ch2 calculations and analysis

7/30/2019 CH2 Calculations and Analysis

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An atom of an element can be described using two numbers:

The atomic number is the smaller of the two numbers , and

it corresponds to the number ofprotons in the nucleus of anatom of that element.

The mass number is the larger of the two numbers and is the total number ofnucleons, both protons and neutrons, in the nucleus. These make up the mass ofthe atom because electrons weigh very little. The Periodic Table is organised in

atomic number order.

The number of neutrons can be found by subtracting the atomic number from the massnumber e.g. for potassium: neutrons = (39 19) = 20

Kpotassium

39

19atomicnumbersymbol

massnumber

Atomic Structure


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End of Structure of AtomsCarry on to Electronic StructureBack to list of topics

Atoms of the same element must have the same number of protonsif they didnt

theyd be different elements !

They must also have the same number of electronsif they didnt the positive and

negative charges wouldnt balance, and theyd be ions not atoms !

Atoms of the same element can have different numbers of neutrons, however.This means that atoms of the same element can have different masses, but all theirother properties are the same.

We call atoms of the same element with different numbers of neutrons (and thereforedifferent masses) ISOTOPES.

Hydrogen has three different isotopes:

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31H H H

Tritium key-fobs

glow in the dark1 proton 1 proton 1 proton

0 neutrons 1 neutron 2 neutrons

also known as Deuterium Tritium


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Now we know the masses of the different atoms in a compound, we can work out

what percentage of mass comes from each different type of atom. We are

calculating % by mass.

For example: what is the percentage by mass, of carbon in carbon dioxide ?

Arof carbon = 12

Mrof carbon dioxide (CO2) = 44

% by mass of carbon = 12 x 100 = 27.3%44

Sometimes there is more than one atom of the element we are interested in, for

example to work out the percentage of oxygen in MgCO3:

Arof oxygen = 16Mrof magnesium carbonate = 84 (24 + 12 + 3 x16)

% by mass of oxygen = 3 x 16 x 100 = 57.1%84

% Composition (% by mass)


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If we know the percentages by mass (i.e. the composition) of a compound, then we

can use this to work out its empirical formula (or simplest formula).

Remember: The EMPIRICAL formula is the simplest ratio of the numbers of atomspresent which may not be the same as the MOLECULAR formula, which is the

actual numbers of atoms present in the compound.

e.g. substance molecular formula empirical formulawater H2O H2O same

ethene C2H4 CH2 differentglucose C6H12O6 CH2O different

A substance contains 27.3% carbon and 72.7% oxygen what is its empirical

formula ?

carbon oxygen% 27.3 72.7 write the percentages of each elementAr 12 16 divide each by the atomic mass of that elementRatio 2.275 : 4.544 divide by smallest (2.275) to get whole numbersRatio 1 : 1.997 round up/down nearly whole numbersRatio 1 : 2 write formula CO2

Empirical formulae


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Note: If you are given the percentage of one element in a compound containing twoelements, you should be able to work out that the percentage of the second

element is 100% take away the percentage of the first.

Note: The method works exactly the same if you have masses rather thanpercentages of the different elements. Just write the masses down rather thanpercentages, and follow the same method.

Note:Round nearly whole numbers such as 3.002 or 1.997 to whole numbers.BUT if you get a half or nearly half e.g. 2.5 or 2.499 then DONT round it to a

whole number, instead DOUBLE the ratios for ALL the elements in the compound

e.g. A sample of 24.0g of iron oxide is found to contain 16.8g of iron. Work out its

empirical formula.

iron oxygenmass 16.8 7.2 write the masses of each elementAr 56 16 divide each by the atomic mass of that elementRatio 0.3 : 0.45 divide by smallest (0.3) to get whole numbersRatio 1 : 1.5 double both, DONT round !Ratio 2 : 3 write formula Fe2O3

mass of oxygen= total mass (24g) mass of iron (16.8g)


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Consider this reaction: 2 H2 + O2 2 H2O

The equation tells us that two hydrogen molecules will react with one oxygen

molecule to make water. We need to measure out two hydrogen molecules for

each oxygen molecule but they are too small to count out individually !

Lets say we want to react 32g of oxygen. What mass of hydrogen should we

weigh out to mix with it ?

Chemists measure the AMOUNT OF substances in moles. (A mole is just a verylarge number of something, like a dozen only much, much bigger). Look again at

the equation at the top of the page. It tells us that TWO MOLES of hydrogen will

react with ONE MOLE of oxygen. Well get TWO MOLES of water made.

Fortunately we CAN measure out moles of something:

One mole is defined as the amount of substance you have when you weigh out therelative formula mass of a substance, in grammes.

Oxygen molecules (O2) have an Mr= 32. So we weigh out 32g of oxygen, and

weve got one mole of oxygen. Now we need two moles of hydrogen. The Mrof

hydrogen molecules (H2) = 2, so its is 2g per mole. Well need 4g of hydrogen to

react with our 32g of oxygen.

Reacting Masses


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One mole .

of carbon weighs 12g

of iodine crystals (I2)

weighs 254g

of sodium metal: 23g

of magnesium weighs 24g

of water weighs 18g

and occupies 18cm3


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We can calculate how many moles, given a mass, using:

moles = mass

Mr**use Ar for single atoms

e.g. how many moles of carbon dioxide do we have if we have 660g ?

1: work out Mrof carbon dioxide (CO2) = 12 + 16 + 16 = 44

2: moles = mass = 660 = 15 molesMr 44

We can also convert numbers of molesback to mass, using: mass = moles x Mr*

e.g. what is the mass of 0.3 moles of chlorine molecules (Cl2) ?

1: work out Mrof chlorine molecules = 35.5 + 35.5 = 71

2: mass = moles x Mr = 0.3 x 71 = 21.3 g


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Remember that chemical equations tell us how many moles of each reactant

produce how many moles of each product. Combined with our ability to convert

masses to moles and vice versa we can now handle questions which ask us about

what masses of reactants to use, or what mass of products would be produced.

e.g. Calcium carbonate is attacked by hydrochloric acid. Carbon dioxide gas is one

of the products. The equation is: CaCO3 + 2 HCl CaCl2 + H2O + CO2

What mass of hydrochloric acid is needed to react with 5g of calciumcarbonate ?

Step 1: convert mass of calcium carbonate to moles of calcium carbonate

Mrof CaCO3 = 40 + 12 + 16 + 16 + 16 = 100

moles = mass Mr = 5g 100 = 0.05 moles of CaCO3

Step 2: use equation to see how many moles of hydrochloric acid will be needed

equation shows 1 mole of CaCO3react with 2 moles of HClWe have 0.05 moles CaCO3 so we need 2 x 0.05 = 0.10 moles HCl

Step 3: convert moles of hydrochloric acid to mass

Mrof HCl = 1 + 35.5 = 36.5

mass = moles x Mr = 0.10 x 36.5 = 3.65 g of HCl


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The steps are always the same. Youll have something you know the mass of, and

something you are trying to work out the mass of

Step 1: convert mass of known to moles of known

Step 2: use chemical equation to convert moles of known to moles of unknown

Step 3: convert moles of unknown to mass of unknown

Another example: What mass of sodium do I need in order to make 186g of

sodium oxide ? 4 Na + O2 2 Na2O

In this example, our known is the sodium oxide. We know its mass, and can work

out its Mr = 23 + 23 + 16 = 62 (Note that the 2 in front of Na2O isnt part of itsformula so isnt part of the relative formula mass !)

Step 1: convert mass of Na2O to moles of Na2O

moles = mass

Mr = 186

62 = 3 moles

Step 2: the equation tells us we need 2 moles of Na for every 1 mole of Na2O so

well need 6 moles of Na ifwe want 3 moles of Na2O

Step 3: convert moles of Na to mass of Na

mass = moles x Ar(Na is a single atom) = 6 x 23 = 138g


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Using moles we can calculate how much product we expect to be made from our

reactants. In reality we dont always manage to collect that much. The amount we

actually end up with is called the yield.

Reasons for not getting as much as wed expect include:

not all the reactants reacted, some may remain unreacted

side reactions - different products beside the ones we expected may be made

some reactions are reversible, and do not go to completion

product may have been lost in removing and purifying it

It is very important for industry to know how efficiently reactions are working. We

calculate the % yield:

% yield = mass of product actually obtained (g) x 100

mass of product expected (g)

The mass of product actually obtained is simply weighed, once the reaction iscomplete and the product has been separated out, purified and dried etc.

The mass of product expected is worked out using a mole calculation similar to

those on the previous pages.

% Yield


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Example:

Sulphur trioxide reacts with water to form sulphuric acid according to the

equation SO3 + H2O H2SO4

80g of sulphur trioxide are bubbled through water. The mass of sulphuric acid

produced was 63.7g What was the % yield in this reaction ?

We need to calculate the mass of product expected:

Step 1: convert mass of SO3 to moles of SO3 (Mr= 80)

moles = mass Mr = 80 80 = 1 mole of SO3

Step 2 equation tells us 1 mole SO3 makes 1 mole H2SO4

Step 3 convert 1 mole of H2SO4 to mass of H2SO4 (Mr= 98)

mass = moles x Mr

= 1 x 98 = 98 grammes expected

Now we can work out the % yield:

% yield = actual mass of product x 100 = 63.7 x 100 = 65%

expected mass of product 98


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Identifying and Analysing Substances

We can either identify substances using their chemical

reactions in a series ofchemical tests.

Alternatively we can use instrumental methods.

The development of modern instrumental methods have

been made possible by advances in electronics and

computing, and sensor technologies.

Advantages: highly accurate

rapid analysis

sensitive (very small quantities can be

analysed e.g. in forensics)

Disadvantages: usually very expensive

requires specialists to use

results may need to be compared with a

library of previous results before a match

can be found


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Chemical Analysis can be used to identify additives in foods.e.g. Paper chromatography can be used to separate different substances suchas the colouring additives in food.

The solubility of each additive determines how fast it will travel up the

chromatography paper when carried along by the solvent, so the components endup being separated, as illustrated below.

Paper chromatography


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How to use chromatography:

A baseline is drawn in pencil on thechromatography paper and a small spot of

the unknown is placed alongside spots ofknown substances which may be present

(these are called references). Each spot islabelled in pencil.

The paper is then dipped in the solvent

(usually water) and left while the solventslowly soaks up to the top of the paper,

separating the mixture as it goes.

Each spot in the finished chromatogram is a

different substance (although sometimes

substances with similar solubilities produce

spots which are overlapping).

A match is found when one of the spots in

the unknown sample is the same colourand at the same height as a reference spot.

baseline

Unknown A B C


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pure reference

samples of additives

foods with

unknown

additives In this example twounknown foods have

been analysed to see

which colouring

additives are present.

The yellow Brightiewas found to contain

sunburst yellow only.

The orange Brightie

was found to contain

sunny yellow and solaryellow.

Using chromatography


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Instrumental methods can also be used to separate mixtures and to identify thecomponents in the mixture.

e.g. Gas chromatography can separate the components in a mixture and massspectrometry can identify what each of these components is. Used together likethis, the method is referred to as GC-MS.

GC-MS is increasingly used for detection of

illegal narcotics, and may eventually supplant

drug-sniffing dogs. It is also commonly used in

forensics to find drugs and/or poisons in

biological specimens of suspects, victims, or thedeceased.

a GC-MS instrument


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The sample is injected into the instrument, and passes through the column, where it

is separated into the different components. Each component emerges from the

capillary at a different time and is then identified in the mass spectrometer.


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The output from the GC instrument shows a peak for each substance present in the

sample. The size of the peak tells how much of that substance is present.

Here the GC is beingused to see how muchpropanone (acetone) ispresent in a urinesample. The presence ofpropanone can indicate

diabetes.


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The output from the Mass Spectrometer can be used to tell us the Relative

Molecular Mass (Mr) of each substance that has been separated from the mixture.

The peak furthest to the right in graph from the mass spectrometer is called the

molecular ion peak, and has a value which corresponds to Mr.

, C3H6O The relativemolecularmass forpropanone= 58

molecular

ion peak

ch2 calculations and analysis

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