Chem1031 Week 1

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CHEM1031:Higher Chemistry 1A

Dr Jason Harper Dalton 223, x54692

[email protected]

Week 1 - Atoms, Inorganic Nomenclatureand Chemical Calculations

Text Books

Blackman, A., Bottle, S., Schmid, S., Mocerino, M., andWille. U, Chemistry, John Wiley and Sons, 2008.Aylward, G.H. and Findlay, T.J.V. SI Chemical Data,(6th ed.).

CHEM1011/CHEM1031 Course Pack, sold at the UNSWBookshop.

Learning Objectives• Name the constituent parts of an atom, together with theirrelative masses and charges.• Calculate numbers of protons, neutrons, electrons in atomsof a particular element.• Name simple inorganic compounds and write the formulaefor simple compounds from their name.• Write and balance simple chemical equations.• Calculate molecular weight from chemical formula.• Calculate % by mass of each element in a compound anddetermine empirical formula from % by mass.• Calculate concentration of solutions in various units.• Calculate yield in a chemical reaction, determine thelimiting reagent.

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Learning Objectives• Name the constituent parts of an atom, together with theirrelative masses and charges.• Calculate numbers of protons, neutrons, electrons in atomsof a particular element.

These are covered in sections 1.2-1.4 of Blackman et al.

A starting point ... and an updateDalton’s Atomic Theory

• All matter consists of tiny particles: “ATOMS”• Atoms of one element can neither be subdivided nor changedinto atoms of another element.• All atoms of the same element are identical in size, mass andother properties.• Atoms of one element differ in mass and other properties fromthe atoms of other elements.• Chemical combination is the union of atoms of differentelements, the elements combine in simple, whole number ratioswith each other.

Dalton’s Atomic Theory• Atoms of one element can neither be subdivided nor changedinto atoms of another element.

(The second point neglects nuclear fission and fusion, not thenunderstood.)

The first part may be reworded- the smallest unit of anelement is the atom.- it is made up of two parts:a) The nucleus(protons and neutrons)b) Electrons in ‘shells’ aroundthe nucleus.

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The different components of the atom are not the samemass and they don’t take up the same amount of space.

Proton mass = 1.673 x 10-27 kgNeutron mass = 1.675 x 10-27 kgElectron mass = 9.109 x 10-31 kg

The electron cloud is muchlarger than the nucleus.(For hydrogen, which hasone proton in the nucleus,it is 100,000 times.)

*Think about what the ‘outside world’ sees.

So what does this mean?i) The size of the atom’s electron cloud defines the atom’s size.

ii) The weight of the atom’s nucleus (that is, how many protons andneutrons are present) defines the atom’s weight.

iii) Together these two define the atom’s average density.

But what determines how many protons, electrons and neutronsthere are?

Let’s start with the easy parts.Need to consider that these species are chargedCharge on a proton = +1.602 × 10-19 CCharge on a neutron = 0Charge on an electron = -1.602 × 10-19 C

For a neutral atom,number of protons = number of electrons

This says nothing about the number of neutrons (which we willcome to later).

If the two values are not the same, the species is charged.Anion: Negatively charged species, protons < electronsCation: Positively charged species, protons > electrons

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Fundamental property of an element

There are the same number of protons in each atom.

A way to remember what defines an elementConsider the following argument:

For two atoms to react they must come into contact.

This means that the electron clouds must interact.

Different number of electrons would mean different interactions(and hence reactions/reactivity).

Since atoms are neutral, different number of electrons correlates to

different number of protons.

Atoms with different numbers of protons will display differentreactivity ... which is the a way of defining an element.

Fundamental property of an element

There are the same number of protons in each atom.

Why define it in terms of protons rather than electrons then?Because electrons may be gained or lost by an atom to form an ion.

The number of protons in a nucleus is the atomic numberfor that element.

This is all consistent with Dalton’s theory, particularly ...

“Atoms of one element differ in mass and otherproperties from the atoms of other elements.”

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So what about neutrons?

Neutrons do not have charge but do have mass.

If a neutron was added to the nucleus of an atom, it would beheavier but the number of electrons would still be the same.

There would be little change to the reactivity but some changes to

properties related to mass.

Isotopes

• Atoms with the same number of protons (and hence electrons)but different number of neutrons are called isotopes.• They have different mass (and hence slightly different physicalproperties).• They have very similar chemical properties and are both thesame element.• So Dalton was incorrect when he proposed

“All atoms of the same element are identical in size,mass and other properties.”

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Was he right about size though?

“All atoms of the same element are identical in size, mass and otherproperties.”

Remember from earlier ...“The size of the atom’s electron cloud defines the atom’s size.”

Two isotopes of the same element have the same number ofprotons and the same number of electrons, just not the samenumber of neutrons.Hence their electron clouds would be the same size.

So Dalton was right!

So how do we represent these symbolically?

A = symbol of elementZ = number of protons (atomic number)M = mass number (protons plus neutrons)n = charge on the system (zero for an atom - see next slide for use)

Note that sometimes the atomic number is left off - it is consideredto be implied by the elemental symbol.

MAn±

Z

C126Examples: Has 6 protons, 6 neutrons, 6 electrons

Br3579

Has 35 protons, 44 neutrons, 35 electrons

A couple of things on this representation

Different elements can be identified by different symbols andatomic numbers.Different isotopes can be identified by different mass numbers.

Charged species (ions) can also be identified this way.Remember that a positive charge means more protons thanelectrons and vice versa.

MA

± n

Z

Example: C6

13 2+

Has 6 protons, 7 neutrons, 4 electrons

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Isotopes

• Some elements haveonly one isotope, othershave many.• Some are stable, somebreak down(radioactivity).

Isotopic examplesHydrogen has three isotopes.

They differ by the number of neutrons present.These isotopes are sometimes referred to as protium, deuterium andtritium but they are all still the same element.

The compounds formed from the different isotopes have slightlydifferent properties, all based on the fact that they have differentmass:

Boiling points: H2 (-253°C); D2 (-250°C) ‏Melting points: H2 (-259°C); D2 (-255°C) ‏Densities: H2 (0.1 g/ml); D2 (0.2 g/ml) ‏

H 1 1 H

2 1 H

3 1

Are they useful?You have probably all heard of carbon dating techniques.Literally, this measures the amount of the isotope of carbon with amass number of 14 (8 neutrons, 6 protons) relative to thecorresponding amount of carbon-12.

It works because carbon-14 is radioactive and breaks down. Aliving organism will stop taking in carbon-14 from the environment(either as carbon dioxide for plants, through ingestion by animals)so this ratio decreases with time.

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Are they useful?But plants actually fractionate (change the ratio of isotopes) incarbon dioxide they take out of air, and not to the same extent.This difference is tiny but it means that different plant steroids havea different isotopic ratio (carbon-14 to carbon-13 to carbon-12);this is useful for drug testing!

(Steroids 2006, 71, 364)

Learning Objectives• Name simple inorganic compounds and write the formulaefor simple compounds from their name.

This is covered to some degree in section 2.3of Blackman et al.

So what’s a compound then?

A compound is made up of two or more elements combined in afixed ratio.

Later in the course you’ll learn about the different types ofbonding involved (Week 4).

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Naming Inorganic Compounds• Inorganic - without carbon.(Organic nomenclature is an entire section later in the course!)• For simple systems made up of one metallic and one non-metallic element

• the first part of the name is the metal,• is followed by a space and• the second part of the name is the non-metal with the lastsyllable changed to -ide,• with no account made of the ratio of the number of atomspresent.

• The same argument goes for all salts, made up of cation and ananion - this will be discussed shortly when we consider namingcomplex anions.

Some examples

NaBr sodium bromide

K2S potassium sulfide

MgO magnesium oxide

BaCl2 barium chloride

Li3N lithium nitride

Note again, there is no indication of the ratio of the elements -barium chloride rather than barium dichloride.This is because in these cases there is only one ratio.

What happens when more than one ratio is possible?This occurs particularly in compounds where there are more thanone non-metallic element present.

For example, there are two combinations of phosphorus andchlorine.

PCl3 PCl5

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What happens when more than one ratio is possible?Compounds that can have several different ratios of elementsuse number prefixes - mono, di, tri, tetra, penta, hexa

So for the previous examples.

phosphorus trichloride phosphorus pentachloride

Some Questions and AnswersQuestion: What if there was more than one phosphorus atom inthe formula, say it was P2Cl5?Answer: Then phosphorus would also get a prefix - and thatcompound (which doesn’t actually exist) would be diphosphoruspentachloride.

Question: Why does the phosphorus part of the name go first inthe previous examples?Answer: This relates to a more general form of the namingconvention ‘metal goes first’. Actually, it relates to theelectronegativity (tendency of an atom to attract electrons) - aconcept which will be discussed further later in the course.

Question: So how do we use it now?Answer: Electronegativity increases as you move to the top righthand side of the periodic table.

So phosphorus is less electronegative than chlorine.

So how do you name the compounds SO2 and SO3?

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Some examplesSulfur is less electronegative than oxygen so it forms the firstpart of the name.

SO2 sulfur dioxide

SO3 sulfur trioxide

This can also happen with some metalsThough for these cases, nomenclature dictates a change in theway the metallic element is described.

FeCl2 ferrous chloride

FeCl3 ferric chlorideHg2Cl2 mercurous chlorideHgCl2 mercuric chloride

This is a bit cumbersome - is there a better way?Yes, using Oxidation Numbers (ONs)

Why do some elements combine in different ratios?

• The concept of valence is often used.• This is the simple observation that each atom typegenerally has a customary ‘combining power’.• e.g. NaCl, CaCl2, FeCl3, Na2O, CaO, Fe2O3 implyvalencies of

• ‘1’ for Na and Cl• ‘2’ for Ca and O• ‘3’ for Fe (in this case - it can also be ‘2’)

• Relates to position in periodic table - which in turn is related toelectronic configuration and electronegativity.

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So how are oxidation numbers (ON) assigned?

• free element ON = 0 (e.g. F2, O2, S8, Ca, C60) ‏• monatomic ion ON = charge on that ion (e.g. Cl–, Na+, Fe3+) ‏• Fluorine’s ON always –1 (except in the element F2) ‏• Oxygen’s ON almost always –2; exceptions peroxides (H2O2)and superoxides.• Hydrogen’s ON is +1, except in metal hydrides (e.g. NaH),when H is –1• Main group metal in a compound:

ON same as + group number (e.g. Na+1 in NaCl) ‏• Non-metals in compounds:

ON usually (group number –18)* ‏• The sum of oxidation numbers is equal to the charge on eitherthe compound or the ion; this allows determination of theoxidation numbers of the other elements present.

Let’s look at previous examples.For phosphorus trichloride, the ON for P = +3For phosphorus pentachloride, the ON for P = +5This can be used in naming, with the oxidation of the variablevalence element placed in roman numerals after the elementname.

phosphorus(III) chloride phosphorus(V) chloride

This is sometimes called ‘STOCK’ nomenclature.

Let’s look at previous examples.Using the same argument, for the sulfur examples previously …

SO2 sulfur(IV) oxide

SO3 sulfur(VI) oxide

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Let’s look at previous examples.It also greatly simplifies the metallic cases we saw before.

FeCl2 iron(II) chloride (ferrous chloride)‏FeCl3 iron(III) chloride (ferric chloride)‏

Hg2Cl2 mercury(I) chloride (mercurous chloride) ‏HgCl2 mercury(II) chloride (mercuric chloride)‏

MnCl2 manganese(II) chloride (manganous chloride)‏

ON are particularly useful when consideringpolyatomic anions containing oxygen

• these are used as per the more electronegative portion earlier.• name depends on the oxidation state of the central atom.• higher oxidation number – name ends in -ATE• lower oxidation number – name ends in -ITE

SO42– sulfate ion - S(VI) CaSO4 calcium sulfate ‏

SO32– sulfite ion - S(IV) (NH4)2SO3 ammonium sulfite

NO3– nitrate ion - N(V) Mg(NO3)2 magnesium nitrate

NO2– nitrite ion - N(III) Mg(NO2)2 magnesium nitrite

PO43– phosphate ion - P(V) Na3PO4 sodium phosphate

PO33– phosphite ion - P(III) Na3PO3 sodium phosphite

CO32– carbonate ion - C(IV)‏ CuCO3 copper carbonate

CO22– “carbonite ion” - does not exist.

The same goes for polyatomic oxyhalides

• Though because more oxidation states are possible, thenomenclature is extended.

ON = +1 hypochlorite ion ClO–

ON = +3 chlorite ion ClO2–

ON = +5 chlorate ion ClO3–

ON = +7 perchlorate ion ClO4–

• Bromine and iodine series have same formulæ and nameformats, with ‘brom’ or ‘iod’ in place of ‘chlor’.

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There are also lots of organic anions

• But the two we’d like you to know are …

Acetate - CH3CO2- Oxalate - C2O4

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ONs are also used for the protonated forms - Acids!

• A correct way of naming would be to use the same form asdiscussed previously.

HCl - hydrogen chlorideH2CO3 - hydrogen carbonate

• May also be named as acids if liquid (remember HCl is a gas!).-ate ions = -ic acid, e.g. H2SO4 = sulfuric acid

-ite ions = -ous acid, e.g. H2SO3 = sulfurous acid

• When needed, specify number of protons attached.HPO4

2– = monohydrogenphosphate ionH2PO4

– = dihydrogenphosphate ionHCO3

– = hydrogencarbonate ion HSO4

– = hydrogensulfate ion

Something you may have noticed.

• Small central atom can fit around it a max. of three O atoms.

• Bigger central atom can fitaround it a max. of four O atoms.

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Learning Objectives• Write and balance simple chemical equations.

This is covered in sections 3.1-3.2 of Blackman et al.

Equations and what they representThere are two types of chemical equation.

Physical Change - the chemical composition does not change.This is usually a change of state (solid, liquid, gas)

H2O(s) H2O(l)

Chemical Change - Change in chemical composition occurs.Product(s) have new different properties.

2H2(g) + O2(g) 2H2O(g)

Parts of a chemical equation

• Chemical species are defined by a combination of atoms.• The state symbols show the physical state of the species, whichcan affect the reaction rate, and its energy change.• Coefficients are used to ensure that there are the same numberof each atom on each side - matter is neither made nor destroyed.• A coefficient of one is generally not shown.• In balancing an equation the numbers within each chemicalspecies CAN NOT be changed. This would be changing theidentities of the chemicals involved.

P4(s) + 5O2(g) P4O10(s)‏

Chemical species

State symbolsCoefficients

1 1

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So how do your write and balance an equation?

Stepwise!• Write reactants on left, products on right (this is convention).• Include states of matter - (s), (l), (g) or (aq)• Change coefficients to give the same count for each element on bothsides - do not change subscript numbers!

• Balance elements that appear in one species on both sides,• Balance elements that appear in several species on both sides.

• For reactions in water, add H2O if needed to balance oxygen atoms.• For reactions in water, add H+ if needed to balance hydrogen atoms.• Give nett ionic equations if possible – convey essential change.

(A nett ionic equation leaves out components that arepresent both in the starting materials and the products … see later.)

So how do your write and balance an equation?

But to be able to do this, you need to some basic skills …• Name translations.• An understanding of the nature of solution.

The first part we’ve done … the second is where we are now …

The key point to remember is that when dissolved in water, saltsand acids tend to dissociate. That is, they exist as ions, surroundedby a shell of water molecules.

Hold on a second …

Question: Why do these compounds dissociate? Aren’t they stableas they are? We’ve always been told that in an ionic solid, the ionsbond together strongly, due to opposite charges (e.g. Na+ and Cl–).

Answer: Yes the bonds are strong. But there is compensation- water molecules form strong bonds with the ions formed; this is called solvation. (you will cover this later in terms of enthalpy)- there is an increased degree of randomness, with moredegrees of freedom for the solvated ions. (you will cover this later in terms of entropy)

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An example

"solid sodium hydroxide is reacted with hydrochloric acid toproduce a sodium chloride solution and water"

Write reactants on left, products on right (this is convention).

H+ + Cl– + NaOH Na+ + Cl– + H2O

An example


Include states of matter - (s), (l), (g) or (aq)

H+(aq) + Cl–(aq) + NaOH(s) Na+(aq) + Cl–(aq) + H2O(l)

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An example


Change coefficients to give the same count for each element on bothsides - do not change subscript numbers!


Already done!

An example


Give nett ionic equations if possible – convey essential change.


H+(aq) + NaOH(s) Na+(aq) + H2O(l)

X X

An example

”solid sodium oxide reacts with waterto give a sodium hydroxide solution"

Write reactants on left, products on right (this is convention).

Na2O + H2O Na+ + OH– ‏

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An example


Include states of matter - (s), (l), (g) or (aq)

Na2O(s) + H2O(l) Na+(aq) + OH–(aq) ‏

An example


Change coefficients to give the same count for each element on bothsides - do not change subscript numbers!

• Balance elements that appear in one species on both sides,• Balance elements that appear in several species on both sides.

Na2O(s) + H2O(l) 2Na+(aq) + 2OH–(aq) ‏

Hint: I find it easiest to leave elements that appear free on one sideuntil last. They are easiest to manipulate!

An example


Give nett ionic equations if possible – convey essential change.

Na2O(s) + H2O(l) 2Na+(aq) + 2OH–(aq)

Already done!

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OK, that’s great, where from here?

• Balancing an equation is step 1.• This gives you the stoichiometry of a process.• For example, the stoichiometric ratio of the reagents in ourprevious case was 1:1.

Na2O(s) + H2O(l) 2Na+(aq) + 2OH–(aq)

• But that doesn’t tell you anything about how much of a substancereacts with a given amount of another substance. That’s the nextstep …

Learning Objectives• Calculate molecular weight from chemical formula.

This is covered in sections 3.3 of Blackman et al.

To start, how heavy is an atom?

• Well, not very given …Proton mass = 1.673 x 10-27 kgNeutron mass = 1.675 x 10-27 kgElectron mass = 9.109 x 10-31 kg

• A uranium-235 atom has a mass of only about 4 x 10-25 kg!

• Typical units are insufficient - a new unit,the atomic mass unit (symbol u) is defined to help here.

= 1.660538 x 10-27 kg• It is defined as one twelfth the mass of an atom of 12C.• Some people still use the old term for this; the Dalton.

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Why is an atomic mass unit lessthan the mass of a proton?

proton mass = 1.673 x 10-27 kgatomic mass unit = 1.660538 x 10-27 kg

It’s about the definition - one twelfth the mass of an atom of 12C.

When two or more particles combine to form a nucleus, somemass is converted to energy. Therefore the mass of nucleus isless than the sum of the mass of the number of particles in it.

(This is not consistent between elements (or even isotopes of anelement) and is the basis of energy release in nuclear fusion andfission. For more details see Blackman et al., Chapter 26.)

Atomic masses - and averaging them.

• For each isotope of an element you can measure the atomicmass using a spectrometer.• When dealing with a bulk sample, which is a mixture ofisotopes, it is better to consider the weighted average usingknown proportions of the isotope.• This gives you the ‘average atomic mass’.

* It’s actually a point of argument (in chemical circles at least)what the average atomic masses of elements should be (based onthe proportions of isotopes).

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An example

Copper occurs naturally as a mixture of two isotopes: 63Cu(abundance 69.09%) and 65Cu (30.91%). Their atomic massesare 62.930 amu and 64.928 amu, respectively. Calculate theaverage atomic mass of copper.

An example

Whenever dealing with percentages, a useful trick is to consider100 of whatever the items are.Here, consider you have 100 atoms of natural copper.Of these, 69.09 atoms will be 63Cu, of mass

= 69.09 x 62.930 amu.Similarly the mass of the 65Cu atoms will be

= 30.91 x 64.928 amu.Thus, the total mass of all 100 atoms …

= [69.09 x 62.930 + 30.91 x 64.928] amu.= 6354.8 amu.

Thus, the average atomic mass of one atom= 6354.8 /100= 63.55 amu.

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Translating this to compounds

• The molecular mass of a compound is simply the sum of theaverage atomic masses of the atoms present (taking into accounttheir number).

• So, once again, using previous examples.

Molecular mass PCl3 = (30.97) + 3 x (35.45) = 137.3 uMolecular mass PCl5 = (30.97) + 5 x (35.45) = 208.2 u

What about systems like NaBr,which aren’t molecules?

• Some compounds form extended lattices and aren’t molecules.• The term formula mass is used to define the mass of the atomsrepresented by the formula as written.• It is determined in the same way.

• So, here the formula mass for NaBr= 22.99 + 79.90 = 102.9 u

So what can you do from here?

Learning Objectives• Calculate % by mass of each element in a compound anddetermine empirical formula from % by mass.

This is covered in section 3.4 of Blackman et al.

From now on, we will talk about it as mass fraction (expressedas %), because it’s a more straightforward way to express it.

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What is this?

• Literally what it sounds like• mass - so deal with this quantity• fraction - the ratio of the quantity for the component ofinterest divided by the total.• expressed as % - x 100.

• So what is the mass fraction of hydrogen in hydrogen chloride,expressed as a percentage?

The average atomic mass of hydrogen is 1.008 uThe molecular mass of hydrogen chloride is

(1.008 + 35.45) = 36.46 uThe mass fraction of H is therefore 1.008/36.46 x 100 = 2.76%

This clearly works for more complicated systems

Cysteine is an amino acid with the molecular formula C3H7NO2S.Calculate the mass fraction (as %) for each element.

Check the working yourselfbut …C - 29.74%H - 7.06%N - 11.56%O - 26.41%S - 26.47%

How is this useful?

Being able to calculate a mass fraction as a % from a molecularformula is not useful in itself but is very useful for determiningmolecular formula of a substance.

The mass fraction can be calculated by combustion analysis(burning!) and analysing the amount of products. It is anexperimental technique and the result is an empirical formula,which gives the simplest ratio of the elements present.

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Calculating an empirical formula

It is literally the reverse of determining the mass fraction.Consider an example: What is the empirical formula of acompound which has a mass fraction of C of 92.3% and a massfraction of H of 7.7%?

Start by considering the ratio by mass:C : H = 92.3 : 7.7

Then convert it to amount of substance by dividing by the averageatomic masses of each species.

C : H = (92.3/12.01) : (7.7/1.008) = 7.69 : 7.64 = 1 : 1 (note: rounding often needed)

So the empirical formula is C1H1.

This can be harder

What is the empirical formula of a compound which has a massfractions of C, H, N and O of 67.3%, 6.98%, 4.62% and 21.1%,respectively?

Ratio by mass:C : H : N : O = 67.3 : 6.98 : 4.62 : 21.1

Ratio by amount of sustanceC : H : N : O = 5.60 : 6.92 : 0.33 : 1.32

= 17 : 20.97 : 1 : 4 = 17 : 21 : 1 : 4

So the empirical formula is C17H21NO4.This is cocaine!

A warning: Empirical formula only gives a ratio

Note that empirical formula only gives a ratio of the elementspresent. Many compounds may have the same empirical formula.Two examples are below.

Benzene has molecular formula C6H6 but empirical formula C1H1.Acetylene has molecular formula C2H2 but empirical formulaC1H1.In order to calculate the molecular formula you need themolecular mass.

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Calculating a molecular formula - method 1

If you have the molecular mass and the empirical formula, it isquite straightforward. Again, an example.

What is the molecular formula of a compound with an empiricalformula C1H1 and a molecular mass of (a) 26 u; (b) 78 u ?

First as the empirical formula is C1H1 then the molecular formulawill be (C1H1)n. As a result the molecular mass will be

(12.01 + 1.008)n = 13.018 n(a) 13.018 n = 26 therefore n = 26/13.018 = 2 (note rounding!)

so molecular formula is (C1H1)2 = C2H2(b) 13.018 n = 78 therefore n = 78/13.018 = 6 (note rounding!)

so molecular formula is (C1H1)6 = C6H6

Calculating a molecular formula - method 2

If you have the molecular mass and the mass fraction, you canskip the empirical formula stage.

What is the molecular formula of a compound with mass fractionsof C, H, N and O of 67.3%, 6.98%, 4.62% and 21.1%,respectively, and a molecular mass of 303.5 u?

For each element, you can calculate the mass per molecule andhence the number of atoms of that element.

C - 67.3% of 303.5 = 204.2 u No. of atoms C = (204.2/12.01) = 17H - 6.98% of 303.5 = 21.17 u No. of atoms H = (21.17/1.008) = 21N - 4.62% of 303.5 = 14.01 u No. of atoms N = (14.01/14.01) = 1O - 21.1% of 303.5 = 64.01 u No. of atoms O = (64.01/16.00) = 4

So the molecular formula is C17H21NO4(Note that here it is the same as the empirical formula!)

How does all of this translate to real world masses?

The atomic mass unit is convenient - but we’re unlikely to be ableto measure something out in it. How do we convert it tosomething more practical - again, it’s a new unit.

An atomic mass unit is defined as one twelfth the mass of an atomof carbon-12

A mole is the amount of substance of a system which contains asmany elementary entities as there are atoms in 0.012 kilogram ofcarbon-12.When the mole is used, the elementary entities must be specifiedand may be atoms, molecules, ions, electrons, other particles, orspecified groups of such particles.

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Amount of substance

So the mole measures how much substance is there - an amount.

The definitions are convenient. The mass of a particle (molecule,atom) in atomic mass units converts to the mass of one mole ofthose particles in grams.

e.g. 1 mole of Fe (55.84 u) weighs 55.84 g

Another way of expressing this is as a molar mass (M)- the molar mass of Fe is 55.84 g mol-1

One mole of a substance contains a lot of entities.

One mole of a substance contains the same number of entities asthere are atoms in 0.012 kilogram of carbon-12.

This corresponds to a number you’ve probably all heard of thatbeing Avogadro’s Number = 6.022 x 1023.

Something to try - work out how many entities are present in 1 gof a variety of substances (Hg, ethanol (C2H5OH), NaCl).

What do you measure to determinethe amount of a substance?

• If it’s a solid - measure a mass and divide by the molar mass.

• If it’s a liquid - measure a volume, convert to a weight usingdensity (g cm-3) and divide by the molar mass.

• If it’s a gas - a number of ways we’ll talk about next week butyou can even just measure a mass and divide by the molar mass.

• What about a substance (the solute) dissolved in a liquid (thesolvent) to form a solution … for this we need to understandconcentration.

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Learning Objectives• Calculate concentration of solutions in various units.

This is covered in section 3.6 of Blackman et al.

There are different ways to measure concentration

BUT … in each case, it can be related to a property that measuresthe amount of substance.

In the following examples, how to calculate the concentration of asolution from the amount of substance in different ways will beshown.

You are strongly encouraged to try the reverse; to calculate theamount of substance in an amount of solution of knownconcentration.

Note the UNITS in each case. It is often expressed as cA or [A].

One example in lots of ways

Consider:“A concentrated solution of NH3 in water contains 256 g of NH3per litre of solution; the solution has a density of 0.880 g ml–1”

1. Mass fraction as a percentage.

This is (mass of solute)/(mass of solution) x 100

In this case, if we consider 1 l of solution (256 g) / (0.880 g ml-1 x 1000 ml) x 100

= 29.0 %

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2. Mole fraction (χ)

This is (amount of solute)/(amount of solute + amount of solvent)

In this case, if we consider 1 l of solutionAmount of ammonia = 256 g = (256)/(14.01 + 3 x 1.008)

= 15.1 molesAmount of water = 624 g = (624)/(16.00 + 2 x 1.008) = 34.7 moles

Mole fraction = (15.1)/(15.1 + 34.7) = 0.303



3. Molality.

This is (amount of solute)/(mass solvent)

In this case, if we consider 1 l of solution amount of ammonia = 15.1 moles (from before)

Mass of solution = 880 g = 0.880 kg, Mass of solvent = 0.624 kg

Molality of ammonia in this solution = 15.1/0.624 = 24.1 mol kg-1



4. Molarity.

This is (amount of solute)/(volume of solution)

In this case, if we consider 1 l of solution amount of ammonia = 15.1 moles (from before)

Molarity of ammonia in this solution = 15.1/1 = 15.1 mol l-1 (M)

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A note …

Remember that some species dissociate in water. As a result, youmay be able to express the concentration in terms of theindividual species.

For example, if CaBr2 were dissolved in water to give a solutionwith a concentration of of 0.5 M.

This solution would contain Ca2+(aq) at a concentration of 0.5 Mand Br-(aq) at a concentration of 1 M.

Learning Objectives• Calculate yield in a chemical reaction, determine the limitingreagent.

This is covered in sections 3.6 of Blackman et al.

All that we have been talking about up until here leads to this.

Why? Because we need to be able to determine how much (ina quantity we measure - g) of a reactant will react with anotherreagent.

Let’s make ethanol.

Ethanol can be made industrially by hydrating ethene. Thereaction is as below - the stoichiometric ratio is 1:1.

C2H4(g) + H2O(g) C2H5OH(g)

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Ethanol can be made industrially by hydrating ethene. Thereaction is as below - the stoichiometric ratio is 1:1.

C2H4(g) + H2O(g) C2H5OH(g)


But what happens if there are more molecules of water … thenthey’re leftover at the end.

Here water is referred to as being in excess and the ethene isreferred to as being the limiting reagent.


In the previous case, all of the ethene would react, leaving theexcess water. The amount of ethanol is therefore determined bythe amount of ethene (not the amount of water) - you consider thelimiting reagent.

This allows you to calculate a theoretical yield; by firstdetermining the amount of each reagent, working out the limitingreagent and hence the theoretical amount of ethanol formed.

For example, what is the theoretical yield of ethanol (in g) if 50 gof water and 270 g of ethene were reacted together.

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First - the equation …C2H4(g) + H2O(g) C2H5OH(g)

So the stoichiometric ratio is 1:1 - one molecule of water reactswith one molecule of ethene.

50 g of water = 50/18.016 = 2.78 moles of water270 g of ethene = 270/28.052 = 9.62 moles of ethene

Therefore, in this case water is the limiting reagent and ethene isin excess. The amount of product is determined by the amount ofwater.

Amount of product = 2.78 moles = 2.78 x (46.07) = 128 g

Different yields

Note that this is a theoretical yield - it assumes that- no material is lost in isolation of the product,- no side reactions consume ethene, and- the equilibrium lies far to the direction of the products.

In practice, the first is the major loss of yield.

The actual yield is how much is isolated.

The percentage yield = actual yield/theoretical yield x 100.

Note that this is always ≤ 100%.

Another example

A sample of solid magnesium (2.05 g) was added to a solution ofhydrochloric acid (50.0 ml, 2.00 M) to produce a magnesiumchloride solution and hydrogen gas. The hydrogen gas wascollected and 0.084 g was isolated.

(a) Determine the limiting reagent.(b) Calculate the theoretical yield of hydrogen gas.(c) Calculate the percentage yield of hydrogen gas.

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Another example

First - the equation …Mg(s) + 2H+(aq) Mg2+(aq) + H2(g)

So the stoichiometric ratio is 1:2 - one atom of magnesium reactswith two hydrogen ions.

2.05 g of Mg = 2.05/24.31 = 0.0843 moles of Mg50 ml of 2.00 M HCl(aq) corresponds to

50 x 10-3 x 2.00 = 0.100 moles of H+

Because the stoichiometric ratio is 1:2, the H+ is the limitingreagent.

Here, 0.100 moles of H+ react with 0.050 moles of Mg (leaving0.0343 unreacted) to give 0.050 moles of H2(g).

Another example

Hence the theoretical yield of H2(g) is 0.050 moles or0.050 x (2 x 1.008) = 0.101 g

Therefore, the percentage yield= (0.084/0.101) x 100= 83%

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Chem1031 Week 1