Chapter 3 Stoichiometry The study of the quantities of materials produced and consumed in chemical reactions Relative Atomic Mass- atoms are small and.

Slides:



Advertisements
Similar presentations
Stoichiometry “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.
Advertisements

Stoichiometry Chapter 3. Atomic Mass 1961: Atomic Mass is based on 12 C 1961: Atomic Mass is based on 12 C 12 C is assigned a mass of EXACTLY 12 AMU 12.
Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1 Chemical Stoichiometry Stoichiometry - The study of quantities of materials consumed.
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.
Chapter 3: STOICHIOMETRY Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions.
Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1 Chapter 3 Stoichiometric Atomic Masses, Mole concept, and Molar Mass (Average atomic.
Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1 Chemical Stoichiometry Stoichiometry - The study of quantities of materials consumed.
Percent Composition, Empirical and Molecular Formulas Courtesy
Chapter 3 Stoichiometry Stoichiometry: The study of quantities of materials consumed and produced in chemical reactions Atomic Masses: Are determined by.
Mass Relationships in Chemical Reactions Chapter 3.
Chapter 3 Percent Compositions and Empirical Formulas
Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations.
Stoichiometry Quantitative nature of chemical formulas and chemical reactions Chapter 3 (Sections )
Stoichiometry Atomic Mass Atomic Mass Molar Mass of an Element Molar Mass of an Element Molecular Mass Molecular Mass Percent Composition Percent Composition.
The S-word Stoichiometry Chapter 3. By definition: 1 atom 12 C “weighs” 12 amu On this scale 1 H = amu 16 O = amu Atomic mass is the mass.
1 Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations.
Chapter 3 Stoichiometry Deals with quantities of materials consumed and produced in chemical reactions.
Chapter 3 Stoichiometry. Chemical Stoichiometry Stoichimetry from Greek “measuring elements”. That is “Calculation of quantities in chemical reactions”
Mass Relationships in Chemical Reactions Chapter 3.
Mass Relationships in Chemical Reactions Chapter 3.
Chapter 3 Stoichiometric
The Mole 1 dozen = 12 1 gross = ream = mole = 6.02 x 1023.
Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1 Chemical Stoichiometry Stoichiometry - The study of quantities of materials consumed.
1 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.
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.
 Objective: Understand molecular formulas and balancing equations.  Before: Introduction to molecular formulas  During: Discuss molecular formulas.
Chapter 9 - 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.
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.
Mass Relationships in Chemical Reactions Chapter 3 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Mass Relationships in Chemical Reactions Chapter 3 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
The Mole Standards Standards The Mole 1 dozen = 1 gross = 1 ream = 1 mole = x There are exactly 12 grams of carbon-12 in one mole.
Percent Composition, Empirical and Molecular Formulas Courtesy
Stoichiometry and the mole Chapter 8 What is stoichiometry?  Quantitative aspects of chemistry  Stoicheon Greek root (element)  Metron Greek root(
Empirical and Molecular Formulas. Formulas  molecular formula = (empirical formula) n  molecular formula = C 6 H 6 = (CH) 6  empirical formula = CH.
Stoichiometry is… Greek for “measuring elements” Defined as: calculations of the quantities in chemical reactions, based on a balanced equation. There.
Stoichiometry Calculations with Chemical Formulas and Equations.
Chapter 3: Calculations with Chemical Formulas and Equations MASS AND MOLES OF SUBSTANCE 3.1 MOLECULAR WEIGHT AND FORMULA WEIGHT -Molecular weight: (MW)
Percent Composition, Empirical and Molecular Formulas Courtesy
1 dozen =12 1 gross =144 1 ream =500 1 mole = 6.02 x (Avogadro’s Number) A mole is a Very large Number.
“Chemical Quantities” Yes, you will need a calculator for this chapter!
Percent Composition, Empirical and Molecular Formulas.
Percent Mass, Empirical and Molecular Formulas. Calculating Formula (Molar) Mass Calculate the formula mass of magnesium carbonate, MgCO g +
Moles and Formula Mass.
Chemical Stoichiometry
Moles and Formula Mass.
Chemistry 200 Fundamentals D Chemical Composition.
Moles and Formula Mass.
Percent Composition, Empirical and Molecular Formulas
Chemical Stoichiometry
Moles and Formula Mass.
The Mole 1 dozen = 12 1 gross = ream = mole = 6.02 x 1023
Stoichiometry “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.
Chapter 3: Stiochiometry
Percent Composition, Empirical and Molecular Formulas
Moles and Formula Mass.
Moles and Formula Mass.
Chemistry 100 Chapter 6 Chemical Composition.
Stoichiometry “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.
Stoichiometry “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.
Percent Composition, Empirical and Molecular Formulas
Moles and Formula Mass.
Stoichiometry “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.
Stoichiometry “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.
Unit 3: Stoichiometry The Mole, Molar Mass, % Composition, Balancing Chemical Equations, Limiting Reagent, Percent Yield.
Chemical Reactions, Chemical Equations, and Stoichiometry
Moles and Formula Mass.
Stoichiometry- Mass Relationships in Chemical Reactions
Chemical Stoichiometry
Presentation transcript:

Chapter 3 Stoichiometry

The study of the quantities of materials produced and consumed in chemical reactions Relative Atomic Mass- atoms are small and as chemists we deal with samples of matter that contain huge numbers of atoms The number of atoms in a sample is determined by its average atomic mass

Atomic Masses The work of Dalton, Gay-Lussac, Lavoisier, Avogadro and Berzelius produced information about atomic masses Modern system of atomic mass instituted 1961 – Based on Carbon-12 – It is assigned a mass of exactly 12 amu – Masses of all other atoms are relative to carbon- 12 standard

Atomic Masses Mass Spectrometer- most accurate method to compare the masses of atoms An analytical technique that produces spectra of the masses of the molecules in sample Spectra are used to determine the elemental composition of the matter Samples can be solids, liquids or gases

Atomic Masses

Mass Spectrometer Ionization- atoms/molecules are passed through a beam of high speed electrons- creates cations Acceleration-Electric field is applied to cations Deflection- applied magnetic field changes path of ion depending on its mass (most massive deflected the least) Detection-Ions hit the detector plate, gives accurate values of their relative masses

Atomic Masses Atomic Mass of carbon is amu Natural carbon is a mixture of isotopes 12 C, 13 C, and 14 C Natural carbon is composed of 98.89% 12 C and 1.11% 13 C, the amount of 14 C is negligible – 12 C has a mass of 12 amu and 13 C has a mass of amu Calculate the average atomic mass

The Mole

The number equal to the number of carbon atoms in exactly 12 grams of pure 12 C One mole of something consists of 6.02 x units of that substance (Avogadro’s number) The mass of 1 mole of an element is equal to its atomic mass in grams

Calculations with Moles: Converting moles to grams How many grams of lithium are in 3.50 moles of lithium? 3.50 mol Li6.94 g Li 1 mol Li = 45.1 g Li

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

Calculations with Moles: Using Avogadro’s Number How many atoms of lithium are in 3.50 moles of lithium? 3.50 mol 6.02 x atoms 1 mol =2.07 x atoms

Calculations with Moles: Using Avogadro’s Number How many atoms of lithium are in 18.2 g of lithium? 18.2 g Li 1 mol Li 6.94 g Li 6.02 x atoms Li 1 mol Li = 1.58 x atoms Li

Molar Mass Mass in grams of one mole of the compound Also referred to as molecular weight Calculate the molar mass of methane CH 4

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

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

Formulas 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.  molecular formula = (empirical formula) n [n = integer]  molecular formula = C 6 H 6 = (CH) 6  empirical formula = CH

Formulas (continued) Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio). 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). 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. 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. 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? 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? 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? 2. Divide the molecular mass by the mass given by the emipirical formula. 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? 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. Multiply the empirical formula by this number to get the molecular formula. 3(12.01 g) + 5(1.01) + 2(16.00) = g (C 3 H 5 O 2 ) x 2 = C 6 H 10 O 4

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:

Calculating Masses of Reactants and Products 1.Balance the equation. 2.Convert mass or volume to moles, if necessary. 3.Set up mole ratios. 4.Use mole ratios to calculate moles of desired substituent. 5.Convert moles to mass or volume, if necessary. 1.Balance the equation. 2.Convert mass or volume to moles, if necessary. 3.Set up mole ratios. 4.Use mole ratios to calculate moles of desired substituent. 5.Convert moles to mass or volume, if necessary.

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

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