stoichiometry & solution stoich chapter 3 and 4.2/4.6

61
Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Upload: bruce-sanders

Post on 28-Dec-2015

228 views

Category:

Documents


1 download

TRANSCRIPT

Page 1: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Stoichiometry & Solution Stoich

Chapter 3 and 4.2/4.6

Page 2: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Chemical Reactions/Equations

Reactants are listed on the left Products on the right Atoms are neither created nor

destroyed Shows what state of matter the

compounds are in: Solid (s), Liquid (l), Gas (g), or dissolved in water (aqueous solution) (aq)

Page 3: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Chemical Equations

Hydrogen burns and reacts with oxygen in the air to form water. write the chemical reaction:

H2 (g)+ O2 (g) H2O (l) What’s wrong with the above equation? Due to the Law of Conservation of

Mass, it balances out to:

2H2 (g)+ O2 (g) 2H2O (l)

Page 4: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Balancing Equations

Determine what reaction is occurring (sometimes it helps to write it in word form)

Write the unbalanced (skeleton) eqn. Balance the equation by inspection, by

adding coefficients (usually works best going from left to right)

Include phase information

Page 5: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Balancing Equations

Al2(SO4)3 + Ca(OH)2 Al(OH)3 + CaSO4

H3PO4 + NaOH Na3PO4 + HOH

Page 6: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Types of Reactions

Knowing the basic types of chemical reactions, helps you to predict the products of many reactions

Combination Reaction:

2Mg (s) + O2 (g) 2MgO (s) Decomposition Reaction:

CaCO3 (s) CaO (s) + CO2 (g)

Page 7: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Combustion Reactions: Rapid reactions that produce a flame Most involve O2 (from air) as a reactant

When Hydrocarbons (CxHy) react with O2 they produce CO2 and H2O.

C3H8(g) + 5O2 (g) 3CO2 (g) + 4H2O(g)

Page 8: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

When an air bag deploys, sodium azide (NaN3) decomposes, rapidly releasing nitrogen gas and sodium.

1. What type of RXN took place?

2. Write a balanced equation

Page 9: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Balancing Combustion RXNs

first start with those elements that occur in the fewest chemical formulas.

Try a few:

CH4 + O2 CO2 + H2O

C2H5OH + O2 CO2 + H2O

Page 10: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

How much sand?

Page 11: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

3 Ways to Measure Matter

By COUNT – 1 million grains of sand By MASS – 1,000 grams of sand By VOLUME – 100 liters of sand

Page 12: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

ATOMIC MASS:

What is the atomic mass of Hydrogen? 1.01 a.m.u.

What is the atomic mass of Oxygen? 15.999 16.0 a.m.u.

Page 13: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

= +

SO3 1 S atom 3 O atomsYou can calculate the mass of a molecule by adding the atomic masses of the atoms making up the molecule. (32.1 + 16 + 16 + 16 = 80.1 amu)

Page 14: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

What is the atomic mass of Water (H2O)? 2H – 2 (1.0) = 2.0 a.m.u. 1O – 1(16.0) =16.0 a.m.u.

18.0 a.m.u

What is the atomic mass of Ca(NO3)2? 1 Ca – 1(40.1) = 40.1 a.m.u 2 N – 2 (14.0) = 28.0 a.m.u. 6 O – 6 (16.0) = 96 a.m.u.164.1 a.m.u.

Formula Weights

Page 15: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Percentage Composition

The percentage by mass contributed by each element in the substance

Calculating any percentage is just like calculating your grade in a class (“the part, divided by the whole, multiplied by 100”)

Calculate the percentage by mass of each element in Ca(NO3)2

Page 16: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Calculate the percentage composition of Oxygen in glucose (C6H12O6)

Page 17: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Problem Solving Hints

1. Analyze the Problem

2. Develop a plan for solving the problem

3. Solve the problem

4. Check the solution

Page 18: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

The Mole

Page 19: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

What is a MOLE? A mole is a quantity equal to Avogadro’s

Number (6.02 x 1023) 6.02 x 1023 particles (atoms or molecules)

– depending on what you are looking at. A mole of anything contains the same

number of “things” as a mole of anything else.

one mole is set by defining one mole of carbon 12 atoms to have a mass of exactly 12 grams.

Page 20: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Molar Mass

The mass in grams of one mole of a compound

It is numerical equivalent to what it’s atomic weight was in a.m.u.’s

Molar Mass of Water (H2O)? 18 grams/mole

Molar Mass of Calcium Nitrate? 161.4 grams/mole

Page 21: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

divide

multiply

Multiply by 22.4

Divide by 22.4

Use 6.02 x 1023

Page 22: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Practice Problems

How many moles are in 5g of Copper? How many grams in 3.2 moles of Oxygen? How many atoms in .350 moles of

Sodium? How many moles in 7.2 x 1024 atoms of

gold? How many glucose molecules in 5.23

grams of glucose, C6H12O6? How many atoms of Oxygen?

Page 23: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Determining the Empirical Formula of a Compound

1. Determine the percentage of each element in your compound

2. Treat % as grams, and convert grams of each to moles of each element

3. Find Smallest whole number ratio (divide the larger number by the smaller one)

4. If ratio is not all whole numbers, multiply each by an integer to make all whole number ratio

Page 24: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Example 1

Mercury forms a compound with chlorine that is 73.9% mercury and 26.1% chlorine.

1. Convert to 73.9 g Hg and 26.1 g Cl

2. Convert to moles of each element

3. Find Smallest whole number ratio (divide the larger number by the smaller one)

Page 25: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Example 2

Ascorbic acid (Vitamin C) contains 40.92% C, 4.58% H, and 54.5% O by mass. What is the empirical formula of ascorbic acid?

C:H:O = 3 (1:1.33:1) = 3:4:3

C3H4O3

Page 26: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Determining the molecular Formula

Find the empirical formula mass. Divide the known molecular mass by

the empirical formula mass, deriving a whole number, n.

Multiply the empirical formula by n to derive the molecular formula.

Page 27: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Example 1

The empirical formula of ascorbic acid is C3H4O3

1. The empirical formula mass is 88.0 amu.2. The experimentally determined

molecular weight is 176 amu.3. Therefore the molecule has twice the

mass (176/88 = 2.00) and must have twice as many of each atom for a molecular formula of C6H8O6

Page 28: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Example 2

Mesitylene, has an empirical formula of C3H4. The experimentally determined molecular weight of this substance is 121 amu. What is the molecular formula of mesitylene?

Page 29: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Combustion Analysis to determine empirical formulas

Used to figure out how much Carbon and Hydrogen were in the original sample of the Hydrocarbon

Page 30: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Example 1

Isopropyl alcohol (rubbing alcohol) is composed of C, H, and O. Combustion of .255g of isopropyl alcohol produces 0.561 g CO2 and 0.306g H2O. Determine the empirical formula of isopropyl alcohol.

1. Calculate the number of grams of C present in the CO2 and grams of H present in H2O using the mole concept and dimensional analysis.

Page 31: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

2. Calculate the mass of O in the final sample by subtracting the mass of C and H in the sample from the total sample mass: Mass of O = mass of sample – (mass of C + mass of H)

3. Then calculate the number of moles of C, H, and O in the sample

4. Find the lowest whole number ratio

Page 32: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Example 2

Caproic acid, which is responsible for the foul odor of dirty socks, is composed of C, H, and O atoms. Combustion of a 0.225g sample of this compound produces 0.512g CO2 and 0.209 g H2O. What is the empirical formula of caproic acid?

It has a molar mass of 116 g/mol. What is its molecular formula?

Page 33: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Stoichiometric Calculations:Quantitative info from balanced EQNs

1. Balance the chemical equation2. Convert grams of reactant or product

to moles.3. Compare moles of the known to moles

of the desired substance (use a ratio derived from the coefficients in the balanced equation.)

4. Convert from moles back to grams if required.

Page 34: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Chemical Calculations

Mole to Mole

Mole to Gram

Gram to Gram

g g

m m

m m

g

m m

Page 35: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

g g

m m

Page 36: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Mole-Mole Calculations

How many moles of ammonia (NH3) are produced when 0.60 moles of nitrogen reacts with hydrogen?

Step 1: Write Chemical EQN N2 + H2 NH3

Step 2: Balance EQN N2 + 3H2 2NH3

Page 37: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Step 3: Find known & unknown, then calculate

Mole Ratio: 1 mol N2

2 mol NH3

Unknown: ? mol NH3

Known: 0.60 mol N2

0.60 mol N2 x 2 mol NH3 = 1.2 mol NH3

1 mol N2

Page 38: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Gram-Gram Calculations

Calculate the number of grams of NH3 produced by the reaction of 5.40 g of hydrogen with an excess of nitrogen.

N2 + 3H2 2NH3

5.40 g ? grams

g g

m m

g g

m m

1

2

3

Page 39: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

5.40 g H2 x 1 mol H2 = 2.70 mol H22.0 g H2

Step 2 : Get a mole ratio fromthe equation.

3 mol H22 mol NH3

then 2.7 mol H2? mol NH3

Step 3 : Solve for the unknown number of moles. (2.7 x 2) / 3 = 1.8 moles NH3

Step 1 : Change grams to moles.

Page 40: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Step 4 : Change moles of theunknown to grams.

1.8 mole NH3 x 17.0 g NH3 = 30.6 g NH3

1 mol NH3

Page 41: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Examples

How many grams of water are produced in the oxidation of 1.00 g of glucose, C6H12O6?

Propane, C3H8, is a common fuel used for cooking and home heating. What mass of O2 is consumed in the combustion of 2.75 moles of propane?

Page 42: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Examples

K2PtCl4(aq) + NH3(aq) Pt(NH3)2Cl2 (s)+ KCl(aq)

what mass of Pt(NH3)2Cl2 can be produced from 65 g of K2PtCl4 ?

How much KCl will be produced? How much from 65 grams of NH3?

Page 43: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Making Chocolate Chip Cookies

Ingredients in Kitchen (I have a BIG kitchen): 40 lbs of butter 2 lbs of salt 1 gallon of vanilla extract 80 lbs of chocolate chips 200 lbs of flour 150 lbs of sugar 10 lbs baking soda 2 eggs

What’s going to determine

how many cookies I can make?

Page 44: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Limiting Reactants

limits or determines the amount of product that can be formed in a reaction.

The reactant that isn’t used up is called the excess reagent

To determine, book says use ratio method and I.C.E. chart, I’ll show you a different method… both work.

Page 45: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

To determine the limiting reagent requires that you do two stoichiometry problems.

Figure out how much product each reactant makes.

The one that makes the least is the limiting reagent

Page 46: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Using an I.C.E. chart

Convert both reactants into moles… suppose we had 10 moles of H2 and 7 moles of O2.

2H2 + O2 2H2O

Initial

Quantities

Change

(reaction)

Expected

Quantities

10 mol 7 mol 0 mol

-10 mol -5 mol +10 mol

0 mol 2 mol 10 mol

Page 47: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Example

Ammonia is produced by the following reaction

N2 + H2 NH3

What mass of ammonia can be produced from a mixture of 100. g N2 and 500. g H2 ?

How much unreacted material remains?

Page 48: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Success of Reaction

The amount of stuff you make is the yield.

The theoretical yield is the amount you would make if everything went perfect.

The actual yield is what you make in the lab.

Page 49: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Percent Yield

% yield = Actual x 100% Theoretical

% yield = what you got x 100%what you could have got

Page 50: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Examples

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. (actual, theoretical, percent)

Page 51: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Precipitation Reactions

Page 52: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Occur when certain pairs of oppositely charged ions attract each other so strongly that they form an insoluble ionic solid.

Can determine if a precipitate will form by following certain guidlines

Page 53: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Solubility Guidelines for Ionic Compounds

Previous picture: Only 1.2 x10-3 mol of PbI2 dissolves in a liter of water.

We will consider it insoluble if the solubility is less than 0.01 mol/L

Refer to Table 4.1 (and Ksp values on pg 1045) in text for solubility guidelines for common ionic compounds in water

Dateless Dudes:

Ammonium, Nitrate, Acetate, and Alkali Metals

Page 54: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Ksp = Solubility-Product Constants pg. 1045 in text book The smaller the number, the more likely it

will precipitate If there is no insoluble product, the

reaction does not occur. Exchange (Metathesis) Reactions are

another name for double replacement reactions… (+ and - ions switch partners)

Page 55: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Simple Solubility Rules

1. Most nitrate slats are soluble

2. Most salts containing the alkali metal ions and ammonium are soluble

3. Most chloride, bromide, and iodide salts are soluble. (exceptions are salts containing the ions Ag+, Pb+2, and Hg+2)

Page 56: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

4. Most sulfate (SO4-2) salts are soluble.

(notable exceptions are BaSO4, PbSO4, HgSO4, and CaSO4)

5. Most OH- salts are only slightly soluble. The important soluble ones are NaOH and KOH. Ba, Sr and Ca are marginally soluble.

6. Most S-2, CO3-2, CrO4

-2 and PO4-3 are

slightly soluble.

Page 57: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Ionic and Net Ionic Equations

1. Write out the molecular equation:Pb(NO3)2(aq)+ 2 KI(aq) PbI2 (s) + 2KNO3(aq)

2. Write out the complete ionic equation and Identify and cancel out spectator ions:

3. Write out the Net Ionic equation:

Page 58: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Example

An aqueous solution of silver nitrate reacts with an aqueous solution of potassium chloride. A precipitate is produced.

1. Determine the precipitate using Ksp.

2. Write the Molecular, Ionic, and Net Ionic equations for the above reaction

Page 59: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Solution Stoichiometry and Chemical Analysis

Molarity = moles of solute / Liters of solution.

M1V1 = M2V2 We can take the molarity and volume of

a solution to find out the moles of a solution. And then use dimensional analysis to determine the moles or grams of another reactant or product.

Page 60: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Example 1

How many grams of Ca(OH)2 are needed to neutralize 25.0 mL of 0.100 M HNO3?

1. Use molarity and volume of HNO3 to convert to moles of HNO3

Balance equation:

HNO3 + Ca(OH)2 H2O + Ca(NO3)2

1. Convert moles of HNO3 to moles of Ca(OH)2 and then to grams of Ca(OH)2

Page 61: Stoichiometry & Solution Stoich Chapter 3 and 4.2/4.6

Example 2

How many grams of NaOH are needed to neutralize 20.0 mL of 0.150M H2SO4 solution?

NaOH + H2SO4 H2O + Na2SO4