# If you remember, we left off this part of the story with defining the mass of one 12

Save this PDF as:

Size: px
Start display at page:

Download "If you remember, we left off this part of the story with defining the mass of one 12"

## Transcription

1 Chapter 11 Chemical Calculations For the past several weeks we have been working on our qualitative understanding of first atoms, then molecules, and finally chemical reactions. In this chapter we enter a new phase: we begin to put numbers to everything and begin building up our quantitative tools so we can determine actual amounts of chemicals to be used in reactions The Mole To start building up on quantitative understanding of chemistry we have to go back to Chapter 2 where we discussed the mass of the atom. If you remember, we left off this part of the story with defining the mass of one 12 atom of 6C as being 12 atomic mass units or 12 u. The numbers associated with all the other atoms was then a relative number that told how much larger or smaller the mass of another atoms was. And then we added the complication that most elements in nature consist of a number of isotopes, and that the numbers we write on the periodic table represent a weighted average of all the isotopes. Now that we know how molecules are built, let s extend our concept of atomic masses to molecular mass 12 If the mass of one atoms of C is 12 units, then the molecule CH 4 which has one C atoms and 4 H atoms is goin to be 12+(4x1) = 16 u, etc. Key Concept: A molecular mass is the sum of the atomic masses of all the atoms in a molecule Let s generalize this one more step A formula mass is the sum of the atomic masses of all the atoms in an atom, molecule or ion. Again we are talking relative numbers one molecule of CH 4 is 16u and is 16u/12u more massive than one atom of C But dealing with ratios is tricky and abstract. I want something concrete and absolute to work with. So I can weigh a chemical out on a balance and know how much material I have. The key to this absolute chemical scale is the mole

2 2 Key concept: The quantity of a substance whose mass in grams is numerically equal to the formula mass of a substance is called a mole (abbreviated mol) -And- The molar mass of a compound is the number of grams need to make up one mole of a compound. Example: CH 4 Formula mass = units ( (1.008)) st 1 definition above: grams CH 4 = 1 mole CH4 nd 2 definition above Molar mass of CH = grams of CH 4 4 But I probably don t have to tell you this; you have probably had this a zillion times in high school, just maybe not presented in this manner. The above equations, linking the molar mass of a substance in grams with the mole is a key conversion factor that we will use over and over in chemical calculation and fits right in with our Dimensional Analysis that you learned in Chapter 1 Example 1: I have 20 grams of CH 4 How many moles is that Given want grams moles Unit conversion Grams X Moles = moles Grams grams CH 4 = 1 mole CH4 Plug & chug grams CH X 1 mole CH =1.25 moles CH g CH 4

3 3 Example 2: I need.450 moles of NaCl for an experiment. How many grams do I need to weigh out? 1 mole NaCl = g NaCl Given moles NaCl Conversion moles NaCl x Need Grams NaCl Grams NaCl = grams NaCl Moles NaCl Plug & Chug.450 moles NaCl x grams NaCl = 26.3 g NaCl 1 mol NaCl Clicker question: gram to mole conversion Ba(OH) 2 show 4 equations each with a different answer have them pick out correct equation 11-2 Avogadro s number 12 So our definition of a mole was that 12g of C is 1 mole of C But what is it really Now that we know that the C is composed of atoms of C, 12 how many atoms of C are in that 1 mole or 12 g of C? Key Concept: 23 1 mole of a thing is 6.022x10 things. 23 So g of CH 4 contains 6.022x10 molecules of CH and g of NaCl contains 6.022x10 g of NaCl(s) or 6.022x10 Na ions 23 - and 6.022x10 Cl ions And now we have another conversion factor 23 1 mole of stuff = 6.022x10 items of stuff A mole is a lot of stuff but I think you can handle it. Example calculations: I have a 10 pound lead weight that I use for an anchor on my boat. How many moles are in that anchor? How many atoms? 1 lbs= g 1 mole Pb = g

4 4 Given lbs Want Moles Set up lbs x grams x moles =Mole lbs grams Plug & Chug 10.0 lbs Pb x g Pb x 1 mole Pb = 21.9 moles Pb 1 lbs Pb g Pb If I wanted to go all the way to atoms Given lbs Want Atoms Set up lbs x grams x moles x atoms = atoms lbs gram Mole Plug & Chug lbs Pb x g Pb x 1 mole Pb x 6.022x10 atoms = 1.32x10 atoms Pb 1 lbs Pb g Pb 1 mole Pb Clicker question Start with Giga molecules of Calcium Chloride (written not formula) Give 3 wrong and 1 correct equation to convert to grams 11-3 Simplest (Empirical) Formulas Back in chapter 2 we talked about analyzing a compound using mass percent where mass % = (mass of element A)/(total mas of all elements in a compound) X100% This is one of the simplest and oldest ways that chemists use to analyze compounds. In this section you will learn how to go from a mass % to an empirical formula But first, what is an empirical formula? Key Concept: An Empirical formula is the simplest possible formula for a compound. It expresses the ratio of one atom to another, but it does not convey the total amount of each element in a compound

5 5 Example: Benzene and acetylene have the same empirical formula CH. This says that there is 1 C atom for every H atom in both of these compound The actual molecular formula of Benzene is C6H 6 and that of acetylene is C2H2 so you can see how the actual molecular formula is a multiple of the empirical formula The way McQuarrie explains how to go from a mass % to an empirical formula makes sense, but he hasn t really boiled it down to a simple step-by-step procedure for you to follow. So let s try doing it this way: Key Procedure: Calculating Empirical Formulas from mass % 1. Calculate the numbers of grams of each element that would occur if the sample was 100 g. We do this to convert from % to g in one easy step. 2. Calculate the Number of moles for each element in a 100 gram sample 3. Find the element with the smallest number of moles, and then divide the number of moles for each element by that smallest number. - This should result in small whole numbers that will represent the subscripts for the elements in the empirical formula - If you do not have whole numbers, multiply all the numbers in the equation by integers until you get whole numbers for all coefficient in the equation Example calculation: Let s try an example, hydrazine is 87.42% nitrogen and 12.58% hydrogen, what is its empirical formula? 1. The first step is to convert % to grams assuming you have a 100 gram sample 100g X.8742 = g N 100 g X = g H 2. Second step - convert from grams to moles g N/14.01 g/mol = 6.24 moles N g H / g/mol = moles H 3. Divide through by the smallest number of moles moles H/ 6.24 moles of N So 2 moles of H 1 mole of N means is stoichiometically equivalent to

6 Three things to notice. A. Always put the largest number over the smallest. If you have more than 2 elements choose the smallest overall and use that as the denominator for all calculations B. You will usually get nice whole numbers, but sometimes may end up with ratios like 1.5/1. In this case you need to multiply both the numerator and the denominator by a single number that will give you all nice whole numbers. In this case 2(1.5)=3 and 2(1) =2, so the actual ratio is 3/2 C. The resulting formula is called the empirical formula. It is not the molecular formula you have the ratio right but the overall numbers can be factors of this NH, (N H, N H etc.) A second example (if you want it) An unknown liquid has 12.5% H, 27.5% C and 50% O, what is the empirical formula of this compound? H : 100g x 12.5% = 12.5g C: 100g x 37.5% = 37.5g O: 100g x 50% = 50.0g H: 12.5g x 1 mole/1.008 g = 12.5 mole C: 37.5g x 1 mole/12.01g = mol O: 50 g x 1 mole/16.00 g = /3.125 = 4 units of H 3.125/3.125 = 1 unit of C 3.125/3.125 = 1 unit of O 4 moles of H 1 mole of C 1 mole of O Empirical formula = CH O Clicker question? Long an complicated break into steps? 4 The book does some additional problems where you burn a compound in a gas and look at the resulting oxide or nitride. I think I will skip this Using the empirical formula to determine the atomic mass of an element in a compound While this may be a fairly standard lab technique, its not one that we do so I will skip this section 6

7 11-5 Molecular Formulas So if we can only get an empirical formula from a mass %, how do we get an actual molar formula? We need one more piece of information, the molar mass. You have seen that the empirical formula is the simplest possible molecular formula. You can think of it as a lowest common denominator. The actual molecular formula is some multiple of this: Key Equation molecular formula = empirical formula x N molecular mass = empirical mass x N Where N is some whole number Example: A compound is 85.7% C and 14.3% H and has a molar of 42. What is the molecular formula of this compound? There are a couple of ways to do this. First the long way gc in 100g sample = 85.7g gh in 100g sample = 14.3 g mole C = 85.7/12.01 = 7.14 moles C mole H = 14.3/1.008 = moles H Dividing by mole C 2 moles of H Empirical formula = CH 2 ; Empirical mass = 14 Molecular mass = empirical mass x N 42 = 14x N 42/14=N N=3 Molecular formula = empirical formula x N = (CH 2) x 3 =C H 3 6 I like the long way because you see that there are three empirical units in the molecular unit. 7

8 8 Now the short way Multiply the molar mass by the % 42 g/mole x 85.7%/100% = 36.0 g/mol C 42 g/mol x 14.3%/100% = 6.00 g/mol H Divide the g/mol fo reach element by the atomic mass g/mol C /12 g/atom = 3 atoms/mole = C 3 6 g/molh / 1g/atom H = 6 atoms/mol = H 6 Molecular formula = C H 3 6 Key calculation Given % composition and molar mass be able to determine the molecular formula. This can be done by two different procedures choose the one that you like best Combustion Analysis This is another fairly standard technique that is used in many labs, but I think we will skip for now Coefficients in Chemical Equations Back in chapter 3 you learned how to balance chemical equations. In this process you came up with a set of balancing coefficients, or numbers in front of every molecule that told you how many of each molecule you needed so that all the atoms that appeared in the reactant molecules were also present in the product molecules so that the mass of every atom was conserved These coefficients are also the key to our stoichiometry problems. For example let s start with the balanced equation for the burning butane: 2C4H 10(l) + 13O 2(g) 8CO 2(g) + 10H2O(g) The equation says that 2 molecules of C H 13 molecules O 8 molecules CO 10 molecules H O Which the same as 2 moles of C H 13 moles O 8 moles CO 10 moles H O This gives you a wealth of conversion factors. A cigarette lighter might have 5 grams of butane in it. How much O 2 is required to burn this amount of butane? First, how many moles of butane do I have What is the Molecular mass of butane? 4(12.01) + 10(1.01) = 58.24

9 9 5 g x 1 mole/58.24 g = moles Now from the above reaction we know that 2 moles 13 moles of Oxygen (or we also say is equivalent to). Now just as we did unit conversions, lets set this up as a proportionality 2 moles butane = 13 moles oxygen 2 moles butane/13 moles oxygen = 1 -or- 13 moles oxygen/2 mole butane =1 Now with these conversions, how do we convert moles of butane to moles of oxygen? moles butane X 13 moles oxygen=.5580 moles of oxygen 2 moles of butane Since O 2 has a molecular mass of 32.0, our weight of O 2 is.5580 moles O 2 x 32g O 2 = grams O2 1 mol O 2 Out of curiosity, lets continue this to calculate the amount of water made in this reaction. From our chemical reaction we know that 2 moles of butane 10 moles of water, so we have the proportion 2 mol butane/10 mol water = 1 -or- 10 mol water/2 mol Butane =1 If we have mol butane x 10 mol water =.429 moles of water 2 mol butane Molar mass of water= so.429 mole water x g water = grams of water 1 mole water Key Procedure: Stiochiometric calculations 1. Balance the equation 2. Convert the mass of given substance to moles 3. Use balanced equation to set up molar conversion factors 4. Use appropriate conversion to calculate the moles of substance you want 5. Use molecular mass of substance to convert moles into grams of substance.

10 Stoichiometry Figure 11.7 Puts the above procedure into a nice flow chart. But I can t seem to grab it from the computer Let s try this in another example; the thermite reaction. Used in rockets, underwater welding and incendiary bombs Fe2O 3 (s)+al(s) Fe(l) + Al2O 3(s) (Unbalanced) (review names iron(iii) oxide and Aluminum oxide (why no (III)?) If I want to generate 15 grams of Fe, how much Al and how much Fe2O 3 is needed Step 1. balance reaction Fe2O 3 (s)+2al(s) 2 Fe(l) + Al2O 3(s) Fe 2 2 Al 2 2 O 3 3 Fe2O 3 (s)+2al(s) 2 Fe(l) + Al2O 3(s) Step 2. Convert 15 g Fe to moles 15 g Fe X 1 mole Fe = moles Fe g Fe Step 3 Convert moles Fe to moles Al mol Fe X 2 mol Al = mole Al; 2 mol Fe Step 4 Convert mole Al to grams Al mole Al x g Al = 7.12 g Al 1 mol Al Step 3 for Fe O : Converting moles Fe to moles Fe O mol Fe X 1 mol Fe O = mol Fe O 2 mol Fe Step 4 Convert moles Fe2O 3 to g Fe2O mol Fe O X g Fe O = 21.5 g Fe O Mole Fe O

11 Stoichiometry without chemical equations Using the principal of conservation of mass, it is sometimes possible to do a stoichiometric calculation without a chemical equation! For example say I am going to react 10 g of sodium in water to make NaOH If NaOH is the only product of this reaction that contains sodium, then the principle of conservation of mass says that all the sodium I start off with has to appear in the NaOH product. So Na NaOH So, not even knowing the reaction I can do my stoichiometric calculation: 10g Na x 1 mole Na 23 g Na =.435 mole Na.435 mole Na x 1 mole NaOH =.435 mol NaOH 1 mole NaOH.435 mol NaOH x 40 g NaOH = 17.4 g NaOH 1 mol NaOH Limiting Reactant When two or more substances react the mass of the product is determined by the limiting reactant. Let s go back to the thermite reaction. When we left that problem we had determined that 7.12 g of Al and 21.5 g of Fe2O 3 were required to make 15 of Fe using this reaction What would happen if we used 15 g of Al and 21.5 g of Fe2O 3, would we make any more product? NO. Adding excess of one reactant doesn t give you more product, because the other reactant limits the amount of product that can be made. Key Concept: In a chemical reacion with more that one reactant, usually one reactant, called the limiting reactant, will limit the amount of product made in the reaction. Other reactants, that are present in excess are called excess reactants. You will often be given problems like this, where you are give the gram amounts of two reactants and you need to figure out which one is the limiting reactant and which one is the excess reactant.

12 12 So let me give you a procedure to figure out which is which Key Procedure and definition: Finding the limiting reactant 1. Pick a product and stick with it for the rest of the problem 2. Calculate how many moles of this product you can get from reactant A alone 3. Calculate how many moles of this product you can get from reactant B alone C, D, E??? The reactant that give you the SMALLEST product is the limiting reactant. The other reactants are all excess reactants. The amount of product obtained with this limiting reactant is the Theoretical Yield Let s try this with a variation of our butane lighter problem. If I start the butane lighter with 0.5 grams of butane, and place it in a 5 gallon jug and seal the jug, which will run out first, the air in the jug or the lighter fuel? We already calculated that 0.5 grams of butane is moles of butane. And we know from the balanced equation 2C H (l) + 13O (g) 8CO (g) + 10H O(g) That 2 moles butane 13 moles O 8 moles CO 10 moles H O Now I will skip ahead a few steps on how to go from 5 gallons of air to moles of O, and just give you the final answer that 5 gallons of air contains.187 moles of 2 O 2 Step 1 pick a product I will pick CO 2 Step 2 calculate amount for product given by reactant A.0086 m butane x 8 mol CO =.0344 moles CO 2 mol butane Step 3 calculate amount for product given by reactant B.187 m O 2 x 8 mol CO 2 =.115 mole CO2 13 m O Butane is the limiting reagent because it makes the smallest amount of product. O is the excess reagent because you have an excess amount of it. 2

13 Which product do you choose? It doesn t matter, and the result will be the same. However, on a test, the next question is usually how much is your expected yield of X so if you chose the reagent that answers that question, you are one step closer to answering that question Percentage Yield In the above section I talked about the theoretical yield. That the amount of yield you calculate from the limiting reactant. You actual yield almost never matches this theoretical value. Why? Reaction fails Reaction not complete Side reaction that gives other products You may lose some product by accident Chemists like to keep track of how close their yield is to the theoretical with a calculation called the % Yield Key Calculation: % yield = (actual yield/theoretical yield) x 100 % Example: Going back to the previous example. If our actual yield was.0301 moles of CO and our theoretical yield was.0344 mole of CO, what is our % yield?.0301/.0344 x 100 = 87.5% Note 1: You can calculate % yield based on grams of product or moles of product. It doesn t matter which one you use they should give you the same number of you do the calculation correctly. Note 2: You might think that the % yield should always be less than 100% but guess again. Sometimes it is >100%. How can that be? Sometimes when you purify your product, you accidentally purify some Kaka with it. Now the weight you record is higher than it should be because of the added Kaka

### 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

### 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

### 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

### 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

### Molecular Formula: Example

Molecular Formula: Example A compound is found to contain 85.63% C and 14.37% H by mass. In another experiment its molar mass is found to be 56.1 g/mol. What is its molecular formula? 1 CHAPTER 3 Chemical

### 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

### Mass and Moles of a Substance

Chapter Three Calculations with Chemical Formulas and Equations Mass and Moles of a Substance Chemistry requires a method for determining the numbers of molecules in a given mass of a substance. This allows

### 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

### 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

### Simple vs. True. Simple vs. True. Calculating Empirical and Molecular Formulas

Calculating Empirical and Molecular Formulas Formula writing is a key component for success in chemistry. How do scientists really know what the true formula for a compound might be? In this lesson we

### Chapter 3 Calculation with Chemical Formulas and Equations

Chapter 3 Calculation with Chemical Formulas and Equations Practical Applications of Chemistry Determining chemical formula of a substance Predicting the amount of substances consumed during a reaction

### 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?

### 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

### 2 Stoichiometry: Chemical Arithmetic Formula Conventions (1 of 24) 2 Stoichiometry: Chemical Arithmetic Stoichiometry Terms (2 of 24)

Formula Conventions (1 of 24) Superscripts used to show the charges on ions Mg 2+ the 2 means a 2+ charge (lost 2 electrons) Subscripts used to show numbers of atoms in a formula unit H 2 SO 4 two H s,

### 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

### 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

### 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 and Chemical Reactions. Moles and Chemical Reactions. Molar mass = 2 x 12.011 + 6 x 1.008 + 1 x15.999 = 46.069 g/mol

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

### Chemical Reactions. Chemical Equations. Mole as Conversion Factor: To convert between number of particles and an equivalent number of moles:

Quantities of Reactants and Products CHAPTER 3 Chemical Reactions Stoichiometry Application of The Law of Conservation of Matter Chemical book-keeping Chemical Equations Chemical equations: Describe proportions

### 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

### 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

### Chemistry I: Using Chemical Formulas. Formula Mass The sum of the average atomic masses of all elements in the compound. Units are amu.

Chemistry I: Using Chemical Formulas Formula Mass The sum of the average atomic masses of all elements in the compound. Units are amu. Molar Mass - The mass in grams of 1 mole of a substance. Substance

### Chemical calculations

Chemical calculations Stoichiometry refers to the quantities of material which react according to a balanced chemical equation. Compounds are formed when atoms combine in fixed proportions. E.g. 2Mg +

### Chapter 3. Stoichiometry: Ratios of Combination. Insert picture from First page of chapter. Copyright McGraw-Hill 2009 1

Chapter 3 Insert picture from First page of chapter Stoichiometry: Ratios of Combination Copyright McGraw-Hill 2009 1 3.1 Molecular and Formula Masses Molecular mass - (molecular weight) The mass in amu

### 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

### 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

### 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.

### Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page 1. A chemical equation. (C-4.4)

Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page 1 1. 2. 3. 4. 5. 6. Question What is a symbolic representation of a chemical reaction? What 3 things (values) is a mole of a chemical

### 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

### Chapter 3. Mass Relationships in Chemical Reactions

Chapter 3 Mass Relationships in Chemical Reactions This chapter uses the concepts of conservation of mass to assist the student in gaining an understanding of chemical changes. Upon completion of Chapter

### 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

### 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.

### 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.

### 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.

### Chem 1100 Chapter Three Study Guide Answers Outline I. Molar Mass and Moles A. Calculations of Molar Masses

Chem 1100 Chapter Three Study Guide Answers Outline I. Molar Mass and Moles A. Calculations of Molar Masses B. Calculations of moles C. Calculations of number of atoms from moles/molar masses 1. Avagadro

### 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

### Simple vs. True Calculating Empirical and Molecular Formulas

17 Calculating Empirical and Molecular Formulas OBJECTIVE Students will learn to calculate empirical and molecular formulas and practice applying logical problemsolving skills. LEVEL Chemistry NATIONAL

### 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

### 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

### Chapter 1 The Atomic Nature of Matter: Selected Answersc for Practice Exam.

Chapter 1 The Atomic Nature of Matter: Selected Answersc for Practice Exam. MULTIPLE CHOICE 50. 5.80 g of dioxane (C 4 H 8 O 2 ) is how many moles of dioxane? 0.0658 mol 0.0707 mol 0.0725 mol d. 0.0804

### 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

### 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

### 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

### 1. P 2 O 5 2. P 5 O 2 3. P 10 O 4 4. P 4 O 10

Teacher: Mr. gerraputa Print Close Name: 1. A chemical formula is an expression used to represent 1. mixtures, only 3. compounds, only 2. elements, only 4. compounds and elements 2. What is the total number

### 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,

### 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

### Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations

CHE11 Chapter Chapter Stoichiometry: Calculations with Chemical Formulas and Equations 1. When the following equation is balanced, the coefficients are. NH (g) + O (g) NO (g) + H O (g) (a). 1, 1, 1, 1

### Chapter 3. Molecules, Compounds, and Chemical Composition

Chapter 3 Molecules, Compounds, and Chemical Composition Elements and Compounds Elements combine together to make an almost limitless number of compounds. The properties of the compound are totally different

### 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?

### The Mole Concept. A. Atomic Masses and Avogadro s Hypothesis

The Mole Concept A. Atomic Masses and Avogadro s Hypothesis 1. We have learned that compounds are made up of two or more different elements and that elements are composed of atoms. Therefore, compounds

### 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 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

### 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

### Chemistry B11 Chapter 4 Chemical reactions

Chemistry B11 Chapter 4 Chemical reactions Chemical reactions are classified into five groups: A + B AB Synthesis reactions (Combination) H + O H O AB A + B Decomposition reactions (Analysis) NaCl Na +Cl

### 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

### IB Chemistry 1 Mole. One atom of C-12 has a mass of 12 amu. One mole of C-12 has a mass of 12 g. Grams we can use more easily.

The Mole Atomic mass units and atoms are not convenient units to work with. The concept of the mole was invented. This was the number of atoms of carbon-12 that were needed to make 12 g of carbon. 1 mole

### 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

### The Mole, Avogadro s Number, and Molar Mass

The Mole, Avogadro s Number, and Molar Mass Example: How many atoms are present in 2.0 kg of silver? (1 amu = 1.6605402x10-24 g) Example: How many molecules are present in 10. mg of smelling salts, (NH

### 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

### 10.3 Percent Composition and Chemical Formulas. Chapter 10 Chemical Quantities Percent Composition and Chemical Formulas

Chapter 10 Chemical Quantities 101 The Mole: A Measurement of Matter 102 Mole-Mass and Mole-Volume Relationships 103 Percent Composition and Chemical Formulas 1 CHEMISTRY & YOU What does the percent composition

### Moles Lab mole. 1 mole = 6.02 x 1023. This is also known as Avagadro's number Demo amu amu amu

Moles I. Lab: Rice Counting II. Counting atoms and molecules I. When doing reactions chemists need to count atoms and molecules. The problem of actually counting individual atoms and molecules comes from

### 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

### MOLES, MOLECULES, FORMULAS. Part I: What Is a Mole And Why Are Chemists Interested in It?

NAME PARTNERS SECTION DATE_ MOLES, MOLECULES, FORMULAS This activity is designed to introduce a convenient unit used by chemists and to illustrate uses of the unit. Part I: What Is a Mole And Why Are Chemists

### Appendix D. Reaction Stoichiometry D.1 INTRODUCTION

Appendix D Reaction Stoichiometry D.1 INTRODUCTION In Appendix A, the stoichiometry of elements and compounds was presented. There, the relationships among grams, moles and number of atoms and molecules

### Practice questions for Chapter 8

Practice questions for Chapter 8 2) How many atoms of nickel equal a mass of 58.69 g? (Refer to the Periodic Table.) A) 1 B) 28 C) 58.69 D) 59 E) 6.02 x 1023 Answer: E Section: 8.1 Avogadro's Number 6)

### 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

### 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

### CHAPTER 3 MASS RELATIONSHIPS IN CHEMICAL REACTIONS

CHAPTER 3 MASS RELATIONSHIPS IN CHEMICAL REACTIONS This chapter reviews the mole concept, balancing chemical equations, and stoichiometry. The topics covered in this chapter are: Atomic mass and average

### Calculations with Chemical Reactions

Calculations with Chemical Reactions Calculations with chemical reactions require some background knowledge in basic chemistry concepts. Please, see the definitions from chemistry listed below: Atomic

### Chapter 3: STOICHIOMETRY: MASS, FORMULAS, AND REACTIONS

Chapter 3: STOICHIOMETRY: MASS, FORMULAS, AND REACTIONS Problems: 3.1-3.8, 3.11, 3.14-3.90, 3.103-3.120, 3.122-3.125, 3.128-3.131, 3.134, 3.137-36.138, 3.140-3.142 3.2 THE MOLE Stoichiometry (STOY-key-OM-e-tree):

### 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

### 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

### 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

### 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

### 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

### 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

### Calculations with Chemical Formulas and Equations

Chapter 3 Calculations with Chemical Formulas and Equations Concept Check 3.1 You have 1.5 moles of tricycles. a. How many moles of seats do you have? b. How many moles of tires do you have? c. How could

### Chapter 6: Chemical Composition

C h e m i s t r y 1 2 C h 6 : C h e m i c a l C o m p o s i t i o n P a g e 1 Chapter 6: Chemical Composition Bonus: 17, 21, 31, 39, 43, 47, 55, 61, 63, 69, 71, 77, 81, 85, 91, 95, 97, 99 Check the deadlines

### 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

### 2 The Structure of Atoms

CHAPTER 4 2 The Structure of Atoms SECTION Atoms KEY IDEAS As you read this section, keep these questions in mind: What do atoms of the same element have in common? What are isotopes? How is an element

### 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

### Chemistry Stoichiometry Lesson 8 Lesson Plan David V. Fansler

Chemistry Stoichiometry Lesson 8 Lesson Plan David V. Fansler The Arithmetic of Equations Objectives: Interpret balanced chemical equations in terms of interacting moles, representative particles, masses,

### Jenn Maeng Lesson overview

Jenn Maeng Lesson overview Subject: Chemistry Grade: 10-12 Topic: Stoichiometry Concepts: Stoichiometric Conversions Essential How do we quantify changes in systems? questions: Objectives Students will

### 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

### Chapter 8 How to Do Chemical Calculations

Chapter 8 How to Do Chemical Calculations Chemistry is both a qualitative and a quantitative science. In the laboratory, it is important to be able to measure quantities of chemical substances and, as

### 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

### 11-1 Stoichiometry. Represents

11-1 Stoichiometry What is stoichiometry? Calculations that relate the quantities of substances. It is the study of quantitative (measurable amounts) relationships in chemical reactions and equations.

### 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

### 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

### Mole Concept and Stoichiometry

1 Mole Concept and Stoichiometry Concept Mole Concept Introduction This is our common experience that when we go to market to buy something, a few things we always get in definite numbers. For example,

### 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

### Chapter 3: ex. P2O5 molecular mass = 2(30.97 amu) + 5(16.00 amu) = amu

Molecular Mass and Formula Mass for molecular compounds: the molecular mass is the mass (in amu) of one molecule of the compound molecular mass = atomic masses of elements present Chapter 3: ex. P2O5 molecular

### Lecture Notes Chemistry E-1. Chapter 3

Lecture Notes Chemistry E-1 Chapter 3 http://inserbia.info/news/wp-content/uploads/2013/05/tamiflu.jpg http://nutsforhealthcare.files.wordpress.com/2013/01/tamiflu-moa.jpg The Mole A mole is a certain

### 7-5.5. Translate chemical symbols and the chemical formulas of common substances to show the component parts of the substances including:

7-5.5 Translate chemical symbols and the chemical formulas of common substances to show the component parts of the substances including: NaCl [salt], H 2 O [water], C 6 H 12 O 6 [simple sugar], O 2 [oxygen

### Unit 9 Stoichiometry Notes (The Mole Continues)

Unit 9 Stoichiometry Notes (The Mole Continues) 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

### Experiment 8: Chemical Moles: Converting Baking Soda to Table Salt

Experiment 8: Chemical Moles: Converting Baking Soda to Table Salt What is the purpose of this lab? We want to develop a model that shows in a simple way the relationship between the amounts of reactants

### CONSERVATION OF MASS During a chemical reaction, matter is neither created nor destroyed. - i. e. the number of atoms of each element remains constant

1 CHEMICAL REACTINS Example: Hydrogen + xygen Water H + H + + - Note there is not enough hydrogen to react with oxygen - It is necessary to balance equation. reactants products + H + H (balanced equation)

### CHEM 101/105 Numbers and mass / Counting and weighing Lect-03

CHEM 101/105 Numbers and mass / Counting and weighing Lect-03 Interpretation of Elemental Chemical Symbols, Chemical Formulas, and Chemical Equations Interpretation of an element's chemical symbol depends