Great works are performed not by strength by perseverance.

- Samuel Johnson-

CHEMISTRY UNIT 3!

CLASS 3 – 04/02/08

OTHER TYPES OF QUESTIONS:

Calculate the grams analyte of the precipitate for the following: i) P in Ag3PO4 ii) Bi2S3 in BaSO4

Trick: Look for the common element. Make sure that the elements are equivalent in number!!!

ANOTHER EXAMPLE:

Phosphate in a 0.2711 g sample was precipitated giving 1.1682 g of (NH4)2PO4.12 MoO3

i) Find percentage P ii) Find the percentage P2O5 in the sample.

YET ANOTHER:

Manganese in a 1.52 g sample was precipitated as Mn3O4 weighing 0.126 g.

i) Find percentage Mn2O3 ii) Find percentage of Mn in the sample

AND ANOTHER……

A mixture containing only FeCl3 and weighs 5.95 g. The chlorides are converted to hydroxides and ignited to Fe2O3 and Al2O3 . The oxide mixture weighs 2.26 g. Calculate:i) the percentage Fe ii) the % of Al in the sample.

A 0.4960 g sample of a CaCO3 is dissolved in an acidic solution. The calcium is precipitated as CaC2O4 .H2O and the dry precipitate is found to weigh 0.6186 g.

What is the percentage of CaO in the sample?

GRAVIMETRIC ANALYSIS – EXTRA THINGS!

VERY accurate process if done correctly (helped determine atomic weight!)

Things to do to get 100% of precipitate;

- Choose compound that precipitates to large crystals that are easy to collect with the filter paper not small

- NB: the LARGER the crystal the better...

WHY ARE LARGE CRYSTALS BETTER?

Because they have a LOWER surface area therefore they cannot absorb contaminants.

The will be easily collected with filter

Ok... but how do we make large crystals?

HOW TO MAKE LARGE CRYSTALS...

Stir solution and add reactants SLOWLY Make the solution hot to increase collision

rate Keep at low pH The solutions should be diluted – the excess

added isn’t meant to be ½ a jar but a few drops.

NOTE: Adding too much excess of reactant can also increase the contamination as the free ions will interact!

WAYS TO IMPROVE YOUR GRAVIMETRIC ANALYSIS!

Use distilled water (deionized) this way it will prevent contamination of the surface of your precipitate.

Filter in a vacuum – a vacuum will prevent the filter paper from raising and your precipitate being lost.

Use a precipitate that absorbs LESS impurities such as barium sulfate, AgCl, BaCrO4 and Fe2O3

ERRORS IN GRAVIMETRIC ANALYSIS AND THEIR CONSEQUENCES...

Not enough precipitate formed; underestimate

Heat solution, stir, low pH

Precipitate too soluble: underestimate Choose a better precipitate that is easier to

...collect ie AgCl

UNDERESTIMATE

GENERAL GAS EQUATION:

The ideal gas law is PV = nRT

P – pressure in kPaV – Volume in Ln – moleR – constant (8.314J/mol.K)T – Temperature (kelvin)

How to get Kelvin?

Kelvin = º C + 273

ERRORS IN GRAVIMETRIC ANALYSIS AND THEIR CONSEQUENCES...

ContaminationOverestimate

Some chemists do an extra reaction to get rid of known contaminant ions – follow ideal conditions

Not dry enough: Overestimate Water is a heavy molecule that will affect the

weight

OVERESTIMATE

ASSUMPTIONS IN IDEAL GAS LAW...

Just because its a formula doesn’t mean its perfect.

It is only a generalization!

ASSUMPTIONS (8.314 for constant can change)

1) Particles are independent: This law does NOT take into account attraction between individual gas particles2) Small number of particles and small sizes of molecules: Otherwise the distances between particles are different and the energy of collision is affected.

Careful: They don’t generally cover this at school but it can come up!!!

MORE ASSUMPTIONS

Elastic CONDITIONS: This means that no energy is lost in the collision. If it is inelastic then the collision energy is converted to a different form

i.e. Light, heat Temperature is high: If it is too low then

there will be a GREATER PROBABILITY of interaction between molecules (i.e. Attraction)

Please note: Pressure = force/area.

ASSUMPTION EXPLANATION FLOW CHART:

KINETIC ENERGY

Kinetic Energy of a particle: You can also be asked about this... The formula for this is simply:

KE = ½ x mass x velocity² or KE = ½mv²

PRESSURE AND CONVERSIONS:

The pressure unit that the ideal gas law needs is kPa

You don’t always get kPa in the question LEARN TO CONVERT!!!

1 atm = 101.325 kPa 1 atm = 760 torr = 760 mmHg

Tip: The BEST way to do conversions is turn into atm...and then work from there.

GAS LAW EXAMPLES:

Charlie was blowing up a balloon for his sister. He blew about 5L of carbon dioxide into it. Then he measured the pressure which was exactly 3.72 atm. The temperature was a pleasant 42ºC.

How many mol. of carbon dioxide did he need to sacrifice?

How many CO2 molecules?

Calculate the number of oxygen atoms that are in the balloon.

OTHER UNITS OF VOLUME...

Your not only going to get units like L or mL for volume.

‘Old fashion’ units 1 cm³ = 1 mL

1 dm³ = 1 L 1 m³ = 1000 L

MORE GAS EQUATIONS:

Before we said....a mole has a mass. A mole can also have volume.

There are only two different types of conditions:

- STP - SLC

Chemist can choose to work at SLC or STP.

They can’t just choose random condition as the temperature will affect the calculations.

VOLUME SLC

Standard conditions All lab work done at 25ºC 1 atm. of pressure 1 mole of gas at these condition occupy

24.5L!

VSLC = 24.5n

This means that when you some mol. Of substance you can sub it and you get the amount it occupies!

VOLUME AT STP

Conditions: 1 atm 22.5 L for 1 mole 0 ºC

VSTP = 22.5n

This is less popular in science as it is not practical.

Sort of like this...

IMPORTANT NOTE!!!!!!!!!!!!!!!!!!!!

Remember how we just covered the assumptions of the ‘ideal gas law’?

We weren’t suppose to do experiments in cold temperatures as this slows down the ‘elasticity’

Right?

NO... This is different...here WE KNOW the temperature and the formulas have been made based on temp... before we USED A CONSTANT!

EXAMPLES...

What volume will 20.0 g of Argon occupy at STP?

HARDER... SIMILAR TO VCAA

3i) 0.190 g of a gas occupies 250.0 mL at STP.ii) What is its molar mass? iii) What gas is it?

THE COMBINED GAS LAW

Sometimes we have initial conditions and then these are changed.

Therefore we need to use TWO EQUATIONS.

Generally PV = nRT

If you know that something will be constant then make the INTIAL reaction equal the FINAL reaction WITH THE FACTOR THAT IS CONSTANT!!!!

FOR EXAMPLE:

If I initially have a gas at a pressure of 12 atm a volume of 23 L and a temperature of 200K and I raise the pressure to 14 atm and the temperature to 300K. What is the new volume of the gas?

PRACTICE:

YOU MOVE IN THE DIRECTION OF YOUR DOMINANT THOUGHTS

Want to succeed in VCE?

Then STOP worrying about being ‘not good enough’ in an area... Instead know that as long as you work hard you WILL be rewarded!

EMPIRICAL FORMULA

It is a chemical formula in the simplest ratio.

Ie/

Glucose’s [C6H12O6 ] empirical formula is CH2O

Whenever you work out a chemical equation you end up getting the EMPIRICAL FORMULAR.

WAY TO WORK OUT THE EMPIRICAL FORMULA: 1) If you are given a % assume that you

have 100g.

Ie/ A compound X is made up of 54% of carbon, 40% hydrogen and 6% chlorine...

Therefore the compound is made of 54g of carbons, 40g of hydrogen and 6 gram of chlorine.

2) Convert your masses into a mole value by n = m/Mr

MORE ON EMPIRICAL FORMULA 3) Divide all the values by the

SMALLEST mole value.

4) Convert all the values into whole numbers.

5) These values are the ratio of your elements in the chemical formula.

Ie/ C : H : O x : y : zCXHYOZ

Whole numbers

AN EXAMPLE:

What is the empirical formula for a compound if an 8.1 g sample contains 4.9 g of magnesium and 3.2 g of oxygen?

MOLECULAR FORMULA

When we get the empirical formula this does NOT always give us the actual formula.

Ie/ If we were to do glucose we would get CH2O as the empirical formula. This tells us absolutely nothing – we wouldn’t know that it is glucose!

Sometimes in the question you are given the MOLECULAR FORMULA. This tells you the molar mass of the actual unknown compound.

MOLECULAR FORMULA:

When you are given the molecular formula you must:

Molecular formula / Empirical (Mr)

Then using the ratio above you multiply the atoms by that number!

EMPIRICAL FORMULAS...DON’T HAVE TO BE WHAT YOU EXPECT...

You may have to use empirical formulas for:

Hydrates Compounds which do not directly give you % Compounds that have been in some way

manipulated. Proteins

HARDER QUESTIONS:

A 3.245 g sample of titanium chloride was reduced with sodium to titanium metal. After the resultant sodium chloride was washed out, the residual titanium metal was dried, and weighed 0.819g. What is the empirical formula of this titanium chloride?

ONE MORE....I LOVE QUE’S

A certain compound was known to have a formula which which could be represented as [PdCxHyNz](ClO4)2.

Analysis showed that the compound contained 30.15% carbon and 5.06% hydrogen. When converted to the corresponding thiocynate [PdCxHyNz](SCN)2 the analysis was 40.46% carbon and 5.94% hydrogen.

Calculate the values of x, y and z.

SOME MOLECULAR ONES...

A compound has the following composition. C= 40.0% H=6.67% and O=53.3%. Its molecular weight is 60. Determine the molecular formula

Do you want that score? WAKE UP!

LAST ONE

One of the earliest methods for determining the molecular weight of proteins was based on chemical analysis. A haemoglobin preparation was found to contain 0.335% iron.

i) If the haemoglobin molecule contains one atom of iron, what is its molecular weight?

ii) If it contains 4 atoms of iron, what is its molecular weight?

THIS WEEK PLEASE:

Do ALL the set questions ASK if you have issues Read the notes again

NEXT WEEK: Concentrations Acids/Bases Volumetric Analysis (Part 1)