Homework and test date

Please finish the questions: "The Ideal Way" and "Some Ideal Questions"

You need to revise for your test on Quantative chemistry on Thursday 6th Oct.

Notes on The Gas Laws

Boyle’s Law:     Pressure is inversely proportional to the volume   P x V = constant

                          P₁V₁=P₂V₂ 

You need to know the graphs, pressure against volume and pressure against (1/V)

Charle’s Law:    Volume is proportional to the temperature. V/T = constant

                            V₁/T₁ = V₂/T₂

                                      You need to know the graph of Volume against temperature (⁰C) and how it can be    

                             extrapolated to give a value of absolute zero. Should be -273 ^oC (0K)

The Ideal Gas Law and the gas constant:

The volume of a gas depends on:

·         The temperature

·         The number of moles

·         The pressure

The volume of the gas particles themselves is insignificant compared to the space between them. Therefore, it doesn’t matter which gas you are referring to, one mole of gas at a fixed temperature and pressure will occupy the same volume. One mole of any gas at 1atm and 273K will occupy 22.4 dm³ (litres).

So the gas constant for one mole of a gas under standard conditions (STP, 1atm, 273K) can be easily worked out. However, its value depends on if you use atmosphere (atm) for pressure or pascals (Pa) or KPa.

For 1atm, 273 K. Note: if you are using atm use dm³  as your volume units

PV/T = constant        = 1 x 22.4/273   = 0.082 atm dm³ k^-1 mol^-1

1 atm = 101317 Pa  Note: 1Pa = 1 N/m². Remember pressure is force/area. So you have to use m³ for the units of volume if you are using pascals and dm³   if you are using kPa.

PV/T = constant     = (101325 x 0.0224)/273  = 8.314 Pa m³ k^-1 mol^-1   

Note: 1 N/m = J  so it can be written as: 8.314 J K^-1 mol^-1

So for n moles PV/T = nR    where R is the gas constant.

So PV = nRT

Unit conversions:

1atm = 101325Pa = 101.325kPa = 760 mm Hg

Homework

Homework PG

Complete the uncertainty and % error calculations for your titration experiment.

Make sure that you have tried the virtual labs on Boyle’s and Charle’s law. Write notes about these gas laws

Boyle's Law

Charle's Law

Error, Uncertainty and Significant figures

Error and uncertainty in practical work

The error is the difference between the result obtained and the generally accepted 'correct' result found in the data book or other literature. If the 'correct' result is available it should be recorded and the percentage error calculated and commented upon in your conclusion. Without the 'correct ' value no useful comment on the error can be made.

The percentage error is equal to:

the difference between the value obtained and the literature value    x 100
                              the literature value

Uncertainty occurs due to the limitations of the apparatus itself and the taking of readings from scientific apparatus. For example during a titration there are generally four separate pieces of apparatus, each of which contributes to the uncertainty.

When making a single measurement with a piece of apparatus then the absolute uncertainty and the percentage uncertainty can both be stated relatively easily. For example consider measuring 25.0 cm³ with a 25 cm³ pipette which measures to + 0.1 cm³. The absolute uncertainty is 0.1 cm³ and the percentage uncertainty is equal to:

                        0.1  x 100   =  0.4%

                       25.0

If two volumes or two masses are simply added or subtracted then the absolute uncertainties are added. For example suppose two volumes of 25.0 cm³ + 0.1 cm³ are added. In one extreme case the first volume could be 24.9 cm³ and the second volume 24.9 cm³ which would give a total volume of 48.8 cm³. Alternatively the first volume might have been 25.1 cm³ which when added to a second volume of 25.1 cm³ gives a total volume of 50.2 cm³. The final answer therefore can be quoted between 48.8 cm³ and 50.2 cm³, that is, 50.0 cm³ + 0.2 cm³.

 

When using multiplication, division or powers then percentage uncertainties should be used during the calculation and then converted back into an absolute uncertainty when the final result is presented. For example, during a titration there are generally four separate pieces of apparatus, each of which contributes to the uncertainty.

e.g. when using a balance that weighs to + 0.001 g the uncertainty in weighing 2.500 g will equal

                               0.001 x 100 = 0.04%

                               2.500

Similarly a pipette measures 25.00 cm³ + 0.04 cm³.

The uncertainty due to the pipette is thus 0.04 x 100 = 0.16%

                                                                  25.00

Assuming the uncertainty due to the burette and the volumetric flask is 0.50% and 0.10% respectively the overall uncertainty is obtained by summing all the individual uncertainties:

Overall uncertainty = 0.04 + 0.16 + 0.50 + 0.10 = 0.80% ~ 1.0%

Hence if the answer is 1.87 mol dm^-3 the uncertainty is 1.0% or 0.0187 mol dm^-3

The answer should be given as 1.87 + 0.02 mol dm^-3.

 

If the generally accepted ‘correct’ value (obtained from the data book or other literature) is known then the total error in the result is the difference between the literature value and the experimental value divided by the literature value expressed as a percentage. For example, if the ‘correct’ concentration for the concentration determined above is 1.90 mol dm^-3 then:

the total error = (1.90 – 1.87)  x 100  = 1.6%.

                             1.9

Significant figures

Whenever a measurement of a physical quantity is taken there will be uncertainty in the reading. The measurement quoted should include the first figure that is uncertain. This should include zero if necessary. Thus a reading of 25.30^oC indicates that the temperature was taken with a thermometer that is accurate to + or - 0.01^oC. If a thermometer accurate to only + or - 0.1^oC was used the temperature should be recorded as 25.3^oC.

Zero can cause problems when determining the number of significant figures. Essentially zero only becomes significant when it comes after a non-zero digit (1,2,3,4,5,6,7,8,9).

000123.4        0.0001234                                1.0234         1.2340

              zero not a significant figure               zero is a significant figure

              values quoted to 4 sig. figs.              values quoted to 5 sig. figs.

Zeros after a non-zero digit but before the decimal point may or may not be significant depending on how the measurement was made. For example 123 000 might mean exactly one hundred and twenty three thousand or one hundred and twenty three thousand to the nearest thousand. This problem can be neatly overcome by using scientific notation.

            1.23000 x 10⁶  quoted to six significant figures

            1.23 x 10⁶        quoted to three significant figures.

Calculations.

1. When adding or subtracting it is the number of decimal places that is important.

e.g.      7.10 g          +          3.10 g          =         10.20 g

          3 sig. figs.                3 sig. figs.                4 sig. figs.

This answer can be quoted to four significant figures since the balance used in both cases was accurate to + or - .01g.

2.  When multiplying or dividing it is the number of significant figures that is important. The number with the least number of significant figures used in the calculation determines how many significant figures should be used when quoting the answer.

e.g. When the temperature of 0.125 kg of water is increased by 7.2^oC the heat required =

       0.125 kg  x  7.2^oC  x 4.18 kJ kg^{-1 o}C^-1 = 3.762 kJ.

Since the temperature was only recorded to two significant figures the answer should strictly be given as 3.8 kJ.

In practice the IB does not tend to penalise in exams if the number of significant figures in an answer differs by one from the correct number (unless the question specifically asks for them) but will penalise if they are grossly wrong.

 Source: Old file, however I think it's from the Geoff Nuess book, Chemistry Companion

Homework

Homework for Tuesday:

Write up the lab

Method – Use the past tense. You should include the names of the apparatus that you used: Burette, volumetric pipette (25.0 cm3), burette, phenolphthalein indicator solution etc

Results- Use a clear table with units and uncertainties

Calculations- Show the step by step calculations in order to calculate the concentration of the unknown

Remember the step by step guide:

Step1. Write a balanced equation

Step2. Calculate the number of moles in the solution with a known concentration (the standard solution)

Step3. Deduce from the ratio in the equation the number of moles in the solution with unknown concentration. The sulphuric acid in this case

Step4. Calculate the concentration of the unknown

Step5. Remember that the solution of unknown concentration was diluted by a factor of four. Therefore, you need to multiply by four to get the concentration of the original acid solution

Evaluation

You need to list the sources of error and evaluate each one in terms of how it could have affected your result

You need to suggest realistic improvements that relate to the weaknesses that you have identified

The Gas Laws

Here is a good site where you can carry out virtual experiments in order to find out the relationship between pressure and volume and between volume and temperature. You can put the data in excel and plot the appropriate graphs: pressure against 1/V  and Volume against T (kelvin). To see the relationships.

The Gas laws virtual site

This is also a good site
Gas Laws

Homework

The first four questions from the concentrate on concentration package.

Moles and titrations

Moles and titration questions

A titration is a means by which an acid or base of known concentration and volume is reacted with a base or acid of unknown concentration. An indicator is used to tell us when the reaction is complete so that we know the volume that is required and the concentration can be calculated.

You have to be able to calculate concentrations and to be able to use moles effectively

Remember:   The number of moles  =  the mass / the mass of one mole

A molar  solution is a solution with one mole dissolved in a litre (mol/L). Usually we use mol dm^-3

E.g. a 2.0M solution means 2.0 moles per litre i.e. 2.0 moles in a 1000cm³  (2.0 mol dm^-3)

If you know the volume and the concentration you can work out the number of moles by:

Number of moles = concentration  x  volume in litres

E.g. 20.0 cm³ of a 0.25 mol dm^-3 solution has 0.25 x 0.020 = 5 x 10^-3 moles

By looking at the mole ratio in the equation you can see how many moles of the reactants and products you have, for example if there are 5.0 x10^-3   moles of sodium hydroxide how many moles of sulphuric acid would react with it?  To do this you need to write down a balanced equation

2NaOH  +  H₂SO4  →  Na₂SO₄ + 2H₂0

Therefore the number of moles of sulphuric acid that would react with 5.0 x 10^-3  moles of sodium hydroxide is 5.0 x 10^-3 / 2   = 2.5 x 10^-3

To work out concentrations you simply use:

concentration = number moles/ volume in litres

Remember 1 litre is 1,000 cm³

Therefore if there are 2.5 x10^-3 moles sulphuric acid in 40cm³ of solution then the concentration = 2.5 x10^-3  / 0.04 = 0.0625 mol dm^-3

Examples

1.   If 15.0 cm³  of 2M hydrochloric acid reacted exactly with 40.0 cm³ of potassium hydroxide. Calculate the concentration of the alkali

Remember there are four simple steps:

• Write a balanced equation
• Calculate the number of moles of the solution with a known concentration
• Determine the number of moles of the unknown solution by using the mole ratio
• Calculate the concentration of the unknown solution

                               c=n/v        n=cxv         v =n/c   

  

The volume must be in litres. If it’s in cm³   then divide by a 1,000

2.  If 25.0 cm³  of  0.50M sulphuric acid reacted with exactly 12.5 cm³  of  sodium hydroxide, calculate the concentration of the alkali.

Experiment to determine the formula of magnesium oxide

You can put your knowledge of moles to work by carrying out an experiment using magnesium ribbon in order to determine the formula of the oxide. Make sure that you follow all the safety procedures for examples wear goggles when using the Bunsen Burner. In addition make sure that you accurately record your data (with inaccuracies) and follow the criteria indicated on the link to practical assessment. You will need to hand in your raw data, processed data and conclusions & evaluations.

Assessment Criterea

Relative Atomic Mass

This is the weighted mean of all the naturally occurring isotopes of an element, relative to 1/12 of the mass of carbon 12.

The mass number is equal to the number of protons and neutrons in the nucleus of a particular atom. Remember that protons and neutrons make up almost all of the mass of an atom. The relative mass of an electron is approximately 1/2000 of that of a proton or neutron and so the mass that they contribute is negligible.

What is the actual mass of a proton?

Homework

Finish the three pages on naming that you have started.

The Mole

A mole is the amount of pure substance containing the same number of chemical units as there are atoms in exactly 12 grams of carbon-12 (i.e., 6.023 X 10²³). This allows us to count atoms, molecules and ions and therefore, allows us to be to use the correct amounts for a reaction. Remember that a balanced equation equation is a ratio in terms of moles so that the number of atoms on the left hand side of the equation is the same as the right hand side. In a reaction the atoms are not created or destroyed, they are just rearranged, old bonds are broken and new ones made.

Common ion names

Monatomic anions:

Cl⁻ chloride

S²⁻ sulfide

P³⁻ phosphide

Polyatomic ions:

NH₄⁺ ammonium

H₃O⁺ hydronium

NO₃⁻ nitrate

NO₂⁻ nitrite

ClO⁻ hypochlorite

ClO₂⁻ chlorite

ClO₃⁻ chlorate

ClO₄⁻ perchlorate

SO₃²⁻ sulfite

SO₄²⁻ sulfate

HSO₃⁻ hydrogen sulfite (or bisulfite)

HCO₃⁻ hydrogen carbonate (or bicarbonate)

CO₃²⁻ carbonate

PO₄³⁻ phosphate

HPO₄²⁻ hydrogen phosphate

H₂PO₄⁻ dihydrogen phosphate

CrO₄²⁻ chromate

Cr₂O₇²⁻ dichromate

BO₃³⁻ borate

AsO₄³⁻ arsenate

C₂O₄²⁻ oxalate

CN⁻ cyanide

SCN⁻ thiocyanate

MnO₄⁻ permanganate

From Wikipedia

Homework

Write down something about the influence of chemistry over the last century and why an understanding of chemistry is essential in order to solve the World's future problems.

Welcome to IB Chemistry

This is a demanding and complete course which requires hard work and commitment. The rewards from this are not only an excellent understanding of chemistry and its application to the world around us but also an extremely well recognized course for universities worldwide.

The course is assessed by exams (multiple choice, structured questions, options        questions and practical work. All details of the course can be found on the web: http://ibchem.com/IB/ibsyllabus-2009.htm

Examination action verbs

As with all examinations, it is important to know what the examiners require. The IBO releases a list of key words and what they mean when referring to exam questions.

For example the word "Draw" is taken to mean "represent by a series of pencil lines and labels (unless specifically told not to do so)" This leaves the students clear as to the actual requirements of the questions. A full list of these so-called action verbs can be found here

---

Practical (Internal) Assessment

Criterion	marks available	marks
Design	two grades out of 6 are to be presented	12 marks
Data collection and processing	two grades out of 6 are to be presented	12 marks
Conclusions and Evaluation	two grades out of 6 are to be presented	12 marks
Manipulative skills	1 grade out of 6 will be presented	6 marks
Personal skills	1 grade out of 6 will be presented	6 marks
	Total =	48 marks

Every year a sample of the internal assment work is sent by the teacher in charge for external moderation. In this process a third part (the moderator) tries to decide whether the teacher has applied the criteria correctly in his/her assessment. More about moderation.

Final grades

The final grade awarded for an IB subject is from 1-7. The schools receive a breakdown of the grade achieved in each part of the exam.

The requirements to achieve the IB diploma are fairly complex and may be found here: IB diploma award requirements

Previous Knowledge and Understanding

Background Knowledge in Chemistry

A lot of you have come from different backgrounds and have done different science courses. Therefore, it would be very helpful for me to know what you have studied before. Please complete the table below and honestly indicate the level in each topic. For each of the topics that you have indicated that you have some idea or completely understand, please give me some key points (underneath) of what you know. Thanks MrO

Topic	Not Covered	Have some idea	Completely understand
Stoichiometry/Moles
Atomic Theory
Periodicity
Bonding
Energetics
Kinetics
Equilibrium
Acids and Bases
Oxidation and Reduction
Organic Chemistry

Topic                                                                      Key Points that I know and understand