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University of Nizwa

College of Arts and Science

Department of Biological Sciences and Chemistry

Introduction to General Chemistry

CHEM 107

Hassan Nimir BSc, MSc, PhD, MRSC.

Title: Associate professor of Inorganic and Biological

Inorganic Chemistry.

Room: 5B-2: office hours: Sat. Mon and Wed: 8-10

4 Parts

Contents: part3

CH1: Chemical foundation: units of measurements;

uncertainty in measurement; precision and

accuracy; dimensional analysis; Density;

classification of matter.

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CH2: Atoms, molecules and ions: structure of the

atom (the modern view); molecules and ions; the

periodic table; naming simple compounds.

CH3: Stoichiometry: Atomic masses; molar mass;

percent composition of compounds, determining

the formula of a compound; chemical equations;

balancing chemical equations; stoichiometric

calculations: amounts of reactants and products;

calculations involving a limiting reagent

(reactant)and percent yield.

Chapter one: Units and measurements:

Table 1.1: The fundamental SI units

Physical quantity

Name of unit Abbreviation

Mass kilogram Kg Length meter m Time Second s Temperature Kelvin K

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Electric current ampere A Amount of substance

mole mol

Luminous intensity

candela cd

The International system (le Systeme International

in French) or the SI system. See table 1.1

Prefixes are used to change the size of the unit see

table 1.2

Table 1.2: The prefixes used in the SI System

Prefix Symbol Meaning Exponential Notation

Giga G 1,000,000,000 109 Mega M 1,000,000 106 Kilo k 1,000 103 Hector h 100 102 Deka da 10 101 - - 1 100 Deci d 0.1 10-1

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Centi c 0.01 10-2 Milli m 0.001 10-3 Micro µ 0.000001 10-6 Nano n 0.000000001 10-9 Pico p 0.000000000001 10-12

Volume has no SI unit; however, it is derived from

length.

1 liter = 1dm3 = 1000 cm3 = 1000 ml.

Precision and accuracy in measurements:

Accuracy refers to the agreement of a particular

value with the true value.

Precision refers to the degree of agreement among

several measurements of the same quantity. It

reflects reproducibility of a given measurement.

Example1:

Suppose we weight a piece of brass five times as

shown below: this result show excellent precision.

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Weighing Result/g 1 2.486 2 2.487 3 2.485 4 2.484 5 If, we assume that the true mass of the piece of brass is close to 2.486 g. Which is the average of the five results?

2.488

The above result is also accurate.

However the measurements are accurate or not this

is depends on the absence or present of a

systematic error.*

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*A systematic error is type of error that occurs

in the same direction each time; it is either

always high or always low.

Example 2:

Check the accuracy of the graduated cylinder: A

student filled the cylinder to the 25.00ml mark

using water delivered from a burette and then

read the volume delivered.

Table 3: results of five trials by the graduated

cylinder:

Trial Volume shown by Graduated cylinder/ml

Volume shown by the burette/ml

1 25 26.54 2 25 26.51 3 25 26.60 4 25 26.49 5 25 26.57 Average 25 26.54

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Remarks:

1. Good precision of graduate cylinder means the

student has good technique.

2. Note the average 26.54 ml, which means that

the graduated cylinder is not very accurate. The

result is low in each measurements (systematic

error)

Dimensional analysis: unit factor method

Converting from one unit to another:

1. Use the equivalence statement that relates the

two units. E.g. 1kg= 2.205 lb. (lb= pound)

2. Derive the appropriate unit factor by looking at

the direction of the required change (to cancel

the unwanted units)

3. Multiply the quantity to be converted by the

unit factor to give the quantity with the desired

units.

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Example 3:

A pencil is 17.80 cm long. What is its length in

inches?

Equivalence statement:

1. 2.54 cm = 1 in

2. Two unit factor are possible 2.54cm/1in and

1in/2.54cm

3. Look at the direction of the required change in

this case from cm to inch. Then the cm must

cancel.

4. 17.80cmx1in/2.54cm = 7.00 in.

Example 4: a student has entered 10.0 km run. How

long is the run in miles?

Statements:

1km=1000m

1m=1.094yd

1760yd=1mi

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Therefore we combine all steps to cancel unwanted

units and leave the wanted unit only.

10.0kmx1000m/1kmx1.094yd/1mx1mi/1760yd =

6.22 mi

Density: The mass of a substance per unit volume of

the substance.

Density = mass/volume

Density is a property that is often used by chemists

to identify a substance. Also density has many other

uses such the car’s lead storage battery should has

a density of 1.30g/ml as well as determine the

amount of antifreeze agent. The densities of

common substances are listed in table 1.5 as shown

below in this chapter.

Example5: if a 25 ml of a compact disc (cd) cleaning

fluid, has a mass of 19.625g at 20°C. Identify the

main component of the cd cleaner from the

compound listed below?

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compound Density in g/ml at 20°C

Chloroform 1.492

Diethyl ether 0.714

Ethanol 0.789

Isopropyl alcohol 0.785

Toluene 0.867

To identify the substance, we must determine its

density.

d = mass of liquid/volume;

d = 19.625g/25.00ml = 0.785 g/ml

The liquid is probably isopropyl alcohol.

Table 1.5: densities of various substances at 20 °C.

Substance Physical state Density g/ml

Oxygen Gas 0.00133

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Hydrogen Gas 0.000084

Ethanol Liquid 0.789

Benzene Liquid 0.880

Water Liquid 0.9982

Magnesium Solid 1.74

Sodium chloride Solid 2.16

Aluminum Solid 2.70

Iron Solid 7.87

Copper Solid 8.96

Silver Solid 10.50

Lead Solid 11.34

Mercury Hg Solid 13.60

Gold Solid 19.32

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Classification of matter:

Matter is anything occupying space and having

mass. Matter exists in three states: solid, liquid and

gas.

A solid is rigid; it has a fixed volume and shape. A

liquid has a definite volume but no specific shape; it

takes the shape of its container. A gas has no fixed

volume or shape; it takes on the shape and volume

of its container.

Liquid and solid are slightly compressible, while

gases are highly compressible.

The solid particles are locked into rigid positions

and are close together they don’t move they can

only vibrate. In the liquid state molecules/particles

are still close together but can move around to

some extent, while, the molecules of a gas are far

apart and move freely and randomly.

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Separation of mixtures:

Most matter around us consists of mixtures of pure

substances. Mixtures are either, homogenous

(having visibly indistinguishable parts) or

heterogeneous (having visibly distinguishable

parts). A mixture has variable composition compare

to a pure substance with constant composition.

Mixtures can be separated into pure substances by

physical methods. The common ways to separate

mixture are: Simple distillation and fractional

distillation used to separate liquid/liquid or

liquid/dissolved solid e.g. desalination of seawater

(removal of dissolved mineral in seawater),

filtration: used to separate liquid/solid mixture.

Chromatography: employ a system with two phases

(states) of matter; a mobile phase (liquid/or gas)

and a stationary phase (solid). The separation

occurs because the components of the mixture

have different affinities (attraction or solubility) for

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the two phases and thus move through the system

at different rates. Common chromatography

techniques include: paper chromatography; thin-

layer chromatography; gas chromatography; liquid

chromatography; etc…

Example6 separation of ink by paper

chromatography:

1. A line or a drop of the ink (mixture of dyes) is

placed at one end of a sheet of porous paper

such as the filter paper. The dipped into a liquid

( the mobile phase)

2. The paper is the stationary phase acts as a wick

to draw up the liquid that travel up.

3. The component in the mixture (dye in this case)

with the weakest attraction for the paper

travels faster than the components that cling to

the paper. See fig on page 29 in Zumdahl.

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Chapter two

Atoms, molecules and ions

Structure of the atom (the modern view);

molecules and ions; the periodic table; naming

simple compounds.

The modern view of atomic structure:

The simplest view of the atom is that it is

spherical in shape and consists of a tiny nucleus

(diameter of around 10-13 cm) in the middle. The

nucleus contain protons, which are positively

charged and equal in magnitude to the negatively

charged electrons and neutrons, which have

virtually the same mass as proton but no charge.

The electrons move around the nucleus. The

chemistry of an atom mainly results from the

number and arrangement of its electrons in the

valence shell (level).

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See table for a view of an atom and of the mass

and charge of the components of the atom, page

53 Zumdahl, 6th ed.

Two striking things about the nucleus are its small

size compared to the overall size of the atom and

its extremely high density. The tiny nucleus

accounts for almost all the atom’s mass. The idea

is Similar, to ball (nucleus-size), and stadium

overall atom size.

Any atom represented by specific symbol. The

atomic number Z (number of protons) is written

as a subscript and the mass number A (the total

number of protons and neutrons) is written as

superscript.

Molecules and ions:

A molecule is a neutral compound consists of a

group of two or more atoms chemically combined

(through either covalent or ionic type of bonds)

and has a definite chemical composition, which

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can be translated into a chemical formula. E.g.

the water molecule formed when, two atoms of

hydrogen combined to one oxygen atom through

covalent bond and has the chemical formula H2O.

Examples of covalent molecules are: hydrogen

molecule H2, oxygen molecule O2, ammonia

molecule NH3, methane molecule CH4, and

hydrochloric acid molecule HCl.

More information about a molecule is given by its

structural formula, in which the individual bonds

are indicated by lines. Structural formula doesn’t

usually represent the actual shape of the

molecule. The structural formula of a molecule

can be represented in many ways in two or three

dimensional shape (2D, 3D).

Ions: is an atom or a group of atoms combined

together and has a net positive or negative

charge. A positive ion is called cation and the

negative ion is called anion. Ionic compound such

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as sodium chloride (table salt) consists of sodium

ions Na+ and chloride ions Cl-, combined together

through ionic bonds which take place through

electrostatic attractions between cations (formed

when a metal for example lost its valence

electron(s) to reach the stable configuration of

the nearest Noble gas) and anions (formed when

a non-metal gain specific number of electron(s) to

reach the octet in the outermost shell). The

particles of an ionic compound arranged in a

crystal lattice which usually in the solid state.

Example of ionic compounds: NaCl, NH4NO3, KBr

etc….

See part 1 for more details in the ionic and

covalent compound and their properties.

Introduction to the periodic table: see part 1:

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Naming simple compounds:

Compounds consist of either binary ions or

polyatomic ions and they are many types

depending on the cation oxidation number.

Binary compounds are compounds composed two

particles (species) they can be of covalent or ionic

origin.

1. Binary ionic compounds (type I): contain a

cation always written first in the formula and an

anion. The following rules apply.

i. The cation always named first and the anion

second.

ii. A monatomic cation takes the element

name. E.g. Na+ called sodium in the

compound.

iii. A monatomic anion is named by taking the

root of the element name and adding the –

ide at the end. E.g. Cl- is called chloride.

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For common monatomic cations and anions,

see table in page 62 Zumdahl.

Examples of naming binary ionic compounds are

illustrated below.

Compound Ions present Name

NaCl Na+, Cl- Sodium chloride

KI K+, I- Potassium iodide

CaS Ca2+, S2- Calcium sulphide

Li3N Li+, N3- Lithium nitride

CsBr Cs+, Br- Cesium bromide

MgO Mg2+, O2- Magnesium oxide

Home work 1:

Name each of the following binary compounds:

a. RbF b. AlCl3 c. LiH.

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2.Binary ionic compounds type II: some metals

have different oxidation numbers e.g. Iron form Fe2+

and Fe3+, copper from Cu1+ and Cu2+ cations. The

binary compounds of type II were named similar to

type I, plus indicating the charge on the cation

(oxidation number) in roman numerical between

brackets. E.g. FeCl2 named Iron(II) chloride and

FeCl3 is Iron(III) chloride. See common type II

cations in page 63 Zumdahl.

Naming type II, compounds: H.W.

Give the systematic name of the following:

1. (CuCl) 2. (HgCl2) 3. (Fe2O3) 4. (PbI2) 5. (MnO2).

3. Binary covalent compounds type III:

Are binary covalent molecules, formed between

two non-metals. They named very similarly to

binary ionic compounds. Prefixes are used to

denote the numbers of atoms present (mono-1, di-

2, tri-3, tetra-4, penta-5 and hexa-6). The prefix

mono- was never used for the first element. E.g. for

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CO, the name is carbon monoxide and NOT

monocarbon monoxide.

NO2 = Nitrogen dioxide; N2O = Dinitrogen monoxide

(common name is Nitrous Oxide); N2O4 = Dinitrogen

tetraoxide etc, see page 68.

H.W.: Name each of the following compounds?

i. CO2. ii. SO3. iii. PCl5. iv. SF6. V. SO2.

Naming ionic compounds with polyatomic ions:

see table on page 67, Zumdahl, for common

polyatomic ions.

Polyatomic ions are assigned special names that

must be memorized to name the compounds

containing them.

Note that several series of anions contain an atom

of a given element and different atoms of oxygen

atoms. These anions called oxyanions. When there

are two groups the one with the small number of

oxygen end in – ite and with larger end with –ate.

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E.g. sulfite SO32-, and sulfate SO4

2-. For more than

two oxyanions in a series, hypo-(less than) and per-

(more than) are used as prefixes to indicate the

member with the fewest and largest number of

oxygen see table oxyanions of chlorine.

H.w:

Give the systematic name of each of the following:

Na2SO3, KHPO4, Fe(NO3)3, KBrO3, CsClO4.

See rules for naming binary compounds in Figure

2.23 page 69, Zumbdal 6th Ed.

Chapter three

Stoichiometry

Stoichiometry (pronounced stoy-ke-om-etry):

Atomic masses; molar mass; percent composition of

compounds, determining the formula of a

compound; chemical equations; balancing chemical

equations; stoichiometric calculations:

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Amounts of reactants and products; calculations

involving a limiting reagent (reactant) and percent

yield.

In this chapter, the quantities of materials

consumed and produced in chemical reactions will

be explained.

Atomic mass:

The modern system of atomic masses is based

on C-12 (carbon twelve) as the standard. C-12 is

assigned a mass of exactly 12 atomic mass unit

(amu), and the masses of all other atoms are

given relative to this standard.

The most accurate method currently available

for comparing the masses o atoms involves the

use of the mass spectrometer, see page 82,

Zumdahl. For example, when 12C and 13C are

analysed in a mass spectrometer, the ratio of

their masses is found to be

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Mass13C/mass12C = 1.0836129

Since, the mass of 12C is exactly 12 amu, and

then on this same scale, the mass of 13C is:

Mass of 13C = (1.0836129) x (12 amu) =

13.003355 amu.

The masses of other atoms were determined in

similar way. The atomic mass (atomic weight)

for each element in the periodic table is given in

the front cover of most chemistry books.

Average atomic masses:

Most elements occur in nature as mixtures of

isotopes (atoms of same element have same Z

but different A, or same number of protons but

different number of neutrons); thus atomic

masses are usually average values.

In the table the atomic mass of carbon is 12.01

amu, why?

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Because natural carbon is a mixture of three

isotopes, the atomic mass of carbon in that

table is actually an average value reflecting the

average of the isotopes composing it.

This average is computed as follow:

Natural carbon is composed of 98.89%12C atoms

and 1.11% 13C and a negligibly small amount of

0% of 14C. Using the masses of 12C (exactly 12

amu) and 13C (13.003355 amu) the average

atomic mass of natural carbon is calculated as

follow:

Summation of all isotopes masses multiplied by

their natural abundances.

(Mass of each isotope X its % abundance).

12 amuX98.89% +13.0034X1.11%

= (0.9889X12 amu + 0.0111X13.0034 amu) =

12.01 amu

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Calculations of an average mass of an element:

1. When a sample of natural copper vaporized

and injected into mass spectrometer, the

abundance of 63Cu is 69.09% and that of

65Cu is 30.91%. If the mass values of 63Cu and 65Cu are 62.93 amu and 64.93 amu,

respectively. Calculate average mass of

copper.

Solution: Assume that the number of copper

atoms is 100 atoms.

(69.09 atoms)(62.93amu/atom)+

30.91(64.93amu/atom) = 6355 amu

The average mass of one copper atom is:

6355 amu/100 atoms = 63.55 amu/atom

This mass value is used in doing calculations

involving the reactions of copper and is the

value in the cover of most chemistry books.

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The mole:

The SI definition of the mole (mol) is the

amount of a substance that contains as many

entities as there are in exactly 12 g of carbon-

12. Techniques such as Mass Spectrometry,

which count atoms very precisely, have been

used to determine this number as

6.02214X1023. This number is called

Avogadro’s number.

As a rule one mole of any substance is

equivalent to 6.022X1023 units (atoms, ions,

molecules, electrons, particles) of that

substance.

How do we use the mole in chemical

calculations?

Since Avogadro’s number is defined as the

number of atoms in exactly 12 g of carbon-12.

Also a sample of 12.01 g of natural carbon

contains the same number of atoms as 4.003

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g of natural helium. Both samples contain 1

mole of atoms (6.022X1023).

Thus the mole of an element is also defined as

the sample of natural element with a mass

equal to the element’s atomic mass expressed

in grams.

Important relations for calculations:

1mole of an element atoms = atomic mass in

grams Ar in grams

1mole of molecules = Molecular (molar) mass

in grams, Mr.

1mole of any particles (atoms, molecules..etc)

contain = 6.022X1023 particles.

The table below showed a comparison of

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1 mole samples of various elements:

Element No. of atoms Mass of sample

Aluminum 6.022X1023 26.98

Copper 6.022X1023 63.55

Iron 6.022X1023 55.85

Sulfur 6.022X1023 32.07

Iodine 6.022X1023 126.90

Mercury 6.022X1023 200.60

Examples:

1. Compute the mass in grams of a sample of

americium (Am, Ar = 243) containing 6

atoms.

1 mole of Am atoms = 243 g

6.022X1023 atom of Am = 243 g

6 atoms of Americium = X g

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Therefore, the mass of 6 atoms of Am = X g

= 6 atomX243g/6.022X1023atom= 2.42X10-21 g

2. Compute both the number of moles of

atoms and the number of atoms in 10 g Al

sample?

Solution:

1 mole of Al atoms = Ar of Al in g

1 mole of Al atoms = 26.98 g

X mole of Al atoms = 10 g

Therefore, no of mole of atoms of Al in 10

g= X mol = 10 gX1 mole/26.98 g = 0.371 mol

Al atom

Also,

1 mol Al contain = Avogadro’s no

Therefore,

1mol Al atom= 6.022X1023 Al atom

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0.371 mol Al = X atom

The number of Al atom in 0.371 mol (10g) =

X atom = 0.731 mol Al X 6.022X1023 Al

atom/ 1 mol Al atom

= 2.23X1023 atoms

Alternatively use the appropriate factor:

10g Al X 1 mol Al/26.98g Al = 0.371 mol Al

atoms.

The number of atoms in 10.0g (0.731mol) of

aluminum is:

0.731 mol Al X(6.022X1023 atoms)/(1mol Al)

= 2.23X1023 atoms

Example3:

A silicon chip used in microcomputer has a

mass of 5.68mg. How many silicon atoms

are present in the chip?

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Solution:

Strategy: convert from:

mg → g → mol → atom of Si

5.65mg Si X 1g Si/1000mg Si = 5.68X10-3g Si

5.68X10-3g SiX1mol Si/28.09g Si=

2.02X10-4mol Si

2.02X10-4mol SiX6.022X1023 atoms/1mol Si

= 1.22X1020 atom

Always check to see if your answer is

sensible.

Calculating the number of moles and mass:

Example4:

Cobalt (Co) is a metal that is added to steel

to improve its resistance to corrosion.

Calculate both the number of moles in a

sample of cobalt containing 5.00X1020

atoms and the mass of the sample?

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Solution:

5.00X1020atoms CoX1mol Co/6.022X1023

atoms Co

=8.30X10-4 mol Co

Since the mass of 1 mole of cobalt is 58.93g,

the mass of 5.00X1020 atoms can be

determined as follows:

8.30X10-4 mol CoX58.93g Co/1mol Co

=4.89X10-2g Co

Molar mass:

A substance's molar mass is the mass in

grams of 1 mole of the substance.

Traditionally, the term molecular weight

has been used.

Example: 1. Calculate the molar mass of

Calcium carbonate?

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2. A certain sample of calcium carbonate

contains 4.86 moles. What is the mass in

grams of this sample? What is the

mass of the carbonate ions present?

Answers: molar mass of calcium carbonate =

100.09 g How is that!!

Mass of 4.86 mol = 4.86 mol CaX100.09/1mol

=486 g

Mass of carbonate in 4.86 mol =

4.86X60.10/1mol = 292 g carbonate.

Percent Composition of Compound:

The mass % of a compound is obtained

from its formula by comparing the mass of

each element present in 1 mole of the

compound to the total mass of 1 mole.

Example: Penicillin F has the formula

C14H20N2SO4. Compute the mass percent of

each element? The molar mass of Penicillin

F = 312.40 g

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mass % of C = 168.10 g C/312.40 X100 =

53.81%

mass % of H = 20.16/312.40X100 = 6.453 %

mass % of N = 28.02/312.40X100 = 8.969%

mass % of S = 32.07/312.4X100% = 10.27%

mass % of O = 64.00/312.4X100% = 20.49%

Determining the formula of a compound:

Empirical formula: EF

It is the smallest (simplest), whole –number

ratio of the elements in a compound.

Molecular formula: MF

It indicates the exact numbers of atoms in

the formula of the compound (molecule).

It could be computed if we know the molar

mass of the compound and it’s EF.

MF = (EF)n where n = 1, 2, 3, ...

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Empirical formula calculations:

Step1: calculate the mass percent of each

element in the unknown compound.

Step2: convert the above masses of the

elements to numbers of atoms. The easiest

way to do this is to work with 100g sample

of the compound.

Step3: calculate the number of moles of

each element in its mass.

Step4: to find the smallest whole-number

ratio of atoms in the unknown compound

(EF), divide each of the mole values by the

smallest mole value.

Step5: to calculate the MF: use the relation:

MF = (EF)n where n is an integer n=1,2,3,..

Find n = MF mass/EF mass

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Empirical formula mass is needed

Sometime MF=EF that is when n=1.

Example: Determine the empirical and

molecular formulas for a compound that is

composed of carbon, hydrogen and

nitrogen. If 0.1156g of this compound

burned in oxygen, 0.1638g CO2 and 0.1676g

of H2O were obtained.

Solution:

The molar mass of CO2 = 44.01g/mol

And the molar mass of water H2O =

18.02g/mol

The mass of C present in 0.1638 carbon

dioxide= 0.1638 g CO2X12.01/44.01 =

0.04470g C and the mass % of C =

0.04470/.01156X100% = 38.67% C

The mass of H present in 0.1676g of H2O =

0.1676g H2OX2.016g H/18.02 = 0.01875g H

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The mass % of H in the sample =

0.01875/0.1156X100% = 16.22% H

The mass % of N = 45.11% N

Step2: The mass of each element in100 g of

the unknown compound is:

C =38.67g, H=16.22g and N=54.11g

Step3: the number of moles of each

element: 38.67/12.01 = 3.220 mol C

For H 16.22/1.008 = 16.09 mol H

And for N 45.11/14.01 = 3.219 mol N

Step4: divide by the smallest number of

mole, 3.2190 mol

C = 3.22/3.219 = 1

H= 16.09/3.219 = 5

N = 3.219/3.220 = 1

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The empirical formula = CH5N the EF mass =

31.06g/mol

The molecular formula = (CH5N)n

If the molar mass of the unknown

compound is = 62.12 g/mol

n = 62.12/31.06 = 2

Then the MF is (CH5N)2 = C2H10N4

Home work:

A white powder is analysed and found to

contain 43.64% Phosphorus and 56.36%

oxygen by mass. The compound has a molar

mass of 283.88 g/mol. What are the

compound Empirical and molecular

formulas? ANS. EF = P2O5 and MF= P4O10

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Chemical equations:

A chemical change involves a reorganization

of atoms in one or more substances. In a

chemical reaction, atoms are neither

created nor destroyed. All atoms present in

the reactants must be accounted for among

the products.

Example1:

When hydrochloric acid in aqueous solution

is added to solid sodium hydrogen

carbonate, the products carbon dioxide gas,

liquid water and sodium chloride solution

are formed. Write a complete balanced

chemical equation?

Solution:

HCl(aq)+ NaHCO3(s) → CO2(g) + H2O(l) + NaCl(aq)

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Example2:

Ethanol burns in oxygen to give carbon

dioxide and water vapour. Write a balanced

chemical equation to represent the

reaction?

The unbalanced equation is:

C2H5OH(l) + O2(g) → CO2(g) + H2O(g)

Most chemical reactions balanced by

inspection, that is, by trial and error. It is

always best to start with the most

complicated molecules (those containing

the greatest numbers of atoms).

Notice the C and H atoms are not balanced.

The most complicated molecule is ethanol.

Begin by balancing the products that

contain the atoms of C and H as in C2H5OH.

Since C2H5OH contains 2C atoms and 6 H

atoms, place the coefficient 2 before CO2 to

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balance the C atoms and place the

coefficient 3 before H2O:

C2H5OH(l) + O2(g) →2 CO2(g) + 3H2O(g)

Last, balance the O atoms by placing the

coefficient 3 before the oxygen molecule.

C2H5OH(l) + 3O2(g) →2CO2(g) + 3H2O(g)

Summary of writing and balancing the

chemical equation for any reaction:

1. Determine what reaction is occurring.

What are the reactants, the products and

the physical state involved?

2. Write the unbalanced equation that

summarises the reaction described in

step1 above.

3. Balance the equation by inspection,

starting with the most complicated

molecule(s). Determine what coefficients

are necessary so that the same number

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of each type of atom appears on both

reactant and product side. Don’t change

the identities (formulas) of any of the

reactants or products.

H.W.

At 1000 °C, ammonia gas, NH3, reacts

with oxygen gas to form gaseous nitrogen

monoxide and water vapour. Balance the

equation for the reaction?

Stoichiometric calculations involving

amounts of reactants and products in

chemical reactions:

Steps:

Balance the equation for the reaction

Convert the known mass of the

reactant/or product given to moles

Use the balanced equation to set up

the appropriate mole ratios.

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Use the appropriate mole ratios to

calculate the number of moles of the

desired reactant or product.

Convert from moles back to grams if

required by the problem.

Example: Solid lithium hydroxide is used

to purge (absorb) carbon dioxide from the

air in space shuttle cabin, by forming solid

lithium carbonate and liquid water. What

mass of gaseous carbon dioxide can be

absorbed by 1.00 kg of lithium hydroxide?

Solution:

2LiOH(s) + CO2(g) → Li2CO3(s) + H2O(l)

The molar mass of LiOH = 23.95 g/mol

Mole of LiOH in 1.00kg = 1000g/23.95 =

41.80 mol of LiOH

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Since we want to determine the amount

of CO2 that reacts with the given amount

of LiOH (41.80 mol), the appropriate mole

ratio is from the balanced chemical

equation is: 1 mol CO2/2 mol LiOH

Step4:

The mole of CO2 neeaded to react with

the given mass of LiOH is:

41.80X1/2 = 20.90 mol CO2

And the mass of CO2 is:

20.90X44.00 = 0.920 kg

Thus 920 g of carbon dioxide will be

absorbed by 1.00 kg of LiOH.

Home work:

Baking soda (NaHCO3) is often used as an

antacid. It neutralizes excess hydrochloric

acid secreted by the stomach:

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NaHCO3(s)+ HCl(aq)→ NaCl(aq) + CO2(aq) +H2O(l)

Milk of magnesia, which an aqueous

suspension of magnesium hydroxide, is also

used as an antacid:

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

Which is the more effective antacid per

gram?

Answer:

1 g of NaHCO3 will neutralises 1.19X10-2 mol

HCl.

However, 1g Mg(OH)2 will neutralises

3.42X10-2 mol HCl.

Milk of magnesia is a better antacid per

gram than baking soda. Show how is that?

48

Calculations involving a limiting reactant:

The limiting reactant (reagent) is the one

that is consumed first and thus

determines how much product can be

formed.

Example:

Consider the ammonia synthesis reaction:

N2(g) + 3H2(g) → 2NH3(g)

Assume that 5 moles of N2 and 9 moles of

H2 are placed in a flask. Is this

stoichiometric mixture of reactants, or

will one of them will be consumed

(limiting reactant) before the other runs

out (excess)?

Solution:

From equation the molar ratio between

reactants is 3hydrogen: 1nitrogen. In the

flask the ratio is 9/5 = 1.8/1 which is less

49

than the stoichiometric 3:1 ratio required

by the balanced equation. Hydrogen is

the limiting reactant and it will runs out

first leaving excess Nitrogen.

Example2: Suppose 25 kg of Nitrogen and

5.00kg of hydrogen are mixed and

reacted to form ammonia. Identify the

limiting reagent? Calculate the mass of

ammonia produced?

Solution: The balanced equation is:

N2(g) + 3H2(g) → 2NH3(g)

Moles of reactant:

Moles of N2=

25kgX1000g/1kgX1molN2/28.0gN2

=8.93X102 mol N2.

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Moles of H2=

5kgX1000g/1kgX1molH2/2.016gH2

=2.48X103 mol H2. (What we have)

The number of moles of H2 that will

exactly react with (8.93X102) mol N2 is:

8.93X102X3/1 = 2.68X103 mol H2(needed)

Therefore, hydrogen is the limiting

reactant, and we must use the amount of

hydrogen to compute the quantity of

ammonia produced.

2.48X103 mol H2X 2molNH3/3molH2

=1.65X103 molNH3

Converting mole of ammonia to mass

gives:

1.65X103 molNH3X17.0gNH3/1molNH3 =

2.80X104 g NH3= 28.0kg NH3.

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Calculating percent yield:

The percent yield =

Actual yield/theoretical yieldX100%

Where the actual yield is the amount of

product actually obtained in a given

experiment, and the theoretical yield, is

the maximum amount of product

calculated when the limiting reactant is

completely consumed (based on the

limiting reactant).

For example: if the amount of ammonia

actually obtained from the above

example is 21 kg. Calculate the percent

yield?

Percent yield of ammonia =

Actual yield/theoretical yieldX100 =

21/28X100% = 75%.