Chapter Three:

Stoichiometry: Calculations with Chemical Formulas and Equations

Chapter Three:

Stoichiometry: Calculations with Chemical Formulas and Equations

Overview

Chemical Equations Patterns/Reactions Atomic/Molecular Weights Moles/Molar Mass Empirical/Molecular Formulas Quantitative Relationships Limiting Reactants/Theoretical Yields

Chemical Equations

chemical ‘sentences’– reactants and products described by formulas or

symbols combined with “punctuation”

2 H2(g) + O2(g) 2 H2O(l)

reactant formulas

product formula

coefficients physical state

‘react to form’

“atoms can be neither created nor destroyed”– all equations must be ‘balanced’ with the same number of

atoms on both sides of the reaction arrow

H2O + O2 H2O2

unbalanced

2 H & 3 O 2 H & 2 O

2H2O + O2 2H2O2

balanced

4 H & 4 O = 4 H & 4 O

unbalanced

balanced

two formula units

one formula unit

H2O O2 H2O2

Examples CH3OH(l) + O2(g) CO2(g) + H2O(l)

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

HBr(aq)+ Ba(OH)2(aq) H2O(l) + BaBr2(aq)

2 2 43

22 2

2 2

Patterns of Chemical Reactivity

Because elements are grouped by chemical properties, their reactions can also be grouped: – alkali metals and water

2K(s) + 2H2O(l) 2KOH(aq) + H2(g) specific

2M(s) + 2H2O(l) 2MOH(aq) + H2(g) general

– Combustion in air

C3H8(g) + O2(g) CO2(g) + H2O(l) 3 45

CxHy + O2(g) CO2(g) + H2O(l)

specific

general

hydrocarbon

– Combination Reactions

2Mg(s) + O2(g) 2MgO(s)

X + Y XY

specific

general

– Decomposition Reactions

CaCO3(s) CaO(s) + CO2(g)

XY X + Y

specific

general

Name the Reaction

PbCO3(s) PbO(s) + CO2(g) decomposition

C(s) + O2(g) CO2(g) combination

2NaN3(s) 2Na(s) + 3N2(g)

decomposition

2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(l) combustion

Atomic and Molecular Masses

Amu scale – defined by assigning the mass of 12C as 12 amu exactly

– 1 amu = 1.66054 x 10-24 g– 1 g = 6.02214 x 1023 amu

Average Atomic Masses– 12C 98.892% abundant 13C 1.1108% abundant

(0.98892)(12 amu) + (0.01108)(13.00335 amu) = 12.011

amu

atomic mass

Formula and Molecular Masses– sum of all atomic masses in the formula of an

ionic or molecular compound

vitamin C C6H8O6

6 x 12.0 = 72.0 amu

8 x 1.0 = 8.0 amu

6 x 16.0 = 96.0 amu

176.0 amu

formula mass of vitamin C (often called molecular mass)

Percentage Composition Calculate the percent mass that each type

of atom contributes to a molecule– % X = (no. X atoms)(X amu) x 100

formula mass cmpd

– C6H8O6

% C = (6)(12.01amu) x 100 = 40.94% C

176.0 amu% H = (8)(1.01amu) x 100 = 4.59% H

176.0 amu

% O = (6)(16.00 amu) x 100 = 54.55% O

176.0 amu

The Mole We can measure masses in amu but how do we

relate that to mass in grams? We define a quantity of atoms – a mole – which has the same mass in grams as the mass of the element in amu.

So how many atoms does it take to make, say, 1.00 g of H?

1.0 g H x 1 atom H 6.0 x 1023 atoms of H

1.7 x 10-24 g H

12.0 g C x 1 atom C 6.0 x 1023 atoms of C

2.0 x 10-23 g C

a mole

Avogadro’s Number

6.02214 x 1023 units/mole

– No. of atoms per mole of an element– No. of molecules per mole of molecular

cmpd.– No. of formula units per mole of ionic cmpd.– No. of cows per mole of cows

Memorize this number & what it means!Memorize this number & what it means!

1 C atom = 12 amu 1 mole C atoms = 12 g

1 Mg atom = 24 amu 1 mole Mg atoms = 24 g

1 CO molecule = 28 amu 1 mole CO molecules = 28

g

1 NaCl fm. unit = 58 amu 1 mole NaCl fm.units = 58g

Molar Mass From this information we can define something

called the molar mass (MM) of an atom (or molecule or formula):

from the equality: 1 mole C = 12.0 g C

we define the molar mass of a substance

12.0 g C = MM or Molar Mass1 mole C (Atomic Mass)

(Molecular Mass)(Formula Mass)conversion

factor

Problems

Practice Ex. 3.9:– How many mole in 508 g of NaHCO3?

Given: MM = 84.02 g/mol NaHCO3 508 g NaHCO3 508 g NaHCO3 x 1 mole = 6.05 mole NaHCO3

84.02 g NaHCO3

– How many formula units of NaHCO3?

Given: 6.02 x 1023 form. units/mole NaHCO3

6.05 mole NaHCO3 x 6.02 x 1023 fm. units = 3.6 x 1024 fm. 1 mole unitsNaHCO3

Molar Mass converts between moles and grams of a substance

Avogadro’s number converts between moles of a substance and atoms (or molecules or formula units) of that substance

These are very important conversion factors, know & understand them!

Problems How many moles of vitamin C are contained in 5.00 g of

vitamin C? C6H8O6 176.0 g/mol

17.5 mg of cocaine (C17H21NO4) per kg of body weight is a lethal dose. How many moles is that? How many molecules?

In 25 g of C12H30O2 THC (tetrahydrocannibinol) how many moles are there? How many molecules are there? How many C atoms are there?

How many moles of O are contained in 1.50 moles of C6H5NO3?

How many grams of nitrogen are contained in 70.0 g of C6H5NO3? How many atoms?

Calculate the number of H atoms in 50.0 mg of acetominophen, C8H9O2N.

Determination Empirical Formulas

simplest ratio of atoms

– change g of each element to moles or

– assume 100 grams of substance & change the % of each element to moles

– change the mole ratio of atoms to the simplest ratio by dividing by the smallest number of moles

Practice Ex. 3.12:– 5.325 g methyl benzoate contains 3.758 g C, 0.316 g H,

1.251 g O. Determine empirical formula.

3.758 g C x 1 mole = 0.313 mol C

12.01 g

0.316 g H x 1 mole = 0.313 mol H

1.01 g

1.251 g O x 1 mole = 0.0782 mol O

16.00 g

C0.313H0.313O0.0782

C4H4O

Determination of Molecular Formulas

actual ratio of atoms

– determine the empirical formula

– divide the actual molar mass by the empirical formula mass to get ‘n’

– multiply the mole ratio in the empirical formula by ‘n’

Practice Ex. 3.13:– Ethylene glycol is composed of 38.7% C, 9.7% H &

51.6% O by mass. Its true molar mass is 62.1 g/mol. What are the empirical and molecular formulas?

38.7 g C x 1 mole = 3.23 mole C

12.0 g

9.7 g H x 1 mole = 9.60 mole H 1.01 g

51.6 g O x 1 mole = 3.22 mole O 16.0 g

C3.23H9.60O3.22

CH3Oempirical formula

n = 2

molecular formula

C2H6O2

Formulas from Combustion Data

Formulas determined from products of combustion products– Menthol is composed of C, H, and O. A 0.1005 g

sample of menthol is combusted, producing 0.2829 g of CO2 and 0.1159 g H2O. What is the empirical formula?

CxHyOz + O2 CO2 + H2O0.1005 g 0.2829 g 0.1159 g

– Calculate moles CO2 & C; moles H2O & H

0.2829 g CO2 x 1 mol x 1 mol C = 0.00643 mol C 44.0 g 1 mol CO2

0.1159 g H2O x 1 mol x 2 mol H = 0.0129 mol H

18.0 g 1 mol H2O

total mass of C + H = 0.0902 g

mass of O = 0.1005 g - 0.0902 g = 0.0103 g O x 1 mol = 16.0 g

6.44 x 10-4 mol O

total mass of all C, H & O

C0.00643H0.0129O0.000644

C10H20O

0.00643 mol C 0.0129 mol H 6.44 x 10-4 mol O

If the MM is 156 g/mol, what is the molecular formula?

n=1 therefore molecular formula is C10H20O

(empirical formula mass 156 g/mol)

Quantitative Stoichiometry

Determination of quantities from balanced chemical reaction equations– mole ratios from balanced chemical equation

convert between species

– if quantities are given for more than one reactant, the limiting reactant must be determined

– Given the following balanced equation:

1Mg(OH)2 + 2HCl 1MgCl2 + 2H2O

– Calculate the number of moles of HCl required to react completely with 0.42 mol of Mg(OH)2

0.42 mol Mg(OH)2 x 2 mol HCl = 0.84 mol HCl 1 mol Mg(OH)2

The mole ratio comes from the balanced chemical equation

– How many grams of MgCl2 can be produced?

0.42 mol Mg(OH)2 x 1 mol MgCl2 x 95.3 g MgCl2 1 mol Mg(OH)2 1 mol

= 40.0 g MgCl2

Theoretical Yield -- maximum amount that can be produced

General Sequence of Conversion:

grams of reactant

moles of reactant

MM reactant

moles of product

mole ratio

grams of product

MM product

Practice Ex. 3.14:

How many grams of O2 can be prepared from 4.50 g of KClO3?

2KClO3 2KCl + 3O2

4.50 g KClO3 x 1 mol x 3 mol O2 x 32.0 g O2 = 1.76 g O2

122.6 g 2 mol KClO3 1 mol

Limiting Reactant

given a non-stoichiometric amount of both reactants, you will have to determine which is the limiting reagent or reactant

example: you have 10 bicycle frames and 16 bicycle wheels and you need to put them together to produce as many bicycles as possible, how many bicycles can be produced, what is the limiting “reagent”, and how much excess “reagent” do you have left over?

—Balanced ‘Equation’

1 (mole) frame + 2 (moles) wheels 1 (mole) bicycles [10 (moles) frames] [16 (moles) wheels] [8(moles) bicycles]

– Limiting Reactant -- will produce the least amount of product

10 mol frames x 1 mol bicycles = 10 bicycles 1 mol frames

16 mol wheels x 1 mol bicycles = 8 bicycles 2 mol wheels

limiting reactant

Practice Ex. 3.16:

A mixture of 1.5 mol of Al and 3.0 mol of Cl2 react. What is limiting & how many moles of AlCl3 are formed?

2Al(s) + 3Cl2(g) 2AlCl3(s) 1.5 mol 3.0 mol

1.5 mol Al x 2 mol AlCl3 = 1.5 mol AlCl3 2 mol Al

3.0 mol Cl2 x 2 mol AlCl3 = 2.0 mol AlCl3 3 mol Cl2

1.5 mol