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Chapter 3Stoichiometric

Atomic Masses, Mole concept, and Molar Mass (Average atomic mass).

Number of atoms per amount of element.Percent composition and Empirical formula

of molecules.Chemical equations, Balancing equations,

and Stoichiometric calculations including limiting reagents.

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Chemical Stoichiometry

Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions.

Copyright©2000 by Houghton Mifflin Company. All rights reserved.

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By definition: 1 atom 12C “weighs” 12 amu

On this scale

1H = 1.008 amu

16O = 16.00 amu

Atomic mass is the mass of an atom in atomic mass units (amu)

Micro Worldatoms & molecules

Macro Worldgrams

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Atomic Masses

Elements occur in nature as mixtures of isotopes

Carbon = 98.89% 12C

1.11% 13C

<0.01% 14C

Carbon atomic mass = 12.01 amu

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Natural lithium is:

7.42% 6Li (6.015 amu)

92.58% 7Li (7.016 amu)

7.42 x 6.015 + 92.58 x 7.016100

= 6.941 amu

Average atomic mass of lithium:

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The Mole

The number equal to the number of carbon atoms in exactly 12 grams of pure 12C.

1 mole of anything = 6.022 1023 units of that

thing

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Avogadro’s number equals

6.022 1023 units

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Molar Mass

A substance’s molar mass (molecular weight) is the mass in grams of one mole of the compound.

C=12 O=16

CO2 = 44.01 grams per mole

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Molar mass is the mass of 1 mole of in gramseggsshoes

marblesatoms

1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g

1 12C atom = 12.00 amu

1 mole 12C atoms = 12.00 g 12C

1 mole lithium atoms = 6.941 g of Li

For any element

atomic mass (amu) = molar mass (grams)

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1 amu = 1.66 x 10-24 g or 1 g = 6.022 x 1023 amu

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Measuring Atomic Mass

Figure 3.1: (left) A scientist injecting a sample into a mass spectrometer. (right) Schematic diagram of a mass spectrometer.

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Spectrum

Most Abundant Isotope

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Do You Understand Molecular Mass?

How many H atoms are in 72.5 g of C3H8O ?

1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O

1 mol H = 6.022 x 1023 atoms H

5.82 x 1024 atoms H

1 mol C3H8O molecules = 8 mol H atoms

72.5 g C3H8O1 mol C3H8O

60 g C3H8Ox

8 mol H atoms

1 mol C3H8Ox

6.022 x 1023 H atoms

1 mol H atomsx =

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Percent composition of an element in a compound =

n x molar mass of elementmolar mass of compound

x 100%

n is the number of moles of the element in 1 mole of the compound

C2H6O

%C =2 x (12.01 g)

46.07 gx 100% = 52.14%

%H =6 x (1.008 g)

46.07 gx 100% = 13.13%

%O =1 x (16.00 g)

46.07 gx 100% = 34.73%

52.14% + 13.13% + 34.73% = 100.0%

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Figure 3.5: A schematic diagram of the combustion device used to analyze substances for carbon and hydrogen.

Determining Elemental Composition(Formula)

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The masses obtained (mostly CO2 and H2O and sometimes N2)) will be used to determine:

1. % composition in compound

2. Empirical formula

3. Chemical or molecular formula if the Molar mass of the compound is known or given.

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Example of Combustion

Combust 11.5 g ethanol

Collect 22.0 g CO2 and 13.5 g H2O

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1mol CO244gCO2 1mol C 12g C

22gCO2 mc

mc =

Collect 22.0 g CO2 and 13.5 g H2O

12 16 x 2

44gCO2

22gCO2 x 12gC

% C in CO2 collected

= 6g C

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1mol C 12g Cnmol C 6g

6g x 1molCnc =

12 gC

= 0.5 mol

Convert g to mole:

Repeat the same for H from H2O

1mol H2O 18g H2O 2 mol H 2g H

13.5g H2O nmol H

2x13.5nmol H = = 1.5 mol H

18 mol H Faster H but still need O

mH

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Empirical formula C0.5H1.5O0.25

Divide by smallest subscript (0.25)

Empirical formula C2H6O

2x13.5mH = = 1.5 g H

18mO = 11.5g – mC – mH

= 11.5 – 6 – 1.5 = 4g

m 4nO = = = 0.25 mol O

MM 16

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OR for C

mC 12.01

1. Fraction of C in CO2 = = =

MMCO2 44.01

2. Mass of C in compound = mass of CO2 x Fraction of C

mC

3. % C in compound = x 100

msample

4. If N then % N = 100 - % C - % H

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ThenEmpirical Formula

Using the previously calculated % in compound:

% in gram

a. Number of mole of C =

Atomic mass of C

% in gram

b. Number of mole of H =

Atomic mass of H

a b cThen divide by the smallest number:

smallest smallest smallest

: :

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NoteIf results are : 0.99 : 2.01 : 1.00Then you have to convert to whole numbers:

1 : 2 : 1

CH2N

If results are : 1.49 : 3.01 : 0.99Then you have to multiply by 2:

3 : 6 : 2

C3H6N2

Hence, empirical formula is the simplest formula of a compound

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Formulasmolecular formula = (empirical formula)n

[n = integer]

molecular formula = C6H6 = (CH)6

empirical formula = CH

Then

Molecular Mass

= n

Empirical Mass

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Figure 3.6: Examples of substances whose empirical and molecular formulas differ. Notice

that molecular formula = (empirical formula)n, where n is a integer.

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Chemical Equations

Chemical change involves a reorganization of the atoms in one or

more substances.

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Chemical Equation

A representation of a chemical reaction:

C2H5OH + O2 CO2 + H2O

reactants products

Unbalanced !

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Chemical Equation

C2H5OH + 3O2 2CO2 + 3H2O

The equation is balanced.

1 mole of ethanol reacts with 3 moles of oxygen

to produce

2 moles of carbon dioxide and 3 moles of water

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Calculating Masses of Reactants and Products

1. Balance the equation.

2. Convert mass to moles.

3. Set up mole ratios.

4. Use mole ratios to calculate moles of desired substituent.

5. Convert moles to grams, if necessary.

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Methanol burns in air according to the equation

2CH3OH + 3O2 2CO2 + 4H2O

If 209 g of methanol are used up in the combustion, what mass of water is produced?

grams CH3OH moles CH3OH moles H2O grams H2O

molar massCH3OH

coefficientschemical equation

molar massH2O

209 g CH3OH1 mol CH3OH

32.0 g CH3OHx

4 mol H2O

2 mol CH3OHx

18.0 g H2O

1 mol H2Ox =

235 g H2O

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2CH3OH + 3O2 2CO2 + 4H2O

2 mol 4 mol

2x(12+4+16) g 4x(2+16) g

209 g m

209 x 4(2+16)m =

2(12+4+16)

OR

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Limiting Reactant

The limiting reactant is the reactant that is consumed first, limiting the

amounts of products formed.

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Solving a Stoichiometry Problem

1. Balance the equation.

2. Convert masses to moles.

3. Determine which reactant is limiting.

4. Use moles of limiting reactant and mole ratios to find moles of desired product.

5. Convert from moles to grams.

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6 green used up6 red left over

Limiting Reagents

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Limiting Reactant Calculations

What weight of molten iron is produced by 1 kg each of the reactants?

Fe2O3(s) + 2 Al(s) Al2O3(s) + 2 Fe(l)1 mol 2 mol6.26mol 18.52

Ratio:

0.160 > 0.108

Limiting Excess

The 6.26 mol Fe2O3 willDisappear first

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Theoretical Yield is the amount of product that wouldresult if all the limiting reagent reacted. Its amount isCalculated using the balanced equation.

Actual Yield is the amount of product actually obtainedfrom a reaction. It is usually given.

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Percent Yield

Actual yield = quantity of product actually obtained

Theoretical yield = quantity of product predicted by stoichiometry

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Percent Yield Example

14.4 g excess

Actual yield = 6.26 g

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Sample Exercise

Titanium tetrachloride, TiCl4, can be made by combining titanium-containing ore (which is often impure TiO2) with carbon and chlorine -

TiO2(s) + 2 Cl2(g) + C(s) TiCl4(l) + CO2(g)

If one begins with 125 g each of Cl2 and C, but plenty of titanium-containing ore, which is the limiting reagent in the reaction? What quantity of TiCl4 can be produced?

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Virtual Laboratory Project

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Practice Example 1

A compound contains C, H, N. Combustion of 35.0mg of the compound produces 33.5mg CO2 and 41.1mg H2O. What is the empirical formula of the compound?

Solution:1. Determine C and H, the rest from 33.5mg is N.2. Determine moles from masses.3. Divide by smallest number of moles.

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Practice Example 2Caffeine contains 49.48% C, 5.15% H, 28.87% N and 16.49% O by mass and has a molar mass of 194.2 g/mol. Determine the molecular formula.

Solution:1. Convert mass to moles.2. Determine empirical formula.3. Determine actual formula.

C8H10N4O2

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Practice Example 3

Nitrogen gas can be prepared by passing gaseous ammonia over solid copper(II) oxide at high temperatures. The other products of the reaction are solid copper and water vapor. If a sample containing 18.1g of NH3 is reacted with 90.4g of CuO, which is the limiting reactant? How many grams of N2 will be formed.

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Practice Example 4

Methanol can be manufactured by combination of gaseous carbon monoxide and hydrogen. Suppose 68.5Kg CO(g) is reacted with 8.60Kg H 2(g). Calculate the theoretical yield of methanol. If 3.57x104g CH3OH is actually produced, what is the percent yield of methanol?

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Practice Example 5

SnO2(s) + 2 H2(g) Sn(s) + 2 H2O(l)

a) the mass of tin produced from 0.211 moles of hydrogen gas.

b) the number of moles of H2O produced from 339 grams of SnO2.

c) the mass of SnO2 required to produce 39.4 grams of tin.d) the number of atoms of tin produced in the reaction of 3.00

grams of H2.

e) the mass of SnO2 required to produce 1.20 x 1021 molecules of water.

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QUESTIONBromine exists naturally as a mixture of bromine-79 and bromine-81 isotopes. An atom of bromine-79 contains: 1) 35 protons, 44 neutrons, 35 electrons. 2) 34 protons and 35 electrons, only. 3) 44 protons, 44 electrons, and 35 neutrons. 4) 35 protons, 79 neutrons, and 35 electrons. 5) 79 protons, 79 electrons, and 35 neutrons.

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ANSWER1) 35 protons, 44 neutrons, 35 electrons. Section 2.5 The Modern View of Atomic Structure: An Introduction (p. 54) The 79 in bromine-79 is the mass number (the number of protons plus neutrons). The atomic number for bromine is 35. The number of neutrons is 79 – 35 = 44. The number of electrons equals the number of protons.

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QUESTIONThe atomic mass of rhenium is 186.2. Given that 37.1% of natural rhenium is rhenium-185, what is the other stable isotope?

1) 183

75Re 4)

181

75Re

2) 187

75Re 5)

190

75Re

3) 189

75Re

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ANSWER

2) 187

75Re

Section 3.1 Atomic Masses (p. 82) (0.371 185) + (0.629 M) = 186.2. Solve for M (the isotopic mass). Since neutrons are slightly heavier than protons, this calculation is only an approximation, but is accurate enough to determine the correct answer.

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QUESTION

Naturally occurring iron contains 5.82% 54

26Fe,

91.66% 56

26Fe, 2.19%

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26Fe, and 0.33%

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26Fe.

The respective atomic masses are 53.940 amu, 55.935 amu, 56.935 amu, and 57.933 amu. Calculate the average atomic mass of iron.

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ANSWER55.85 amu Section 3.1 Atomic Masses (p. 83) (.0582 53.940 amu) + (.9166 55.935 amu) + (.0219 56.935 amu) + (.0033 57.933 amu) = 55.85 amu

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QUESTIONWhat is the mass of one atom of copper in grams? 1) 63.5 g 2) 52.0 g 3) 58.9 g 4) 65. 4 g 5) 1.06 10

–22 g

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ANSWER

5) 1.06 10–22

g Section 3.2 The Mole (p. 87) A mole of copper atoms has a mass of 63.55 g. The mass of 1 copper atoms is

63.55 g/(6.022 1023

) = 1.06 10–22

g.

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ANSWER3) 46.07 Section 3.3 Molar Mass (p. 90) The molar mass is the sum of masses of all the atoms in the molecule.

2 12.01 + 6 1.008 + 1 16.00 = 46.07

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QUESTIONWhat is the molar mass of ethanol (C2H5OH)? 1) 45.07 2) 38.90 3) 46.07 4) 34.17 5) 62.07

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QUESTIONFor which compound does 0.256 mole weigh 12.8 g? 1) C2H4O 2) CO2 3) CH3Cl 4) C2H6 5) None of these

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ANSWER3) CH3Cl Section 3.3 Molar Mass (p. 90) The molar mass has units of g/mol. (12.8 g/0.256 mol) = 50.0 g/mol. The molecule with the closest molar mass is CH3Cl.

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QUESTIONPhosphorus has the molecular formula P4 and sulfur has the molecular formula S8. How many grams of phosphorus contain the same number of molecules as 6.41 g of sulfur? 1) 3.10 g 2) 3.21 g 3) 6.19 g 4) 6.41 g 5) None of these

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ANSWER1) 3.10 g Section 3.3 Molar Mass (p. 90) 6.41 g of S8 converts to 0.0250 mol. Therefore 0.0250 mol of phosphorus is needed to have the same number of atoms as sulfur. 0.0250 mol P4

(123.88 g/mol P4) = 3.10 g.

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QUESTIONHow many grams are in a 6.94-mol sample of sodium hydroxide? 1) 40.0 g 2) 278 g 3) 169 g 4) 131 g 5) 34.2 g

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ANSWER2) 278 g Section 3.3 Molar Mass (p. 91) The molar mass of sodium hydroxide, NaOH, is 22.99 g/mol + 16.00 g/mol + 1.008 g/mol = 40.00 g/mol. Convert to grams: 6.94 mol (40.00 g/mol) = 278 g.

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QUESTIONHow many grams of potassium are in 12.5 g of K2CrO7? 1) 2.02 g 2) 8.80 g 3) 4.04 g 4) 78.2 g 5) 25.0 g

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ANSWER

3) 4.04 g Section 3.4 Percent Composition of Compounds (p. 94) The molar mass of K2CrO7 is 2 39.10 + 52.00 + 7 16.00 = 242.2. The mass fraction of potassium is (2 39.10)/242.2 = 0.3229. 0.3229 12.5 g = 4.04 g.

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QUESTIONIn balancing an equation, we change the __________ to make the number of atoms on each side of the equation balance. 1) formulas of compounds in the reactants 2) coefficients of compounds 3) formulas of compounds in the products 4) subscripts of compounds 5) none of these

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ANSWER

2) coefficients of compounds Section 3.7 Balancing Chemical Equations (p. 104) Changing the subscripts of a compound or the addition of a compound to a chemical equation changes the reaction. Only the coefficients can be adjusted without altering the reaction.

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QUESTIONSuppose the reaction

Ca3(PO4)2 + 3H2SO4 3CaSO4 + 2H3PO4

is carried out starting with 103 g of Ca3(PO4)2 and 75.0 g of H2SO4. How much phosphoric acid will be produced? 1) 74.9 g 2) 50.0 g 3) 112 g 4) 32.5 g 5) 97.6 g

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ANSWER

2) 50.0 g Section 3.9 Calculations Involving a Limiting Reactant (p. 113) This is a limiting reactant problem. The first step is determining which reactant will run out first.

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ANSWER (continued)

Use the coefficients of the balanced reaction to determine the amount of phosphoric acid that will be produced by the reaction of 103 g of Ca3(PO4)2 and 75.0 g H2SO4. Which ever produces the lesser amount is the limiting reactant.

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QUESTIONWhich of the following compounds has the same percent composition by mass as styrene, C8H8? 1) Acetylene, C2H2 2) Benzene, C6H6 3) Cyclobutadiene, C4H4 4) -ethyl naphthalene, C12H12 5) All of these

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ANSWER

5) All of these Section 3.4 Percent Composition of Compounds (p. 93) The ratio of C to H in C8H8 is 1:1. This is the same ratio found for each of the compounds, so all have the same percent composition by mass.

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QUESTIONThe empirical formula of styrene is CH; its molar mass is 104.1. What is the molecular formula of styrene? 1) C2H4 2) C8H8 3) C10H12 4) C6H6 5) None of these

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ANSWER

2) C8H8 Section 3.5 Determining the Formula of a Compound (p. 96) The mass of CH is 13.018. 13.018 divides into 104.1 about 8 times. Therefore there are 8 CH groups in this compound. C8H8 is the molecular formula.

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QUESTIONBalanced chemical equations imply which of the following? 1) Numbers of molecules are conserved in

chemical change. 2) Numbers of atoms are conserved in

chemical change. 3) Volume is conserved in chemical change. 4) 1 and 2 5) 2 and 3

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ANSWER

2) Numbers of atoms are conserved in chemical change.

Section 3.7 Balancing Chemical Equations (p. 104) This is another way of stating the Law of Conservation of Mass.

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QUESTIONHow many of the following statements are true concerning chemical equations?

a. Coefficients can be fractions. b. Subscripts can be fractions. c. Coefficients represent the relative

masses of the reactants and/or products. d. Changing the subscripts to balance an

equation can only be done once. e. Atoms are conserved when balancing

chemical equations.

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QUESTION (continued)

1) a 2) b 3) c 4) d 5) e

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ANSWER

2) b Section 3.7 Balancing Chemical Equations (p. 104) Coefficients can be fractions, but usually fractions are multiplied by the appropriate number to make them whole numbers. Atoms are conserved according to the Law of Conservation of Mass.

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QUESTIONDetermine the coefficient for O2 when the following equation is balanced in standard form (smallest whole number integers)

C4H10(g) + O2(g) CO2(g) + H2O(g) 1) 4 2) 8 3) 10 4) 13 5) 20

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ANSWER

4) 13 Section 3.7 Balancing Chemical Equations (p. 104) O2 should be balanced last since it contains only one type of element and balancing it will not cause an imbalance in another element.

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QUESTIONWhat is the subscript of aluminum in the formula of aluminum phosphate? 1) 1 2) 2 3) 3 4) 4 5) 0

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ANSWER

1) 1 Section 3.7 Balancing Chemical Equations (p. 104) The phosphate ion has a charge of –3, while the aluminum ion takes a charge of +3. Therefore the formula of aluminum phosphate is AlPO4.

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QUESTIONHow many grams of Ca(NO3)2 can be produced by reacting excess HNO3 with 7.40 g of Ca(OH)2? 1) 10.2 g 2) 16.4 g 3) 32.8 g 4) 8.22 g 5) 7.40 g

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ANSWER

2) 16.4 g Section 3.8 Stoichiometric Calculations: Amounts of Reactants and Products (p. 108) The balanced equation is Ca(OH)2 + 2HNO3

Ca(NO3)2 + 2H2O. Water is the other product for an acid/base reaction.

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QUESTIONThe limiting reactant in a reaction: 1) is the reactant for which there is the least

amount in grams. 2) is the reactant which has the lowest

coefficient in a balanced equation. 3) is the reactant for which there is the most

amount in grams. 4) is the reactant for which there is the fewest

number of moles. 5) none of these

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ANSWER

5) none of these Section 3.9 Calculations Involving a Limiting Reactant (p. 113) The limiting reactant in a reaction is the reactant that produces the least number of grams of any product.

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QUESTIONThe molecular formula always represents the total number of atoms of each element present in a compound.

1) True 2) False

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ANSWER

1) True Section 3.5 Determining the Formula of a Compound (p. 98) The molecular formula contains all the information about the compound.