition of water in the hydrate, (CaSO4 · 2 H2O ) Example 13: Calculate the % composition of water in the hydrate, (CaSO4 · 2 H2O ) Ca 1 x 40.08 g = 40.08 g Cl 2 x 35.45 g = 70.90 g 36.02g/147.00 g H2O 2 x 18.02g = 36.02 g 24.50% H2O Copyright © Cengage Learning. All rights reserved

Emprical Formula From percent composition, you can determine the empirical formula. Empirical Formula the lowest ratio of atoms in a molecule. Based on mole ratios.

Calculating Empirical Formulas If you are given a % composition, assume you have 100 grams and switch the % unit to g. If you are given the actual number of grams of each element, use those instead Convert the grams of each element to moles of each element. Divide each mole amount by the smallest mole amount. If your answers are within .1 of whole number, round up or down and these numbers become your subscripts. If your answers are not within .1 of a whole number, you must find the smallest number that you can multiply by all the mol amounts you have just calculated to get them to be all whole numbers or within .1 of a whole number. © 2009, Prentice-Hall, Inc.

Calculating Empirical Formulas The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA. © 2009, Prentice-Hall, Inc.

Calculating Empirical Formulas Assuming 100.00 g of para-aminobenzoic acid, C: 61.31 g x = 5.105 mol C H: 5.14 g x = 5.09 mol H N: 10.21 g x = 0.7288 mol N O: 23.33 g x = 1.456 mol O 1 mol 12.01 g 14.01 g 1.01 g 16.00 g © 2009, Prentice-Hall, Inc.

Calculating Empirical Formulas Calculate the mole ratio by dividing by the smallest number of moles: C: = 7.005  7 H: = 6.984  7 N: = 1.000 O: = 2.001  2 5.105 mol 0.7288 mol 5.09 mol 1.458 mol © 2009, Prentice-Hall, Inc.

Calculating Empirical Formulas These are the subscripts for the empirical formula: C7H7NO2 © 2009, Prentice-Hall, Inc.

Calculating Empirical Formulas Example 14 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. © 2009, Prentice-Hall, Inc.

Calculating Empirical Formulas Example 15 A compound has the following percent composition: 31.9%K, 29.0 % Cl and 39.1%O. What is the empirical formula? © 2009, Prentice-Hall, Inc.

Extra Practice A substance has the following composition by mass: 60.80 % Na ; 28.60 % B ; 10.60 % H. What is the empirical formula of the substance? © 2009, Prentice-Hall, Inc.

Empirical To Molecular Formulas Empirical is lowest ratio. Molecular is actual molecule. Need Molar mass. Ratio of empirical to molar mass will tell you the molecular formula. Must be a whole number because...

Example 16 A compound is made of only sulfur and nitrogen. It is 69.6% S by mass. Its molar mass is 184 g/mol. What is its formula?

Exercise The composition of adipic acid is 49.3% C, 6.9% H, and 43.8% O (by mass). The molar mass of the compound is about 146 g/mol. What is the empirical formula? C3H5O2 What is the molecular formula? C6H10O4 Assume 100.0g. 49.3 g of carbon is 4.105 mol C (49.3/12.01). 6.9 g of hydrogen is 6.845 mol H (6.9/1.008). 43.8 g of oxygen is 2.7375 mol O (43.8/16.00). The ratio of C:H:O is 1.5:2.5:1 (C: 4.105/2.7375 = 1.5; H: 6.845/2.7375 = 2.5). Multiplying each by 2 to get a whole number becomes a ratio of 3:5:2. The empirical formula is therefore C3H5O2. The molar mass of the empirical formula is 73.07 g/mol, which goes into the molar mass of the molecular formula 2 times (146/73.07). The molecular formula is therefore C6H10O4. Copyright © Cengage Learning. All rights reserved

Sample is burned completely to form CO2 and H2O Pure O2 in CO2 is absorbed Sample is burned completely to form CO2 and H2O H2O is absorbed Combustion Analysis

How To: Use the mass of CO2 to covert to moles of CO2 to moles of C to grams of C. Use the mass of H2O and convert to moles of H2O to mole H to g H Subtract the combined mass of C and H from the mass of the sample to get the mass of oxygen. Convert it to moles. Simplify the mole ratios of C, H and O to get the empirical formula

Example 17 A 0.2000 gram sample of a compound (vitamin C) composed of only C, H, and O is burned completely with excess O2 . 0.2998 g of CO2 and 0.0819 g of H2O are produced. What is the empirical formula?

Chemical Equations Are sentences. Describe what happens in a chemical reaction. Reactants ® Products Equations should be balanced. Have the same number of each kind of atoms on both sides because ...

Balancing Chemical Equations Must have the same number of each kind of atom on both sides of the equation must exist __ CH4 + __ O2  __ CO2 + __ H2O 1 C 1 4 H 2 2 O 3

Balancing Chemical Equations Add coefficients to make atoms of each element equal on both sides 2 CH4 + 4 O2  2 CO2 + 4 H2O 2 C 2 8 H 8 8 O 8

Example 18 __ Ca(OH) 2 + __ H3PO4  __ H2O + __ Ca3(PO4) 2 __ Cr + __ S8  __ Cr2S 3 16 Cr + 3 S8  8 Cr2S 3 __ KClO 3  ___ KCl + __ O 2 2 KClO 3  2 KCl + 3 O 2

Example 18 cont. Translate this and then balance: Solid iron(III) sulfide reacts with gaseous hydrogen chloride to form solid iron(III) chloride and hydrogen sulfide gas Fe2S3 + HCl  FeCl3 + H2S 1 Fe2S3 + 6 HCl  2 FeCl3 + 3 H2S

Steps to solve problems Stoichiometry Given an amount of either starting material or product, determining the other quantities. Steps to solve problems 1. Write balanced chemical equation 2. Convert quantities of known substances into moles Use coefficients in balanced equation to calculate the number of moles of the sought quantity 4. Convert moles of sought quantity into desired units  

Meaning of Coefficients 2H2 + O2 ® 2H2O 2 dozen molecules of hydrogen and 1 dozen molecules of oxygen form 2 dozen molecules of water. 2 x (6.02 x 1023) molecules of hydrogen and 1 x (6.02 x 1023) molecules of oxygen form 2 x (6.02 x 1023) molecules of water. 2 moles of hydrogen and 1 mole of oxygen form 2 moles of water.

Conversion Factors to USE! 1 mol = 6.022 x 1023 particles molar mass: 1 mol = molar mass (g) mole ratio: ? Mol A = ? Mol B molar volume: 1 mol = 22.4 L @STP

Mole Ratio: conversion factor between 2 different amounts in a balance equation 2 Al2O3 ® 4Al + 3O2 every time we use 2 moles of Al2O3 we make 3 moles of O2 2 moles Al2O3 3 mole O2 or 3 mole O2 2 moles Al2O3

Mole to Mole Conversions 1 step process using a mole ratio to convert from moles of substance A to moles of substance B How many moles of O2 are produced when 3.34 moles of Al2O3 decompose? 2 Al2O3 ® 4Al + 3O2 3.34 moles Al2O3 3 mole O2 = 5.01 moles O2 2 moles Al2O3

4 Types of Problems Example 19 Mole- Mole How many moles of sulfur trioxide can be produced if 8.1 moles of oxygen react with sulfur. 2S + 3 O2  2 SO3

4 Types of Problems Mole-Mass: 2 step process using a mole ratio to convert from moles of substance A to moles of substance B and then the molar mass conversion factor to convert between moles of B to grams of B No example

4 Types of Problems Mass-Mole: 2 step process using the molar mass conversion factor to convert from mass of A to moles of A and then the mole ratio to convert from moles A to moles of B Example 20 One way of producing oxygen O2 involves the decomposition of potassium chlorate into potassium chloride and oxygen gas. A 25.5 g sample of potassium chlorate is decomposed. How many moles of O2 are produced? 2 KClO 3  2 KCl + 3 O 2

Example 20 A 25.5 g sample of potassium chlorate is decomposed. How many moles of O2 are produced? 2KClO 3  2 KCl + 3 O 2 1 mol KClO 3 32.00 g O 2 25.5 g KClO 3 1 mol O 2 122.55 g KClO 3 2 mol KClO 3 1 mol O 2 =.312 mol O2 =

4 Types of Problems Mass-Mass: 3 step process using the molar mass conversion factor to convert from mass of A to moles of A, then a mole ratio used to convert from moles A to moles of B, then a molar mass conversion factor to convert between moles of B to mass of B Example 21 If 209 g of methanol are used up in the combustion, what mass of water is produced? 2CH3OH + 3O2  2CO2 + 4H2O

Periodic Table Balanced Equation Periodic Table Mass g A MolesA MolesB Mass g B Decide where to start based on the units you are given and stop based on what unit you are asked for

Example 21 If 209 g of CH3OH are used up in the combustion, what mass of water is produced? 2CH3OH + 3O2 ® 2CO2 + 4H2O 1 mol CH3OH 18.02 g H2O 209 gCH3OH 4 mol H2O 32.05 gCH3OH 2 molCH3OH 1 mol H2O =235 g H2O =

We can also change Liters of a gas to moles At STP 0ºC and 1 atmosphere pressure At STP 22.4 L of a gas = 1 mole

Example 22 If 6.45 grams of water are decomposed, how many liters of oxygen will be produced at STP? 2H2O ® 2H2 + O2 1 mol H2O 1 mol O2 22.4 L O2 6.45 g H2O 2 mol H2O 1 mol O2 18.02 g H2O = .0178 L O2

Limiting Reagent If you are given one dozen loaves of bread, a gallon of mustard and three pieces of salami, how many salami sandwiches can you make? The limiting reagent is the reactant you run out of first. The excess reagent is the one you have left over. The limiting reagent determines how much product you can make

Limiting Reagent Steps to determine the LR (limiting reagent) Pick a product to convert to Convert each reactant to the selected product in moles or grams The reactant that gives the smallest answer will be the limiting reactant Once you determine the LR, you should always start with the amount given of it to answer any other parts

How much excess reagent? Use the limiting reagent to find out how much excess reagent you used Subtract that from the amount of excess you started with

Cu is Limiting Reagent = 19.0 g Cu2S Example 23: If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be formed? 2Cu + S ® Cu2S Cu is Limiting Reagent 1 mol Cu 1 mol Cu2S 159.16 g Cu2S 10.6 g Cu 63.55g Cu 2 mol Cu 1 mol Cu2S = 13.3 g Cu2S = 13.3 g Cu2S 1 mol S 1 mol Cu2S 159.16 g Cu2S 3.83 g S 32.06g S 1 mol S 1 mol Cu2S = 19.0 g Cu2S

Short Cut to Finding the Limiting reagent 1. Take each reactant amount and convert to moles. If already in moles move to the next step. 2. Divide each mole amount by the coefficient of that substance from th e balanced chemical equation. 3. Whichever is the smallest number will identify the limiting reagent (LR) 4. Use the original amount of the reactant from the problem to begin your conversion process to determine either how much excess reagent you have left over or how much product you will form.

Example 24 10.0 g Al reacts with 35.0 g Cl2 gas to produce AlCl3. Which reactant is the limiting? Which is in excess? How much product is produced? 2Al + 3 Cl2 ® 2AlCl3 1 mol Al = .3706 mol Al/2 = .1853 mol 10.0 g Al 26.98 g Al 1 mol Cl2 35.0 g Cl2 = ..4937 mol Cl2/3 = .1646 mol 70.90 g Cl2 Cl2 is LR Al is ER

10. 0 g Al reacts with 35. 0 g Cl2 gas to produce AlCl3 10.0 g Al reacts with 35.0 g Cl2 gas to produce AlCl3. Which reactant is the limiting? Which is in excess? How much product is produced? 2Al + 3 Cl2 ® 2AlCl3 1 mol Cl2 2 mol AlCl3 133.33 g AlCl3 35.0 g Cl2 3 mol Cl2 1 mol AlCl3 70.90 g Cl2 Smallest Amount  = 43.9 g AlCl3

Yield How much you get from an chemical reaction

Type of Yield The theoretical yield is the amount you would make if everything went perfect. The actual yield is what you make in the lab.

% yield = what you got x 100% what you could have got Percent Yield % yield = Actual x 100% Theoretical % yield = what you got x 100% what you could have got

Example 25 Aluminum burns in bromine producing aluminum bromide. In a laboratory 6.0 g of aluminum reacts with excess bromine. 50.3 g of aluminum bromide are produced. What are the three types of yield.

Density of a Gas D = M /V for a gas the units will be g / L We can determine the density of any gas at STP if we know its formula. To find the density we need the mass and the volume. If you assume you have 1 mole than the mass is the molar mass (PT) At STP the volume is 22.4 L.

Examples Find the density of CO2 at STP. D= 44.02 g/22.4 L = 1.97 g/L You Try: Find the density of CH4 at STP.