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Chapter 4


Chemistry study notes: Chapter 4

Thursday, 25 April 2013

11:15 AM

 

Unit 4.1 THE MASS OF AN ATOM

 

Atoms are so small they are difficult to comprehend, they are far to small to be weighted directly

 

Eg, the mass of hydrogen is 0.000000000000000000000000000167g not only is this a highly impractical but the existence of isotopes also adds difficultly in determining the weight of an element/atom.

 

Machine generated alternative text: TABLE 4.1.1 ISOTOPES OF HYDROGEN No. of No. of No. of Name Mass (g) protons neutrons electrons Hydrogen 1 0 1 1.674 X 10-24 Deuterium 1 1 1 3.348 X 10-24 Tritium 1 2 1 5.023 X 10-24

 

The question remains, what exactly does the mass of an element mean. The fact that each isotope is present in different amounts also increases the difficulty of answering this question.

Unit 4.2 RELATIVE ATOMIC MASS

The problems presented in the pervious section have been over come by a relative atomic scale. 

  • It is important to realise that relative scales have no units because they are simply a comparison of one quantity to another.

 

  • The advantage of relative scales is that very large or very small numbers cab be compared easily.

 

Carbon-12 was the isotope of carbon chosen to generate the relative scale of atomic masses.  It was given a mass of exactly 12 units.

 

Carbon was choses because:

 

  • Carbon is very cheap and is widely available.
  • It is relatively easy to isolate and purify this isotope.
  • Carbon is not toxic in any way.  

 

Carbon was allocated a mass of 12 units rather than 1 because as this number mirrored the mass number of the isotope and protons and neutrons are the basic building blocks of atoms the relative atomic mass would closely parallel the number of these fundamental particles in the element.

 

Machine generated alternative text: = 12 exactly each ivig atom is approximtiy twice the mss of aC atom ‘2C = 12 exactly figure 4.2.2 What are the masses of these aton when compared to carbon-i 2? /D\ each ‘H atom has a mass of approximately 1

We can now formally define the relative mass of an isotope of a particular element.

 

Machine generated alternative text: The relative isotopic mass (RIM) of an isotope of a particular element is defined as the mass of an atom of the isotope on a scale such that one atom of the carbon-12 isotope (‘2C) has a mass of 12 units exactly.

 

To be useful we need a relative scale of masses atoms of elements rather than just their isotopes. To achieve this the weighed mean or average of the masses of the isotopes and their percentage abundances is calculated. The percentage abundance and isotopic masses can be determined using a mass spectrometer.

 

Machine generated alternative text: sa mp le gas electron beam Figure 4.2.3 Schematic diogram of the major components of a muss spectrometer. - detecting screen ion filament beam accelerating plates

 

The mass spectrometer

 

A mass spectrometer is a complex instrument. The underlying principle of its operation is that moving charged particles will be affected in their movement as they pass through a magnetic field. The degree to which these particles are deflected from there original path will depend on their mass and charge.

 

  • The sample to be analysed is passed into a tube in a vapour state.
  • The atoms or molecules are then ionised (stripped of electrons) by a high energy beam of electrons
  • Positively charged ions are then accelerated along the tube and passed through slits to control the direction and velocity of their motion.
  • The stream of ions is then passed into a very strong magnetic field.
  • The result is that the ions are deflected through a curved path toward a detecting device that measures the location and number of particles that collide with it.
  • As the lighter particles are deflected more than the heavier ones, it is relatively easy to determine the mess/charge ratio and this can determine which elements and isotopes are present and what amounts of each.
     

Machine generated alternative text: ionisation and m.s acceleration separation detection spectrum _Ø*CZ I—[1—II dLi applied electric ions are particles ate fie4ds ionise separated by sorted by atoms, then a magnetic mass/charge accelerate ions field ratio

 

In simple elemental mass spectra the number of peaks recorded indicates the number of isotopes of the element present and their mass. The height of each peak represents how much of that element is present in the sample.

 

Eg.

Machine generated alternative text: copper 4.9 63 65 mass/c harg ratio

 

Machine generated alternative text: 11.1 d_______ — 11.1+4.9 160an 11.1+4.9 16.0 To convert these abundance fractions to percentage abundances, we simply multiply by 100: 69.49 and = 3O.69 16.0 1 16.0 1

 

Calculating relative atomic mass

 

 

Machine generated alternative text: The r1ative atomic mass (RAM) of an element is defined as the weighted mean of the masses of the naturally occurring isotopes, on the scale in which the mass of an atom of the carbon-12 isotope (‘2C) is 12 units exactly. The symbol for RAM is A.


To find the RAM of any element X, we multiply the relative isotopic mass of each naturally occurring isotope by its abundance fraction and add each other these values.

Eg.
 

Machine generated alternative text: Worked example 1 Use the data provided to determine the relative atomic mass of magnesium. Isotope Relative isotopic mass Percentage abundance 24Mg 23.99 78.70 Mg 24.99 10.13 Mg 25.98 11.17 Solution RlMx%abundanœ) A(Mg)= ioo 23.99x 78.70 24.99 x 10.13 25.98 xll.17 — 100 ‘ 100 + 100 = 18.88 • 2.53 + 2.90 = 24.31

 

Machine generated alternative text: Worked example 3 Gallium has two naturally occurring isotopes: Ga with a relative isotopic mass of 68.93 and 7tGa with a relative isotopic mass of 70.92. Given that the relative atomic mass of gallium is 69.72, determine the percentage abundances of each isotope. Solution Let the percentage abundance of the lighter isotope be x%, then the abundance of the other isotope must be (100 —x). Then: RIM X % abundance) Ar(Ga) = 100 = 69.72 6893xx 70.92x(100—x) ... + =69.72 100 100 6972 = 68.93x + 70.924100 — x) .-. 6972 = 68.93x + 7092 — 70.92x 6972— 7092 = 68.93x — 70.92x :—120=—1.99x .x=60.30 The percentage abundance ofGa is 60.30% and of 7tGa is 39.70’L Note: Relative atomic masses for all elements are provided I» appendix 3 and are also provided on standard periodic tables. Students are not expected to commit relative atomic mass data to memory though you will most likely find that some of the more common elements’ values will be memorised as you solve the problems associated with this section of the course.

 

Relative molecular mass and relative formula mass

 

Just as RAM refers to atoms relative molecular mass (RMM) refers to molecules. To determine the RMM, merely at together all the RAM of atoms present in the molecule.

Mr is the symbol.

 

Machine generated alternative text: Worked example 1 Determine the relative molecular mass of the following molecules. a Water(H.20) b Methane (CH4) c Sulfuric acid (112504) Solution a Mr(FLO) = 2 x 10 + 16.0 = 18.0 b M,.CH4 = 12.0 + 4 x 1.0 = 16.0 c MjI-L,SOj=2x 1.0+32.li-4x 16.0=98.1



RFM or relative formula mass is exactly the same thing only it is used for ionic compounds which cannot be described as molecules.

Unit 4.3 THE MOLE CONCEPT

 

Machine generated alternative text: A mole is dofined as the amount of substanœ which contains as many elementary particles as there are atoms in precisely 12 grams of the carhon-12 isotope.


The mole is defined in terms of the number of atoms present in 12 grams of Carbon-12 isotope.

The number required for this is 6.02 x 1023 This is known as Avogadro's constant symbol(NA)

 

  • If we counted 6.02 x 1023 atoms of oxygen it would have a mass of 16 grams.
  • It is important to recognise that a mole is simply a number just as a dozen is 12.

 

Machine generated alternative text: The number of elementary particles in one mole is known as Avogadro’s constant where: = 6.02 X 10 mol’

 

B y defining relative atomic masses and the mole in terms of the same reference, an atom of the carbon-12 isotope, an important link is made such that the mass of one mole of substance is equivalent to its relative atomic mass, measured in grams.
 

Machine generated alternative text: The mass of one mole of any substance is known as its molar mass (symbol M), where the molar mass is equal to the relative atomic (or molecular or formula mass measured in units of grams per mole (g mol-’). Fre 4.3.4 iile mie male al ojbon-12 has a mass of 121 the saine cinowit of cqIper has a mass of 63.5 i The numbei of atoms of each element ûhetd. 6.02 x 10 atoms *i %• ••á•’ — . — r , 63/Q”\ 1 mole otCu=63.5g 1 mole of ‘2C = 12g exactly

 

Calculations involving the mole

 

Using the idea of molar mass, we can determine the number of mole of a particular substance in a given mass.

 

Machine generated alternative text: n = where n = number of mole M m = mass (g) M = molar mass (g mol’)

This formula can be transposed

 


 

Whenever performing calculations involving quantities data you should always write the relevant formula and record the information provided in terms of the appropriate symbols.

 

Machine generated alternative text: Worked example I Determine the number of mole present in: a 32.0 g of oxygen atoms b 11.0 g of carbon dioxide molecules (CO2). Solution a n = -,wheren =?,m. = 32.0 g,M(O= 16.0 gmol’ = 2.00 mol b n = .!.where n = ?,rn = 11.0 g, M(CO)= 44.0 g moi1 :. n(C02) = = 0.250 mol 440

 

Knowing the number of particles present in one more of substance is 6.02 X 1023, it is possible to calculate the number of particles present in any given substance.

 

Machine generated alternative text: n = whore n = number of mole N A N = number of particles NA = Avogadro’s constant


 

Machine generated alternative text: Worked example 4 Calculate the number of partides present in each of the following. a Atoms of neon in 3.00 mole of Ne b Molecules of sulfur trioxide in 0.125 mole of SO3 c Oxygen atoms in 0.0855 mole of magnesium orthophosphate (Mg3( PO4)2) Solution a N = n X NA, where n(Ne) = 3.00 mol, NA = 6.02 X 1023 molt N(Ne) = 3.00 X 6.02 X 10 = 1.81 X 1O b N=n X NA, where n(S03)= 0.125 mol.NA= 6.02 X 10n mol NS03 = 0.125 x 6.02 X 10 = 7.53 X 10 c N = n X NA, where n(Mg3( PO4)2) = 0.0855 mol, NA = 6.02 X 10 mol’ :. N(Mg3(P04)2) = 0.0855 X 6.02 X lO = 5.15 x lO :.MO)=8x5.15x10=4.12x10

 

Unit 4.4 EMPIRICAL AND MOLECULAR FORMULAS

One of the many useful applications of the mole concept is to assist us in the determination of the formula of newley synthesised compounds. While this is a long process, one of the first steps is to determine the elements precent and their percentage composition by mass.

Determining % composition by mass of a compound


To determine percentage composition by mass of a compound, divide the molar mass of each element present by the molar mass of the whole compound and express it as a percentage.

Alternatively, divide the mass of each element of each element present by the mass of the whole sample and express as a percentage.

 

Machine generated alternative text: Worked example I Determine the percentage composition by mass of each element in the following compounds. a Sodium chloride b Potassium nitrate. Solution M(Na) 100 23.0 100 a %NainNaCl= X — - X =39.3% M(NaC1) 1 58.5 1 M(Na) 100 > .!Š?Š =607% %C1mNaCI= X - 1 M(NaC1) 1 58.5 (or simply 100 —39.3 = 60.7%) M(E) 100 39.1 100 b%KinKNOa= X X 387% M(KNOa) 1 101.1 1 M(N 100 14.0 100 %NinKNO3= X X 138% MKN03 riorr -1—S . 3xM(O) 100 48.0 100 % O in KNO3 = MKNO3 x —r- = 101.1 X 47.5% (or simply 100— 38.7 — 13.8 = 47.5%)


Determining empirical formulas

 

Machine generated alternative text: The empirical formula of a compound is the simplest whole-number ratio of atoms of different elements in the compound. The molecular formula of a compound is the actual number of atoms of different elements covalently bonded in a molecule.


If the % composition by mass is known, it is possible to determine the empirical formula of the compound. It provides information about the ratio at which atoms are bonded with out showing the actual number of atoms present.

 

Machine generated alternative text: To determine the empirical formula of a compound, take the following steps. 1 Write the elements present in the compound. 2 Underneath each element write its percentage composition (or mass proportion). 3 Divide the percentage (or mass) by the relative atomic mass of the relevant element and calculate this value. 4 Divide each ratio obtained by the smallest quotient to obtain a whoh— number ratio. 5 Express this ratio as an empirical formula.

 

Machine generated alternative text: Worked example 1 A compound of carbon and hydrogen is analysed and found to consist of 7&09 by mass carbon and 25.0% by mass hydrogen. Determine the empirical formula of the hydrocarbon. Solution C: H Elements in compound 75.0%: 25.0’) Percentage composition 75.0 25.0 . . —: — Divide by RAM 12.0 1.00 625 : 25.0 Determine ratio 6.25 25.0 . . . ; — Sunphfy ratio 6.25 6.25 1:4 The empirical formula of the hydrocarbon is CH4.

 

Determining molecular formulas

 

The molecular formula is defined as the actual number of atoms found in each molecule. The empirical formulas and molecular formulas must have the same ratios of atoms.

Eg.

Machine generated alternative text: TABLE 4.4.1 A COMPARISON OF EMPIRICAL AND MOLECULAR FORMULAS Compound Molecular formula Empirical formula Ethene C,H4 CH, Propene C3H6 CH, Butene C4H9 CH, Pentene CH CH, Water H,O H20 Carbon dioxide CO, CO7 Hydrogen peroxide H», HO Glucose CSH,O6 CH,O Ethanoic acid C,H40, CH,O To deWrmin the molecular formula of a compound, you must know the empirical formula and the molar mass i or relative molecular mass) of the compound. The molar mass of the empirical formula will be in direct ratio to that of the actual molecular formula.

 

Machine generated alternative text: Worked example I Given that the empirical formula of a hydrocarbon is CH and its molar mass is 26 g mol’, determine its molecular formula. Solution M(CH = 13g mol ‘and Mcompound) = 26g mol . Ratio of M(compound) : Mempirical formula) = 26: 13 = 2: 1 So the molecular formula is C2H2.

 

To find Molecular formula:

 1. Calculate or find empirical formula.
 2. Calculate the RMM or RFF of compound.
 3. Compare this to given actual mass of Molecule.
 4. Determine how many times you need to multiply molar mass of empirical to get molecular mass and multiply each number in empirical formula by this.

 

Machine generated alternative text: Worked example 2 Hydrazine is a highly reactive substance used principally as a rocket fuel. When it is analysed, its percentage composition is determined to be 87.5% nitrogen and 12.5% hydrogen and its molar mass as 32 g mol’. Determine the molecular formula of hydrazine.

 

Machine generated alternative text: Solution First calculate the empirical formula of hydrazine. N:H 87.5%: 12.5% 87.5 12.5 14.0 1.00 6.25: 12.5 6.25 12.5 6,25 ‘ 6.25 1:2 The empirical formula of hydrazine is NH2. M(NH2) 16g mol and M(compound = 32 g mol Ratio of M(œmpound : Mempirical formula) = 32: 16 = 2: 1 So the molecular formula of hydrazme is N2H4.

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