 # Chapter 3 Stoichiometry. Stoichiometry  Stoichiometry is just a long word for changing units in chemistry  If you can do Dimensional Analysis, you can.

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Chapter 3 Stoichiometry

Stoichiometry  Stoichiometry is just a long word for changing units in chemistry  If you can do Dimensional Analysis, you can do stoichiometry  Chemists need a convenient method for counting accurately the number of atoms, molecules, or formula units in a sample of a substance.

The Mole  The mole (mol) is the SI base unit used to measure the amount of a substance.  A mole of anything (atoms, particles, ions, etc.) contains 6.022 x 10 23 representative particles.  We call this number Avogadro’s Number  1 mol = 6.022 x 10 23 particles  Just like….1 dozen = 12 anything

Mole – Representative Particle Calculations  Calculate the number of atoms in 3.50 mol of copper

Mole – Representative Particle Calculations  Try the following: How many moles of MgO are in 9.72 x 10 23 molecules of MgO?

Mole – Mass Relationship  The mass in grams of one mole of any pure substance is called its molar mass.  The molar mass of any element is numerically equal to its atomic mass and has the units g/mol.  The molar mass comes from the periodic table

Calculating Molecular Mass What is the molecular mass of (C 3 H 5 ) 2 S?

Mole – Mass Calculations What is the mass of 4.21 moles of iron (III) oxide (Fe 2 O 3 )?

Mole – Mass Calculations  Try the following: How many moles of Ca(OH) 2 are in 325 grams?

Mass – Particle Conversions  How many atoms of gold are in 25.0 g of gold?

Mass – Particle Conversions  Try the following: How many grams of H 2 O are in 9.03 x 10 23 atoms of water?

Section 3.5 and 3.6 % Composition, Empirical Formulas, & Molecular Formulas

% Composition % composition = (part / whole ) x 100  When calculating the % composition, you are calculating the % of each element in a compound

% Composition  Calculate the % Composition of MgO

% Composition  Try the following: Calculate the % Composition of iron (III) oxide (Fe 2 O 3 ).

Empirical & Molecular Formulas  ________________– the smallest whole number ratio of elements  ________________– the true number of elements in a compound

Empirical Formula  What is the empirical formula for H 2 O 2 ?  What is the empirical formula for C 6 H 12 O 6 ?

Steps for Calculating the Empirical Formula 1. List your givens 2. Change % to grams 3. Change grams to moles 4. Divide everything by the smallest number of moles 5. Write your formula

Empirical Formula Problem  Calculate the empirical formula of a compound containing 40.05 % S and 59.95 % O.

Empirical Formula Problem  Calculate the empirical formula for a compound containing 48.64 g C, 8.16 g H, and 43.20 g O.

Steps for Calculating Molecular Formula 1. Calculate the empirical formula 2. Get the molecular mass of the empirical formula that you just determined 3. Divide the experimentally determined molecular mass (given) by the molecular mass of the empirical formula 4. You will get a whole number 5. Multiply everything in the empirical formula by this number

Molecular Formula Problem  Calculate the molecular formula of a compound containing 40.68% C, 5.08% H, and 54.25% O with an experimentally determined molecular weight of 118.1 g/mol

Molecular Formula Problem

C2H3O2C2H3O2C2H3O2C2H3O2  Molecular Mass =  EDMM / EFMM =  Molecular Formula

Molecular Formula Problem  Try the following: Calculate the molecular formula of a compound containing 57.84 g C, 3.64 g H, and 38.52 g O with an experimentally determined molecular mass of 249.21 g/mol

Section 3.9 and 3.10 Stoichiometric Calculations and Limiting Reactant

Mole – Mole Relationship 4 Fe + 3O 2  2Fe 2 O 3 You can determine the Mole Ratios from the balanced equation! mol Fe / mol O 2 mol Fe / mol O 2 mol Fe / mol Fe 2 O 3 mol Fe / mol Fe 2 O 3 mol O 2 / mol Fe 2 O 3

Mole – Mole Relationship How many moles of Fe 2 O 3 will form from 5.0 mol of Fe? 4 Fe + 3O 2  2Fe 2 O 3

Mole – Mole Relationship Try the following: How many moles of Mg will be needed to form 2.5 mol of MgO? How many moles of Mg will be needed to form 2.5 mol of MgO? Mg + O 2  MgO

Mass – Mole Relationship How many g of NaCl will be produced from 1.25 mol of chlorine gas reacting with sodium? 2Na + Cl 2  2NaCl Mole Ratio Mass Ratio

Mass – Mole Relationship

Mass – Mass Relationships Ammonium nitrate decomposes into dinitrogen monoxide gas and water. Determine that amount of water produced if 25.0 g of ammonium nitrate decomposes. NH 4 NO 3  N 2 O + 2 H 2 O Mole Ratio Mass Ratio

Mass – Mass Relationships

Limiting Reactant  If you are given a loaf of bread, a bottle of mustard, and four slices of salami, how many salami sandwiches can you make?

“Limiting” Reagent  The ____________is the reactant you run out of first.  The ____________is the one you have left over.  The limiting reagent determines how much product you can make

How do you find out?  Do as many stoichiometry problems as there are givens.  The one that makes the least product is the limiting reagent.

 If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be formed? 2Cu + S  Cu 2 S Mole Ratio Mass Ratio

 If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be formed? 2Cu + S  Cu 2 S Mole Ratio Mass Ratio

If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be formed? 2Cu + S  Cu 2 S 10.6 g of Cu produces 13.3 g of Cu 2 S. 3.83 g of S produces 19.0 g Cu 2 S.  ___ is the Limiting Reagent and only ____ of Cu 2 S are produced.

Another example AgCl + NaNO 3  AgNO 3 + NaCl If 10.1 g of AgCl and 2.87 g of NaNO 3 are reacted, how many grams of NaCl will be produced?

Still another example 4KBr + CS 2  2K 2 S + CBr 4 If 14.3 g of KBr are reacted with 21.7 g of CS 2 how much CBr 4 will be produced? Which reactant is the Limiting Reagent?

Chemical Yield Chemical Yield   The amount of product made in a chemical reaction. There are three types: 1.______________the amount of Product that should be made (from calculations) 2. ___________- the amount of Product formed in the laboratory (always given)

Chemical Yield Chemical Yield 3. ___________– a percentage / ratio between the actual yield and the theoretical yield. % Yield =   % yield tells us how “efficient” a reaction is.   % yield can not be bigger than 100 %.

Example #1  According to your calculations the theoretical yield for the production of NaCl is 13.6 grams. In the laboratory your actual yield is 11.8 grams of NaCl. What is the percent yield?

Example In the laboratory 6.78 g of copper is actually produced when 3.92 g of Al are reacted with excess copper (II) sulfate. 2Al + 3 CuSO 4  Al 2 (SO 4 ) 3 + 3Cu  What is the actual yield?  What is the theoretical yield?  What is the percent yield?

Types of Reactions  Synthesis A + B  AB  Decomposition AB  A + B  Single Replacement AB + C  AC + B  Double Replacement AB + CD  AD + CB  Combustion X (normally a hydrocarbon) + O 2  CO 2 + H 2 O

Try the following: 1. H 2 + O 2  2. H 2 O  3. Zn + H 2 SO 4  4. HgO  5. KBr +Cl 2  6. AgNO 3 + NaCl  7. Mg(OH) 2 + H 2 SO 3 

4.1: Water The Common Solvent

Shape of Water  Bent or V-shaped  Angle of 105°

Polarity of Water  ______ Bonds are covalent because they are sharing electrons  They are not sharing the electrons equally because the oxygen has a greater attraction for electrons  This creates a slight negative charge on the oxygen and a slight positive charge on the hydrogen

Polarity of Water Polar Molecule- a molecule that has an uneven dispersion of charge

Polarity of Water  Because of water’s polarity, it can dissolve many substances  ___________- the interaction between the dissolving molecules and water molecules  Positive ends of water molecules (close to hydrogen atoms) are attracted to anions  Negative ends of water molecules (close to oxygen atoms) are attracted to cations

Hydration

Hydration  When an ionic compound dissolves in water, the compound breaks apart into cations and anions Ex: Ca(NO 3 ) 2 (s) Ca 2+ (aq) + 2NO 3 - (aq)  Ionic compounds do not all dissolve completely because of attractions between different ions and attraction between ions and water molecules.

Dissolving Nonionic  Water can dissolve other substances, that do not create ions in solution  Many molecules that have an O-H bond like water, dissolve in water because of their polarity  Nonpolar substances (like fat molecules) do not dissolve in water  Rule: “____________________”

4.2: Aqueous Solutions Strong and Weak Electrolytes

Solutions  ________- what is being dissolved  ________- what it is being dissolved in  Arrhenius thought that conductivity came from the presence of ions  the level of conductivity depends directly on how many ions are present in solution

Strong electrolytes  conducts electricity very efficiently  all solute particles break apart into ions  includes:

Weak electrolytes  conducts electricity only slightly  a few solute particles create ions  includes:

Nonelectrolytes  does not conduct electricity  no solute particles create ions  still must dissolve  includes: soluble but nonionic compounds (ex. sugar, ethanol)

Electrolytes

4.3 Composition of Solutions

Molarity  (M) moles of solute per volume of solution  Calculate the molarity of a solution prepared by dissolving 1.56 g of gaseous HCl in enough water to make 26.8 mL of solution.

Molarity  Give the concentration of each type of ion in a 0.50 M Co(NO 3 ) 2 solution:

Standard Solutions  solution whose concentration is known

Dilutions  when water is added to a standard solution to decrease the molarity to a desired level

Dilution

4.4 Types of Chemical Reactions

Precipitation Reactions  when a solid forms from mixing liquids

Acid-Base Reactions  neutralization

Redox Reactions  when oxidation numbers change

4.5 Precipitation Reactions

Precipitation Reactions  When 2 solutions are mixed and an insoluble substance forms  ____________- this insoluble substance  To figure out what the solid is, we need to know what individual species are present in the solution after the mixture occurs

Example 1. Write the reactants as they exist in solution. We add potassium chromate and barium nitrate together and a yellow solid forms. What is it?

Example 2. Find the possibilities for the solid.  Since K 2 CrO 4 and Ba(NO 3 ) 2 are the beginning solutions, they cannot be the yellow solid.  The only other options are:

Example 3. Look at the solubility rules to figure out if either of these are soluble.  Since ______ is soluble, the solid must be: ________  So equation must be:

4.6 Describing Reactions in Solution

3 Types of Equations  ___________ - overall reaction but not the actual species in solution  ______________ – represents all strong electrolytes as ions  ____________ – Only species that undergo change are included, no spectators

Example Aqueous potassium chloride is added to aqueous silver nitrate. Molecular Equation:  Check solubility rules to figure out what symbols to use after products. KCl(aq) + AgNO 3 (aq)  KNO 3 (aq) + AgCl(s)

Example Complete Ionic Equation:  Break all strong electrolytes apart into ions  Leave solids, gases, liquids, and weak electrolytes alone.

Example Net Ionic Equation:  Cancel out all spectator ions: species that are identical on both sides of the equation  Must have same charges, states, formulas, etc.

4.7 Stoichiometry of Precipitation Reactions

Stoichiometry  Determine limiting reactant 1 st  Always write down species present  Use volume and molarity to find moles

Steps 1. Identify species present 2. Write balanced net ionic equation 3. Calculate moles of reactants 4. Determine limiting reactant 5. Calculate moles of product 6. Convert to grams or other unit

Example Sodium sulfate and lead (II) nitrate are mixed and a precipitate forms. Find the mass and identity of precipitate.

Example 1. Identify species present Molecular Equation: Complete Ionic Equation:

Example 2. Write balanced net ionic equation Spectator ions: sodium and nitrate Complete Ionic Equation:

Example 3. Calculate moles of reactants Sodium Sulfate: Lead (II) Nitrate:

Example 4. Determine the limiting reactant Limiting reactant is SO 4 2- since extra Pb 2+ is available. OR Limiting reactant is SO 4 2- since there is not enough SO 4 2-

Example 5. Calculate moles of product  Always start with LR 6. Convert to grams

Example 2 Calculate the mass of solid NaCl that must be added to 1.50 L of a 0.100 M AgNO 3 solution to precipitate all the Ag + ions in the form of AgCl.

Example 2  Write balanced net ionic equation  Find moles of Ag + that need to be precipitated

Example 2  Find the moles of Ag+ needed  Find grams of NaCl needed

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