Chapter 1 : Matter 1.1 Atoms and molecules # I dentify and describe protons, electrons and neutrons as sub-atomic particles. # Define proton number, nucleon number and isotopes and write isotope denotation. # Define relative atomic mass and relative molecular mass based on the C-12 scale. Objectives : LECTURE 1
What is matter? 1. Matter is made up of elements, compounds and mixtures 2. The basic building block of matter is called atom . 3. An element consists of same kind of atoms e.g. Fe Anything that occupies space and possesses mass is called matter. Example : air, earth, animals, trees, atoms,… The three states of matter are solid, liquid and gas
3. A compound is made up of 2 or more kinds of atoms chemically bonded in a definite proportion e.g. NaCl, CO 2 . 4. A compound may consist of molecules or ions. E.g. NH 3 , K + Br - . 5. A molecule is made up of 2 or more atoms of the same or different element/s bonded covalently to form a discrete unit.e.g. O 3 , H 2 O. 6. An ion is an atom/group of atoms either gained electron/s (anion) or lost electron/s(kation). E.g. S 2- , Cu + , SO 4 2- .
Matter is made up of very tiny particles called atoms.
. Atom is spherical and consists of a nucleus surrounded with a cloud of moving –ve electrons (e)
THE FUNDAMENTAL PARTICLES OF ATOMS
. Nucleus of an atom is small, dense & consists of +ve proton/s (p) and neutral neutron/s(n)
Atoms consist of protons, electrons and neutrons.
nucleus electron cloud
Name charge mass/amu 1. Proton, p 1+ 1.007277 2. Electron, e 1- 0.0005480 3. Neutron, n 0 1.0008605
Nucleus = p + n.
Mass of proton mass of neutron
m e : m p = 1: 1836
Mass of atom mass of nucleon (p + n).
Properties of subatomic particles
PROTON NUMBER AND NUCLEON NUMBER
Total number of protons is called proton number, Z ( or atomic number)
Total number of protons and neutrons is called nucleon number, A (or mass number)
Z determine the identity of an element.
Symbol notation of atom :
A X where A= nucleon/ mass no.
= Z + n
Z = atomic/ p no. = p
No. of e - = Z– (charge of atom/ion)
e.g. 23 Na + 11 Total charge on the ion = 1+ nucleon no., A = No. of (p+ n) = 23 proton no., Z = no. of p = 11 no. of n in nucleus of = A - Z = 23 – 11= 12 No. of e - = Z – charge = 11- (+1) = 10 Find: the no. of I) proton, ii) neutron, iii) electron?
Example 1.1.2: Give the number of protons, neutrons and electrons in each atom of the following species: (a) b) 2+ 2+ 2 2 No. of Symbol charge 2- 1 10 9 8 2+ 2 78 120 80 atom e n p
For neutral atom, number of protons = number of electrons
For +ve ion, number of electron is less than number of protons.
For –ve ion, number of electrons is more than number of protons.
ISOTOPES <>Atoms with the same proton number but different in nucleon numbers. Hence same chemical properties but different physical properties. E.g :12 14 13 6 6 6 C Ex. What is the no. of (I) p ii) n iii) e in a) 27 Al 3+ b) 31 P 3- ? 13 15 C C
Ex. 2 Give the isotopes’ names of hydrogen and oxygen.and write their respective atomic denotation. Ans: H: protium, deuterium and tritium O : oxygen- 16, oxygen-17 and oxygen -18 1 2 3 H H H 1 1 1 16 17 18 O O O 8 8 8
Complete the following table: 40 18 ----- Ca ---- ----- f. 80 125 82 e. 23 26 56 d. 32 18 *32 16 c. 24 Mg 2+ 12 b. 35 18 18 17 35 17 35 Cl - 17 a. Atomic mass No. of e - No. of n No. of p A Z Symbol notation No.
40 18 20 20 40 20 40 2+ ----- Ca ---- 20 ----- f. 207 80 125 82 207 82 207 Pb 2+ 82 e. 56 23 30 26 56 26 56 Fe 3+ 26 d. 32 18 16 16 *32 16 32 S 2- 16 c. 24 12 - 2 = 10 12 12 24 12 24 Mg 2+ 12 b. 35 18 18 17 35 17 35 Cl - 17 a. Atomic mass No. of e No. of n No. of p A Z Symbol notation No.
RELATIVE MASS <>Isotope Carbon-12 as a reference or standard for comparing the masses of other atoms. Definition: Relative atomic mass => mass of 1atom of an element(a.m.u.) 1/12 mass of 1 atom of C-12 isotope(a.m.u.) e.g. Relative atomic mass of Na = mass of 1atom Na 1/12 mass of 1 atom 12 C isotope = 23 a.m.u ./1.0 a.m.u = 23
Relative molecular mass Definition: Sum of all the relative atomic masses of all atoms present in a molecule. e.g. : Molecular formula of sulphuric acid is H 2 SO 4 , determine its relative molecular mass. [ RAM: H=1.00; S=32.0; O=16.0] Ans: RMM H 2 SO 4 = 2X1.00 + 32.0 + 4X16.0 = 98.0
1. For each of the substances:: i) Na 2 CO 3 ii. MgO. [ Mg=24.0; O= 16.0; Na=23.0; C=12.0] Find its relative molecular mass Exercise: .2. .Determine the relative atomic mass of an element X if the ratio of the atomic mass of X to carbon-12 is 0.75 3. The relative atomic mass of chlorine atom is 35.5 Explain why this no. is not a whole no.. 4. Relative atomic mass of bromine atom is 80. What is the mass of Br compared to carbon-12 atom? Ans: 9.00 Ans: 6.7
The mass spectrometer is used to : i. determine the atomic mass and molecular mass of substances. ii. reveal the presence of isotopes of elements and its relative abundance. MASS SPECTROMETRY Who invented the mass spectrometer?
Vaporisation chamber Acceleration chamber (Electric field) Magnetic field Ionization chamber Ion detector vacuum Figure : Diagramatic representation of Mass Spectrometer A B
Vaporisation Chamber Sample of the element is vaporised into gaseous atom Ionisation Chamber A gaseous sample (atom or molecule) is bombarded by a stream of high-energy electrons that are emitted from a hot filament. Collisions between the electrons and the gaseous atom (or molecule) produce positive ions by dislodging an electron from each atom or molecule M + e M + + e + e
Acceleration Chamber (Electric field) The positive ions are accelerated by an electric field towards the two oppositely charged plates. The electric field is produced by a high voltage between the two plates. The emerging ions are of high and constant velocity Magnetic field The positive ions are separated and deflected into a circular path by a magnet according to its mass/charge ( m/e) ratio . Positive ions with small m/e ratio are deflected most and appear near A. Ions with large m/e ratio are deflected least and appear near B
Ion detector The numbers of ions and types of isotopes are recorded as a mass spectrum . In practice, the ion detector is kept in a fixed position. The magnetic field is varied so that the positive ions of different masses arrive at the detector at different times
Mass spectrum : The horizontal axis = m/e ratio or nucleon number or isotopic mass or relative atomic mass of the ions entering the detector. The vertical axis= the abundance or detector current or relative abundance or ion intensity or percentage abundance of the ions. The height is proportional to the amount of each isotope present.
Relative abundance 63 8.1 9.1 0 24 25 26 m/e The mass spectrum of magnesium shows that naturally occurring magnesium consists of three isotopes: 24 Mg, 25 Mg and 26 Mg. The height of each line is proportional to the abundance of each isotope. In this example, 24 Mg is the most abundance of the three isotopes
A r Mg = (24 a.m.u. x 63) + (25 a.m.u. x 8.1) + (26 a.m.u x 9.1) (63 + 8.1 + 9.1) = 24.33 A r = where Q = the abundance of an isotope of the element = the percentage of the isotope found in the naturally occurring element m = the relative isotopic mass of the element General formula for A r
. The relative atomic mass of an element of
= ( % abundance X 1 x isotopic mass X 1 +
% abundance X 2 x isotopic mass X 2 ) ÷
sum of % abundance X 1, X 2
Mass spectrum of an atom shows the
presence of isotopes
by representative lines with lengths
the abundance of their ions
Eg.: Boron comprises of 2 isotopes, i.e. 14.6 % 10 B dan 85.4 % 11 B with their isotopic masses of 10.01294 dan 11.00931 respectively. Calculate the relative atomic mass of B. Ans: RAM Boron = % 10 B x A r + % 11 B x A r 100 % = 14.6 x 10.01294 + 85.4 x 11.00931 % 100% = 10.8
1. The atomic masses of 6 Li and 7 Li are 6.0151 amu and 7.0160 amu respectively. What is the relative abundance of each isotope if the relative atomic mass of lithium is 6.941 amu? Ans:(92.95%, 7.05 %) 2. Naturally occurring iridium, Ir is composed of 2 isotopes and in the ratio of 5:8. The relative isotopic mass of and are 191.021 and 193.025 respectively. Calculate the relative atomic mass of iridium. Ans:(192.254) Exercise your mind:
Objectives: #1 . Write chemical formulae of ionic compounds #2 Define mole in term of mass #2. Relate the no. of mole with Avogadro constant #3. perform calculation involving converting : i. Mole to mass ii. mass to mole #4. Relate the number of particles to mass through Avogadro ‘s constant, N A LECTURE 2
Chemical formulae of ionic compounds: Ionic compounds consists of : Cation(+ve ion) and anion(-ve ion) chemically bonded together by ionic bond. types : I. between monoatomic cation and anion e.g K + Cl - 2 . Monoatomic cation and polyatomic anion e.g. Mg 2+ CO 3 2- 3. Polyatomic cation and monoatomic anion e.g. ( NH 4 +) 2 S 2- 4. Polyatomic cation and anion e.g. . NH 4 + NO 3 - 5. Other combinations.
IUPAC names of some common anions & caions -ve Ion +ve Ion Chromate CrO 4 2- Chromium(III) Cr 3+ Carbonate CO 3 2- Manganese (II) Mn 2+ Sulphate SO 4 2- Manganese(IV) Mn 4+ Permanganate MnO 4 - Iron(III) Fe 3+ Nitrate NO 3 - Iron(II) Fe 2+ Hydroxide OH - Aluminium Al 3+ Nitride N 3- Barium Ba 2+ Sulphide S 2- Ammonium NH 4 + Bromide Br - Potassium K +
IUPAC names of some common anions & caions -ve Ion +ve Ion Phosphate PO 4 3- Lead(IV) Pb 4+ Oxide O 2- Lead(II) Pb 2+ Sulphite SO 3 2- Tin(IV) Sn 4+ Cyanide CN - Zinc Zn 2+ Nitrite NO 2 - Barium Ba 2+ Chlorite ClO 2 - Cobalt (II) Co 2+ Chlorate ClO 3 - Silver Ag + peroxide O 2 - Copper (I) Cu + Chloride Cl - Hydrogen H +
Writing chemical formulae of ionic compounds e.g. Potassium sulphite: . K + SO 3 2- 2- 1+ X2 =2 x2 K 2 SO 3 Steps: : i. write symbol of cation followed by that of anion. ii. Combine the 2 , make sure their total charge = 0 e.g. 2K + + SO 3 2- = (2+) + (2-) = 0 Hence the chemical formula for potassium sulphite is K 2 SO 3
Try write chemical formula for:
a. Calcium hydrogen carbonate
b. Aluminium oxide
c. Sodium nitride
d. Potasium dichromate
e Barium sulphide
f. Potassium permanganate
2. Name the following compounds using IUPAC Nomenclature: i. K 2 CrO 4 ii. Na 2 Cr 2 O 7 iii. Ca(OH) 2 iv. ZnSO 4 v. Al 2 (CO 3 ) 3 vi. HNO 3 vii. MgS viii. MgSO 4 .7H 2 O
The amount of substance which contains equal amount of elementary entities ( atoms, molecules, ions etc.) as there are 12 C atoms in 12.00g of 12 C .
In 12.00 g of 12 C, there are 6.022045 x 10 23 atoms which is 1 mole of the substance carbon-12
Therefore 1 mole is just a unit to represent an amount equal to 6.022045 x 10 23
1 mole = 6.022045 x 10 23 = Avogadro constant , N A
MOLE: 12.00g of carbon-12 will have the same number of atoms as that of 1 H atoms in 1.00 g of hydrogen-1, i.e. 1 mole of any substance consists of 6.022045 x 10 23 = N A particles ( atoms, ions, molecules) of it.
Exercise 1. Which substance has more atoms?
1 mole of Na or 1 mole of B?
0.11 mole of Mg or 0.72 mole of Cl?
Symbol used for mole is n
MOLAR MASS: Is the mass of 1 mole of an element or compound. E.g. 1 mole of copper,Cu , contains 6.022045 x 10 23 atoms and weighs 63.5 g . What is the molar mass of Cu? Ans: 63.5 g mol -1 Unit used = g mol -1 Mass of Cu atom = 63.5 amu ( from mass spectrometry) 1 mole Cu atoms ≡ 6.022 x 10 23 atoms mole -1 Cu x 63.5 a.m..u/atom Cu Since 1 a.m.u. = 1.6605 x 10 -24 g (from Std.atomic mass of 12 C scale) Hence molar mass of Cu, M Cu =( 3.824 x 10 25 a.m.u. mol -1 x 1.66505 x 10 -24 g / 1 a.m.u.) = 63.5 g mol -1 molar mass is taken as the mass with value of the relative atomic mass , A r or molecular mass, M r in unit g mole -1
MOLAR MASS OF ATOM (a) 30.97 g mole -1 Exercise 2: (a) What is the molar mass of phosphorus with Its relative atomic mass of 30.97? (b) 0.50 mole of K weighs 19.50 g. What is its molar mass? (b) 39.0 g mole -1 MOLAR MASS OF a molecule/compound Is the sum of atomic masses of the elements in 1 mole of the molecule or compound. e.g. Find the molar mass of Cl 2 Ans: Molar mass of Cl 2 , M Cl = 2 x M Cl = 2 x 35.5 g mole -1 = 70.0 g mole -1 2
MOLAR MASS OF a molecule/compound Exercise 3: Find the molar mass of SO 2. [Given A r of : S = 32.0, O = 16.0] Ans : 64.0 g mole -1 Relate mass to mole of a substance Relate mole to mass to become quantity we can measure, ie. That can be weighed. Mass (g) = moles(n) x molar mass(g mole -1 ) i.e. m = n x M
e.g.1. . What is the mass of 2.50 moles of iron, Fe? [ Fe = 56.0] From m = n x M, Hence mass of Fe, m = 2.50 moles x 56.0 g mole -1 = 140.0 g e.g. 2 : How may mole of N 2 contained in 1.40 g of N 2 (g)? . From m = n x M, I.e. n = m / M Ans: Mole of N 2 , n = 1.40 g / 28.0 g mole -1 = 0.050 mole
Exercise 4: [Given A r : C = 12.00 ; O = 16.0, Ca = 40.0]
(a) Find the mass of 0.25 mole of CO.
(b) State the mass of 1.50 mole of Ca.
(c ) Find the amount(moles)
of element in 54.0g of Al.
of ion in 4.8 g of Mg 2+
Ans: ( a) 7.00 g (b) 60.0 g ( c) I) 0.2 mole ii) 0.2 mole
no. of particles = n x N A e.g. 1. Find the no. of ion in 0.80 mole of O 2- Ans: No. of ion O 2- = 0.80 mole x 6.022 x 10 23 ions mole -1 = 4.48 x 10 24 ions Relate mole to no. of elementary particles e.g. Find the amount(moles) in 6.022 x 10 24 atoms of Na? Ans: n = 6.022 x 10 24 atoms = 10.00 moles 6.022 x 10 23 atoms mole -1
Exercise 5: Find the no. of i) atoms in 1.20 mole of S. ii) Molecules in 40.0 g of CO 2
7.23 x 10 23
5.47 x 10 23
Exercise 6: How many moles are there in 2.50 x 10 26 of NO 2 molecules? What is the mass of NO 2 ? [ N= 14.0; O = 16.0} Ans : 4.15 x 10 2 mole ; 19090 g
Summary/review #1 Mole is an amount of substance containing the same no. of elementary. particles as that of 12 C atoms in 12.00g of 12 C. #2 molar mass of different substances contain the same no. of elementary particles. e.g. the no. of : Na atoms in 23.0 g = CO 2 molecules in 44.0g #3 Molar mass Ξ 1 mole Ξ 6.022 x 10 23 ( i.e.N A ) #4 Molar atomic mass obtained by expressing the A r in g mole -1 #5 Molar molecular mass = sum of molar atomic masses of all elements in molecule.
Relate mole to mass and no. of particles. Summary/review # 1 Convert mole to mass: m = n x M #2 Convert mass to mole: n = m / M #3 Convert mole to no. of elementary particles: no. of particles = m x N A #4 Convert no of elementary particles to mole: n = no. of particles / N A #5 Relate mole to no. of elementary particles No. of particles(atom/ ion/molecule) = n x N A #6 Relate no. of elementary particles to mole n = No. of particles / N A
THE MOLE CONCEPT Objectives: #1 . Perform calculation involving converting i. mass to No. of elementary particles ii. no. of particles to mass #2. Relate mole with molar volume of gases #4. Relate volume of a gas to quantity in moles at : i. STP ii. room condition LECTURE 3
Convert mass to No. of elementary particles mass mole No. of particles e.g. How many molecules are found in 34.0 g of NH 3 ? Ans: molar mass of NH 3 = 14.0 + 3.00 = 17.0 g mole -1 moles of NH 3 , n = m / M = 34.0 g / 17.0 g mole -1 = 2.00 mole No. of NH 3 molecules = 2.00 mole x 6.022 x 10 23 molecules mole -1 = 1.20 x 10 24 molecules i.e. No. of particles = m x N A M
Convert no. of particles to mass No. of particles mole mass i.e. m = no. of particles x M N A e.g. What is the mass of 1.807 x 10 25 formula units of sodium hydroxide, NaOH ? Ans: molar mass, M NaOH = 23.0+16.0+1.00 = 40.0 g mole -1 Hence mass of NAOH, m NaOH = 1.807 x 10 24 f-units x 40.0 g mole -1 6.022 x 10 23 f-units mole -1 = 1200 g
How many molecules are there in 3.65 g of
hydrogen chloride, HCl ?
2. How may oxygen atoms in 20.0 g of SO 3
3. Find the mass of 1.25 x 10 22 Ca 2+ ions ?
4. Which is heavier , 6.022 x 10 30 H 2 molecules or
1.205 x 10 21 atom Na?
5. What mass of sulphur dioxide, SO 2 contains the
same no. of molecules as are in 1.00 g of ammonia?
#1. 6.022 x 10 22 #2. 3x 1.506 x 10 23 #3. 0.830g #4. H 2 #5. 3.77 g
Relate mole with molar volume of gases Molar volume of any gases at STP, V m = 22.4 L mole -1 STP =standard temperature and pressure Where T = 273.15 K P = 1 atm @ 760 mm Hg Amount of gas at STP, n = V gas (L) V m (22.4 L mole -1 ) e.g. How many moles are there in 6.5 L oxygen at STP ? n = 6.5 L O 2 x 1 mole O 2 = 0.29 mole 22.4 L O 2
Relate volume with molar volume of gases Molar volume of any gases at STP, V m = 22.4 L mole -1 STP =standard temperature and pressure Where T = 273.15 K P = 1 atm @ 760 mm Hg Amount of gas at STP, n = V gas (L) Vm (22.4 L mole -1 ) e.g. How many moles are there in 6.5 L oxygen at STP ? n = 6.5 L O 2 x 1 mole O 2 = 0.29 mole 22.4 L O 2
Relate volume of a gas to quantity in moles Molar volume of any gases at STP, V m = 22.4 L mole -1 STP =standard temperature and pressure Where T = 273.15 K P = 1 atm @ 760 mm Hg At STP V gas (L) = n x V m (22.4 L mol -1 ) e.g. Find volume of 0.25 mole of HCl gas at STP ? ANS: V HCl = 0.25 mol HCl x 22.4 L mol -1 = 5.6 L
Exercise/e.g: Calculate the mass of 4.80 L of hydrogen gas at room conditions. Solution: using V gas (L) = n x V m (24 L mol -1 ) n = ( 4.80 / 24 ) mol mass of hydrogen = ( 4.80 / 24 ) mol x 2.00 g mol -1 = 0.40 g Molar volume of any gases at room conditions, V m = 24L mole -1 Room temperature & pressure Where T = 25 0 [email_address] K P = 1 atm @ 760 mm Hg
A balloon filled with gas at STP occupies a volume of 2.24 L.
Calculate the amount of hydrogen gas. [Ans: 0.1 mole.]
2. Calculate the volume of 24 x 10 23 molecules of gas at STP.
[Ans: 89.6 L ]
3. A sample of CO 2 has a volume of 56 mL at STP. Calculate: (a) the no. of moles of CO 2 molecules: (b) the no. of CO 2 molecules ( c) the no. of oxygen atoms in the sample ( d) the mass of CO 2 4. For a sample of chlorine gas with a mass of 71.0g, Find: a) the amount( mole) of its molecules b) its volume at STP c) its volume at room conditions. Ans: 3 (a) 0.0025 b) 1.5 x 10 21 c) 3.0 x 10 21
Summary/review #1 Relate mass to No. of elementary particles : No. of particles = m x N A M where m = mass of substance M = its molar mass; NA = 6.022 x 1022 #2 Relate mole with molar volume of gases Amount of gas at STP, n = Vgas (L) V m (22.4 L mol -1 ) Amount of gas at room conditions, n = Vgas (L) V m (24 L mol -1 )
THE MOLE CONCEPT Objectives: #1 . Define i. Empirical formula ii. Molecular formula of a chemical compound. #2. Determine the empirical formula and molecular formula of a compound from it’s a) mass composition, b) combustion data. #3 Determine the % composition by mass of a chemical compound.
Chemical Formula Empirical Formula indicates which elements are present and the simplest whole-number ratio of their atoms in a molecule. Molecular Formula shows the exact number of atoms of each element in the smallest unit of a substance. molecula r formula = (empirical formula) n Molecules Empirical formula Molecular Formula n Water H 2 O H 2 O 1 Hydrogen peroxide HO H 2 O 2 2 Benzene CH C 6 H 6 6 Ethyne CH C 2 H 2 2
Which 2 compounds have same empirical formula and molecular formula ? Give 2 molecules that might have different molecular formulae but the same empirical formula 1. Calculating the empirical formula from the masses of constituents E.g. 18.3 g sample of hydrated compound contained 4.0 g of calcium, 7.1 g of chlorine and 7.2 g of water only. Calculate its empirical formula.
Empirical formula = CaCl 2 . 4H 2 O Constituent Ca Cl H 2 O Mass/g 4.0 7.1 7.2 Amount/mol = 0.10 = 0.20 = 0.40 Simplest ratio of relative amount = 1.0 = 2.0 = 4.0 1 2 4
Exercise 50.0 g of phenol. A general disinfectant, has 38.29g C, 3.21 g H ang 8.50 g O. Determine its empirical formula. Ans: C 6 H 6 O Calculating the empirical formula from percentage composition by mass Example : Ascorbic acid (vitamin C) cures scurvy and may help prevent the common cold. It is composed of 40.92% carbon, 4.58% hydrogen and 54.50% oxygen by mass. The molar mass of ascorbic acid is 176 g mol 1 . Determine its empirical formula and molecular formula
Empirical formula = C 3 H 4 O 3 n = = = 2 Molecular formula = (C 3 H 4 O 3 ) n = (C 3 H 4 O 3 ) 2 = C 6 H 8 O 6 Element C H O Mass/g 40.92 4.58 54.50 Amount/mol = 3.41 = 4.58 = 3.41 Simplest ratio = 1 = 1.33 = 1 1x3 1.33x3 1x3 3 4 3
Allicin is the compound responsible for the characteristic smell of garlic. An analysis of the compound gives the following percent composition by mass: C: 44.4%; H: 6.21%; S: 39.5%; O: 9.86%. Calculate its empirical formula. What is its molecular formula given that its molar mass is about 162 g mol 1 .
2) Determine the formula of a mineral with the following mass composition: Na=12.1%, Al=14.2%, Si=22.1%, O=42.1%, H 2 O=9.48%.
Calculating the empirical formula from elemental analysis data (Combustion data) E.g 1.00 g sample of compound A was burnt in excess oxygen producing 2.52 g of CO 2 and 0.443 g of H 2 O. Determine the empirical formula of the compound 1 mol CO 2 contains 1 mol C & 1 mol H 2 O 2 mol H = 0.688
the mass of oxygen = m sample – (m C + m H ) = 1.00 g – (0.688 g + 0.0492 g) = 0.263 g O Element C H O Mass/g 0.688 0.0492 0.263 Amount/mol = 0.0573 = 0.0492 = 0.0164 Simplest ratio of relative amount = 3.49 = 3.00 = 1.00 3.49x2 3.00x2 1.00x2 7 6 2
Empirical formula = C 7 H 6 O 2 Exercise 1. 0.535 g sample of thiophene, a carbon-hydrogen-sulphur compound used in manufacturing of pharmaceuticals, is burned completely in excess oxygen to yield 1.119 g CO 2 , 0.229 g H 2 O, and 0.407 g SO 2 . Determine empirical formula of thiopene.
Percentage composition from formula For an element in a compound = atoms of element x RAM x 100 Relative formula mass e.g. Calculate the % composition of hydrogen in C 12 H 22 O 11 Where RFM C 12 H 22 O 11 = 342.0 ANS: % C = 22(1.0) X 100 = 6.4 % 342.0 Exercise: Calculate the % of nitrogen, by mass, in Ca(NO 3 ) 3 Ans: 17.1