lect 2

Bromine has two naturally occurring isotopes. One of them, Bromine-79 has a mass of 78.9183 and an abundance of 50.69%. What must be the abundance and the mass of the other, Bromine-81?

If Br-79 accounts for 50.69% the other must account for 100-50.69 = 49.31%

From the periodic table we see that the averaged Br weight is 79.904. Thus: 79. 904 = (0.5069 x 78.9183) + (0.4931 x Br-81)

Example

Counting Atoms

Recall that we must account for all of the atoms in a chemical reaction.

Physically counting atoms is impossible. We must be able to relate measured mass to

numbers of atoms. using atoms by the gram

Avogadros number The mole is an amount of substance that

contains the same number of elementary entities as there are carbon-12 atoms in exactly 12 g of carbon-12.

NA = 6.02214199 x 1023 mol-1

Combining Several Factors in a CalculationMolar Mass, the Avogadro Constant, Percent Abundance.

Potassium-40 is one of the few naturally occurring radioactive isotopes of elements of low atomic number. Its percent natural abundance among K isotopes is 0.012%. How many 40K atoms do you ingest by drinking one cup of whole milk containing 371 mg of K?

Want atoms of 40K, need atoms of K, Want atoms of K, need moles of K,

Want moles of K, need mass and M(K).

Example

Convert strategy to plan

mK(mg) x (1g/1000mg) mK (g) x 1/MK (mol/g) nK(mol)

Convert mass of K(mg K) into moles of K (mol K)

Convert moles of K into atoms of 40K

nK(mol) x NA atoms K x 0.012% atoms 40K

nK = (371 mg K) x (10-3 g/mg) x (1 mol K) / (39.10 g K) = 9.49 x 10-3 mol K

atoms 40K = (9.49 x 10-3 mol K) x (6.022 x 1023 atoms K/mol K)

x (1.2 x 10-4 40K/K) = 6.9 x 1017 40K atoms

Chapter 3 Molecules Only the noble gases exist in nature as single atoms

The atoms of all other elements combine to form MOLECULES

e.g. H2 a DIATOMIC molecule H2O a POLYATOMIC molecule

In a neutral atom or molecule # protons = # electrons

Molecules and Molecular Compounds

Molecular Structural and Empirical Formulas

Molecular Mass

Molecular Mass/Molecular Weight

Determine the molecular mass of H2O:

molecular mass =

sum of atomic mass of each atom in the molecule (in amu)

1 molecule H2O = 2.016 amu + 15.999 amu = 18.015 amu

2 atoms H x 1.008 amu 1 H atom

= 2.016 amu

1 atom O x 15.999 amu 1 O atom

= 15.999 amu

molecular mass

Molar Mass of Molecules mass in grams of 1 mole of the molecule

What is the molar mass of H2O?

1 mole H2O = 2.016 g for two moles of H + 15.999 g for one mole of O = 18.015 g/mol

molar mass

Molecular Mass: mass of one molecule 18.015 amu Molar Mass: mass of one mole of molecules 18.015 g/mol

Molecular mass

HOO

H

H

HO

H

OHOHH

H

OH

Molecular formula C6H12O6 Empirical formula CH2O

Glucose

6 x 12.01 + 12 x 1.01 + 6 x 16.00

Molecular Mass: Use the naturally occurring mixture of isotopes,

= 180.18

Exact Mass: Use the most abundant isotopes,

6 x 12.000000 + 12 x 1.007825 + 6 x 15.994915 = 180.06339

Molecules and Molecular Compounds

Picturing Molecules Molecules occupy three dimensional space.

Molecular Compounds

Some Molecules H2O2 CH3CH2Cl P4O10

CH3CH(OH)CH3 HCO2H

Inorganic Molecules

P4

s8

Ions

When an atom or molecule loses electrons, it becomes positively charged. For example, when Na loses an electron it becomes Na

+. Positively charged ions are called cations.

When # protons # electrons, the species has a net charge and is called an ION.

When an atom or molecule gains electrons, it becomes negatively charged. For example when Cl gains an electron it becomes Cl-.

Negatively charged ions are called anions. An atom or molecule can lose more than one electron.

Ionic compounds Atoms of almost all elements can gain or lose

electrons to form charged species called ions.

Compounds composed of ions are known as ionic compounds.

Metals tend to lose electrons to form positively charged ions called cations.

Non-metals tend to gain electrons to form negatively charged ions called anions.

Oxidation States Metals tend to lose electrons.

Na Na+ + e-

Non-metals tend to gain electrons.

Cl + e- Cl-

Reducing agents Oxidizing agents

We use the Oxidation State to keep track of the number of electrons that have been gained or lost by an element.

Determination of Empirical Formula Combustion Analysis

Step 1: Convert the mass of each product into moles of C, H and S.

An Example of Determining the Empirical and Molecular Formulas of a Compound from Combustion Analysis Data. Complete combustion of a 1.505 g sample of an unknown compound consisting of C, H and S yields 3.149 g CO2, 0.645 g H2O and 1.146 g of SO2. What is the empirical formula for the unknown?

3.149 g CO2 x 1mol/44.010g CO2 x 1mol C/1mole CO2 = 0.07155 mol C 0.645 g H2O x 1mol/18.02g H2O x 2mol H/mol H2O = 0.0716 mol H 1.146 g of SO2 x 1 mol/64.06 SO2 x 1mol S/mol SO2 = 0.01789 mol S

Step 2: Write a tentative formula C0.07155 H0.0716 S0.01789 = C4H4S

Example Complete combustion of a 1.505 g sample of an unknown compound yields 3.149 g CO2, 0.645 g H2O and 1.146 g of SO2. What is the empirical formula for the unknown? Step 1: Convert the mass of each product into moles of C, H and S. 0.07155 mol C, 0.0716 mol H, 0.01789 mol S

By the way: 0.07155 mol C = 0.859 g 0.0716 mol H = 0.0716 g 0.01789 mol S = 0.574 g

1.505 g

Determining the Empirical and Molecular Formulas of a Compound from Its Mass Percent Composition.

Step 1: Determine the mass of each element in a 100g sample.

Example 2

When an unknown compound is decomposed into its constituent elements, it is found to contain 71.65% Cl, 24.27% C, and 4.07% H by mass. What is the empirical formula for the unknown?

71.65 g Cl, 24.27 g C and 4.07 g H in the sample.

Step 2: Convert masses to amounts in moles (molar mass).

Step 3: Write a tentative formula.

Step 4: Convert to small whole numbers.

1 mol Cl 35.45 g Cl

71.65 g Cl x = 2.021 mol Cl

24.27 g C x 1 mol C 12.01 g C

= 2.021 mol C

4.07 g H x 1 mol H 1.008 g H

= 4.04 mol H

Cl2.021C2.021H4.04

Cl1C1H2 or ClCH2

Possible molecular formulas: Cl2C2H4 or Cl3C3H6 or Cl4C4H8 etc.

Example 2

Step 4: Convert to small whole numbers. Cl1C1H2 or ClCH2

We experimentally determine the molar mass to be 98.96 g/mol. Determine the molecular formula of the unknown.

Empirical formula mass is 49.48 g/mol This is half of the actual molar mass so the correct formula is

Possible molecular formulas: Cl2C2H4 or Cl3C3H6 or Cl4C4H8 etc.

Cl2C2H4

Example 2

Empirical Formula

Burn 0.115 g of a hydrocarbon, CxHy, and produce 0.379 g of CO2 and 0.1035 g of H2O. What is the empirical formula of CxHy?

CxHy + some oxygen 0.379 g CO2 + 0.1035 g H2O

First, recognize that all C in CO2 and all H in H2O is from CxHy.

1. Calculate moles of C in CO2 8.61 x 10-3 mol C

2. Calculate moles of H in H2O 1.149 x 10 -2 mol H


Now find ratio of mol H/mol C to find values of x and y in CxHy.

1.149 x 10 -2 mol H/ 8.61 x 10-3 mol C = 1.33 mol H / 1.00 mol C = 4 mol H / 3 mol C Empirical formula = C3H4


An Example from Test 1 2004 A 0.157 g sample of an unknown material undergoes combustion analysis. From this,

0.213 g of CO2, 0.0310 g of H2O, and 0.0230 NH3 was produced. The remainder of the material is oxygen. What is the simplest empirical formula for this compound? The concept: CxHyNzOw is converted into CO2, H2O, NH3 and there is Ow that is not directly measured

Moles CO2 = 0.213 g / 44.02 g/mol = 0.00484 moles Moles C = 0.00484 moles Mass C = 0.00484x12.011 = 0.05812 g % mass C = 0.05812 / 0.157 = 37.0%

Moles H2O = 0.0310 g / 18.01 g/mol = 0.00172 moles Moles H = 2x0.00172 = 0.00344 moles Mass H = 0.00344x1.0079 = 0.00347 g % mass H = 0.00347 / 0.157 = 2.21%

Moles NH3 = 0.0230 g / 17.03 g/mol = 0.00135 moles Moles N = 0.00135 moles Mass N = 0.00135x14.0067 = 0.00189 g % mass N = 0.00189 / 0.157 = 12.0%

% mass C = 0.05812 / 0.157 = 37.0% % mass H = 0.00347 / 0.157 = 2.21% % mass N = 0.00189 / 0.157 = 12.0% % mass O = 100 (37.0 + 2.21 + 12.0) = 48.79%

assume 100 g: moles C = 37.0/ 12.011 = 3.08 moles moles H = 2.21 / 1.0079 = 2.19 moles moles N = 12.0 / 14.0067 = 0.856 moles moles O = 48.79 / 15.994 = 3.05 moles Ratio: C 3.58 H 2.54 N 1.00 O 3.05 C18H13N5O18

CHEMICAL REACTIONS Chapter 4

How do elements combine?

Compounds - two or more elements in a specific ratio

Molecules - a discrete group of atoms elements or compounds or ions

Ions - positively or negatively charged atom or group of bonded atoms

Chemical Equations Chemical Equations depict the kind of

reactants and products and their relative amounts in a reaction.

4 Al(s) + 3 O2(g) 2 Al2O3(s) Some key features: stoichiometric coefficients the letters (s), (g), and (l) are the physical

states of compounds.

Chemical Equations 4 Al(s) + 3 O2(g) 2 Al2O3(s) This equation means: 4 Al atoms + 3 O2 molecules gives 2 molecules of Al2O3 4 moles of Al + 3 moles of O2 gives 2 moles of Al2O3

A trivial example:

Nitrogen monoxide + oxygen nitrogen dioxide

Step 1: Write the reaction using chemical symbols.

NO + O2 NO2

Step 2: Balance the chemical equation.

2 1 2

Balancing Equations

Balancing Equations Never introduce extraneous atoms to balance.

Never change a formula for the purpose of balancing an equation.

NO + O2 NO2 + O

NO + O2 NO3 X

Balancing Equation Strategy Balance elements that occur in only one

compound on each side first.

Balance free elements last. Balance unchanged polyatomics as groups. Fractional coefficients are acceptable and can

be cleared at the end by multiplication.

Please practice on your own. Tonnes of exercises are available at the end of the chapter

lect 2

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