· Web viewCHEM 101 – CHAPTER 3 STOICHIOMETRY OF FORMULAS AND EQUATIONS Stoichiometry – the study of the quantitative aspects of formulas and reactions Defining the Mole Mole

CHEM 101 – CHAPTER 3

STOICHIOMETRY OF FORMULAS AND EQUATIONS

Stoichiometry – the study of the quantitative aspects of formulas and reactions

Defining the MoleMole (mol) – the SI unit for the amount of substance

Defined as the amount of a substance that contains the same number of entities as the number of atoms in 12 g of carbon-12o This number is called Avogadro’s number

One mole (1 mol) contains 6.022 x 1023 entities 1 mol of carbon-12 contains 6.022 x 1023 atoms

and has a mass of 12 g

Examples (elements)… (The mass in atomic mass units (amu) of one atom of an element is the same numerically as the mass in grams (g) of 1 mole of atoms of the element.)

Sulfuro 1 atom has a mass of 32.07 amuo 1 mol [6.022 x 1023 atoms] has a mass of 32.07 g

Irono 1 atom has a mass of 55.85 amuo 1 mol [6.022 x 1023 atom] has a mass of 55.85 g

Examples (compounds)… (The mass in atomic mass units (amu) of one molecule of a compound is the same numerically as the mass in grams (g) of 1 mole of molecules of the compound.

Water (H2O)o 1 molecule of water has a mass of 18.02 amuo 1 mol [6.022 x 1023 molecules] has a mass of 18.02 g

Sodium Chloride (NaCl)o 1 molecule of NaCl has a mass of 58.44 amuo 1 mol [6.022 x 1023 molecules] has a mass of 58.44 g

Two key points about the mole unit…

(1) The mole lets us relate the number of entities to the mass of a sample of those entities.

(2) The mole maintains the same numerical relationship between mass on the atomic scale (atomic mass units, amu) and mass on the macroscopic scale (grams, g)

Determining Molar MassMolar mass – the mass per mole of the entities (atoms, molecules, or formula units) of a substance and has the units of grams per mole (g/mol)

For elements… look up the atomic mass on the periodic tableo For monatomic elements the molar mass is the

periodic table value in grams per mole Molar mass of neon is 20.18 g/mol Molar mass of gold is 197.0 g/mol

o For molecular elements you would need to know the formula to determine its molar mass

Molar mass for O2 (2 x 16.00 g/mol = 32.00 g/mol)

Molar mass for S8 (8 x 32.07 g/mol) = 256.6 g/mol)

For compounds… the molar mass is the sum of the molar masses of the atoms in the formulao Example… SO2

Molar mass SO2 = (1 x molar mass of S) + (2 x molar mass of O)

Molar mass SO2 = (1 x 32.07 g/mol) + (2 x 16.00 g/mol) = 64.07 g/mol

o Example… K2S Molar mass K2S = (2 x molar mass of K) + (1 x

molar mass of S) Molar mass K2S = (2 x 39.10 g/mol) + (1 x 32.07

g/mol) = 110.27 g/mol

Converting Between Amount, Mass, and Number of Chemical EntitiesConverting from amount (mol) and mass (g)

1.25 mol of carbon 12.01 g carbon = 15.0 g C1 mole of carbon

Converting from mass (g) to amount (mol)

34.6 g carbon 1 mol carbon = 2.88 mol C12.01 g carbon

Converting from amount (mol) to number of entities

1.25 mol of carbon 6.022 x 1023 entities = 7.53 x 1023 entities1 mol carbon

Converting from number of entities to amount (mol)

8.13 x 1025 entities

1 mol carbon = 1.35 x102 mol carbon

6.022 x 1023 entities

Sample Problem 3.1 p. 86Calculating the Mass of a Given Amount of an Element0.0342 mol Ag 107.9 g Ag = 3.69 g Ag

1 mol AgFollow-Up Problem 3.1315 mg C 1 g C 1 mol C = 0.0262 mol C

1000 mg C 12.01 g C

= 2.62 x 10-2 mol C

Sample Problem 3.2 p. 86Calculating the Number of Entities in a Given Amount of an Element2.85 x 10-3 mol Ga 6.022 x 1023 Ga

atoms= 1.72 x 1021 Ga atoms

1 mol Ga

Follow-Up Problem 3.2 p. 879.72 x 1021 nitrogen molecules

1 mol nitrogen molecules

2 mol nitrogen atoms

= 3.23 x 10-2 mol of nitrogen atoms

6.022 x 1023 nitrogen molecules

1 mol nitrogen molecules

Sample Problem 3.3 p. 87Calculating the Number of Entities in a Given Mass of an Element95.8 g Fe 1 mol Fe 6.022 x 1023

Fe atoms= 1.03 x 1024 Fe atoms

55.85 g Fe 1 mol Fe

Follow-Up Problem 3.3 p. 873.22 x 1020 Mn atoms

1 mol Mn atoms

54.94 g Mn = 2.94 x 10-2 g Mn

6.022 x 1023

Mn atoms1 mol Mn atoms

Sample Problem 3.4 p. 88Calculating the Number of Chemical Entities in a Given Mass of a Compound8.92 g NO2 1 mol NO2 6.022 x 1023 NO2

molecules= 1.17 x 1023 NO2 molecules

46.01 g NO2 1 mol NO2

Follow-Up Problem 3.4 p. 881.19 x 1019 formula units NaF

1 mol NaF 41.99 g NaF = 8.30 x 10-4 g NaF

6.022 x 1023 formula units NaF

1 mol NaF

Sample Problem 3.5 p. 89Calculating the Number of Chemical Entities in a Given Mass of a Compound (Part 2)(a)

41.6 g (NH4)CO3

1 mol (NH4)CO3

6.022 x 1023 formula units (NH4)CO3

= 2.61 x 1023 formula units (NH4)CO3

96.09 g (NH4)CO3

1 mol (NH4)CO3

(b)

2.61 x 1023 formula units (NH4)CO3

3 O atoms = 7.83 x 1023 O atoms

1 formula unit (NH4)CO3

Determining Mass Percent from a Chemical Formula

For a molecule…

Mass % of element X =

[(atoms of X in formula) x (atomic mass X (g/mol)] x 100%

molecular or formula mass of compound (amu)

For a mole of compound

Mass % of element X =

[(moles of X in formula) x (molar mass of X (g/mol)] x 100%

mass (g) of 1 mol of compound

Sample Problem 3.6 p. 90Calculating the Mass Percent of Each Element in a Compound from the FormulaMass percent carbon (C)

(a) Calculate mass of carbon in 1 mol C6H12O6

6 mol carbon 12.01 g carbon = 72.06 g carbon1 mol carbon

(b) Calculate mass percent of carbon in C6H12O6

72.06 g carbon = .4000 x 100% = 40.00% carbon180.16 g C6H12O6

Mass percent of hydrogen (H)

(a) Calculate the mass of hydrogen in 1 mol of C6H12O6

12 mol of hydrogen 1.008 g hydrogen = 12.096 g hydrogen1 mol hydrogen

(b) Calculate the mass percent hydrogen in C6H12O6

12.096 g hydrogen = 0.06714 x 100% = 6.714% hydrogen180.16 g C6H12O6

Mass percent of oxygen (O)

(a) Calculate the mass of oxygen in 1 mol of C6H12O6

6 mol of oxygen 16.00 g oxygen = 96.00 g oxygen1 mol oxygen

(b) Calculate the mass percent of oxygen in C6H12O6

96.00 g oxygen = 0.5329 x 100% = 53.29% oxygen180.16 g C6H12O6

Follow-Up Problem 3.6 p. 91Ammonium nitrate – NH4NO3

Formula mass = 80.05 g/mol

(a) Calculate the mass of oxygen in one mole of NH4NO3

2 mol of nitrogen 14.01 g nitrogen = 28.02 g nitrogen1 mol nitrogen

(b) Calculate the mass percent of nitrogen in NH4NO3

28.02 g nitrogen = 0.3500 x 100% = 35.00% nitrogen80.05 g NH4NO3

Determining the Mass of an Element from its Mass Percent

What if we wanted to know how much oxygen was in 15.5 g of NO2?

15.5 g NO2 32.00 g oxygen = 10.8 g oxygen46.01 g NO2

Sample Problem 3.7 p. 91Calculating the Mass of an Element in a Compound16.55 g C6H12O6 72.06 g carbon = 6.620 g carbon

180.16 g C6H12O6

Follow-Up Problem 3.7 p. 9135.8 kg NH4NO3

1000 g NH4NO3 28.02 g nitrogen = 1.25 x 104 g nitrogen

1 kg NH4NO3 80.05 g NH4NO3

Determining the Formula of an Unknown Compound

Empirical Formula – shows the lowest whole number of moles, and thus the relative number of atoms, of each element in the compound.

Molecular Formula – shows the actual number of atoms of each element in a molecule.

Structural Formula – shows the relative placement and connections of atoms in the molecule

Sample Problem 3.8 p. 92Determining an Empirical Formula from Amounts of Elements

(1) Zn0.21P0.14O0.56

(2) Zn0.21P0.14O0.56

0.14 0.14 0.14

(3) Zn1.5P1.0O4.0

(4) Zn3P2O8

Follow-Up Problem 3.8 p. 93

(1) H0.255B0.170

(2) H0.255B0.170 0.170 0.170

(3) H1.5B1.0

(4) H3B2

Sample Problem 3.9 p. 93Determining an Empirical Formula from Masses of ElementsCalculate moles of Na

2.82 g Na 1 mol Na = 0.123 mol Na22.99 g Na

Calculate moles of Cl

4.35 g Cl 1 mol Cl = 0.123 mol Cl35.45 g Cl

Calculate moles of O

7.83 g O 1 mol O = 0.489 mol O16.00 g O

(1) Na0.123Cl0.123O0.489

(2) Na0.123Cl0.123O0.489 0.123 0.123 0.123

(3) Na1.0Cl1.0O4.0

(4) NaClO4

Follow-Up Problem 3.9 p. 932.88 g S 1 mole S = 0.0898 mol S

32.07 g S

0.0898 mol S 2 mol M = 0.0599 mol M3 mol S

3.12 g M = 52.09 g/mol (Chromium)0.0599 mol M

Cr2S3 Chromium (III) Sulfide

Molecular FormulasSample Problem 3.10 p. 94Determining a Molecular Formula from Elemental Analysis and Molar MassStep 1 – Assume 100.0 g of compound to express each mass percent directly as mass (g)

We have 40.0 g C, 6.71 g H, and 53.3 g O

Step 2 – Convert each mass (g) to amount (mol)

40.0 g C 1 mol C = 3.33 mol C12.01 g C

6.71 g H 1 mol H = 6.66 mol H1.008 g H

53.3 g O 1 mol O = 3.33 mol O16.00 g O

Step 3 – Derive the empirical formula

(a) C3.33H6.66O3.33

(b) C3.33H6.66O3.33

3.33 3.33 3.33

(c) CH2O

Step 4 – Determine the compound’s empirical formula mass

Formula mass = (1 x 12.01) + (2 x 1.008) + (1 x 16.00)


Step 5 – Divide the compound’s molar mass by the empirical formula mass to find the whole number multiple

90.08 g/mol / 30.026 g/mol = 3.00

Step 6 – Multiply each subscript in the empirical formula by the multiple found in step 5

C3H6O3

Follow-Up Problem 3.10 p. 95Step 1 – Assume 100.0 g of compound to express each mass percent directly as mass (g)

95.21 g C, 4.79 g

Step 2 – Convert each mass (g) to amount (mol)

95.21 g carbon 1 mol carbon = 7.93 mol carbon12.01 g carbon

4.79 g hydrogen 1 mol hydrogen = 4.75 mol hydrogen1.008 g hydrogen

Step 3 – Derive the empirical formula

(a) C7.93H4.75

(b) C7.93H4.75 4.75 4.75

(c) C1.67H1.00

(d) C5H3

Step 4 – Determine the compound’s empirical formula mass

Formula mass = (5 x 12.01) + (3 x 1.008)


Step 5 – Divide the compound’s molar mass by the empirical formula mass to find the whole number multiple

252.30 g/mol / 63.074 = 4.00

Step 6 – Multiply each subscript in the empirical formula by the multiple found in step 5

C20H12

Sample Problem 3.11 p. 95Determining a Molecular Formula from Combustion DataStep 1 – Determine mass of CO2 collected

Mass of CO2 absorber after combustion

– Mass of CO2 absorber before combustion

85.35 g - 83.85 g = 1.50 g CO2 collected

Step 2 – Determine the mass of H2O collected

Mass of H2O absorber after combustion

– Mass of H2O absorber before combustion

37.96 g - 37.55 g = 0.41 g H2O collected

Step 3 – Determine mass of carbon collected

1.50 g CO2 12.01 g C = 0.409 g C44.01 g CO2

Step 4 – Determine the mass of hydrogen collected

0.41 g H2O 2.016 g H = 0.046 g H18.02 g H2O

Step 5 – Determine the mass of oxygen collected

Mass of oxygen collected

= 1.000 g vitamin C

- mass of carbon collected

- mass of hydrogen collected

Mass of oxygen collected = 1.000 g – 0.409 g – 0.046 g

Mass of oxygen collected = 0.545 g O

Step 6 – Calculate the amount (mol) of each element

0.409 g C 1 mol C = .0341 mol C12.01 g C

0.046 g H 1 mol H = .0456 mol H1.008 g H

0.545 g O 1 mol O = .0341 mol O16.00 g O

Step 7 – Determine the empirical formula of vitamin-C

(a) C.0341H.0456O.0341

(b) C.0341H.0456O.0341 .0341 .0341 .0341

(c) C1.0H1.33O1.0

(d) C3H4O3

Step 8 – Determine empirical formula mass

Formula mass = (3 x 12.01) + (4 x 1.008) + (3 x 16.00)


Step 9 - Divide the compound’s molar mass by the empirical formula mass to find the whole number multiple

176.12 g/mol / 88.06 g/mol = 2.00

Step 10 - Multiply each subscript in the empirical formula by the multiple found in step 9

C6H8O6

Follow-Up Problem 3.11 p. 96Step 1 – Determine mass of carbon collected

0.451 g CO2 12.01 g C = 0.123 g C44.01 g CO2

Step 2 – Determine mass of hydrogen collected

0.0617 g H2O 2.016 g H = 0.00690 g H18.02 g H2O

Step 3 – Determine mass of chlorine collected

0.250 g sample - 0.123 g C - 0.00690 g H = 0.120 g Cl

Step 4 – Calculate mol of carbon

0.123 g C 1 mol C = 0.0102 mol C

12.01 g C

Step 5 – Calculate mol of hydrogen

0.00690 g H 1 mol H = 0.00685 mol H1.008 g H

Step 6 – Calculate mol of chlorine

0.120 g Cl 1 mol Cl = 0.00339 mol Cl35.45 g Cl

Step 7 – Determine empirical formula

(a) C.0102H.00685Cl.00339

.00339 .00339 .00339

(b) C3H2Cl

Step 8 – Calculate the empirical formula mass

Formula mass = (3 x 12.01) + (2 x 1.008) + (1 x 35.45)


Step 9 – Determine the Multiplier

146.99 g/mol / 73.50 g/mol = 2.00

Step 10 – Calculate the molecular formula

C6H4Cl2

Chemical Formulas and Molecular Structure; Isomers

Different compounds can have the same empirical formula as we can observe in table 3.3 (the empirical formula tells us nothing about molecular structure because it is based solely on mass analysis)

Isomers – compounds with the same molecular formula and molar mass, but different properties. They have different structural formulaso Constitutional (structural) isomers – isomers where

the atoms link together in different arrangements

Writing & Balancing Chemical Equations

Macroscopic level… 2.016 g of H2 and 38.00 g F2 react to form 40.02 g HF

Macroscopic level… 1 mol of H2 and 1 mol of F2 react to form 2 mol of HF

Molecular level… 1 molecule of H2 and 1 molecule of F2 react to form 2 molecules of HF

Steps for Balancing Equations

Translating the statemento Reactants – substance(s) on the left side of the arrow

o Products – substance(s) on the right side of the arrowo Arrow – means “yields” or “produces”

Balancing the atomso Start with any metals that you have in the equation and

balance them first (if there is more than one, balance them one at a time)

Inventory the number of atoms on reactant and product side of the equation

Place a balancing (stoichiometric) coefficient in front of the metal in the equation (or the formula containing the metal) to balance the number of metal atoms on both sides of the equation.

Proceed to any other metals (if there are other metals) in the equation and balance them in the same manner.

After the metals are balanced, balance any nonmetals other than hydrogen or oxygen by placing balancing (stoichiometric) coefficients in front of their formulas.

Once you have balanced all other elements, balance hydrogen atoms and finally oxygen atoms.

o Re-inventory the number of atoms of each element on both the reactant and product side of the equation and check to make sure that the coefficients are the smallest possible whole numbers

Sample Problem 3.12 p. 101Balancing Chemical EquationsStep 1 – Translate the statement

C8H18 + O2 CO2 + H2O

Step 2 – Balance the atoms

Balance the carbons (there is 8 on the left and only 1 on the right)… place a balancing coefficient of 8 in front of the compound with carbon in it on the left (CO2)

C8H18 + O2 8 CO2 + H2O

Balance the hydrogens (there are 18 on the left and only 2 on the right)… place a balancing coefficient of 9 in front of the compound that contains hydrogen on the right side of the equation(H2O)

C8H18 + O2 8 CO2 + 9 H2O

At this point in the process, the carbons and hydrogens are balanced (8 carbons and 18 hydrogens on both sides of the equation). When you atom inventory the oxygens, there are now 2 oxygens on the left and 25 on the right (remember the balancing coefficients multiply every element in the formula)

Balancing coefficients have to be whole numbers… so you cannot do the following…

C8H18 + 12 ½ O2 8 CO2 + 9 H2O

To get rid of the ½ in the balancing coefficient, you need to multiply all of the coefficients in the equation by 2. This gives the following result…

2 C8H18 + 25 O2 16 CO2 + 18 H2O

Specify states of matter

2 C8H18 (l) + 25 O2 (g) 16 CO2 (g) + 18 H2 (g)

Follow-up Problem 3.12 p. 101(a)

Na (s) + H2O (l) H2 (g) + NaOH (aq)

Na (s) + H2O (l) H2 (g) + 2NaOH (aq)

Na (s) + 2H2O (l) H2 (g) + 2NaOH (aq)

2Na (s) + 2H2O (l) H2 (g) + 2NaOH (aq)

(b)

CaCO3 (s) + HNO3 (aq) CO2 (g) + H2O (l) + Ca(NO3)2 (aq)

CaCO3 (s) + 2HNO3 (aq) CO2 (g) + H2O (l) + Ca(NO3)2 (aq)

(c)

PCl3 (g) + HF (g) PF3 (g) + HCl (g)

PCl3 (g) + 3HF (g) PF3 (g) + 3HCl (g)

(d)

C3H5N3O9 (l) CO2 (g) + H2O (g) + N2 (g) + O2 (g)

2C3H5N3O9 (l) CO2 (g) + H2O (g) + 3N2 (g) + O2 (g)

2C3H5N3O9 (l) 6CO2 (g) + H2O (g) + 3N2 (g) + O2 (g)

2C3H5N3O9 (l) 6CO2 (g) + 5H2O (g) + 3N2 (g) + O2 (g)

4C3H5N3O9 (l) 12CO2 (g) + 10H2O (g) + 6N2 (g) + O2 (g)

Sample Problem 3.13 p. 102Balancing an Equation from a Molecular Scene

4N2O5 (g) 8NO2 (g) + 2O2 (g)

2N2O5 (g) 4NO2 (g) + O2 (g)

Follow-Up Problem 3.13 p. 1026CO (g) + 3O2 (g) 6CO2 (g)

2CO (g) + O2 (g) 2CO2 (g)

Calculating Quantities or Reactants and ProductsStoichiometrically Equivalent Molar Ratios from the Balanced Equation

In a balanced equation, the amounts (mol) of substances are stoichiometrically equivalent to each other…

A specific amount of one substance is formed from, produces, or reacts with a specific amount of the other.o The quantitative relationships are expressed as

stoichiometrically equivalent molar ratios Use the molar ratios as conversion factors to

calculate amounts

Example… the combustion of propane

C3H8 (g) + 5O2 (g) 3CO2 (g) + 4 H2O

From the balanced equation we know that…

1 mol C3H8 reacts with 5 mol of O2

1 mol C3H8 produces 3 mol of CO2

1 mol C3H8 produces 4 mol of H2O

Therefore, in this reaction…

1 mol C3H8 is stoichiometrically equivalent to 5 mol O2

1 mol C3H8 is stoichiometrically equivalent to 3 mol CO2

1 mol C3H8 is stoichiometrically equivalent to 4 mol H2O

This also means that…

3 mol CO2 is stoichiometrically equivalent to 4 mol H2O 5 mol O2 is stoichiometrically equivalent to 3 mol of CO2

5 mol O2 is stoichiometrically equivalent to 4 mol CO2

If we want to convert between mol of O2 and mol of H2O, we could construct to conversion factors depending on which quantity we are solving for…

Mol H2O 5 mol O2 Mol of O2

4 mol H2O

Mol O2 4 mol H2O Mol H2O5 mol O2

If we wanted to know how many mol of O2 is consumed when 10.0 mol of H2O is produced…

10 mol H2O 5 mol O2 = 12.5 mol O2

4 mol H2O

Sample Problem 3.14 p. 104Calculating Quantities of Reactants and Products: Amount (mol) to Amount (mol)Step 1 – Make sure the equation is balanced

Copper (I) sulfide + oxygen Copper (I) oxide + Sulfur dioxide

Cu2S (s) + O2 (g) Cu2O (s) + SO2 (g)

2Cu2S (s) + 3O2 (g) 2Cu2O (s) + 2SO2 (g)

Step 2 – Convert from mol Cu2S to mol O2

10.0 mol Cu2S 3 mol O2 = 15.0 mol O2

2 mol Cu2S

Follow-Up Problem 3.14 p. 105Step 1 – Make sure the equation is balanced

Iron (III) oxide + Aluminum Aluminum oxide + Iron

Fe2O3 + Al Al2O3 + Fe

Fe2O3 + 2Al Al2O3 + 2Fe

Step 2 – Convert from mol Fe to mol Fe2O3

3.60 x 103 mol Fe 1 mol Fe2O3 = 1.80 x 103 mol Fe2 mol Fe

Sample Problem 3.15 p. 105Calculating Quantities of Reactants and Products: Amount (mol) to Mass (g)Step 1 – Make sure the equation is balanced

Same equation as sample problem 3.14


Cu2S (s) + O2 (g) Cu2O (s) + SO2 (g)

2Cu2S (s) + 3O2 (g) 2Cu2O (s) + 2SO2 (g)

Step 2 – Convert from mol Cu2S to grams SO2

10.0 mol Cu2S 2 mol SO2 64.07 g SO2 = 641 g SO2

2 mol Cu2S 1 mol SO2


*Same equation as follow-up problem 3.14 on page 105




Step 2 – Convert from formula units of Al2O3 to mol Fe

1.85 x 1025 formula units Al2O3

1 mol Al2O3 2 mol Fe = 61.4 mol Fe

6.022 x 1023

formula units Al2O3

1 mol Al2O3

Sample Problem 3.16 p. 105

Calculating Quantities of Reactants and Products: Mass to MassStep 1 – Make sure the equation is balanced

Same equation as sample problem 3.14 and 3.15


Cu2S (s) + O2 (g) Cu2O (s) + SO2 (g)

2Cu2S (s) + 3O2 (g) 2Cu2O (s) + 2SO2 (g)

Step 2 – Convert from kg Cu2O to kg O2

2.86 kg Cu2O

1000 g Cu2O

1 mol Cu2O

3 mol O2 32.00 g O2 1 kg O2 =

1 kg Cu2O

143.10 g Cu2O

2 mol Cu2O

1 mol O2 1000 g O2

= .960 kg O2


*Same equation as follow-up problem 3.14 on page 105




Step 2 – Convert from grams Al2O3 to atoms of aluminum

1 g Al2O3 1 mol Al2O3

2 mol Al 6.022 x 1023

Al atoms= 1.18 x 1022 Al atoms

101.96 g Al2O3

1 mol Al2O3

1 mol Al atoms

Reactions That Occur in a SequenceIn many situations, a product of one reaction becomes a reactant for the next in a sequence of reactions.

For stoichiometric purposes, when the same (common) substance forms in one reaction and reacts in the next, we eliminate it in an overall (net) equation.

Sample Problem 3.17 p. 106Writing an Overall Equation for a Reaction SequenceStep 1 – Write the sequence of balanced equations.

2Cu2S (s) + 3O2 (g) 2Cu2O (s) + 2SO2 (g) [equation 1]

Cu2O (s) + C (s) 2 Cu (s) + CO (g) [equation 2]

Step 2 – Adjust the equations arithmetically to cancel the common substance.

2Cu2S (s) + 3O2 (g) 2Cu2O (s) + 2SO2 (g) [equation 1]

(2x) 2 Cu2O (s) + 2C (s) 4Cu (s) + 2CO (g) [equation 2]

We multiplied equation 2 through by 2 so that we would have equal coefficients in front of the common substance (Cu2O) in both equations

Step 3 – Add the adjusted equations together to obtain the overall balanced equation

2Cu2S (s) + 3O2 (g) + 2C (s) 2SO2 + 4Cu (s) + 2CO (g)

Follow-Up Problem 3.17 p. 107Step 1 – Write the sequence of balanced equations

2SO2 (g) + O2 (g) 2SO3 (g) [equation 1]

SO3 (g) + H2O (l) H2SO4 (aq) [equation 2]

Step 2 – Adjust the equations arithmetically to cancel the common substance.

2SO2 (g) + O2 (g) 2SO3 (g) [equation 1]

(2x) 2SO3 (g) + 2H2O (l) 2H2SO4 (aq) [equation 2]

Step 3 – Add the adjusted equations together to obtain the overall balanced equation

2SO2 (g) + O2 (g) + 2H2O (l) 2H2SO4 (aq)

Example of Reaction Sequence in Biology…

C6H12O6 (aq) + 6O2 (aq) 6CO2 (g) + 6H2O (l)

Cellular respiration (net reaction)… this is the net equation of a 30 reaction process!

Reactions That Involve a Limiting Reactant

2 scoops ice cream + 1 cherry + 50 mL syrup 1 sundae

We can only make 2 sundaes because we run out of syrupo The syrup is our “limiting reactant”o We have excess ice cream and cherries

Sample Problem 3.18 p. 109Using Molecular Depictions in a Limiting Reactant ProblemStep 1 – Make sure you have a balanced equation

Cl2 (g) + 3F2 (g) 2ClF3 (g)

Step 2 – Determine the limiting reactant

3 molecules Cl2 2 ClF3 molecules = 6 molecules ClF3

1 molecule Cl2

6 molecules F2 2 ClF3 molecules = 4 molecules of ClF3

3 molecules F2

F2 is the limiting reactant because it limited the number of ClF3

molecules that can be produced


Step 1 – Make sure you have a balanced equation

B2 (g) + 2AB (g) 2AB2 (g)


3 molecules B2 2 molecules AB2 = 6 molecules AB2

1 molecule B2

4 molecules AB 2 molecules AB2 = 4 molecules of AB2

2 molecules AB

AB is the limiting reactant because it produces the least amount of AB2

Sample Problem 3.19 p. 110Calculating Quantities in a Limiting Reactant Problem: Amount to AmountStep 1 – Make sure you have a balanced equation

Cl2 (g) + 3F2 (g) 2ClF3 (g)


0.750 mol Cl2 2 mol ClF3 = 1.50 mol ClF3

1 mol Cl2

3.00 mol F2 2 mol ClF3 = 2.00 mol ClF3

3 mol F2

Cl2 is the limiting reactant because it produces the least amount of ClF3

Follow-Up Problem 3.19 p. 110Step 1 – Make sure you have a balanced equation

B2 (g) + 2AB (g) 2AB2 (g)


1.50 mol B2 2 mol AB2 = 3.00 mol AB2

1 mol B2

1.50 mol AB 2 mol AB2 = 1.50 mol AB2

2 mol AB

AB is the limiting reactant because it can only form 1.50 mol of AB2

Sample Problem 3.20 p. 111Calculating Quantities in a Limiting Reactant Problem: Mass to MassStep 1 – Make sure you have a balanced equation

2N2H4 (l) + N2O4 (l) 3N2 (g) + 4H2O (g)


100. g N2H4

1 mol N2H4

3 mol N2 28.02 g N2 = 131 g N2

32.05 g N2H4

2 mol N2H4 1 mol N2

200. g N2O4

1 mol N2O4 3 mol N2 28.02 g N2 = 183 g N2

92.02 g N2O4 1 mol N2O4 1 mol N2

N2H4 is the limiting reactant because it would produce less N2



2Al (s) + 3S (s) Al2S3 (s)


10.0 g Al

1 mol Al

1 mol Al2S3 150.15 g Al2S3 = 27.8 g Al2S3

26.98 g Al

2 mol Al 1 mol Al2S3

15.0 g S

1 mol S 1 mol Al2S3 150.15 g Al2S3 = 23.4 g Al2S3

32.07 g S

3 mol S 1 mol Al2S3

Sulfur (S) is the limiting reactant because it produces the least mass of product

Step 3 – Calculate the mass of excess reactant

23.4 g Al2S3

1 mol Al2S3

2 mol Al 26.98 g Al = 8.40 g Al (used)

150.15 g Al2S3

1 mol Al2S3

1 mol Al

10.0 g Al – 8.40 g Al = 1.60 g Al remaining

Theoretical, Actual, and Percent YieldsUp until now we have assumed that…

100% of the limiting reactant becomes product Ideal methods exist for isolating the product We have perfect lab techniques and collect all the product

In reality, these three assumptions are rarely observed, so chemist recognize three types of reaction yields…

Theoretical yield – The amount of product calculated from the molar ratio in the balanced equation.

Side reactions form different products than the main product.

In sample problem 3.20 we looked at this reaction…

2N2H4 (l) + N2O4 (l) 3N2 (g) + 4H2O (g)

A side reaction occurs that produces some NO instead of N2

N2H4 (l) + N2O4 (l) 6NO (g) + 2H2O (g)

Actual yield – amount of product actually obtained

Percent yield – is the actual yield expressed as a percent of the theoretical yield

Percent yield = actual yield / theoretical yield x 100%

Sample Problem 3.21 p. 112Calculating Percent Yield


SiO2 (s) + 3C (s) SiC (s) + 2CO (g)

Step 2 – Calculate the theoretical yield

100.0 kg SiO2

1000 g SiO2

1 mol SiO2

1 mol SiC

40.10 g SiC

1 kg SiC

1 kg SiO2

60.09 g SiO2

1 mol SiO2

1 mol SiC

1000 g SiC

= 66.73 kg SiC [Theoretical yield]

Step 3 – Calculate the percent yield


Percent yield = 51.4 kg / 66.73 kg x 100%

Percent yield = 77.0%


CaCO3 (s) + 2HCl (aq) CaCl2 (aq) + H2O (l) + CO2 (g)

Step 2 – Calculate the theoretical yield

10.0 g CaCO3

1 mol CaCO3

1 mol CO2 44.01 g CO2

= 4.40 g CO2

100.09 g CaCO3

1 mol CaCO3 1 mol CO2

Step 3 – Calculate the percent yield


Percent yield = 3.65 g / 4.40 g x 100%

Percent yield = 83.0%

Yields in Multistep SynthesesInteresting examples…

In a multistep synthesis of a complex compound the overall yield can be surprisingly low, even if the yield for each step is high.o If a 6 step synthesis has a 90% yield for each step…

(.90 x .90 x .90 x .90 x .90 x .90) x 100 = 53.1 % Multistep sequences are common in the laboratory

synthesis of medicines, dyes, pesticides, and many other organic compounds.o The antidepressant Sertraline is prepared in 6 steps…

Atom Economy: A Green Chemistry Perspective on Yield

Green chemistry – development of methods to reduce or prevent the release of harmful substances into the environment and the wasting of energy resources

Atom economy – the proportion of reactant atoms that end up in the desired producto Percent atom economy – represents the efficiency of

a synthesis process

Look at the bottom of page 113 for example…

Fundamentals of Solution StoichiometryExpressing Concentrations in Terms of MolarityA solution consists of…

Solute – the substance that dissolves in the solvent

Solvent – the substance in which the solute(s) dissolve

The relationship between the amount of solute and solvent is expressed as concentration… (the quantity of solute dissolved in a given quantity of solution)

Molarity (M) – expresses the concentration in units of moles of solute per liter of solution

Molarity = moles of solute / liters of solution

Sample Problem 3.22 p. 115Calculating the Molarity of a Solution0.715 mol glycine 1000 mL = 1.44 M glycine

495 mL solution 1 L

Follow-Up Problem 3.22 p. 11584 mL KI 1 L 0.50 mol KI = 0.042 mol KI

1000 mL 1L

Sample Problem 3.23 p. 115Calculating Mass of Solute in a Given Volume of Solution1.75 L .460 mol

Na2HPO4

141.96 g Na2HPO4

= 114 g Na2HPO4

1 L 1 mol Na2HPO4

Follow-Up Problem 3.23 p. 115135 g C12H22O11

1 mol C12H22O11

1 L = 0.120 L

342.296 g C12H22O11

3.30 mol C12H22O11

Preparing and Diluting Molar SolutionsImportant point…

Solution volume not solvent volumeo 1 mol solute + 1 liter solvent to make a 1M solution

would have a volume greater than 1 liter (the solute

adds volume), so the concentration would be less than 1M

Preparing a solution… [0.500 L of 0.350 M nickel (II) nitrate hexahydrate]

1. Weigh the solute – calculate the mass of solute needed

0.500 L 0.350 mol Ni(NO3)2 6H2O

290.82 g Ni(NO3)2 6H2O

= 50.9 g Ni(NO3)2 6H2O

1 L 1 mol

2. Transfer the solid into a 0.500 L volumetric flask with about half of the needed volume of distilled water

3. Dissolve the solid (solute)4. Add enough distilled water to make the final volume

Sample Problem 3.24 p. 117Preparing a Dilute Solution from a Concentrated SolutionMolarity dilute x Volume dilute = Molarity conc x Volume conc

M dilute x V dilute = M conc x V conc

(0.15 M) (0.80 L) = (6.0 M) (V conc)

0.012 L = V conc

To make this solution… place 0.012 L (12 mL) of 6.0 M NaCl in a 1.0 L graduated cylinder and add distilled water (about 780 mL) to the 0.80L (800 mL) mark, and stir thoroughly.

Follow-Up Problem 3.24 p. 117M dilute x V dilute = M conc x V conc

(M dilute) x (500. m3) = (7.50 M) (25.0 m3)

M dilute = 0.375 M

0.375 mol H2SO4

1 L 98.09 g H2SO4 = 3.68 x 10-2 g/mL solution

1 L solution 1000 mL 1 mol H2SO4

Stoichiometry of Reactions in SolutionsSample Problem 3.26 p. 118Calculating Quantities of Reactants and Products for a Reaction in SolutionStep 1 – Make sure you have a balanced equation

Mg(OH)2 (s) + 2HCl (aq) MgCl2 (aq) + 2H2O (l)

Step 2 – Calculate volume of HCl

0.10 g Mg(OH)2

1 mol Mg(OH)2

2 mol HCl

1 L HCl = 3.4 x 10-2 L HCl

58.33 g Mg(OH)2

1 mol Mg(OH)2

0.10 mol HCl


Al(OH)3 (s) + 3HCl (aq) AlCl3 (aq) + 3H2O (l)

Step 2 – Calculate volume of HCl

0.10 g Al(OH)3

1 mol Al(OH)3

3 mol HCl

1L HCl = 3.8 X 10-2 L HCl

78.00 g Al(OH)3

1 mol Al(OH)3

0.10 L HCl

*Al(OH)3 is the most effective antacid

Sample Problem 3.27 p. 119Solving Limiting-Reactant Problems for Reactions in SolutionsStep 1 – Make sure you have a balanced equation

Hg(NO3)2 (aq) + Na2S (aq) HgS (s) + 2NaNO3 (aq)

Step 2 – Calculate mol of HgS formed from Hg(NO3)2

0.050 L Hg(NO3)2

0.010 mol Hg(NO3)2

1 mol HgS = 5.0 x 10-4 mol HgS

1 L Hg(NO3)2 1 mol Hg(NO3)2

Step 3 – Calculate mol of HgS formed form Na2S

0.020 L Na2S 0.10 mol 1 mol HgS = 2.0 x 10-3 mol HgS1 L Na2S 1 mol Na2S

Hg(NO3)2 is the limiting reactant…

Step 4 – Calculate mass of HgS formed

5.0 x 10-4 mol HgS 232.7 g HgS = 0.12 g HgS1 mol HgS


Pb(C2H3O2)2 (aq) + 2NaCl (aq) PbCl2 (s) + 2NaC2H3O2 (aq)

Step 2 – Calculate mol of PbCl2 formed from Pb(C2H3O2)2

268 mL Pb(C2H3O2)2

1 L 1.50 mol Pb(C2H3O2)2

1 mol PbCl2 = 0.402 mol PbCl2

1000 mL

1 L Pb(C2H3O2)2

1 mol Pb(C2H3O2)2

Step 3 – Calculate mol PbCl2 formed from NaCl

130. mL NaCl

1L 3.40 mol NaCl

1 mol PbCl2 = 0.221 mol PbCl2

1000 mL 1 L NaCl 2 mol NaCl

Step 4 – Determine which reactant is limiting?

NaCl is the limiting reactant because it produces the least amount of PbCl2 (0.221mol)

Step 5 – Calculate the mass of PbCl2 formed

0.221 mol PbCl2 278.1 g PbCl2 = 61.5 g PbCl2

1 mol PbCl2

Documents

· Web viewCHEM 101 – CHAPTER 3 STOICHIOMETRY OF FORMULAS AND EQUATIONS Stoichiometry – the study of the quantitative aspects of formulas and reactions Defining the Mole Mole