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Standardized Test Prep
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Chapter 7 Table of Contents
The Mole and Chemical Composition Chapter 7 Table of Contents Section 1 Avogadro’s Number and Molar Conversions Section 2 Relative Atomic Mass and Chemical Formulas Section 3 Formulas and Percentage Composition

Chapter 7 Chapter 7 Bellringer
Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Chapter 7 Bellringer List as many common counting units as you can.

Section 1 Avogadro’s Number and Molar Conversions
Chapter 7 Objectives Identify the mole as the unit used to count particles, whether atoms, ions, or molecules. Use Avogadro’s number to convert between amount in moles and number of particles. Solve problems converting between mass, amount in moles, and number of particles using Avogadro’s number and molar mass.

Avogadro’s Number and the Mole
Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Avogadro’s Number and the Mole The SI unit for amount is called the mole (mol). A mole is the number of atoms in exactly 12 grams of carbon-12. Scientists use the mole to make counting large numbers of particles easier. The number of particles in a mole is called Avogadro’s Number. Avogadro’s number is  1023 units/mole.

Avogadro’s Number and the Mole, continued
Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Avogadro’s Number and the Mole, continued The Mole Is a Counting Unit The mole is used to count out a given number of particles, whether they are atoms, molecules, formula units, ions, or electrons. The mole is just one kind of counting unit: 1 dozen = 12 objects 1 hour = 3600 seconds 1 mole =  1023 particles

Chapter 7 Visual Concepts The Mole

Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Avogadro’s Number and the Mole, continued Amount in Moles Can Be Converted to Number of Particles Counting units are used to make conversion factors. The definition of one mole is 6.022  1023 particles = 1 mol The conversion factor is

Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Avogadro’s Number and the Mole, continued Choose the Conversion Factor That Cancels the Given Units All conversion factors are equal to 1, so you can use them to convert among different units. You can tell which conversion factor to use, because the needed conversion factor should cancel the units of the given quantity to give you the units of the answer or the unknown quantity.

Converting Between Amount in Moles and Number of Particles
Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Between Amount in Moles and Number of Particles

Converting Amount in Moles to Number of Particles
Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Amount in Moles to Number of Particles Sample Problem A Find the number of molecules in 2.5 mol of sulfur dioxide.

Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Amount in Moles to Number of Particles Sample Problem A Solution 2.5 mol SO2  ? = ? molecules SO2 You are converting from the unit mol to the unit molecules. The conversion factor must have the units of molecules/mol. You use  1023 molecules/1 mol.

Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Amount in Moles to Number of Particles Sample Problem A Solution, continued 2.5 mol SO2  ? = ? molecules SO2 2.5 mol SO2 = 1.5  1024 molecules SO2

HW. 3,4 Chapter 7 Sample Problem A, practice pg. 228
How many ions are there in mol of Na+ ions? 2. How many atoms are there in mol of arsenic?

Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Avogadro’s Number and the Mole, continued Number of Particles Can Be Converted to Amount in Moles The reverse calculation is similar to that in Sample Problem A but the conversion factor is inverted to get the correct units in the answer. example: How many moles are 2.54  1022 iron(III) ions?

Converting Number of Particles to Amount in Moles
Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Number of Particles to Amount in Moles Sample Problem B A sample contains 3.01  1023 molecules of sulfur dioxide, SO2. Determine the amount in moles.

Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Number of Particles to Amount in Moles Sample Problem B Solution 3.01 × 1023 molecules SO2  ? = ? mol SO2 You are converting from the unit molecules to the unit mol. The conversion factor must have the units of mol/molecules. You use 1 mol/6.022  1023 molecules.

Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Number of Particles to Amount in Moles Sample Problem B Solution, continued 3.01  1023 molecules SO2  ? = ? mol SO2 3.01  1023 molecules SO2 = mol SO2

Chapter 7 HW. 5 Sample Problem B, practice pg. 229
3. A biologist estimates that there are 2.7 × 1017 termites on Earth. How many moles of termites is this? 4. How many moles do 5.66 × 1025 lithium ions, Li+, equal?

Molar Mass Relates Moles to Grams
Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Molar Mass Relates Moles to Grams Amount in Moles Can Be Converted to Mass The molar mass is the mass in grams of one mole of an element or compound. Molar mass is numerically equal to the atomic mass of monatomic elements and the formula mass of compounds and diatomic elements. The units for molar mass are g/mol. Molar mass can be used as a conversion factor in problems converting between mass and amount.

Chapter 7 Visual Concepts Molar Mass

Molar Mass as a Conversion Factor
Chapter 7 Visual Concepts Molar Mass as a Conversion Factor

Molar Mass Relates Moles to Grams, continued
Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Molar Mass Relates Moles to Grams, continued The Mole Plays a Central Part in Chemical Conversions To convert from number of particles to mass, you must use a two-part process: First, convert number of particles to amount in moles. Second, convert amount in moles to mass in grams. One step common to many problems in chemistry is converting to amount in moles.

Converting Between Mass, Amount, and Number of Particles
Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Between Mass, Amount, and Number of Particles

Converting Number of Particles to Mass
Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Number of Particles to Mass Sample Problem C Find the mass in grams of 2.44  1024 atoms of carbon, whose molar mass is g/mol.

Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Number of Particles to Mass Sample Problem C Solution First part: 2.44  1024 atoms  ? = ? mol Select the conversion factor that will take you from number of atoms to amount in moles. You use 1 mol/6.022  1023 atoms.

Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Number of Particles to Mass Sample Problem C Solution, continued Second part: ? mol  ? = ? g Select the conversion factor that will take you from amount in moles to mass in grams. You use the molar mass of carbon, g C/1 mol.

Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Number of Particles to Mass Sample Problem C Solution, continued = 48.7 g C

Chapter 7 HW. 3,4 Sample Problem C , Practice pg. 231
Find the mass in grams of atoms of copper (molar mass of Cu = g/mol) × 1023 formula units of NaCl (molar mass of NaCl = g/mol)

Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Molar Mass Relates Moles to Grams, continued Mass Can Be Converted to Amount in Moles Converting from mass to number of particles is the reverse of the operation in the previous problem. To convert from mass to number of particles, you must use a two-part process: First, convert mass in grams to amount in moles. Second, convert amount in moles to number of particles.

Converting Mass to Number of Particles
Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Mass to Number of Particles Sample Problem D Find the number of molecules present in 47.5 g of glycerol, C3H8O3. The molar mass of glycerol is g/mol.

Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Mass to Number of Particles Sample Problem D Solution First part: 47.5 g  ? = ? mol Select the conversion factor that will take you from mass in grams to amount in moles. You use the inverse of the molar mass of glycerol:

Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Mass to Number of Particles Sample Problem D Solution, continued Second part: ? mol  ? = ? molecules Select the conversion factor that will take you from amount in moles to number of particles. You use

Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Mass to Number of Particles Sample Problem D Solution, continued = 3.11  1023 molecules

Chapter 7 Sample Problem D, practice pg. 233
Find the number of atoms in 237 g Cu (molar mass of Cu = g/mol). 2. Find the number of ions in 20.0 g Ca2+ (molar mass of Ca2+ = g/mol). 3. Find the number of atoms in 155 mol of arsenic.

Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Molar Mass Relates Moles to Grams, continued Remember to Round Consistently Remember that an answer must never be given to more significant figures than is appropriate. Round molar masses from the periodic table to two significant figures to the right of the decimal point.

Homework Section Review, pg. 233, Qs. 1-13

What is the definition of a mole?
Section 7.1 Review pg. 233 What is the definition of a mole? 2. How many particles are there in one mole? 3. Explain how Avogadro’s number can give two conversion factors. 4. Which will have the greater number of ions, 1 mol of nickel(II) or 1 mol of copper(I)? 5. Without making a calculation, is 1.11 mol Pt more or less than x1023 atoms?

Section 7.1 Review pg. 233 6. Find the number of molecules or ions. a mol Fe3+ 7. Find the number of sodium ions, Na+. b mol Na4P2O7 8. Find the number of moles. c ions Na+ 9. Find the mass in grams. a × 1016 atoms He, 4.00 g/mol

Section 7.1 Review pg. 233 10. Find the number of molecules or ions b. 3.5 g Cu2+, g/mol 11. What is the mass of × 1023 molecules of ibuprofen (molar mass of g/mol)? 12. Find the mass in grams. c × 1019 ions Na+, g/mol 13. Find the number of molecules. a mol H2, 2.02 g/mol

2) Cyclopentane has the molecular formula
1) There are exactly 1000 mg in 1 g. A cup of hot chocolate has 35.0 mg of sodium ions, Na+. One cup of milk has 290 mg of calcium ions, Ca2+. a. How many moles of sodium ions are in the cup of hot chocolate? b. How many moles of calcium ions are in the milk? 2) Cyclopentane has the molecular formula C5H10. How many moles of hydrogen atoms are there in 4 moles of cyclopentane? ____________________ How many atoms of Fe are in the formula Fe3C? How many moles of Fe are in one mole of Fe3C 3) How many moles of NaNO2 are there in a beaker that contains kg of NaNO2 (molar mass of NaNO2 = g/mol)?

Section 2 Relative Atomic Mass and Chemical Formulas
Chapter 7 Bellringer Compare the masses of a roll of pennies and a roll of dimes. Both contain 50 coins. Why are the masses of the rolls different when both rolls contain the same number of coins? If given a roll of mixed coins, what information would you need to determine the mass of the roll?

Chapter 7 Chapter 7 Objectives
Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7 Chapter 7 Objectives Use a periodic table or isotopic composition data to determine the average atomic masses of elements. Infer information about a compound from its chemical formula. Determine the molar mass of a compound from its formula.

Average Atomic Mass and the Periodic Table
Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7 Average Atomic Mass and the Periodic Table Most Elements are a Mixture of Isotopes Isotopes are atoms that have different numbers of neutrons than other atoms of the same element do. 12C, 14C potassium-39, potassium-41 chlorine-35, chlorine-37 234Np, 235Np, 236Np, 237Np, 238Np, 239Np Average atomic mass is a weighted average of the atomic mass of an element’s isotopes. If you know the abundance of each isotope, you can calculate the average atomic mass of an element.

Chapter 7 Visual Concepts Average Atomic Mass

Calculating Average Atomic Mass
Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7 Calculating Average Atomic Mass Sample Problem E The mass of a Cu-63 atom is amu, and that of a Cu-65 atom is amu. Using the data below, find the average atomic mass of copper. abundance of Cu-63 = 69.17% abundance of Cu-65 = 30.83%

Calculating Average Atomic Mass
Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7 Calculating Average Atomic Mass Sample Problem E Solution The contribution of each isotope is equal to its atomic mass multiplied by the fraction of that isotope. contribution of Cu-63: amu × contribution of Cu-65: amu × Average atomic mass is the sum of the individual contributions: (62.94 amu × ) + (64.93 amu × ) = amu

Chapter 7 Sample Problem E Practice p.236 1) Calculate the average atomic mass for gallium if 60.00% of its atoms have a mass of amu and 40.00% have a mass of amu.

Chapter 7 Sample Problem E Practice p.236 2)Calculate the average atomic mass of oxygen. Its composition is 99.76% of atoms with a mass of amu, 0.038% with a mass of amu, and 0.20% with a mass amu.

HW. SectionReview,p.240 # 8,9

Chemical Formulas and Moles
Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7 Chemical Formulas and Moles Formulas Express Composition A compound’s chemical formula tells you which elements, as well as how much of each, are present in a compound. Formulas for covalent compounds show the elements and the number of atoms of each element in a molecule. Formulas for ionic compounds show the simplest ratio of cations and anions in any pure sample.

Chemical Formulas and Moles, continued
Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7 Chemical Formulas and Moles, continued Formulas Express Composition, continued Any sample of compound has many atoms and ions, and the formula gives a ratio of those atoms or ions.

Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7 Chemical Formulas and Moles, continued Formulas Give Ratios of Polyatomic Ions Formulas for polyatomic ions show the simplest ratio of cations and anions. They also show the elements and the number of atoms of each element in each ion. For example, the formula KNO3 indicates a ratio of one K+ cation to one anion.

Understanding Formulas for Polyatomic Ionic Compounds
Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7 Understanding Formulas for Polyatomic Ionic Compounds

Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7 Chemical Formulas and Moles, continued Formulas Are Used to Calculate Molar Masses The molar mass of a molecular compound is the sum of the masses of all the atoms in the formula expressed in g/mol. The molar mass of an ionic compound is the sum of the masses of all the atoms in the formula expressed in g/mol.

Chapter 7 Visual Concepts Formula Mass

Calculating Molar Mass for Ionic Compounds
Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7 Calculating Molar Mass for Ionic Compounds

Calculating Molar Mass of Compounds
Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7 Calculating Molar Mass of Compounds Sample Problem F Find the molar mass of barium nitrate, Ba(NO3)2.

Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7 Calculating Molar Mass of Compounds Sample Problem F Solution Find the number of moles of each element in 1 mol Ba(NO3)2: Each mole has 1 mol Ba, 2 mol N, and 6 mol O. Use the periodic table to find the molar mass of each element in the formula: molar mass of Ba: g/mol molar mass of N: g/mol molar mass of O: g/mol

Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7 Calculating Molar Mass of Compounds Sample Problem F Solution, continued Multiply the molar mass of each element by the number of moles of each element. Add these masses to get the total molar mass of Ba(NO3)2. mass of 1 mol Ba = 1  g/mol = g/mol mass of 2 mol N = 2  g/mol = g/mol + mass of 6 mol O = 6  g/mol = g/mol molar mass of Ba(NO3)2 = g/mol

HW. #3,4 Chapter 7 Sample Problem F Practice p.239
2) Write the formula and then find the molar mass. sodium hydrogen carbonate b. cerium hexaboride

c. magnesium perchlorate
d. aluminum sulfate

e. iron(III) hydroxide f. tin(II) chloride

g. tetraphosphorus decoxide
h. iodine monochloride

HW. SectionReview,p.240 # (1-7),10( d, e), 11( b, d), 12(a, b)
Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7 HW. SectionReview,p.240 # (1-7),10( d, e), 11( b, d), 12(a, b)

Section 7.2 Review, pg. 240 What is a weighted average?
It is an average that takes into account both the value and the frequency of each number involved. 2. On the periodic table, the average atomic mass of carbon is g. Why is it not exactly 12.00? Some atoms of carbon have a mass greater than 12.00, so they raise the average atomic mass. 3. What is the simplest formula for cesium carbonate? Cs2CO3 4. What ions are present in cesium carbonate? Two Cs+ ions for every Co3 2- ion

One nitrogen atom: 3 hydrogen atoms
5. What is the ratio of N and H atoms in NH3? One nitrogen atom: 3 hydrogen atoms 6. What is the ratio of calcium and chloride ions in CaCl2? One calcium ion : 2 chloride ions 7. Why is the simplest formula used to determine the molar mass for ionic compounds An ionic compound is composed of positive and negative ions in an infinite crystal. The simplest formula reflects the specific ratio of positive and negative ions

Section 7.2 Review, pg. 240 8) Calculate the average atomic mass of
chromium. Its composition is: 83.79% with a mass of amu; 9.50% with a mass of 52.94 amu; 4.35% with a mass of amu; 2.36% with a mass of amu. (51.99 amu)

Section 7.2 Review, pg. 240 9) Element X has two isotopes. One has a mass of 10.0 amu and an abundance of 20.0%.The other has a mass of 11.0 amu and an abundance of 80.0%. Estimate the average atomic mass. What element is it? (10.80, boron)

Section 7.2 Review, pg. 240 10. Find the molar mass. d. (NH4)2HPO4 (132.08g/mol) e. C2H5NO ( 75.08g/mol)

Section 7.2 Review, pg. 240 11. Determine the formula, the molar mass, and the number of moles in 2.11 g of each of the following compounds. b. phosphorus trifluoride ( PF3, 87.97g/mol, 2.40E-2 mol) d. mercury(II) bromate ( Hg(BrO3)2, g/mol, 4.62E-3mol)

12. Find the molar mass and the mass of 5
12. Find the molar mass and the mass of mol of each of the following compounds. calcium acetate, Ca(C2H3O2)2 (158.18g/mol, g) b. iron(II) phosphate, Fe3(PO4)2 (357.49g/mol, g)

Section 3 Formulas and Percentage Composition
Chapter 7 Bellringer Brainstorm a list of what you know about percentages.

Chapter 7 Objectives Determine a compound’s empirical formula from its percentage composition. Determine the molecular formula or formula unit of a compound from its empirical formula and its formula mass. Calculate percentage composition of a compound from its molecular formula or formula unit.

Chapter 7 Using Analytical Data
Section 3 Formulas and Percentage Composition Chapter 7 Using Analytical Data The percentage composition is the percentage by mass of each element in a compound. Percentage composition helps verify a substance’s identity. Percentage composition also can be used to compare the ratio of masses contributed by the elements in two different substances.

Chapter 7 Percentage Composition of Iron Oxides

Using Analytical Data, continued
Section 3 Formulas and Percentage Composition Chapter 7 Using Analytical Data, continued Determining Empirical Formulas An empirical formula is a chemical formula that shows the simplest ratio for the relative numbers and kinds of atoms in a compound. An actual formula shows the actual ratio of elements or ions in a single unit of a compound. For example, the empirical formula for ammonium nitrate is NH2O, while the actual formula is NH4NO2.

Chapter 7 Empirical and Actual Formulas

Section 3 Formulas and Percentage Composition Chapter 7 Using Analytical Data, continued Determining Empirical Formulas, continued You can use the percentage composition for a compound to determine its empirical formula. Convert the percentage of each element to g. Convert from g to mol using the molar mass of each element as a conversion factor. Compare these amounts in mol to find the simplest whole-number ratio among the elements.

Percentage Composition
Chapter 7 Visual Concepts Percentage Composition

Determining an Empirical Formula from Percentage Composition
Section 3 Formulas and Percentage Composition Chapter 7 Determining an Empirical Formula from Percentage Composition Sample Problem G Chemical analysis of a liquid shows that it is 60.0% C, 13.4% H, and 26.6% O by mass. Calculate the empirical formula of this substance.

Section 3 Formulas and Percentage Composition Chapter 7 Determining an Empirical Formula from Percentage Composition Sample Problem G Solution Assume that you have a g sample, and convert the percentages to grams. for C: 60.0%  g = 60.0 g C for H: 13.4%  g = 13.4 g H for O: 26.6%  g = 26.6 g O

Section 3 Formulas and Percentage Composition Chapter 7 Determining an Empirical Formula from Percentage Composition Sample Problem G Solution, continued Convert the mass of each element into the amount in moles, using the reciprocal of the molar mass.

The empirical formula is C3H8O.
Section 3 Formulas and Percentage Composition Chapter 7 Determining an Empirical Formula from Percentage Composition Sample Problem G Solution, continued The formula can be written as C5H13.3O1.66, but you divide by the smallest subscript to get whole numbers. The empirical formula is C3H8O.

Chapter 7 Sample Problem G , practice p. 243 Determine the empirical formula for each substance. 1) A dead alkaline battery is found to contain a compound of Mn and O. Its analysis gives 69.6% Mn and 30.4% O.

Chapter 7 Sample Problem G , practice p. 243 2) A compound is 38.77% Cl and 61.23% O.

HW. Practice 3,4; p.243 Section 3 Formulas and Percentage Composition
Chapter 7 HW. Practice 3,4; p.243

Section 3 Formulas and Percentage Composition Chapter 7 Using Analytical Data, continued Molecular Formulas Are Multiples of Empirical Formulas The formula for an ionic compound shows the simplest whole-number ratio of the large numbers of ions in a crystal of the compound. A molecular formula is a whole-number multiple of the empirical formula. The molar mass of any compound is equal to the molar mass of the empirical formula times a whole number, n.

Comparing Molecular and Empirical Formulas
Chapter 7 Visual Concepts Comparing Molecular and Empirical Formulas

Determining a Molecular Formula from an Empirical Formula
Section 3 Formulas and Percentage Composition Chapter 7 Determining a Molecular Formula from an Empirical Formula Sample Problem H The empirical formula for a compound is P2O5. Its experimental molar mass is 284 g/mol. Determine the molecular formula of the compound.

Determining a Molecular Formula from an Empirical Formula
Section 3 Formulas and Percentage Composition Chapter 7 Determining a Molecular Formula from an Empirical Formula Sample Problem H Solution Find the molar mass of the empirical formula P2O5. 2  molar mass of P = g/mol + 5  molar mass of O = g/mol molar mass of P2O5 = g/mol

n (empirical formula) = 2 (P2O5) = P4O10
Section 3 Formulas and Percentage Composition Chapter 7 Determining a Molecular Formula from an Empirical Formula Sample Problem H Solution, continued n (empirical formula) = 2 (P2O5) = P4O10

Chapter 7 Sample Problem H , practice p. 245 1 ) A compound has an experimental molar mass of 78 g/mol. Its empirical formula is CH. What is its molecular formula?

HW. Practice 2,3; p.245

Sample Problem H , practice p. 245
2) A compound has the empirical formula CH2O. Its experimental molar mass is 90.0 g/mol. What is its molecular formula?

Sample Problem H , practice p. 245
3) A brown gas has the empirical formula NO2. Its experimental molar mass is 46 g/mol. What is its molecular formula?

Section 3 Formulas and Percentage Composition Chapter 7 Using Analytical Data, continued Chemical Formulas Can Give Percentage Composition If you know the chemical formula of any compound, then you can calculate the percentage composition. From the subscripts, determine the mass contributed by each element and add these to get molar mass. Divide the mass of each element by the molar mass. Multiply by 100 to find the percentage composition of that element.

Section 3 Formulas and Percentage Composition Chapter 7 Using Analytical Data, continued Chemical Formulas Can Give Percentage Composition CO and CO2 are both made up of C and O, but they have different percentage compositions.

Using a Chemical Formula to Determine Percentage Composition
Section 3 Formulas and Percentage Composition Chapter 7 Using a Chemical Formula to Determine Percentage Composition Sample Problem I Calculate the percentage composition of copper(I) sulfide, Cu2S, a copper ore called chalcocite.

Section 3 Formulas and Percentage Composition Chapter 7 Using a Chemical Formula to Determine Percentage Composition Sample Problem I Solution Find the molar mass of Cu2S. 2 mol  g Cu/mol = g Cu mol  g S/mol = g S molar mass of Cu2S = g/mol

Section 3 Formulas and Percentage Composition Chapter 7 Using a Chemical Formula to Determine Percentage Composition Sample Problem I Solution, continued Calculate the fraction that each element contributes to the total mass by substituting the masses into the equations below and rounding correctly. 79.852% Cu

Section 3 Formulas and Percentage Composition Chapter 7 Using a Chemical Formula to Determine Percentage Composition Sample Problem I Solution, continued 20.15% S

Chapter 7 Sample Problem I, practice p.248 1) Calculate the percentage composition of Fe3C, a compound in cast iron.

Chapter 7 Sample Problem I, practice p.248 2) Calculate the percentage of both elements in sulfur dioxide. 3) Calculate the percentage composition of ammonium nitrate, NH4NO3

Sample Problem I, practice p.248
4) Calculate the percentage composition of each of the following: SrBr c. Mg(CN)2 b. CaSO d. Pb(CH3COO)2

Chapter 7 HW. Section 7.3 Review, p.248 #(1-3), 4c, 6c, 7c

Section 7.3 Review, pg. 248 a. Suppose you know that a compound is 11.2% H and 88.8% O. What information do you need to determine the empirical formula? The molar masses of H and O b. What additional information do you need to determine the molecular formula? The compound’s experimental molar mass 2. Isooctane has the molecular formula C8H18. What is its empirical formula? C4H9

3. What information do you need to calculate the percentage composition of CF4
The molar masses of carbon and fluorine 4. Determine the empirical formula. c. The analysis of a compound shows that it is 27.0% Na, 16.5% N, and 56.5% O.

6. Determine the formula, and then calculate the percentage composition.
c. silver nitrate 7. Calculate the percentage composition. c. iron(III) sulfate, Fe2(SO4)3

Understanding Concepts
Chapter 7 Standardized Test Preparation Understanding Concepts 1. Element A has two isotopes. One has an atomic mass of 120 and constitutes 60%; the other has an atomic mass of 122 and constitutes 40%. Which range below includes the average atomic mass of Element A? A. less than 120 B. between 120 and 121 C. between 121 and 122 D. greater than 122

Chapter 7 Standardized Test Preparation Understanding Concepts 2. Which of the following can be determined from the empirical formula of a compound alone? F. the true formula of the compound G. the molecular mass of the compound H. the percentage composition of the compound I. the arrangement of atoms within a molecule of the compound

Chapter 7 Standardized Test Preparation Understanding Concepts 3. How many ions are in 0.5 moles of NaCl? A  1023 B  1023 C  1023 D  1023

Chapter 7 Standardized Test Preparation Understanding Concepts 4. How many moles of calcium (mass = 40.1) are in a serving of milk containing 290 mg of calcium?

Chapter 7 Standardized Test Preparation Understanding Concepts 4. How many moles of calcium (mass = 40.1) are in a serving of milk containing 290 mg of calcium? Answer: 7.23  10–3 mol

Chapter 7 Standardized Test Preparation Understanding Concepts 5. How is Avogadro's number related to moles?

Chapter 7 Standardized Test Preparation Understanding Concepts 5. How is Avogadro's number related to moles? Answer: Avogadro's number is the number of particles in a mole.

Chapter 7 Standardized Test Preparation Understanding Concepts 6. Antimony has two isotopes. One, amounting to 57.3% of the atoms, has a mass of amu. The other, 42.7% of the atoms, has a mass of amu. What is the average atomic mass of antimony?

Chapter 7 Standardized Test Preparation Understanding Concepts 6. Antimony has two isotopes. One, amounting to 57.3% of the atoms, has a mass of amu. The other, 42.7% of the atoms, has a mass of amu. What is the average atomic mass of antimony? Answer: amu

Chapter 7 Reading Skills Standardized Test Preparation
Read the passage below. Then answer the questions. In 1800 two English chemists, Nicholson and Carlisle, discovered that when an electric current is passed through water, hydrogen and oxygen were produced in a 2:1 volume ratio and a 1:8 mass ratio. This evidence helped to support John Dalton's theory that matter consisted of atoms, demonstrating that water consists of the two elements in a constant proportion. If the same number of moles of each gas occupy the same volume, then each molecule of water must consist of twice as much hydrogen as oxygen, even though the mass of hydrogen is only one-eighth that of oxygen.

Chapter 7 Reading Skills
Standardized Test Preparation Reading Skills 7. Based on this experiment, what is the empirical formula of water? F. HO G. H2O H. H2O8 I. O8H

Standardized Test Preparation Reading Skills 8. How would the experimental result have been different if hydrogen gas existed as individual atoms while oxygen formed molecules with two atoms bound by a covalent bond? A. The result would be the same. B. The ratio of hydrogen to oxygen would be 1:1. C. The ratio of hydrogen to oxygen would be 1:4. D. The ratio of hydrogen to oxygen would be 4:1.

Standardized Test Preparation Reading Skills 9. How does the empirical formula for water compare to its molecular formula?

Standardized Test Preparation Reading Skills 9. How does the empirical formula for water compare to its molecular formula? Answer: Both formulas are the same, H2O.

Interpreting Graphics
Chapter 7 Standardized Test Preparation Interpreting Graphics Use the diagram below to answer questions 10–13.

Chapter 7 Standardized Test Preparation Interpreting Graphics 10. How many moles of oxygen atoms are there in moles of carbon dioxide? F. 66.7 G. 72.7 H I

Chapter 7 Standardized Test Preparation Interpreting Graphics 11. Explain why the percentage of oxygen in carbon dioxide is not twice the percentage of oxygen in carbon monoxide, if there are twice as many oxygen atoms.

Chapter 7 Standardized Test Preparation Interpreting Graphics 11. Explain why the percentage of oxygen in carbon dioxide is not twice the percentage of oxygen in carbon monoxide, if there are twice as many oxygen atoms. Answer: Although there are twice as many oxygen atoms in carbon dioxide, the percentage composition is based on the mass, not the number of atoms.

Chapter 7 Standardized Test Preparation Interpreting Graphics 12. If you did not know the true formulas for carbon monoxide and carbon dioxide, what information would you need beyond what is provided in the illustration in order to calculate them? A. the percentage compositions B. the atomic masses of carbon and oxygen C. the melting and boiling points of each compound D. the number of atoms of each element in the compound

Chapter 7 Standardized Test Preparation Interpreting Graphics 13. How many grams of carbon are contained in grams of carbon dioxide? F G H I

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How to Use This Presentation

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Presentation on theme: "How to Use This Presentation"— Presentation transcript:

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About project

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