2015 prep notes

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AUSTRALIAN CHEMISTRY OLYMPIAD(AChO)

2015PREPARATORY

NOTES0. INTRODUCTION

The AChO Summer School is a 2 week intensive program that aims to present new,

interesting and challenging chemistry to students who have shown a talent in the subject.

Each year, AChO students come from diverse backgrounds. In particular, the high school

curricula vary from state to state. The aim of these preparatory notes is to provide you with

some background information which will help you get into gear for the camp. It is important

that you read these notes and are familiar with the material before arriving in January.

We are also happy to provide guidance if you find any part of these notes unclear. Feel free tocontact us by email:

• Sam Alsop email: [email protected]

• Audrey Lee email: [email protected]

We hope you will spend some time going through these notes and we look forward to seeing

you in the summer.

The Australian Chemistry Olympiad Team

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1. MATHEMATICS

As with all branches of science, the accurate modelling of physical phenomena requires a

mathematical toolkit to aid and concrete the description. You will find that these

mathematical tools pop up frequently during the Summer School, particularly in the physical

chemistry and analytical chemistry topics. For this reason, it is important to read the

following section thoroughly and attempt the problems associated with each section. Along

with what is covered here, it is also expected that students:

• are confident solving simple linear and quadratic equations and inequations

• are able to solve a system of simultaneous equations using elimination and

substitution techniques

• are comfortable with tabulating, graphing and interpreting data (particularly during

laboratory sessions)

1.1 Significant Figures

When you are doing any sort of scientific calculations you need to be aware of significant

figures, decimal places, and how accurately you can report your results.

Significant figures include every figure in the number except zeros that are used to place the

decimal point

Example. The following all have three significant figures

a. 0.465

b. 0.0465

c. 15.0 (this zero is not placing the decimal point but makes the number more accuratethan 15 so is significant)

When performing operations with numbers with different significant figures, one must be

aware of the accuracy with which the answer can be reported. The following table is a

summary of the rules to use:

Operation Rule Example

a + b = c

a – b = c

The answer should be given to the same

number of decimal places as the original

numbers (whichever of a or b has fewer)

53.06 – 52.1 = 1.0

c ba

c ba

=÷

=!

The answer should be given to the same

number of significant figures as the original

numbers (whichever of a or b has fewer)

43 ! 0.01 = 0.4

log b a = c The answer should be given to the same

number of decimal places as significant figures

in the original number a.

log10 11.2 = 1.049

10a = c The answer should be given to the same

number of significant figures as decimal places

in the original number a.

100.42

= 2.6

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Note that multiplication and division take priority over addition and subtraction so would be

reported to the correct number of significant figures.

Rounding of a number should not occur until the final answer. When rounding the following

rules apply for the first number excluded from the answer:

Less than 5 (0-4): nothing will change

Greater or equal to 5 (5-9): the last number that you have in your answer will be

increased by one

Example. Round the following to 2 significant figures:

a. 34.5 = 35

b. 0.09385 = 0.094

c. 1.34 = 1.3

Additional References:

www.chem.tamu.edu/class/fyp/mathrev/mr-sigfg.html

chemed.chem.purdue.edu/genchem/topicreview/bp/ch1/sigfigs.html

Problem Set 1.1:

1.1. Write the following to the correct number of significant figures or decimal places as

required.

a. 32.15-7.98

b.

c. (34.5+12.45)-1.00

d.

e. 10.0)11.932.12( ÷+

1.2 Logarithms

Logarithms are used to solve equations such as 10x = 100 by rearranging the equation to give

log10 100 = x allowing for the solving of x. It is the inverse (reverse action) function of an

exponential function (log10x is the inverse of 10x, log2x for 2

x etc.). While it might seem

redundant for simple propositions like the one above, once x starts to change from a whole

integer it becomes much more important. For example, 10x = 567.

There are two common forms of logarithms, or logs, used in chemistry. The first is log to the

base 10 written as log10 or more simply as log. The other is the natural log, loge or more

simply as ln, where the base e is approximately equal to 2.71828.

For all purposes in chemistry the answer of a log can be produced using a calculator.

However manipulating an equation may become important for simplifying problems. The

following rules should be obeyed and apply to log and ln equally.

04556.0253.0 !

000.5)00.900.10( ÷!

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Logarithm Law

Equation Rearranged equation(associated logarithm law)

Example Calculation (Solve for x)

10y = x log x = y Solve ex = 901

x = ln 901 = 6.804

log x = y 10y = x Solve log x = 1.69

x = 101.69

= 49

log (a ! b) = y log (a) + log (b) = y

(logarithm of a product of

two elements is the sum of

the logarithm of each

element)

Solve log (10 ! x) = 2

log (10) + log (x) = 2

1 + log (x) = 2

log (x) = 1

x = 10

log (a ÷ b) = y log (a) – log (b) = y

(logarithm of a quotient of

two elements is the

difference of the logarithm

of each element)

Solve log (x ÷42) = -0.26

log (x) - log (42) = -0.26

log(x) = -0.26 + log (42)

x = 10-0.26 + log (42)

x = 23

log (ac) = y c ! log (a) = y

(follows from the 3rd

property)

Solve ln (4x) = 19

x ln (4) = 19

x = 19 ÷ ln (4)

x = 17.71

logx x = y y = 1

(logarithm of a number to

the base of itself is 1)

Solve ex = 901

ln(ex) = ln (901)

x ln(e) = ln (901)

x = ln (901) = 6.804

logx1 = y y = 0

(the logarithm of 1 is 0 in

any base y)

Solve log (1 ÷ x ) = 2

log (1) – log (x) =2

log (x) = -2

x = 10-2

= 0.01

log b (x) = loga (x) ÷ loga (b)

(changing the base of a

logarithm from a to b)

Find log3 (7) in terms of logarithms to the

base 10 only.

log3 (7) = log10 (7) ÷ log10 (3)

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www.physics.uoguelph.ca/tutorials/LOG/

www.chem.tamu.edu/class/fyp/mathrev/mr-log.html

Problem Set 1.2.:

1.2 Find y remembering to use the correct number of significant figures.

a. log y = 0.20

b. 10y = 14

c. log y = -5.87

d. log (12 ! 3.00) = y

e. log (y ÷ 3.00) = 0.60

f. log y-1

= 5.60

1.3 Calculus

1.3.1 Functions, Operators and Functionals

Functions are the working objects of basic calculus. A function of a variable, f, maps an input

x (usually a real number) to another number f ( x) as determined by the function. It defines a

one-to-one relationship between one variable and another.

Figure 1.3.1: the action of a function

Many phenomena can be expressed as such relationships. Some examples include:

a) the position of a car on a road could be expressed as a function of time, i.e. x(t )

b) the population of whales in a given region of ocean might be expressible in terms

of a function of the average krill population in that region, i.e. w pop(k pop).

We are most familiar with functions in their graphical form: the set of points ( x, f ( x)) on two

axes. For instance, we have all probably seen graphs of y = f( x) = x2 (a parabola) and

y = f ( x) = sin( x) (the sine wave function).

These functions are called analytic as the action of the function is expressible in some closed

algebraic form. Note that there is no restriction that states functions have to be analytic - we

could imagine a function which was still a one-to-one relationship but that we could not

express with any algebra known to us (consider a random scatter of points). In fact, it is

frequently the case that physical problems involving functions cannot be solved in any

closed/analytic form. In this case a lot more work is required to describe the function. Luckily

for us, most of the functions we'll deal with at the Summer School are smooth-looking

(continuous) and analytic!

We can also have functions of more than one variable. For instance, the population of whales

might depend on both the population of krill and the average temperature. So we write the

function as w pop(k pop, Tavg), in terms of variables k pop and Tavg.

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We have just defined a function as a relationship that maps one real number to another real

number. There are other kinds of relationships, such as:

• An operator maps a function to another function - that is, if you take a function, and

apply an operator, you'll get another function. Lots of operators have a special

notation, a circumflex or 'hat' above them: Oˆ

. Examples are the differential andindefinite integral operators, as explained in the following sections.

• A functional maps a function to real number - that is, if you take a function and

apply a functional, you'll end up with a real number. The simplest example of a

functional is a definite integral, as defined below.

1.3.2 Differentiation

KEY IDEA:

Differentiation is about finding the rate of change of one variable compared to another, or

the rate of change of a (continuous) function f(x) with respect to the variable x.

If we have a graph comparing two variables (e.g. y, x), where y = f(x), then differentiation is

equivalent to finding the gradient of the curve.

The gradient of a straight line function is given by “Rise over Run”. We can express the

gradient as , where is the change in y and x ! the change in x between two

points (see Fig 1). For a graph/function that is curved (for instance for the graph 2x y = ) the

gradient is no longer constant, but is different at each point. We get closer to the true value of

the gradient at a certain point by decreasing the size of x ! and y ! (see Fig 2).

Differentiation finds the true value of the gradient by allowing x ! and to approach (or

get infinitely close to) zero.

Notation:

An infinitely small value of x ! is expressed as “dx”, and y ! as “dy”

The true value of the gradient for any function (at a given point) is given by:

m =" y," x#0lim

" y

" x=

dy

dx

m =" y

" x y !

y !

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In terms of the function defining the y variable, this equivalently becomes:

x

x f x x f

dx

xdf

dx

dy

x !

"!+==

#!

)()()(lim

0

This object, d f /d x, is referred to as the derivative (rate of change) of the function f with respect to the variable x.

Differentiation changes one function into another.

As we said, the value of the gradient (the derivative) can vary from point to point (like in a

parabolic function). For this reason, we expect the action of taking the derivative

(differentiating ) of a function to return another function (an operator relationship). For

instance, to find the derivative of the function2)( x x f = , we start with:

x

x x x

dx

df

x !

"!+=

#!

22

0

)(

lim (1)

Expanding gives:

x

x x

x

x x x

x

x x x x x

dx

df

x

x

x

2

)2(

)2(

)(2

lim

lim

lim

0

0

222

0

=

!+=

!

!!+=

!

"!+!+=

#!

#!

#!

where the last line follows from the fact that " x is approaching 0.

Therefore xdx

df 2= , where f ( x) = x

2.

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Table of Derivatives

Luckily for us, it is unnecessary to apply the formal definition of differentiation every time

we want to find a derivative. In fact, in practical use - no-one does. We use short-cuts. Below

is a tabulated list of common derivatives and how to find them. It is very important to learn

these.

( ) 0=C dx

d

Differentiating a constant gives zero

(i.e. a horizontal line has zero

gradient)

( ) a b ax dx

d =+

Differentiating a linear equation gives

a constant (i.e. the gradient of a

straight line is constant)

( ) 1!= nn nx x dx

d

Differentiating a polynomial lowers its

power by 1, but multiplies the result

by the original power (recall the

derivative of2

x y = )

( ) x x e e dx

d =

Differentiating the exponential of

Euler’s constant e (e = 2.71828…)

gives itself

( )x

x dx

d 1ln =

Differentiating the natural logarithm

gives 1/x

( ) ( ) x x dx

d x x dx

d sincos,cossin !== Differentiating sine gives cosine, anddifferentiating cosine gives thenegative of sine

( ) ( ) ( ))()()lg()( x g dx

d l x f

dx

d k x xkf

dx

d !+!=+

where k, l are constants.

The linearity property:

- the derivative of a scalar multiple of

a function is just the scalar times the

derivative of the function.

- the derivative of the sum of two

functions is the sum of the two

derivatives

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Product Rule (by illustration):

Suppose that two functions x(t) and y(t) change

with time t in Fig 3, e.g. the side lengths of a

rectangle.

Note that

)()()( t yt xt Area = .

Then the change in area:

ydx xdy Aread +=)(

dx dy ! is infinitely small compared with

either of dy , dx and is thus negligible.

Hence:

dt

dx y

dt

dy x

dt

Area d +=

)(

In general:

dx

df g

dx

dg f

dx

fg d +=

)(

where f and g are arbitrary functions. This result is called the product rule and allows one to

calculate more complicated derivatives.

Chain Rule:

When there is a function within a function (e.g. when )ln(sin)( x x z =

, which is equivalentto stating two functional relationships: x x y sin)( = and y y z ln)( = ), then

dx

dy

dy

dz

dx

dz !=

In the given example,x

x x

x y dx

dy

dy

dz

dx

dz

tan

1cos

sin

1cos

1=!=!=!=

Problem Set 1.3.2:

1.3.1. Evaluatedx

dy , using the appropriate rules:

Differential Rules

a. 154 2

!+!= x x y

b. x

e x x y 5cosln !+=

Product Rule

c. x x y ln=

d. x x y cossin=

e. 41 22 +!= x x y

Chain Rule

f. ( )kx y sin=

g. 2

sin x y =

h. ( )33 1+= x y

NB: k is a constant.

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1.3.3 Integration

KEY IDEAS:

1. Integration is the opposite operation of differentiation. It computes an antiderivative.

It takes you from a rate of change function to an explicit relationship between the two

quantities. This means if you have a rate of change of a quantity expressible in functional form f(x) and want to calculate a direct relationship between the quantity and the

variable, you would integrate the rate of change f(x) with respect to x. Since this action

takes a function to a function, it is an operator . We call this taking an indefinite integral .

For example, if you knew the rate of change of position with respect to time (the

velocity), as v(t ) = d x(t )/dt , you would integrate the function v(t) to find the function x(t ).

2. Integration can also find the area under the curve ( x, f(x)) between two points. If you

take the indefinite integral function, F(x), found from Key Idea #1 and compute the

difference in F(x) between two points you will obtain the area (bounded by the axis)

under the original function f(x). Since this takes a function to a number (an area), it is a

functional . We call this taking a definite integral.

How can both these ideas be true? It turns out they're equivalent. This is proved below for

your reference. This is not essential knowledge, however, and may be skipped at your

discretion.

Proof:

Suppose you are trying to estimate the area under a curve )(x f y = by drawing rectangles

under the curve and calculating its area (Fig 4). By decreasing the width ( x ! ) of the

rectangles, you arrive at a better estimate (Fig 5). Obviously, the best estimate is when x ! is

infinitely small (i.e. x ! approaches zero).

Note: if the curve is in below the horizontal axis, the “area” between the zero line and the

curve is negative in value.

Now, let us define )(x F y = which we can use to find the area under the curve of )(x f y = .

Here is how the function works (see Fig 6):

An area bordered on the left by x = A, on the right by x = B, above by )(x f y = , and below

by y = 0 is given by

AREA = )()( AF B F ! (This is AREA (2) in Fig 6)

NB: If the curve )(x f y = lies below y = 0 (i.e. is negative), then the value of this “area” is

also negative

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Using these rules, we can see that AREA (1) is expressed as

AREA (1) = F(A) – F(0),

and the sum of AREA (1) and AREA (2) is given by

AREA (1) + AREA (2) = F(B) – F(0)

Then the area bounded by the infinitely small rectangles we were exploring in Figure 4 and 5

is given by:

x x f x F x x F AREA !="!+= )()()(

As x ! approaches zero, this becomes:

dx x f x F dx x F AREA )()()( =!+=

Note that )()( x F dx x F !+ is simply another way of writing change in y (or dy ) or

)(x dF !

Rearranging:

)()()()(

x f dx

x dF

dx

x F dx x F ==

!+

Thus, the differential of our “area under the curve” function is the original function.

Definite and Indefinite Integrals:

An indefinite integral in the above example is simply )(x F y = , a reverse of differentiation.

F(x) is called the anti-derivative.

Notation:

A definite integral is an expression that finds the area bound by the curve, horizontal axis,and two horizontal axis values (the vertical lines x = a and x = b).

Notation: ! ="b

a

dx x f A F B F )()()(

! = dx x f x F )()(

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Note: In this way (expressed in Figure 5 above), the integral notation is actually a continuous

sum! It sums the product f(x)dx (product of the graphical height of the function f(x) and the

width of each rectangle) continuously between the bounds a and b. This can be expressed

mathematically as:

! " =

#$%=

b

an

nnn

b

a

x x f dx x f )(lim)(

(where xn is the midpoint of the one of the rectangles, n is the number of rectangles, and " xn is

the width of the each rectangle (which decreases as the number of rectangles increases)).

Example. To find the area under the curve2

x y = between 1=x and 4=x , we find the

indefinite integral, which is C x dx x += ! 32

3

1 (see the table below), and then turn it into a

definite integral by substituting the lower and upper bounds (i.e. the “A” and “B”):

213

63

3

1

3

4

3)(

334

1

34

1

2==!="

#

$%&

'=== ( ( x dx x dx x f AREA

B

A

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Table of Indefinite Integrals

Luckily, once again, we do not have to use the formal definition of integration to compute

integrals - we have rules. These rules for computing anti-derivatives are summarised in the

table below. You'll notice that they are simply the reverse of differentiating.

Note that indefinite integrals always contain an integration constant, C. This is because

differentiating a constant gives 0 (previous section), so we are free to choose an arbitrary

constant in the reverse process. The constant is often determined after the integration by using

a physical condition given to us (for instance, the value of the new function at a particular

value of x). This integration constant will always cancel out in definite integrals.

C dx = ! 0 Integrating anything gives an infinitenumber of similar functions, all varying by a constant number C.

C axadx += ! Integrating a constant gives a linear

equation

(a curve with a constant gradient is

obviously any straight line with that

gradient)

C x dx nx nn

+= ! "1

OR From the derivative law!

C e dx e

x x

+= ! From the derivative law!

C x dx x

+= ! ln1

! dx x

1is sometimes written as !

x

dx

C x dx C x dx +=+!= " " sincos;cossin Simple reversals of the differentiationrules

[ ] C dx x g bdx x f adx xbg xaf ++=+ ! ! ! )()()()( Integrals, like derivatives, are linear.

Problem Set 1.3.3:

1.4. Integrate

a. ! ++ dx x x 352

b. ! + xdx x sincos

c. ! x

kdx where k is a constant

d. ! nx

kdx where k, n are constants and n > 1

C n

x dx x

n

n+

+

=

+

! 1

1

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1.3.4 Representation as Operators

We've talked about the fact that both differentiating (finding the derivative) and integrating

(finding the indefinite integral) are operations, so now it's time to make some important

distinctions.

• If we write before a function, it means we take that function and return its

derivative.

• Similarly, if we write after a function and place the integration ( ! ) sign in front of it, it means we take that function and return its integral.

e.g. x xdx

d 2)(

2= , ! =

22 x xdx

This idea of representing a differential operator as is not to be confused with the idea of

the derivative itself,dx

df , which is a well-defined quantity (a ratio of two changes) and is no

longer an operator. The derivative can be manipulated in a normal way, for instance if

dxdf

dx

df

=!

=1

is implied.

We can take the derivative of a function more than once. For example, the second derivative

of a function f is defined as2

2

)(dx

f d

dx

df

dx

d = . For examples, the second derivative of

position with respect to time is acceleration in mechanics. Higher derivatives are defined

similarly.

Note: at no point can you cancel the d letters. Remember that df and dx represent

changes in a particular variable and can be manipulated in the normal algebraic way,

and that the d by itself is meaningful only as an operator.

1.3.5 Separable Differential Equations

These calculus ideas actually underpin most of physics (and hence physical chemistry). The

most common form in which one finds calculus is differential equations. Differential

equations involve relationships between derivatives and functions, and are often incredibly

complicated to solve. They are often unsolvable in terms of analytic functions.

The reason they are so common in physics is because physics describes dynamics, or rates of

change (derivatives). The most obvious exam is Newton's law of motion, which translates

into a simple differential equation.

ma = F,

where F is force, m is mass and a is acceleration

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In differential equation language, this translates to:

2

2

( )( , )

d x t m F x t

dt =

These differential equations also have more obvious chemical applications, particularly inthermodynamics and chemical kinetics.

At Summer School, you will only have to solve a special class of differential equations called

separable differential equations. This means that all variables of one type can be collected

on one side of the equation by algebraic manipulation, and all variables of the other type can

be collected on the other side. Since both sides (infinitesimal changes) are equal everywhere,

both sides can be easily integrated (summed) to give the explicit relationship desired. This

technique is demonstrated in the example below.

Solve kA

dt

dA= (where k is a constant)

Solution:

(1) We separate the two variables, A and t, to either side:

dAA

kdt 1=

(2) We integrate both sides:

! ! = dA A

kdt 1

(3) The desired relationship is reached:

t

+= where C is the constant of integration

(4) The constant of integration is either left arbitrary, or determined by some given

condition (e.g. the value of A when t = 0)

Problem Set 1.3.5:

1.5.1. The concentration of a reagent A (denoted [A]) is found to decrease as a function of

time t via the following relationship:kt

e AA !

= 0][][ (where k is a constant and

0][A is the initial concentration of reagent A). Show that the rate of change of [A]

(i.e.dt

Ad ][) is proportional to the concentration of [A].

1.5.2. The rate of a reaction is found to vary according to the following relation:

2][

][Ak

dt

Ad != . The initial concentration (i.e. t = 0) of the reagent is [A]0

a. By separating variables, prove that kt C A

!=+

!

][

1

b. At t = 3 sec, prove using a definite integral that k AA

3][

1

][

1

03

=!

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1.4 Complex Numbers

If you were asked to solve the equation !! ! !!, what would your response be? If your

answer is anything along the lines of “you can’t take the square root of a negative number”,

then boy, is this the section for you! Read on, brave Summer Scholar, to discover the world of

complex numbers.

1.4.1 The ! in the sky

I’m going to go out on a limb here, and define a number called !, which stands for imaginary.

This number is defined by this simple equation:

! ! !!

! itself is called the imaginary unit.

“You can’t do that!” I hear you say.

Why not? Just as all numbers are inventions, so too is this new one. As long as it fits neatly

into our current system of numbers, why can’t we use it alongside what we already know? In

fact, let’s see what happens when we apply some simple maths and see where that leads us:

!!! !!

!

! !!

!!! !!!

!! !!!! ! !!

!!! !

!!!

!! !!!!! ! !

That’s pretty cool, so if we try multiplying ! by itself lots of times, we always end up going in

a circle of !!!!!!!!!! !!!

!! ! !!!! ! !! !! ! !!

And now, we can take square roots of any number!

Problem Set 1.4.1:

Evaluate the following:

a) !!

b) !!!"

c) !!""

d) !!"

e) !!"

f) ! !!""

Solve the following equations:

g) !!!

!!"

h) !! ! !!"

i) !! ! !! ! !" ! !

1.4.2 Things are going to get more complex

So, things are about to get more complex. How so? Think a moment about the numbers you

use every day. In mathematical jargon, these are called real numbers.

What happens if we put both real numbers and imaginary numbers together?

! ! ! ! !

We can’t actually combine the two into one single number – they are fundamentally different

types of things. The best we can do is write it out just like we have above, with a realcomponent, and an imaginary component. This sum we’ve just created still represents one

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c) !! !! ! ! ! !!

d) !! ! ! ! !! ! !

e) !! !! ! ! ! !

f) ! !! !!!!! ! ! !!!

g) !!! !! ! !! ! !!

h) !!

! ! !!

! ! i) Challenge – What is !!

Hint – Write ! ! ! ! !" and solve !! ! !. Remember that ! and ! are real.

1.4.2 Egad! !’s gone crazy!

So, for Summer School, there’s just one more important piece of info you need to know about

complex numbers. And it’s something called Euler’s Formula.

Perhaps you’ve come across this wacky but devastatingly beautiful equation:

!

!"

! !!

!

Five of the key numbers of mathematics united in one equation. It brings a tear to your eye,

doesn’t it? This can easily be shown using Euler’s Formula, which states:

!!"! !"# ! ! ! !"# !

We must briefly touch on radians in order to use this formula.

What are radians you ask? So everybody’s been familiar with dividing circles into 360° since

primary school. A radian is simply dividing the circle into !! radians. Simple! As an

example, have a look at the following table:

Degrees Radians0 0

30 ! !

60 ! !

90 ! !

180 !

360 !!

Degrees are probably the standard units for angles that you've come across so far. However, it

turns out that mathematically, there are very deep and fundamental reasons why radians are

used, as annoying as they may seem right now! So we'll be using radians when we apply

Euler's formula.

For example, let’s prove our favourite equation by substituting ! ! ! into Euler’s Formula.

!!"! !"# ! ! ! !"# ! !! !"# !"#$ ! ! !"# !"#$!

! !! ! !!! ! !!

Problem Set 1.4.3:

Find exact expressions for the following:

a) !!!"

b) !!" !

c) !!!" !

d) !!"#!" !

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2. ORGANIC NOMENCLATURE

The rules for naming organic compounds are set by an international body called the

International Union for Pure and Applied Chemistry (or IUPAC for short). The basics of

naming and drawing organic molecules will be covered here, to be expanded upon oversummer.

2.1 Basics:

There are three ways of writing formulae of organic compounds:

Structural Formulae: show all atoms, including hydrogen atoms which becomes difficult

when molecules become more complex

Example: 2-chloropropane

C C C

ClH

H

H

H H

H

H

Condensed Structural Formulae: we just write out how many hydrogen atoms are present

without drawing those carbon to hydrogen bonds


CH3

CH

CH3

Cl

Skeletal Formulae – there is a carbon at every vertex and the appropriate number of

hydrogen atoms is assumed to be bonded to each carbon atom. Hydrogen atoms are drawn

when bonded to non-carbon atoms.


Cl

Example: diethylamine

NH

Any organic molecule contains three main components:

• the carbon backbone

• the principal functional group, which influences the properties of the compound, is

referred to in the suffix of the name

• substituents, which are secondary functional groups are named as prefixes

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2.2 The carbon backbone:

The name of the backbone arises from how many carbon atoms are in the longest chain within

the molecule. For an example it is best to look at the alkanes, which are the simplest organic

compounds, containing only carbon and hydrogen with only single bonds between them. The

prefix (in bold) is used throughout molecule naming to describe a chain length with that manycarbon atoms in it.

CH4 methane 1 carbon

C2H6 ethane 2 carbons

C3H8 propane 3 carbons

C4H10 butane 4 carbons

C5H12 pentane 5 carbons

C6H14 hexane 6 carbons

C7H16 heptane 7 carbons

C8H18 octane 8 carbons

We can then add any of a number of substituents to the chain and as you can see, when a

basic carbon chain is added the same prefix appears:

-CH3 methyl- -F fluoro-

-CH3CH2 ethyl- -Cl chloro-

-CH3CH2CH2 propyl- -Br bromo-

isopropyl- -I iodo-

-CH2CH2CH2CH3 butyl- -NO2 nitro-

tert -butyl- -NH2 amino-

-OCH3 methoxy- -OCH2CH3 ethoxy-

At this point, there are a number of rules to introduce:

1. When there is more than one type of substituent, order prefixes in alphabetical

order.

Example: bromochloromethane not chlorobromomethane.

H

CH Cl

Br

CH3

-CH CH3

-C-CH3

CH3

CH3

C

H

HH

H

H3C CH3

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2. When there is more than one of a substituent type, use prefixes di-, tri-, tetra- etc. to

signify the number. These prefixes do not contribute to the alphabetical order.

Example: dibromochloromethane not chlorodibromomethane

H

CBr Cl

Br

3. To define where a substituent is attached, assign numbers to the carbon chain. Start

numbering from the end that will result in the lowest number for the substituent at

the first point of difference.

Example: 1-chloropropane not 3-chloropropane

Cl

4. Put a dash between all numbers and letters, a comma between numbers, and leave no

spaces between letters.

Example: 1,3-dichloropropane

ClCl

5. Make sure that the backbone is the longest chain.

Examples of other Carbon Backbones:

Cycloalkanes: Are alkanes which have joined into a ring. These are named in exactly the

same way as alkanes, with a cyclo- prefix.

Example: 1-chloro-1-methylcyclohexane

Cl

Alkenes: Contain one or more double bonds. Similar rules to those applied to alkanes apply

here also. But you must number only the first carbon that has the double bond, and replace the

-ane of the alkane with -ene.

Example: 1-butene or but-1-ene

Example: 1,3-butadiene or buta-1,3-diene

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The double bond takes priority over halo substituents so number from the end that will give

the lowest number to the alkene.

Example: 4-chloro-1-butene or 4-chlorobut-1-ene

Cl

Alkynes: Contain one or more triple bonds, and have the suffix -yne. The same rules as

alkenes apply, but if both are present then the alkene has higher priority than the alkyne.

Example: but-1-en-3-yne, not but-3-en-1-yne.

2.3 Functional groups:

Alcohols contain the –OH group. Remember because the hydrogen is bonded to the oxygen it

is still drawn. If the alcohol is the principal functional group, the compound is named with an

-ol suffix, and numbering begins at the end which will give the lowest number for the carbon

attached to the –OH.

Example: 1-propanol or propan-1-ol

OH

Example: 3-chloro-1-propanol or 3-chloropropan-1-ol

Cl OH

If there is another group of higher priority, –OH is named as a substituent, with the prefix

hydroxy- and is numbered accordingly. We will see the order of priority later.

Carbonyl compounds contain the R1

O

R2 group. Varying either R 1 or R 2 will then

change the type of carbonyl group. There are a number of different types of carbonyl

compounds:

Aldehyde - when one R group is a –H and the other is a carbon atom bonded to anything else.

These are named with the suffix -al. An aldehyde will only be present at the end of a chain as

it cannot extend in both directions due to the requirement of the hydrogen.

Example: butanal

O

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Ketone - when both R groups are a carbon atom bonded to anything else. These are named

with a suffix -one. Remember that because the C=O is in the middle of the chain, its position

will need to be shown by numbering.

Example

O

2-butanone or butan-2-one

Carboxylic acid: when one R group is an –OH and the other is a carbon atom bonded to

anything else. These are named with the suffix -oic acid. Again this must be at the end of a

chain.

Example

O

OH

butanoic acid

We now come to priorities of functional groups. If a molecule is composed of more than

one functional group, the group of highest priority determines the root of the name (suffix)

and must be included in the longest chain. Any other functional groups are treated as

substituents and are described with the prefix.

i n c r e a s i n g

p r i o r i t y

functional group suffix prefix

carboxylic acid -oic acid

aldehyde -al oxo-

ketone -one oxo-

alcohol -ol hydroxy-

alkene -ene -en-

alkyne -yne -yn-

alkane -ane -an-

Example: 2-hydroxypropanoic acid

OOH

OH

Example: 2-oxopropanoic acid

OO

OH

Example: but-3-enal

O

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2.4 Aromatic compounds:

We now move to an entirely different area of organic chemistry, that of benzene.

Benzene consists of alternating double and single bonds, and can be represented as a

resonance of the above two forms. For simplicity, it is also represented by

A number of compounds of benzene have special names, listed here in order of

decreasing priority (as you read from left to right).

OOH

OH

NH2

OCH3

CH3

benzoic acid phenol aniline anisole toluene

If there are more than two groups attached, we need to define the positions:

Ortho, meta and para can be abbreviated to o, m and p.

Examples:

Cl

Cl

OH

Br

meta-dichlorobenzene p-bromophenol

orthoortho

meta meta

ara

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If there are more than 2 substituents, then we use numbering.

Example:

OH

Cl

Cl

3,4-dichlorophenol

If benzene is a substituent, it is known as the phenyl group, and can be abbreviated to –Ph in

formulae.

Example:

OH

O

or Ph OH

O

3-phenylbutanoic acid


Brown and Poon, Introduction to Organic Chemistry, 3rd Ed, Wiley International

2005, Sections 3.1-3.6, 9.4, 13.3, 14.3 and pg 101-102.

Silberberg, Chemistry The Molecular Nature of Matter and Change, Sections 15.2:“The Structures and Classes of Hydrocarbons”, 15.4: “ Properties and Reactivities of

Common Functional Groups”

Problem Set 2:

2.1 Name the following compounds

a.

b.

c.

d.

e.

O

f.

OOH

g.

O

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h.

Cl

Cl

i.

NH2

OH

j. OH

OO

2.2. A compound has been named 3-chloro-2-ethyl-5-methyl-heptane. What should it

actually be called?

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3. TRANSITION METAL CHEMISTRY

3.1 Introduction:

At Summer School, we will be talking about coordination compounds, also known as metal

complexes, which are known for all metals, but we will mostly be looking at the transition

metals. The molecules and ions which coordinate to the metal centre are called ligands. In

general, we can think of a coordinate bond formed between the metal and a ligand, in which

the ligand donates a pair of electrons to the metal centre. This bonding can quite drastically

change the reactivity of the metal and ligand.

For example, hexacyanidoferrate(III) (also called ferricyanide) comprises of six cyanide

ligands and an iron(III) centre. Usually the cyanide ion is very toxic, however

hexacyanidoferrate(III) has a very low toxicity, due to the strong iron-carbon bonds.

Another point to note is that hexacyanidoferrate(III) is an anion, so there is a positive counter

ion. The counter ion, frequently sodium or potassium for hexacyanidoferrate(III), is not

covalently bound to the metal or ligands and does not drastically change the reactivity.

One thing you may not have seen before is the wedged bonds. These are used to show that a bond is not in the plane of the page. The solid wedged bonds come out of the page and the

dashed bonds go into the page. Here is another diagram of the hexacyanidoferrate(III) ion to

compare with the structure before.

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3.2 How to draw a metal complex:

Metal complexes can have from one to twelve atoms attached to the metal centre so it is

important to be able to clearly show the geometry around the metal centre. The number ofatoms attached to the metal is called the coordination number, often abbreviated to CN.

Coordination numbers of 4 and 6 are the most common and there are multiple different

geometries around the metal centre that can be observed for the same coordination number.

CN = 6

Complexes with coordination numbers of six are usually octahedral, which puts the metal

centre at the centre of an octahedral formed by the atoms attached to it. The

hexacyanidoferrate(III) above is an example of an octahedral complex. This is the most

common way to draw octahedral complexes, however sometimes we draw them as shown

below.

CN = 4

Complexes with a coordination number of four can be either square planar or tetrahedral and

we will discuss why at Summer School. For the purpose of getting you to practice drawing

them, you will be told if a complex is square planar or tetrahedral for the exercises.

Tetrahedral Square Planar (either is acceptable)

3.3 Naming:

There are a number of rules for naming coordination compounds.

1. The cation is names before the anion, as you've seen for naming simple ionic

compounds.

2. The ligands are named (in alphabetical order) before the metal.

3. If the ligand is anionic, replace the final -e with an -o.

4. If the ligand is neutral, use the full name. Do not abbreviate the name.

5. For multiples of the same ligand, use the prefixes below

Fe

NC

NC CN

CN

CN

CN

3-

Br

Ni

Br Br

Br

2-

Pt NH3

NH3

H3N

NH3

2+

Pt

H3N

H3N NH3

NH3

2+

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If the ligand doesn't contain one of the

following prefixes e.g. chlorido

If the ligand does contain one of the following prefixes

e.g. 1,2-ethanediamine

2 di Bis

3 tri Tris

4 tetra tetrakis

5 penta pentakis

6 hexa hexakis

6. If the complex is anionic, add -ate onto the end of the name of the metal. If it ends in -ium

or -um, replace it with -ate. Some metals also go back to their Latin roots.

Metal Anionic Name Metal Anionic Name

Nickel Nickelate Palladium Palladate

Copper Cuprate Iron Ferrate

Lead Plumbate Gold Aurate

Silver Argentate Tin Stannate

Mercury Mercurate Manganese Manganate

7. Write the oxidation number of the metal in Roman numerals after the metal. For example,

Fe(III).

For formulas, there are similar rules.

1. Cation before anion, as for ionic compounds.2. The formula of the complex is in square brackets.

3. Inside the square brackets, the metal centre is first.

4. Ligands are written with the atom attached to the metal first or abbreviated and

ordered alphabetically. If there are multiples of the one ligand, follow with a

subscript. If the ligand comprises of multiple atoms, surround it with round

brackets.

Examples:

tris(1,2-ethanediamine)cobalt(III) nitrate potassium tetrachloridocuprate(II)

[Co(en)3](NO3)3 K 2[CuCl4]

Co

NH2

H2N NH2

NH2

H2N

H2N

NO3

-

NO3-

NO3-

3+

K+

K+

Cl

Cu

ClCl

Cl

2-

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3.4 Common ligands:

There are a number of common ligands, which we will repeatedly see during Summer School.

The most important of these, namely water, ammonia, carbon monoxide and nitrogen

monoxide, are the exceptions to the rule for the names of neutral ligands are unchanged. The

differences are highlighted in the table below.

There are also ligands that can attach to the metal centre at two or more atoms, like

1,2-ethanediamine as before, through its two nitrogen atoms. These are called polydentate

ligands. Ligands which can only bind through one atom are called monodentate. Ligands with

two binding atoms are called bidentate; three tridentate, and so on. These can be contrasted

with ligands like the nitrite ion (NO2 – ) and the thiocyanate ion (SCN

– ), which have a lone pair

on more than one atom in the ion, so they can bind through either, but not both atoms. These

are called ambidentate ligands.

Molecule or ion Ligand name ( - if neutral) Formula (abbreviation) Donor atoms

ammonia ammine NH3 N

water aqua H2O O

carbon monoxide carbonyl CO C

nitrogen monoxide nitrosyl NO N

acetonitrile - NCCH3 N

bromide bromido Br - Br

carbonate carbonato CO32-

O or 2 O (can be

monodentate or

bidentate)

chloride chlorido Cl- Cl

cyanide cyanido CN- C

iodide iodido I- I

nitrite nitrito NO2- N or O

nitride nitrido N3-

N (triply bonded to

the metal)

oxide oxido O2-

O (doubly bonded

to the metal)

thiocyanate thiocyanato SCN- S or N

pyridine -

C5H5 N (py)

N

1,2-ethanediamine

(ethylenediamine)

- H2 NCH2CH2 NH2 (en) 2 N

oxalate (ethanedioate) oxalato

C2O4- (ox) 2 O

2,2'-bipyridine - 2 N

N

N N

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C10H8 N2 (bipy)

triphenylphosphine -

P(C6H5)3, PPh3

P

glycinate glycinato H2 NCH2COO- (gly) 1 N and 1 O

ethylenediaminetetracetate

(EDTA)

ethylenediaminetetracetato

C10H16 N2O8 (EDTA)

2 N and 4 O

Problem Set 3:

Note that some of these complexes have isomers, but we will discuss isomerism in detail at

Summer School.

1. Name and write the formula for each of the following complexes.

a) b)

2. Draw and write the formula for each of the following complexes.

a) diaquabis(2,2'-bipyridine)iron(II) chloride b) diamminedichloridoplatinum(II) – square planar

3. Draw and name each of the following complexes.

a) [RhCl(PPh3)3] – square planar

b) [Zn(gly)2] – tetrahedral

P

O-

N

-

O

O

N-O

O O-

O

O

Cr

H3N

H3N Cl

Cl

OH2

NH3

+

NO3-

Mn

O

O OH2

OH

CO

OHO

ONa+

Na+

2-

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4. THEROETICAL CHEMISTRY / ELECTRONIC

STRUCTURE

In principle, all the chemical properties of matter can be derived from physical laws of

motion. For example, the simplest chemical system – a hydrogen atom – consists of a positively charged nucleus accompanied by a negatively charged electron. The two particles

experience an attractive force and from this, the motion of the electron can be calculated.

From this, chemically important properties like the radius of a hydrogen atom, the ionization

energy and even the colour can be calculated.

As part of Summer School we will attempt to approach chemistry from this bottom-up

approach by considering what occurs at the microscopic level (electronic structure) but also

the physics behind chemical transformations. Both of these concepts originally had their roots

in simple classical physics.

4.1 Basic Mechanics

At Summer School, we will begin by considering the classical (pre-20th century) laws of

physics, particularly classical mechanics and classical wave theory. You should be familiar

with some of the very basic concepts before you arrive. This section will contain no examples

- it just aims to familiarise you with some of the vocabulary we will be using. Most of this is

covered in a standard year 11/12 physics course.

Particles

In classical mechanics, the state of a particle is described by quantities such as position (x),

velocity (v, directional speed, measured in ms-1), acceleration (a, measured in m s-2)momentum ( p, measured in kg m s

-1), energy ( E, measured in J) (kinetic ( K ) and potential

(V )), forces acting on ( F , in N), mass (m, in kg), etc. The important terms are defined in the

table at the end of the section. In mechanics calculations, SI units should be used at all times.

Velocity and momentum describe the direction and magnitude of motion, while energy is a

number which describes the ability of a particle to push against forces (do 'work').

In classical mechanics, objects stay in constant motion (with a constant velocity and

momentum) unless they are acted upon by a force. Forces acting on a particle change the

momentum of a particle, and hence change its motion:

F dt dp =

Common forces include gravity, and electric forces.

• Electrostatic: ! !!

!!!!

!!!!

!!

• Gravitational: ! ! !!!!!

!!

where r is the distance between the objects, qi is the charge of particle i and mi is the mass of

particle i.

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In particular, classical motion obeys Newton's Law:

ma F = ,

Computation of the motion of a complicated system can be difficult, so often we can use the

more general conservation laws to aid us in our calculations:

• Conservation of energy ( E ): total energy in an isolated physical system is constant

over time.

• Conservation of momentum ( p): in a closed system of objects, with no external

forces, momentum is constant over time.

Waves

A wave is a form of energy which is spread out and transmitted through space - a disturbance

or oscillation which propagates through time. There are many different kinds of waves that

we are familiar with:

- Mechanical waves involve deformation or vibration of substances (water waves, sound

waves, etc.)

- Electromagnetic waves (light, X-rays, radio waves, etc.) are the oscillation of electric and

magnetic fields which can carry energy through a vacuum.

Waves are characterised by a frequency ( f , number of cycles per unit time), a wavelength (! ,

the length from crest to crest), an amplitude ( A, the height of the oscillations) and a speed (v =

f ! ) which relate directly to their properties. Waves are subject to many special phenomena,

including diffraction, reflection, refraction and interference, all of which will be discussed at

the Summer School.

Particle

Mechanics

Forces cause

changes in

momentum, and

hence determine

the motion ofobjects.

Potential Energy

is associated with

the location of a

particle (its

interaction withforces).

Kinetic Energy

is associated

with the motion

of a particle.

Momentum

measures how

much an object

is moving

(taking intoaccount mass).

dx

dV t x F !=),( )( xV

2

2

1mv K = mv p =

Properties

of waves

The speed of a wave, v,

is related to frequency

and wavelength.

Frequency counts the number

of oscillations of a wave per

unit time.

Wavelength is

the spatial

distance covered

by a wave in

one oscillation.

v =f ! f !

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4.2 Electronic Structure

It is only since 1911 that chemists have known that atoms consist of electrons orbiting a tiny,

positively charged nucleus. The orbits of the electron are similar to planetary orbits, and can

be labeled by their average distance, angle of inclination etc. However, the laws of quantum

mechanics imply that only a handful of the infinite possible orbits are allowed. We call thesequantised orbits “orbitals” and label them with each of the following:

• Shell Number:

which can be any of 1,2,3,4 … in order of increasing distance from the nucleus.

• Subshell Label:

which can be s, p, d, f, g, h … The nth shell has n subshells, e.g. the second shell has

two subshells: 2s and 2p.

• Direction Indicator:

The “s” subshell has no particular direction and contains only one orbital, whereas the

“p” subshell contains three orbitals (px, py, pz). The “d” subshell contains fiveorbitals; the “f” subshell contains seven and so on.

• Each orbital can hold two electrons, so long as they have opposite spin to each

other. However, it is slightly more energetically advantageous to have electrons

unpaired rather than paired.

Electrons fill the orbitals so as to minimise total energy: the electrons fill the lowest shells

first. We call this configuration the ground state.

The main contribution to the energy is the distance from the nucleus, so the shells are filled in

the order 1 < 2 < 3 < 4.

Next, the subshells are filled in the order: s < p < d < f, but there is a catch. The 4s subshell is

actually lower in energy than the 3d subshell! Likewise, the 5s subshell gets filled before the

4d, and the 5p and 6s subshells both get filled before the 4f.

s p d

5

! ! ! " "

4

! d xy d xz d yz d z2 d x2-y2

! ! ! 3d

3

! p x p y p z

s ! ! !

!

!

h

l

Subshel l

Here is a helpful mnemonic for

remembering the order in which the

orbitals are filled. Write out all the

orbitals as shown, then read them

diagonally starting from the top right

and going to the bottom left.

1s 1s

2s 2p 2s 2p3s 3p3d 3s 3p3d

4s 4p4d4f 4s 4p4d4f

5s 5p5d5f 5s 5p5d5f

6s 6p6d 6s 6p6d

7s 7p 7s 7p

The order of filling orbitals is 1s, 2s,

2p, 3s, 3p, 4s and so on…

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Example The electron configuration of Nickel is 1s22s

22p

63s

23p

64s

23d

8

(as shown in the diagram above).

Problem Set 4:

3.1. Write the ground state electron configuration fora. Fe

b. F –

c. Hg.

3.2. Which neutral elements would have the following ground state electron

configurations

a. 1s22s22p4

b. 1s22s

22p

63s

23p

64s

23d

3

c. 1s22s

22p

63s

23p

64s

23d

104p

1

3.3. Can you work out a formula (in terms of n) for the number of orbitals in the nth shell?

What about the maximum number of electrons allowed in the nth

shell?

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5. STOICHIOMETRY AND LABORATORY

CALCULATONS

At this stage, it is expected you would have a clear understanding of basic stoichiometryincluding use of the formulae:

(g/mol)MassMolar

(g)MassMoles =

(L)Volume(mol/L)ionConcentratMoles !=

Students are also required to perform calculations with correct use of significant figures. (see

Section 1.1)

A recommended revision text is:

Silberberg , Chemistry The Molecular Nature of Matter and Change, Chapter 3:

“Stoichiometry” & Chapter 4: “The Major Classes of Chemical Reactions”.

5.1 Empirical Formula

The empirical formula is given as the simplest whole number ratio of elements in a

compound.

Examples:

Consider acetic acid: CH3COOH. Its empirical formula is C1H2O1 or CH2O.

The empirical formula of Methane CH4 is the same as its molecular formula.

Many ionic compounds can only be expressed in terms of its empirical formula, such as table

salt NaCl.

Determining Empirical Formula:

The composition of a compound of interest may be expressed in mass or as a percentage

mass. As we are only interested in the ratio of elements, both can be treated the same:

Example: Determine the empirical formula of Vitamin C, which has the following

composition: 40.91% Carbon, 4.59% Hydrogen and 54.50% Oxygen.

Step 1. Determine the number of moles per element:

Assume that 100g of the sample is present when percentages are given.

Divide the mass of each element by its molar mass to determine the mole ratio:

C:40.91 g

12.01 g mol –1

=3.406 mol H:4.59 g

1.008 g mol –1=4.55 mol

O:54.50 g

16.00 g mol –1=3.406 mol

Our current preliminary formula is C3.406H4.59O3.406.

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Step 2. Convert to integer ratio:

Divide each by the smallest number of moles present for an element, or by the smallest

subscript. Here, both C and O are present in the least amount.

C: 1=406.3

406.3 H: 34.1=

406.3

55.4 O: 1=

406.3

406.3

Often integer ratios appear at this step. If this is not the case, we multiply to obtain whole

numbers: here, we multiply by 3:

C: 1!3= 3 H: 1.34!3= 4.01! 4 O: 1!3= 3

Thus, the empirical formula of Vitamin C is C3H4O3.

Problem Set 5.1:

4.1.1. Determine the empirical formula of an oxide of nitrogen containing 30.45% N and

69.55% O by mass.

4.1.2. A sample contains 0.80 g C; 0.134 g H; 1.066 g O. Suggest a molecular formula if

its molecular weight is 180.16 g mol –1

4.1.3. A 1.30 g sample of a unknown hydrocarbon undergoes complete combustion (in

oxygen) to produce 4.40 g of CO2 and 2.70 g of H2O. Suggest a molecular formula

if its molecular weight is 30.0 g mol –1

.

5.2 Laboratory Calculations

The aim of this section is to illustrate the expected layout when writing laboratory

calculations.

Standard Titrations:

In the standard titration , the concentration of a solution is determined by reaction with a set

volume of a solution with known concentration.

Example: The concentration of an unknown sodium hydroxide was determined by titration

with 20.00 mL of approximately 0.2 mol L –1

hydrochloric acid which was standardised with

sodium carbonate.

a. The standard was made by dissolving 5.617 g in 500.0 mL. Determine its molarity.

Moles (Na2CO3) = Mass/ Molar Mass

n(Na2CO3) = 5.617 g / 105.99 g mol –1

n(Na2CO3) = 0.05300…mol (Don’t Round Off)

Concentration (Na2CO3) = n (Na2CO3)/ Volume

[Na2CO3] = 0.05300 mol /0.500 L ( [ ] indicates concentration)

[Na2CO3] = 0.1060 mol L –1

(Four sig. figs)

b. Titration of 20.00 mL of Na2CO3 with HCl gave an average titre of 19.71 mL.

Determine the concentration of the acid.

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2H+

(aq) + CO32-

(aq) ! CO2 (g) +H2O(l) (Write a balanced equation)

n(Na2CO3) = 0.1060 mol L –1

" 0.02000 L

= 0.002120 mol L –1

n(HCl) = 2 x n(Na2CO3)

= 0.004240 mol

[HCl] = 0.004240 mol / 0.01971 L= 0.2151 mol L –1

(Again, 4 sig figs)

c. Below are burette readings in the titration with sodium hydroxide. Determine the

volumes of the NaOH titres and calculate its concentration.

V(NaOH)

(mL)

Titre 1 Titre 2 Titre 3 Titre 4

Initial Volume 0.34 0.23 0.45 0.34

Final Volume 27.69 27.57 27.59 27.70

Titre 27.35 27.34 27.14 27.36

Average 27.35 mL

(27.14mL titre is an outlier and not included in the results)

HCl + NaOH ! H2O + NaCl

n(HCl) = 0.02000 L " 0.2151 mol L –1

= 0.004302 mol

n(HCl) = n(NaOH)

[NaOH] = 0.004302 mol / 0.02735 L

= 0.1573 mol L –1

Back Titration:

Sometimes a direct titration is not possible. In a Back Titration the sample is reacted with an

excess amount of reagent with volume and concentration known. The unreacted reagent is

then titrated with a known solution, allowing for compositional analysis of the initial sample.

Note that the limiting reagent runs out before the excess reagent does: i.e. The maximum

number of moles of product is equivalent to the number of moles of limiting reagent.

Exampl e: The calcium carbonate content in rock was analysed by treating the sample with

15.00 mL of 0.912 mol L –1

hydrochloric acid. The sample was back-titrated with

17.77 mL of 0.130 mol L –1

NaOH. Determine the mass of calcium carbonate in the

sample.

Total Amount of HCl

Excess HCl reacted with NaOH

Amount of HCl used by

CaCO3

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39

HCl XS + NaOH! H2O + NaCl (Equation for back titration)

n(NaOH) = 0.130 mol L –1

" 0.01777 L

= 0.00231 mol

n(HCl XS) = 0.00231 mol

n(HCl total) = 0.9120 mol L

–1

"

0.01500 L= 0.01368 mol

n(HCl used) = n(HCl total) – n(HCl XS)

= 0.01368 mol – 0.00231 mol

= 0.0114 mol

2HCl + CaCO3 ! CO2 + H2O + CaCl2 (Equation for HCl reaction with rock)

n(CaCO3) = # n(HCl)

= 0.00569 mol

m(CaCO3) = 0.00569 mol x 100.07 g mol –1

= 0.569 g

Yield:

The theoretical yield is the expected mass of a product in a reaction assuming 100%

conversion has occurred .

Example: a. Determine the theoretical yield of ethyl acetate in the following reaction, given

the following data

OHHO

O

+

O

O

+ H2O

Ethanol Acetic Acid Ethyl AcetateVolume Used

(mL)

5 5

Density (g mL –1

) 0.789 1.044 0.897

Molecular

Weight (g mol –1

)

46.07 60.05 88.105

Mass(ethanol) = Density x Volume

= 5 mL " 0.789 g mL –1

= 3.95 g

n (ethanol) = 3.945 / 46.07

= 0.0856 mol

Mass (acetic acid) = 5 mL x 1.044 g mL –1

= 5.22 g

n (acetic acid) = 5.22 g / 60.05 g mol –1

= 0.0869 mol

Ethanol is the limiting reagent and acetic acid is in excess.

Theoretical Yield = Masscalc (ethyl acetate) (Masscalc is the theoretical mass)

= 0.0856 mol " 88.105 g mol –1

= 7.54 g

a. Determine the percentage yield if 5 mL of ethyl acetate was obtained.

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42

b. Using a direct integral instead: ! ! "=3

0

][

][

2

3

0

][][

1dt k Ad

A

A

A

becomes:

)03(][

1

][

1

03

!!=

!

!

!

k AA

which gives the answer.

1.4.1

a. -1

b. 1

c. !

d. !!

e. !! !

f. !!"!

g. !!!

h. !!! !"

i. !!

!

! !"

!

!

1.4.2

a. !" ! !!

b. !! ! !!

c. !!! !

d. !! ! !!

e. !!" ! !!

f. !! ! !! !

g.!

!!

!

!!

h. ! !

!!

!

!!

i. !!!

!

!!

1.4.3

a. !!

b. !

c.!

!!

!

!!

d. !!

!!

!

!!

Section 2: Organic Nomenclature

2.1 . The compounds are:

a. 2-methylpentane

b. 2-methylpent-1-ene or 2-methyl-1-pentene

c. 1,4-dimethylcyclohexane

d. 3-methylheptane

e. hexan-3-one or 3-hexanone

f. 4-hydroxypentan-2-one or 4-hydroxy-2-pentanone

g. hex-1-en-3-one

h. o-dichlorobenzene or 1,2-dichlorobenzene

i. o-aminophenol or 2-aminophenol

j. 3-oxobutanoic acid

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2.2. 3,6-dimethyl-4-chlorooctane

Section 3: Transition Metal Chemistry

1. a) triammineaquadichloridochromium(III) nitrate, [CrCl2(NH3)3(OH2)]NO3

b) sodium aquacarbonyldihydroxidooxalatomanaganate, Na2[Mn(CO)(OH)2(OH2)(ox)]

2. a) [Fe(bipy)2(OH2)2]Cl2 b) [PtCl2(NH3)2]

3. a) chlorotris(triphenylphosphine)rhodium(I) b) bis(glycinato)zinc(II)

Section 4: Electronic Configuration

4.1 The electron configurations are

a. Fe 1s22s

22p

63s

23p

64s

23d

6

b. F- 1s

22s

22p

6 (note the extra electron)

c. Hg 1s22s

22p

63s

23p

64s

23d

104p

65s

24d

105p

66s

24f

145d

10

(or [Xe] 4f 14

5d10

6s2)

4.2 The elements are (a, Oxygen) (b, Vanadium) (c, Gallium)

4.3 The number of orbitals in the nth shell is the sum of the first n odd numbers, 1 + 3 + 5

+ … + (2n – 1) = n2. The maximum number of electrons is therefore 2 n2.

Section 5: Stoichiometry and Laboratory Calculations

5.1.1. NO2

5.1.2. C6H12O6

5.1.3. C2H6

5.2.1. 0.609M

5.2.2. 0.618g

5.2.3. 40.4%

Fe

N

NN

OH2

N

OH2

2+

Cl-

Cl-

Pt

Cl

Cl NH3

NH3

Rh PPh3

PPh3

Cl

PPh3

NH

Zn

O

O

NH

O

O

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7. ACKOWLEDGEMENTS

These notes were written based on the previous notes compiled by Elizabeth New and

Anthony Phillips. The material in these notes has been collated by Alex Wong, Tristan

Reekie, Dustin Stuart, Anne Trinh, Shannon He, Nicholas Malouf, Trent Wallis and SamSalman. A special thanks is extended to William Jackson for his careful editing.

2015 prep notes

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