Chapter 3

Chapter 3

Mass Relations: Stoichiometry

Atomic number

• # of p+ in nucleus

Mass number

• # of p+ and n0 in nucleus

Isotope Mass # Atomic mass

Oxygen-16 16 15.994

Copper-63 63 62.939

Mass # v. Atomic mass

Avg. Atomic Mass (atomic weight)

• Weighted avg. of atomic masses of naturally occurring isotopes of an element

Isotopes

• Different forms of the same element with different mass

• Same # p+

• Diff. # n0

e.g. : naturally occurring Cu

• 69.17% Cu-63 (atomic mass 62.939)

• 30.83% Cu-65 (atomic mass 64.927)

• (0.6917)(62.939)+(0.3083)(64.927) = 63.546Found on periodic table

Masses of individual atoms

The Mole (mol)

Mole (mol)

• Amt. of a substance that contains same # of particles as # of atoms in exactly 12 g of carbon-12

Avogadro’s number

• Number of particles in exactly one mole of a pure substance (6.02 x 1023)

• Named for Amedeo Avogadro [Lorenzo Romano Amedeo Carlo Avogadro, conte di Quaregna e di Cerreto (1776 - 1856) ]

Molar mass

• Mass in grams of one mole of a pure substance

• Used as a conversion factor, number is taken from the periodic table

• What is the mass (in g) of 2.5 mol of cobalt

Molar mass Co = 58.933g

2.5 mol Co x _________

2.5 mol Co x 58.933 g Co / 1 mol Co

= 147.3325 g Co

= 150 g Co

How many atoms are in 0.000820 g of platinum?

0.000820 g Pt x ________

0.000820 g Pt x I mol Pt/ 195.08 g Pt =0.000004203 mol Pt

0.000004203 mol Pt x 6.02 x 1023 atoms Pt / 1 mol Pt =2.530449047 x 1018 atoms2.53 x 1018 atoms

Percent composition (Mass of element / molar mass cmpd. X 100%)

• e.g. find % composition of Cu2S

• 2 mol Cu and 1 mol S• 2 mol Cu x 63.546g Cu / 1 mol Cu = 127.09 g

Cu• 1 mol S x 32.06g S / 1 mol S = 32.06 g S

molar mass Cu2S = 159.15 g

• 127.09g Cu/ 159.15 g Cu2S x 100% = 79.855% Cu

• 32.06g S/ 159.15g Cu2S x 100% = 20.14% S

Determining simplest formula (Formula showing smallest whole number

ratio of atoms)• e.g. find the simplest formula for a cmpd.

containing 26.56% K, 35.41% Cr, and 38.03% O

• If 100g of cmpd., then: K = 26.56 g, Cr = 35.41 g, O = 38.03 g

(cont.)

• 26.56g K x 1 mol K / 39.098g K = 0.6793 mol K• 35.41g Cr x 1 mol Cr / 51.996g Cr = 0.6810mol

Cr• 38.03g O x 1 mol O / 15.999g O = 2.377 mol O divide by smallest number 0.6793 mol K / 0.6793 = 1.000 mol K 0.6810 mol Cr / 0.6793 = 1.003 mol Cr 2.377 mol O / 0.6793 = 3.499 mol O• 1.000 :1.003 : 3.499 2 : 2 : 7

K2Cr2O7

Chemical Equations and Chemical Reactions

Chemical equation

• Represents (w/ symbols and formulas) the reactions and products in a chemical reaction

• The same # of atoms of each element must appear on each side of the equation

• Use coefficients to balance equation

Word equations

e.g. methane + oxygen

carbon dioxide + water

Formula equations

• CH4(g) + O2(g) CO2(g) + H2O(g) (unbalanced)

• CH4(g) + O2(g) CO2(g) + 2H2O(g)

• CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

Rules

• H2, N2, O2, F2, Cl2, Br2, I2

(ClIF H BrON)

Writing equations

• e.g., Write a formula equation for the reaction between hydrogen gas and fluorine gas to produce hydrogen fluoride gas

• H2(g) + F2(g) HF(g)

• H2(g) + F2(g) 2HF(g)

Balancing equations

• Balance: Al + Fe2O3 Al2O3 + Fe

• 2Al + Fe2O3 Al2O3 + Fe

• 2Al + Fe2O3 Al2O3 + 2Fe

Balance:

• NH3(g) + O2(g) N2(g) + H2O(g)

• 2NH3(g) + O2(g) N2(g) + H2O(g)

• 2NH3(g) + O2(g) N2(g) + 3H2O(g)

• 2NH3(g) + 2O2(g) N2(g) + 4H2O(g)

wrong

• 4NH3(g) + 3O2(g) 2N2(g) + 6H2O(g)

Mole relationships

• H2(g) + Cl2(g) 2HCl(g)• 1 molecule of hydrogen reacts with 1

molecule of chlorine to yield 2 molecules of hydrogen chloride

or• 1 mol H2 reacts with 1 mol Cl2 to yield 2

mol HCl or

(cont.)

• 2g H2 (1 x molar mass) reacts with 71g Cl2 (1 x molar mass) to yield 73g HCl (2 x molar mass)

• I mol H2 : 1 mol Cl2 : 2 mol HCl

• 2g H2 : 71g Cl2 : 73g HCl

Types of chemical reactions

1. Synthesis, 2Mg(s) + O2(g) 2MgO(s)

2. Decomposition, 2H2O(l) 2H2(g) + O2(g)

3. Single replacement, Mg(s) + 2HCl(aq) H2(g) + MgCl2

4. Double replacement, Pb(NO3)2(aq) + 2KI(aq) PbI2(s) + 2 KNO3(aq)

(cont.)

5. Combustion, C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

Specific examples

• Decomposition of metal hydroxides, Ca(OH)2(s) CaO(s) + H2O(g)

• Decomposition of metal chlorates 2KClO3(s) 2 KCl(s) + 3O2(g)

• Replacement of hydrogen in water by a metal, 2K(s) + 2H2O(l) 2 KOH(aq) + H2(g)

Stoichiometry

Molar mass

• Mass in grams of one mole of a pure substance

• Used as a conversion factor, number is taken from the periodic table

• Stoichiometry is the calculation of quantitative (measurable) relationships of the reactants and products in chemical reactions

• Stoichiometric air-fuel ratios of common fuels

Fuel By weight Percent fuel

Gasoline 14.7 : 1 6.8%

Natural Gas 17.2 : 1 5.8%

Ethanol 9 : 1 11.1%

Diesel 14.6 : 1 6.8%

Stoichiometry(mass relationships)

• Start with a balanced equation mole ratio

• 2Al2O3 4Al + 3O2

• 2 mol Al2O3 : 4 mol Al : 3 mol O2

• e.g. 2 mol Al2O3/ 4 mol Al

? Mol Al produced f/ 15.0 mol Al2O3

15.0 mol Al2O3 x 4 mol Al / 2 mol Al2O3 =

30.0 mol Al

4 types of stoichiometry problems

1. Mole – mole

2. Mole – mass

3. Mass – mole

4. Mass - mass

mole - mole

• CO2 + 2LiOH Li2CO3 + H2O

• How many of LiOH are required to

react with 30 mol of CO2

(cont.)

• 30 mol CO2 x 2 mol LiOH / 1 mol CO2

= 60 mol LiOH

Mole - mass

6 CO2 + 6H2O C6H12O6 + 6O2

Given 3.00 mol of water and an excess of carbon dioxide how many grams of glucose will be produced?

• 3.00 mol H2O x 1 mol C6H12O6/ 6 mol H2O

x 180 g C6H12O6/ 1 mol C6H12O6

= 90.0 g C6H12O6

Mass-mole• C + SO2 CS2 + CO

• If 8.00 g of SO2 reacts with an excess of carbon how many moles of CS2 are formed?

• 5C + 2SO2 CS2 + 4CO

• 8.00 g SO2 x 1 mol SO2/ 64.1 g SO2

x 1 mol CS2/ 2 mol SO2

= 0.0624 mol CS2

Mass -mass

• Sn(s) + 2HF(g) SnF2(s) + H2(g)

• How many g of SnF2 is produced from the reaction of 30.00g of HF with an excess of Sn?

30.00g HF x 1 mol HF/ 20.01g HF x

1 mol SnF2/ 2 mol HF x

156.7g SnF2/ 1 mol SnF2 =

117.5 g SnF2

Limiting Reactant

Limiting reactant

• Reactant that limits the amt. of the other reactants that can combine, and the amt. of product formed

Limiting reactant• Silicon dioxide (quartz) reacts with

hydrogen fluoride according to the following reaction:

SiO2 (s) + 4HF(g) SiF4(g) + 2H2O(l)

(cont.)

• If 2.0 mol of HF is combined with 4.5 mol of SiO2, which is the limiting reactant?

2.0 mol HF x 1 mol SiO2/ 4 mol HF =

0.50 mol SiO2

• Therefore 2.0 mol HF requires 0.50 mol of SiO2 to completely react, 4.5 mol SiO2 is more than 0.50 mol, therefore HF is the limiting reactant

Percent yield

• Percent yield = actual yield/ theoretical yield x 100%

• C6H6 + Cl2 C6H5Cl + HCl

• If 36.8 g of C6H6 reacts with an excess of Cl2 the actual yield is 38.8 g of C6H5Cl. What is the percent yield?

(cont.)

• 36.8 g C6H6 x 1 mol C6H6/ 78.1 g C6H6

x 1 mol C6H5Cl/ 1 mol C6H6

x 113 g C6H5Cl/ 1 mol C6H5Cl

= 53.2 g C6H5Cl

(theoretical yield)

(cont.)

percent yield = 38.8 g C6H5Cl/ 53.2 g C6H5Cl

x 100%

= 72.9%

Calculation of Molecular Formulas

• The simplest (empirical) formula of a compound of phosphorus and oxygen was found to be P2O5. Experiment shows that the formula mass of this compound is 283.889g. What is the molecular formula of this compound?

(cont.)• molecular mass P2O5 = 141.945g

283.889g = 1.9999

141.945g

(P2O5) 2 = P4O10