Chapter Three Fe atoms 1 mol Fe When balancing a chemical equation, changing the subscripts changes the identity of the substance.
|
|
|
- Helena Wilcox
- 7 years ago
- Views:
Transcription
1 3. The mole is the SI unit for the amount of a substance. A mole is equal in quantity to Avogadro s number ( ) of particles, or the formula mass in grams of a substance. 3.4 There are the same number of molecules in.5 moles of H O and.5 moles of H. 3.5 There are moles of iron atoms in 1 mole of Fe O 3. The stoichiometric equivalent between Fe and Fe O is mol Fe 1 mol Fe O 3. For the number of iron atoms in 1 mole of Fe O 3 : 1 mol Fe O mol Fe 3 1 mole FeO Fe atoms 1 mol Fe = atoms Fe 3.1 When balancing a chemical equation, changing the subscripts changes the identity of the substance There are three distinct empirical formulas represented AB, AB 3, and A 3 B 8. There are two molecules with the empirical formula AB 3 ; AB 3 and A B 6. There is one A 3 B 8, and there are two with the formula AB ; A 6 B 1 and A 3 B Student B is correct. Student A wrote a properly balanced equation. However, by changing the subscript for the product of the reaction from an implied one, NaCl, to a two, NaCl, this student has changed the identity of the product. When balancing chemical equations, never change the values of the subscripts given in the unbalanced equation :4, 1 mol C:4 mol H mol I molecules of I molecules I =.99 mole I 3.30 (a) atom C: 1 atom O mole C: 1 mole O (c) 1 atom C: atom H (d) 1 mole C: mole H mol O mol O = (4.5 mol CaCO 3 ) 1 mol CaCO3 = 1.8 mol O mol H atoms H atom H = (.31 mol C 3 H 8 ) 1 mol C3H8 1 mol H = atoms H 3.07 g S 3.48 (a) g S = (0.546 mol S) 1 mole S = 17.5 g S (c) g N = (3.9 mol N) g N 1 mole N = 46.1 g N 6.98 g N g Al = (8.11 mol Al) 1 mole N = 19 g Al 3.50 g Na = mol Na.99 g Na atoms Na atoms Na 1 mol Na 1 mol Cr 3.5 mol Cr = 85.7 g Cr = 1.65 mol Cr 5.00 g Cr = g Na 31
2 3.54 Note: all masses are in g/mole (a) Ca(NO 3 ) = 1Ca + N + 6O = (40.078) + ( ) + ( ) = g/mole = g/mol Pb(C H 5 ) 4 = 1Pb + 8C + 0H = (07.) + ( ) + ( ) = 33.4 g/mole (Since the mass of Pb is known exactly.) (c) Na SO 4 10H O = Na + 1S +14O + 0H = (.98977) ( ) + ( ) = g/mole = g/mol (d) Fe 4 [Fe(CN) 6 ] 3 = 7Fe + 18C + 18N = ( ) + ( ) + ( ) = g/mole = g/mol (e) Mg 3 (PO 4 ) = 3Mg + P + 8O = ( ) + ( ) + ( ) = g/mole = g/mol g ZnCl 3.56 (a) g ZnCl = (0.754 mol ZnCl ) 1 mol ZnCl = 103 g ZnCl 1 mol KIO g KIO μg KIO3 μg KIO 3 = (0.194 μmol KIO 3 ) 6 10 mol KIO μ 1 mol KIO g KIO3 = g KIO 3 1 mol POCl g POCl3 (c) g POCl 3 = (0.3 mmol POCl 3 ) 3 10 mmol POCl = g POCl 3 1 mol POCl 3 3 ( ) 13.1 g NH HPO (d) g (NH 4 ) HPO 4 = ( mol (NH 4 ) HPO 4 ) 1 mol ( NH4) HPO 4 = g (NH 4 ) HPO (a) 1 mole Ca(OH) mol Ca(OH) = (9.36 g Ca(OH) g Ca(OH) 1000 g PbSO4 1 mole PbSO4 mol PbSO 4 = (38. kg PbSO 4 ) = 16 mol PbSO 4 1 kg PbSO g PbSO4 1 mole H O (c) mol H O = (4.9 g H O ) g H O = 0.16 mol H O 1 g NaAuCl (d) mol NaAuCl 4 = 4.65 mg NaAuCl 4 1 mol NaAuCl mg NaAuCl g NaAuCl = mol NaAuCl (a) The molar mass of (CH 3 ) N H is 60.1 g/mol. 4.0 g C % C = 60.1 g (CH 3) NH 100% = 40.0% 8.06 g H % H = 60.1 g (CH 3) NH 100% = 13.4% 8.0 g N % N = 60.1 g (CH 3) NH 100% = 46.6% 4 4 3
3 (c) (d) (e) The molar mass of CaCO 3 is g/mol g Ca % Ca = g CaCO3 100% = 40.0% 1.01 g C % C = g CaCO3 100% = 1.0% g O % O = g CaCO3 100% = 48.0% The molar mass of Fe(NO 3 ) 3 is 41.9 g/mol g Fe % Fe = 100% = 3.1% 41.9 g Fe NO ( ) 3 3 % N = 4.03 g N 41.9 g Fe( NO3 ) 3 100% = 17.4% % O = g O 41.9 g Fe( NO3 ) 3 100% = 59.5% The molar mass of C 3 H 8 is g/mol g C % C = 100% = 81.7% g C H g H % H = 100% = 18.3% g C3H8 The molar mass of Al (SO 4 ) 3 is 34. g/mol g Al % Al = 100% = 15.8% 34. g Al SO ( ) 4 3 % S = % O = 96. g S 34. g Al( SO4) g O 34. g Al( SO4) 3 100% = 100% = 8.1% 56.1% 3.68 % N in carbamazepine = % N in carbetapentane = 8.0 g N 36.9 g C15H1NO g N g C0H31NO 3 100% = 11.9% N 100% = 4.0% N Therefore, carbamazepine has a higher percentage of nitrogen (a) CH 3 O HSO 4 (c) C H 5 (d) BH 3 (e) C H 6 O mol C mol C = (0.43 g C) = mol C 1.01 g C 1 mol Cl mol Cl = (.50 g Cl) = mol Cl g Cl 1 mol F mol F = (1.34 g F) = mol F g F Now we divide each of these numbers of moles by the smallest of the three numbers, in order to obtain the simplest mole ratio among the three elements in the compound: 33
4 for C, moles / moles = 1.00 for Cl, moles / moles =.000 for F, moles / moles =.00 Chapter Three These relative mole amounts give us the empirical formula CCl F 3.84 Assume a 100 g sample: 1 mol C mol C = (63. g C) 1.01 g C = 5.6 mol C 1 mol H mol H = (5.6 g H) g H = 5. mol H 1 mol O mol O = (31.6 g O) g O = 1.98 mol O Now we divide each of these numbers of moles by the smallest of the three numbers, in order to obtain the simplest mole ratio among the three elements in the compound: for C, 5.6 moles / 1.98 moles =.66 = (8/3) for H, 5. moles / 1.98 moles =.64 = (8/3) for O, 1.98 moles / 1.98 moles = 1.00 = (3/3) These relative mole amounts give us the empirical formula C 8 H 8 O All of the carbon is converted to carbon dioxide so, g C = (.01 g CO ) 1 mol CO 1 mol C 1.01 g C g CO 1 mol CO 1 mol C = g C 1 mol C mol C = (0.549 g C) 1.01 g C = mol C All of the hydrogen is converted to H O, so g H = (0.87 g H O) 1 mol H O mol H g H 18.0 g HO 1 mol HO 1 mol H = g H 1 mol H mol H = (0.095 g H) = mol H g H The amount of O in the compound is determined by subtracting the mass of C and the mass of H from the sample. g O = 0.8 g g g = g 1 mol O mol O = (0.181 g O) g O = mol O The relative mole ratios are: for C, moles / moles = 4.04 for H, moles/ moles = 8.1 for O, moles / moles = 1.00 The relative mole amounts give the empirical formula C 4 H 8 O 3.90 (a) Formula mass = 1.1 g 34
5 73.6 g/mol 1.1 g/mol = The molecular formula is Na 1 Si 6 O 18 Formula mass = 10.0 g g/mol = g/mol The molecular formula is Na 3 P 3 O 9 (c) Formula mass = g 6.1 g/mol = g/mol The molecular formula is C H 6 O 3.94 From the information provided, the mass of sulfur is the difference between the total mass and the mass of antimony: g S = g compound g Sb = g S To determine the empirical formula, first convert the two masses to a number of moles. 1 mole S mol S = (0.645 g S) g S = mol S 1 mole Sb mol Sb = ( g Sb) g Sb = mol Sb Now, divide each of these values by the smaller quantity to determine the simplest mole ratio between the two elements: For Sb: moles/ moles = mol Sb For S: moles/ moles =.500 mol S Hence the empirical formula is Sb S 5, and the empirical mass is ( Sb) + (5 S) = g/mol. Since the molecular mass reported in the problem is the same as the calculated empirical mass, the empirical formula is the same as the molecular formula NO(g) + O (g) NO (g) 3.10 (a) SO + O SO 3 NaHCO 3 + H SO 4 Na SO 4 + H O + CO (c) P 4 O H O 4H 3 PO 4 (d) Fe O 3 + 3H Fe + 3H O (e) Al + 3H SO 4 Al (SO 4 ) 3 + 3H (a) CaO + HNO 3 Ca(NO 3 ) + H O Na CO 3 + Mg(NO 3 ) MgCO 3 + NaNO 3 (c) (NH 4 ) 3 PO 4 + 3NaOH Na 3 PO 4 + 3NH 3 + 3H O (d) LiHCO 3 + H SO 4 Li SO 4 + H O + CO (e) C 4 H 10 O + 6O 4CO + 5H O 35
6 3.108 (a) 5 mole O mol O = (6 mol C 8 H 18 ) = 80 mol O mole C8H18 (Note: This calculation is limited due to sig figs.) 16 mole CO mol CO = (0.5 mol C 8 H 18 ) = 4 mol CO mole C8H18 18 mole HO (c) mol H O = (8 mol C 8 H 18 ) mole C8H = 70 mol H O 18 5 mole O (d) mol O = ( 6.00 mol CO ) 16 mole CO = 9.38 mol O mole C8H18 mol C 8 H 18 = (6.00 mol CO ) 16 mole CO = mol C 8 H (a) 3 mol C 3 H 8 5 mol O 3.00 g O 1 mol C3H = 500 g O 8 1 mol O 0.1 mol C 3 H 8 3 mol CO g CO 1 mol C3H = 13 g CO 8 1 mol CO (c) 4 mol C 3 H 8 4 mol H O g H O 1 mol C3H = 300 g H O 8 1 mol HO 1 mol N H 7 mol H O 34.0 g H O g H O = (85 g N H 4 ) 3.05 g NH4 1 mol NH4 1 mol HO 4 = 6330 g H O (a) First determine the amount of C H 4 that would be required to react completely with the given amount of H O: 1000 g C g H O = 1.0 kg C H H4 1 mol CH4 1 mol HO 4 1 kg CH g CH4 1 mol CH g HO 1 kg HO = 0.64 kg H O 1 mol HO 1000 g HO Since only kg of H O are supplied, it is the limiting reactant. This can be confirmed by calculating the amount of C H 4 that would be required to react completely with all of the available H O: 1000 g HO 1 mol HO 1 mol CH4 g C H 4 = 0.01 kg H O 1 kg HO 18.0 g HO 1 mol HO 8.05 g CH4 1 kg CH4 1 mol CH g CH4 = kg C H g HO 1 mol HO g C H 5 OH = (0.010 kg H O) 1 kg HO 18.0 g HO 1 mol CH5OH g CH5OH = 6 g C H 5 OH 1 mol HO 1 mol CH5OH 36
7 3.1 (a) First calculate the number of moles of water that are needed to react completely with the given amount of PCl 5 : 4 mole HO mol H O = (0.360 mol PCl 5 ) 1 moles PCl5 = 1.44 mol H O Since this is less than the amount of water that is supplied, the limiting reactant must be PCl 5. This can be confirmed by the following calculation: 1 mol PCl5 mol PCl 5 = (.88 mol H O) 4 mol HO = 0.70 mol PCl 5 which also demonstrates that the limiting reactant is PCl 5. 5 mol HCl g HCl g HCl = (0.360 mol PCl 5 ) 1 mol PCl 5 1 mol HCl = 65.6 g HCl 3.16 Assume there is excess oxygen present and determine the theoretical yield of carbon dioxide. 1 mol CH3OH mol CO g CO g CO = (6.40 g CH OH) = 8.79 g CO 3.04 g CH3OH mol CH3OH 1 mol CO 6.1 g % yield = 100% = 69.6% 8.79g 3 37
Answers and Solutions to Text Problems
Chapter 7 Answers and Solutions 7 Answers and Solutions to Text Problems 7.1 A mole is the amount of a substance that contains 6.02 x 10 23 items. For example, one mole of water contains 6.02 10 23 molecules
Calculating Atoms, Ions, or Molecules Using Moles
TEKS REVIEW 8B Calculating Atoms, Ions, or Molecules Using Moles TEKS 8B READINESS Use the mole concept to calculate the number of atoms, ions, or molecules in a sample TEKS_TXT of material. Vocabulary
Balance the following equation: KClO 3 + C 12 H 22 O 11 KCl + CO 2 + H 2 O
Balance the following equation: KClO 3 + C 12 H 22 O 11 KCl + CO 2 + H 2 O Ans: 8 KClO 3 + C 12 H 22 O 11 8 KCl + 12 CO 2 + 11 H 2 O 3.2 Chemical Symbols at Different levels Chemical symbols represent
Other Stoich Calculations A. mole mass (mass mole) calculations. GIVEN mol A x CE mol B. PT g A CE mol A MOLE MASS :
Chem. I Notes Ch. 12, part 2 Using Moles NOTE: Vocabulary terms are in boldface and underlined. Supporting details are in italics. 1 MOLE = 6.02 x 10 23 representative particles (representative particles
Moles. Balanced chemical equations Molar ratios Mass Composition Empirical and Molecular Mass Predicting Quantities Equations
Moles Balanced chemical equations Molar ratios Mass Composition Empirical and Molecular Mass Predicting Quantities Equations Micro World atoms & molecules Macro World grams Atomic mass is the mass of an
1. How many hydrogen atoms are in 1.00 g of hydrogen?
MOLES AND CALCULATIONS USING THE MOLE CONCEPT INTRODUCTORY TERMS A. What is an amu? 1.66 x 10-24 g B. We need a conversion to the macroscopic world. 1. How many hydrogen atoms are in 1.00 g of hydrogen?
Chapter 3: Stoichiometry
Chapter 3: Stoichiometry Key Skills: Balance chemical equations Predict the products of simple combination, decomposition, and combustion reactions. Calculate formula weights Convert grams to moles and
How much does a single atom weigh? Different elements weigh different amounts related to what makes them unique.
How much does a single atom weigh? Different elements weigh different amounts related to what makes them unique. What units do we use to define the weight of an atom? amu units of atomic weight. (atomic
Part One: Mass and Moles of Substance. Molecular Mass = sum of the Atomic Masses in a molecule
CHAPTER THREE: CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS Part One: Mass and Moles of Substance A. Molecular Mass and Formula Mass. (Section 3.1) 1. Just as we can talk about mass of one atom of
Atomic Masses. Chapter 3. Stoichiometry. Chemical Stoichiometry. Mass and Moles of a Substance. Average Atomic Mass
Atomic Masses Chapter 3 Stoichiometry 1 atomic mass unit (amu) = 1/12 of the mass of a 12 C atom so one 12 C atom has a mass of 12 amu (exact number). From mass spectrometry: 13 C/ 12 C = 1.0836129 amu
10 Cl atoms. 10 H2O molecules. 8.3 mol HCN = 8.3 mol N atoms 1 mol HCN. 2 mol H atoms 2.63 mol CH2O = 5.26 mol H atoms 1 mol CH O
Chem 100 Mole conversions and stoichiometry worksheet 1. How many Ag atoms are in.4 mol Ag atoms? 6.0 10 Ag atoms 4.4 mol Ag atoms = 1.46 10 Ag atoms 1 mol Ag atoms. How many Br molecules are in 18. mol
Chapter 6 Chemical Calculations
Chapter 6 Chemical Calculations 1 Submicroscopic Macroscopic 2 Chapter Outline 1. Formula Masses (Ch 6.1) 2. Percent Composition (supplemental material) 3. The Mole & Avogadro s Number (Ch 6.2) 4. Molar
Formulas, Equations and Moles
Chapter 3 Formulas, Equations and Moles Interpreting Chemical Equations You can interpret a balanced chemical equation in many ways. On a microscopic level, two molecules of H 2 react with one molecule
How To Calculate Mass In Chemical Reactions
We have used the mole concept to calculate mass relationships in chemical formulas Molar mass of ethanol (C 2 H 5 OH)? Molar mass = 2 x 12.011 + 6 x 1.008 + 1 x15.999 = 46.069 g/mol Mass percentage of
Chemistry 65 Chapter 6 THE MOLE CONCEPT
THE MOLE CONCEPT Chemists find it more convenient to use mass relationships in the laboratory, while chemical reactions depend on the number of atoms present. In order to relate the mass and number of
Chapter 3. Chemical Reactions and Reaction Stoichiometry. Lecture Presentation. James F. Kirby Quinnipiac University Hamden, CT
Lecture Presentation Chapter 3 Chemical Reactions and Reaction James F. Kirby Quinnipiac University Hamden, CT The study of the mass relationships in chemistry Based on the Law of Conservation of Mass
Stoichiometry. Lecture Examples Answer Key
Stoichiometry Lecture Examples Answer Key Ex. 1 Balance the following chemical equations: 3 NaBr + 1 H 3 PO 4 3 HBr + 1 Na 3 PO 4 2 C 3 H 5 N 3 O 9 6 CO 2 + 3 N 2 + 5 H 2 O + 9 O 2 2 Ca(OH) 2 + 2 SO 2
Chemical Calculations: The Mole Concept and Chemical Formulas. AW Atomic weight (mass of the atom of an element) was determined by relative weights.
1 Introduction to Chemistry Atomic Weights (Definitions) Chemical Calculations: The Mole Concept and Chemical Formulas AW Atomic weight (mass of the atom of an element) was determined by relative weights.
The Mole Notes. There are many ways to or measure things. In Chemistry we also have special ways to count and measure things, one of which is the.
The Mole Notes I. Introduction There are many ways to or measure things. In Chemistry we also have special ways to count and measure things, one of which is the. A. The Mole (mol) Recall that atoms of
Chapter 4. Chemical Composition. Chapter 4 Topics H 2 S. 4.1 Mole Quantities. The Mole Scale. Molar Mass The Mass of 1 Mole
Chapter 4 Chemical Composition Chapter 4 Topics 1. Mole Quantities 2. Moles, Masses, and Particles 3. Determining Empirical Formulas 4. Chemical Composition of Solutions Copyright The McGraw-Hill Companies,
Calculations and Chemical Equations. Example: Hydrogen atomic weight = 1.008 amu Carbon atomic weight = 12.001 amu
Calculations and Chemical Equations Atomic mass: Mass of an atom of an element, expressed in atomic mass units Atomic mass unit (amu): 1.661 x 10-24 g Atomic weight: Average mass of all isotopes of a given
Element of same atomic number, but different atomic mass o Example: Hydrogen
Atomic mass: p + = protons; e - = electrons; n 0 = neutrons p + + n 0 = atomic mass o For carbon-12, 6p + + 6n 0 = atomic mass of 12.0 o For chlorine-35, 17p + + 18n 0 = atomic mass of 35.0 atomic mass
The Mole Concept. The Mole. Masses of molecules
The Mole Concept Ron Robertson r2 c:\files\courses\1110-20\2010 final slides for web\mole concept.docx The Mole The mole is a unit of measurement equal to 6.022 x 10 23 things (to 4 sf) just like there
Lecture 5, The Mole. What is a mole?
Lecture 5, The Mole What is a mole? Moles Atomic mass unit and the mole amu definition: 12 C = 12 amu. The atomic mass unit is defined this way. 1 amu = 1.6605 x 10-24 g How many 12 C atoms weigh 12 g?
Ch. 10 The Mole I. Molar Conversions
Ch. 10 The Mole I. Molar Conversions I II III IV A. What is the Mole? A counting number (like a dozen) Avogadro s number (N A ) 1 mole = 6.022 10 23 representative particles B. Mole/Particle Conversions
1. What is the molecular formula of a compound with the empirical formula PO and a gram-molecular mass of 284 grams?
Name: Tuesday, May 20, 2008 1. What is the molecular formula of a compound with the empirical formula PO and a gram-molecular mass of 284 grams? 2 5 1. P2O 5 3. P10O4 2. P5O 2 4. P4O10 2. Which substance
Chem 115 POGIL Worksheet - Week 4 Moles & Stoichiometry
Chem 115 POGIL Worksheet - Week 4 Moles & Stoichiometry Why? Chemists are concerned with mass relationships in chemical reactions, usually run on a macroscopic scale (grams, kilograms, etc.). To deal with
Chemical Equations & Stoichiometry
Chemical Equations & Stoichiometry Chapter Goals Balance equations for simple chemical reactions. Perform stoichiometry calculations using balanced chemical equations. Understand the meaning of the term
Calculation of Molar Masses. Molar Mass. Solutions. Solutions
Molar Mass Molar mass = Mass in grams of one mole of any element, numerically equal to its atomic weight Molar mass of molecules can be determined from the chemical formula and molar masses of elements
STOICHIOMETRY UNIT 1 LEARNING OUTCOMES. At the end of this unit students will be expected to:
STOICHIOMETRY LEARNING OUTCOMES At the end of this unit students will be expected to: UNIT 1 THE MOLE AND MOLAR MASS define molar mass and perform mole-mass inter-conversions for pure substances explain
Worksheet # 8 Graham/09 Due
CHE 100 Worksheet # 8 Graham/09 Name Key Due 1. According to the law of definite proportions, if a sample of a compound contains 7.00 grams of sulfur and 3.50 grams of oxygen, then another sample of the
The Mole Concept and Atoms
Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 4 24 September 2013 Calculations and the Chemical Equation The Mole Concept and Atoms Atoms are exceedingly
MOLECULAR MASS AND FORMULA MASS
1 MOLECULAR MASS AND FORMULA MASS Molecular mass = sum of the atomic weights of all atoms in the molecule. Formula mass = sum of the atomic weights of all atoms in the formula unit. 2 MOLECULAR MASS AND
The Mole. 6.022 x 10 23
The Mole 6.022 x 10 23 Background: atomic masses Look at the atomic masses on the periodic table. What do these represent? E.g. the atomic mass of Carbon is 12.01 (atomic # is 6) We know there are 6 protons
CP Chemistry Review for Stoichiometry Test
CP Chemistry Review for Stoichiometry Test Stoichiometry Problems (one given reactant): 1. Make sure you have a balanced chemical equation 2. Convert to moles of the known substance. (Use the periodic
Unit 2: Quantities in Chemistry
Mass, Moles, & Molar Mass Relative quantities of isotopes in a natural occurring element (%) E.g. Carbon has 2 isotopes C-12 and C-13. Of Carbon s two isotopes, there is 98.9% C-12 and 11.1% C-13. Find
Chapter 3! Stoichiometry: Calculations with Chemical Formulas and Equations. Stoichiometry
Chapter 3! : Calculations with Chemical Formulas and Equations Anatomy of a Chemical Equation CH 4 (g) + 2O 2 (g) CO 2 (g) + 2 H 2 O (g) Anatomy of a Chemical Equation CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2
We know from the information given that we have an equal mass of each compound, but no real numbers to plug in and find moles. So what can we do?
How do we figure this out? We know that: 1) the number of oxygen atoms can be found by using Avogadro s number, if we know the moles of oxygen atoms; 2) the number of moles of oxygen atoms can be found
Chapter 1 The Atomic Nature of Matter
Chapter 1 The Atomic Nature of Matter 6. Substances that cannot be decomposed into two or more simpler substances by chemical means are called a. pure substances. b. compounds. c. molecules. d. elements.
CHEM 110: CHAPTER 3: STOICHIOMETRY: CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS
1 CHEM 110: CHAPTER 3: STOICHIOMETRY: CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS The Chemical Equation A chemical equation concisely shows the initial (reactants) and final (products) results of
Moles. Moles. Moles. Moles. Balancing Eqns. Balancing. Balancing Eqns. Symbols Yields or Produces. Like a recipe:
Like a recipe: Balancing Eqns Reactants Products 2H 2 (g) + O 2 (g) 2H 2 O(l) coefficients subscripts Balancing Eqns Balancing Symbols (s) (l) (aq) (g) or Yields or Produces solid liquid (pure liquid)
Chapter 3 Mass Relationships in Chemical Reactions
Chapter 3 Mass Relationships in Chemical Reactions Student: 1. An atom of bromine has a mass about four times greater than that of an atom of neon. Which choice makes the correct comparison of the relative
Stoichiometry. Unit Outline
3 Stoichiometry Unit Outline 3.1 The Mole and Molar Mass 3.2 Stoichiometry and Compound Formulas 3.3 Stoichiometry and Chemical Reactions 3.4 Stoichiometry and Limiting Reactants 3.5 Chemical Analysis
Study Guide For Chapter 7
Name: Class: Date: ID: A Study Guide For Chapter 7 Multiple Choice Identify the choice that best completes the statement or answers the question. 1. The number of atoms in a mole of any pure substance
Stoichiometry. What is the atomic mass for carbon? For zinc?
Stoichiometry Atomic Mass (atomic weight) Atoms are so small, it is difficult to discuss how much they weigh in grams We use atomic mass units an atomic mass unit (AMU) is one twelfth the mass of the catbon-12
Stoichiometry Review
Stoichiometry Review There are 20 problems in this review set. Answers, including problem set-up, can be found in the second half of this document. 1. N 2 (g) + 3H 2 (g) --------> 2NH 3 (g) a. nitrogen
AS1 MOLES. oxygen molecules have the formula O 2 the relative mass will be 2 x 16 = 32 so the molar mass will be 32g mol -1
Moles 1 MOLES The mole the standard unit of amount of a substance the number of particles in a mole is known as Avogadro s constant (L) Avogadro s constant has a value of 6.023 x 10 23 mol -1. Example
Chemical Calculations: Formula Masses, Moles, and Chemical Equations
Chemical Calculations: Formula Masses, Moles, and Chemical Equations Atomic Mass & Formula Mass Recall from Chapter Three that the average mass of an atom of a given element can be found on the periodic
2. The percent yield is the maximum amount of product that can be produced from the given amount of limiting reactant.
UNIT 6 stoichiometry practice test True/False Indicate whether the statement is true or false. moles F 1. The mole ratio is a comparison of how many grams of one substance are required to participate in
CHEMICAL REACTIONS. Chemistry 51 Chapter 6
CHEMICAL REACTIONS A chemical reaction is a rearrangement of atoms in which some of the original bonds are broken and new bonds are formed to give different chemical structures. In a chemical reaction,
F321 MOLES. Example If 1 atom has a mass of 1.241 x 10-23 g 1 mole of atoms will have a mass of 1.241 x 10-23 g x 6.02 x 10 23 = 7.
Moles 1 MOLES The mole the standard unit of amount of a substance (mol) the number of particles in a mole is known as Avogadro s constant (N A ) Avogadro s constant has a value of 6.02 x 10 23 mol -1.
Chem 115 POGIL Worksheet - Week 4 Moles & Stoichiometry Answers
Key Questions & Exercises Chem 115 POGIL Worksheet - Week 4 Moles & Stoichiometry Answers 1. The atomic weight of carbon is 12.0107 u, so a mole of carbon has a mass of 12.0107 g. Why doesn t a mole of
Chapter 3 Stoichiometry
Chapter 3 Stoichiometry 3-1 Chapter 3 Stoichiometry In This Chapter As you have learned in previous chapters, much of chemistry involves using macroscopic measurements to deduce what happens between atoms
CHEMISTRY COMPUTING FORMULA MASS WORKSHEET
CHEMISTRY COMPUTING FORMULA MASS WORKSHEET Directions: Find the formula mass of the following compounds. Round atomic masses to the tenth of a decimal place. Place your final answer in the FORMULA MASS
Name: Teacher: Pd. Date:
Name: Teacher: Pd. Date: STAAR Tutorial : Energy and Matter: Elements, Compounds, and Chemical Equations: 6.5C Differentiate between elements and compounds on the most basic level. 8.5F Recognize whether
Chemical Composition. Introductory Chemistry: A Foundation FOURTH EDITION. Atomic Masses. Atomic Masses. Atomic Masses. Chapter 8
Introductory Chemistry: A Foundation FOURTH EDITION by Steven S. Zumdahl University of Illinois Chemical Composition Chapter 8 1 2 Atomic Masses Balanced equation tells us the relative numbers of molecules
Tuesday, November 27, 2012 Expectations:
Tuesday, November 27, 2012 Expectations: Sit in assigned seat Get out Folder, Notebook, Periodic Table Have out: Spiral (notes), Learning Target Log (new) No Backpacks on tables Listen/Pay Attention Learning
CHAPTER 3 Calculations with Chemical Formulas and Equations. atoms in a FORMULA UNIT
CHAPTER 3 Calculations with Chemical Formulas and Equations MOLECULAR WEIGHT (M. W.) Sum of the Atomic Weights of all atoms in a MOLECULE of a substance. FORMULA WEIGHT (F. W.) Sum of the atomic Weights
= 11.0 g (assuming 100 washers is exact).
CHAPTER 8 1. 100 washers 0.110 g 1 washer 100. g 1 washer 0.110 g = 11.0 g (assuming 100 washers is exact). = 909 washers 2. The empirical formula is CFH from the structure given. The empirical formula
CHEM 120 Online: Chapter 6 Sample problems Date: 2. Which of the following compounds has the largest formula mass? A) H2O B) NH3 C) CO D) BeH2
CHEM 120 Online: Chapter 6 Sample problems Date: 1. To determine the formula mass of a compound you should A) add up the atomic masses of all the atoms present. B) add up the atomic masses of all the atoms
Unit 6 The Mole Concept
Chemistry Form 3 Page 62 Ms. R. Buttigieg Unit 6 The Mole Concept See Chemistry for You Chapter 28 pg. 352-363 See GCSE Chemistry Chapter 5 pg. 70-79 6.1 Relative atomic mass. The relative atomic mass
Stoichiometry. Web Resources Chem Team Chem Team Stoichiometry. Section 1: Definitions Define the following terms. Average Atomic mass - Molecule -
Web Resources Chem Team Chem Team Section 1: Definitions Define the following terms Average Atomic mass - Molecule - Molecular mass - Moles - Avagadro's Number - Conservation of matter - Percent composition
stoichiometry = the numerical relationships between chemical amounts in a reaction.
1 REACTIONS AND YIELD ANSWERS stoichiometry = the numerical relationships between chemical amounts in a reaction. 2C 8 H 18 (l) + 25O 2 16CO 2 (g) + 18H 2 O(g) From the equation, 16 moles of CO 2 (a greenhouse
Unit 10A Stoichiometry Notes
Unit 10A Stoichiometry Notes Stoichiometry is a big word for a process that chemist s use to calculate amounts in reactions. It makes use of the coefficient ratio set up by balanced reaction equations
Atomic mass is the mass of an atom in atomic mass units (amu)
Micro World atoms & molecules Laboratory scale measurements Atomic mass is the mass of an atom in atomic mass units (amu) By definition: 1 atom 12 C weighs 12 amu On this scale 1 H = 1.008 amu 16 O = 16.00
English already has many collective nouns for fixed, given numbers of objects. Some of the more common collective nouns are shown in Table 7.1.
96 Chapter 7: Calculations with Chemical Formulas and Chemical Reactions Chemical reactions are written showing a few individual atoms or molecules reacting to form a few atoms or molecules of products.
Matter. Atomic weight, Molecular weight and Mole
Matter Atomic weight, Molecular weight and Mole Atomic Mass Unit Chemists of the nineteenth century realized that, in order to measure the mass of an atomic particle, it was useless to use the standard
Sample Exercise 3.1 Interpreting and Balancing Chemical Equations
Sample Exercise 3.1 Interpreting and Balancing Chemical Equations The following diagram represents a chemical reaction in which the red spheres are oxygen atoms and the blue spheres are nitrogen atoms.
Stoichiometry. 1. The total number of moles represented by 20 grams of calcium carbonate is (1) 1; (2) 2; (3) 0.1; (4) 0.2.
Stoichiometry 1 The total number of moles represented by 20 grams of calcium carbonate is (1) 1; (2) 2; (3) 01; (4) 02 2 A 44 gram sample of a hydrate was heated until the water of hydration was driven
Chem 31 Fall 2002. Chapter 3. Stoichiometry: Calculations with Chemical Formulas and Equations. Writing and Balancing Chemical Equations
Chem 31 Fall 2002 Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations Writing and Balancing Chemical Equations 1. Write Equation in Words -you cannot write an equation unless you
Chapter 8: Quantities in Chemical Reactions
Ch 8 Page 1 Chapter 8: Quantities in Chemical Reactions Stoichiometry: the numerical relationship between chemical quantities in a balanced chemical equation. Ex. 4NH 3 + 5O 2 4NO + 6H 2 O The reaction
CHEMICAL FORMULA COEFFICIENTS AND SUBSCRIPTS. Chapter 3: Molecular analysis 3O 2 2O 3
Chapter 3: Molecular analysis Read: BLB 3.3 3.5 H W : BLB 3:21a, c, e, f, 25, 29, 37,49, 51, 53 Supplemental 3:1 8 CHEMICAL FORMULA Formula that gives the TOTAL number of elements in a molecule or formula
APPENDIX B: EXERCISES
BUILDING CHEMISTRY LABORATORY SESSIONS APPENDIX B: EXERCISES Molecular mass, the mole, and mass percent Relative atomic and molecular mass Relative atomic mass (A r ) is a constant that expresses the ratio
Percent Composition and Molecular Formula Worksheet
Percent Composition and Molecular Formula Worksheet 1. What s the empirical formula of a molecule containing 65.5% carbon, 5.5% hydrogen, and 29.0% 2. If the molar mass of the compound in problem 1 is
PART I: MULTIPLE CHOICE (30 multiple choice questions. Each multiple choice question is worth 2 points)
CHEMISTRY 123-07 Midterm #1 Answer key October 14, 2010 Statistics: Average: 74 p (74%); Highest: 97 p (95%); Lowest: 33 p (33%) Number of students performing at or above average: 67 (57%) Number of students
IB Chemistry. DP Chemistry Review
DP Chemistry Review Topic 1: Quantitative chemistry 1.1 The mole concept and Avogadro s constant Assessment statement Apply the mole concept to substances. Determine the number of particles and the amount
SYMBOLS, FORMULAS AND MOLAR MASSES
SYMBOLS, FORMULAS AND MOLAR MASSES OBJECTIVES 1. To correctly write and interpret chemical formulas 2. To calculate molecular weights from chemical formulas 3. To calculate moles from grams using chemical
KEY for Unit 1 Your Chemical Toolbox: Scientific Concepts, Fundamentals of Typical Calculations, the Atom and Much More
KEY for Unit 1 Your Chemical Toolbox: Scientific Concepts, Fundamentals of Typical Calculations, the Atom and Much More The Modern Periodic Table The Periodic Law - when elements are arranged according
MASS RELATIONSHIPS IN CHEMICAL REACTIONS
MASS RELATIONSHIPS IN CHEMICAL REACTIONS 1. The mole, Avogadro s number and molar mass of an element. Molecular mass (molecular weight) 3. Percent composition of compounds 4. Empirical and Molecular formulas
602X10 21 602,000,000,000, 000,000,000,000 6.02X10 23. Pre- AP Chemistry Chemical Quan44es: The Mole. Diatomic Elements
Pre- AP Chemistry Chemical Quan44es: The Mole Mole SI unit of measurement that measures the amount of substance. A substance exists as representa9ve par9cles. Representa9ve par9cles can be atoms, molecules,
MOLAR MASS AND MOLECULAR WEIGHT Themolar mass of a molecule is the sum of the atomic weights of all atoms in the molecule. Molar Mass.
Counting Atoms Mg burns in air (O 2 ) to produce white magnesium oxide, MgO. How can we figure out how much oxide is produced from a given mass of Mg? PROBLEM: If If 0.200 g of Mg is is burned, how much
Chemical formulae are used as shorthand to indicate how many atoms of one element combine with another element to form a compound.
29 Chemical Formulae Chemical formulae are used as shorthand to indicate how many atoms of one element combine with another element to form a compound. C 2 H 6, 2 atoms of carbon combine with 6 atoms of
Sample Problem: STOICHIOMETRY and percent yield calculations. How much H 2 O will be formed if 454 g of. decomposes? NH 4 NO 3 N 2 O + 2 H 2 O
STOICHIOMETRY and percent yield calculations 1 Steps for solving Stoichiometric Problems 2 Step 1 Write the balanced equation for the reaction. Step 2 Identify your known and unknown quantities. Step 3
Chapter Three: STOICHIOMETRY
p70 Chapter Three: STOICHIOMETRY Contents p76 Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions. p70 3-1 Counting by Weighing 3-2 Atomic Masses p78 Mass Mass
1. When the following equation is balanced, the coefficient of Al is. Al (s) + H 2 O (l)? Al(OH) 3 (s) + H 2 (g)
1. When the following equation is balanced, the coefficient of Al is. Al (s) + H 2 O (l)? Al(OH) (s) + H 2 (g) A) 1 B) 2 C) 4 D) 5 E) Al (s) + H 2 O (l)? Al(OH) (s) + H 2 (g) Al (s) + H 2 O (l)? Al(OH)
THE MOLE / COUNTING IN CHEMISTRY
1 THE MOLE / COUNTING IN CHEMISTRY ***A mole is 6.0 x 10 items.*** 1 mole = 6.0 x 10 items 1 mole = 60, 00, 000, 000, 000, 000, 000, 000 items Analogy #1 1 dozen = 1 items 18 eggs = 1.5 dz. - to convert
Formulae, stoichiometry and the mole concept
3 Formulae, stoichiometry and the mole concept Content 3.1 Symbols, Formulae and Chemical equations 3.2 Concept of Relative Mass 3.3 Mole Concept and Stoichiometry Learning Outcomes Candidates should be
Concept 1. The meaning and usefulness of the mole. The mole (or mol) represents a certain number of objects.
Chapter 3. Stoichiometry: Mole-Mass Relationships in Chemical Reactions Concept 1. The meaning and usefulness of the mole The mole (or mol) represents a certain number of objects. SI def.: the amount of
CHAPTER 8: CHEMICAL COMPOSITION
CHAPTER 8: CHEMICAL COMPOSITION Active Learning: 1-4, 6-8, 12, 18-25; End-of-Chapter Problems: 3-4, 9-82, 84-85, 87-92, 94-104, 107-109, 111, 113, 119, 125-126 8.2 ATOMIC MASSES: COUNTING ATOMS BY WEIGHING
Chemical Calculations
Moles and Molecules Moles and Chemical Reactions Moles, Chemical Reactions, and Molarity All done as UNIT CONVERSIONS!!! and practice, practice, practice CHM 1010 Sinex 1 Mass moles particles conversions
SCH 4C1 Unit 2 Problem Set Questions taken from Frank Mustoe et all, "Chemistry 11", McGraw-Hill Ryerson, 2001
SCH 4C1 Unit 2 Problem Set Questions taken from Frank Mustoe et all, "Chemistry 11", McGraw-Hill Ryerson, 2001 1. A small pin contains 0.0178 mol of iron. How many atoms of iron are in the pin? 2. A sample
The Mole. Chapter 10. Dimensional Analysis. The Mole. How much mass is in one atom of carbon-12? Molar Mass of Atoms 3/1/2015
The Mole Chapter 10 1 Objectives Use the mole and molar mass to make conversions among moles, mass, and number of particles Determine the percent composition of the components of a compound Calculate empirical
Chemical Proportions in Compounds
Chapter 6 Chemical Proportions in Compounds Solutions for Practice Problems Student Textbook page 201 1. Problem A sample of a compound is analyzed and found to contain 0.90 g of calcium and 1.60 g of
Mole Notes.notebook. October 29, 2014
1 2 How do chemists count atoms/formula units/molecules? How do we go from the atomic scale to the scale of everyday measurements (macroscopic scale)? The gateway is the mole! But before we get to the
UNIT (4) CALCULATIONS AND CHEMICAL REACTIONS
UNIT (4) CALCULATIONS AND CHEMICAL REACTIONS 4.1 Formula Masses Recall that the decimal number written under the symbol of the element in the periodic table is the atomic mass of the element. 1 7 8 12
Chemistry - A Quantitative Science
56 Ba 137.33 1 H 1.0079 2 He 4.0026 5 B 10.811 6 C 12.011 7 N 14.007 8 O 15.999 9 F 18.998 10 Ne 20.180 13 Al 26.982 14 Si 28.086 15 P 30.974 16 S 32.066 17 Cl 35.453 18 Ar 39.948 3 Li 6.941 4 Be 9.0122
A dozen. Molar Mass. Mass of atoms
A dozen Molar Mass Science 10 is a number of objects. A dozen eggs, a dozen cars, and a dozen people are all 12 objects. But a dozen cars has a much greater mass than a dozen eggs because the mass of each
Chapter 1: Moles and equations. Learning outcomes. you should be able to:
Chapter 1: Moles and equations 1 Learning outcomes you should be able to: define and use the terms: relative atomic mass, isotopic mass and formula mass based on the 12 C scale perform calculations, including
Name Date Class STOICHIOMETRY. SECTION 12.1 THE ARITHMETIC OF EQUATIONS (pages 353 358)
Name Date Class 1 STOICHIOMETRY SECTION 1.1 THE ARITHMETIC OF EQUATIONS (pages 353 358) This section explains how to calculate the amount of reactants required or product formed in a nonchemical process.
The Mole and Molar Mass
The Mole and Molar Mass 1 Molar mass is the mass of one mole of a substance. Molar mass is numerically equal to atomic mass, molecular mass, or formula mass. However the units of molar mass are g/mol.