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CHEMISTRYCHEMISTRYCHEMISTRY
Cof atom one of Mass X 121
element of atom one of Mass Amass,atomic lativeRe 12r
C of atom one of Mass x 121
molecule one of Mass
Mr mass, molecular Relative
12
abundance
)abundancemass (isotopic mass atomic Average
e% abundancmass) isotopicabundance (%
)2.93.05.90()u 222.9()u 213.0()u 205.90(
e% abundancmass) isotopicabundance (%
)91.3009.69()u 93.6491.30()u 93.6209.69(
abundancemassisotopic abundance
)85()u 025.1938()u 021.1915(
+ _ _
MATTER1.2 Mole Concept
Learning Outcome
At the end of this topic, students should be able to:
a) Define mole in terms of mass of carbon-12 and Avogadro’s constant, NA
Avogadro’s Number, NA
Atoms and molecules are so small – impossible to count
A unit called mole (abbreviated mol) is devised to count chemical substances by weighing them
A mole is the amount of matter that contains as many objects as the number of atoms in exactly 12.00 g of carbon-12 isotope
The number of atoms in 12 g of 12C is called Avogadro’s number, NA = 6.02 x 1023
Example:
1 mol of Cu contains Cu atoms
1 mol of O2 contains O2 molecules
O atoms
1 mol of NH3 contains NH3 molecules
N atoms
H atoms
6.02 6.02 10102323
33 6.02 6.02 10102323
22 6.02 6.02 101023236.02 6.02 10102323
6.02 6.02 10102323
6.02 6.02 10102323
1 mol of CuCl2 contains Cu2+ ions
Cl- ions
6.02 1023
2 6.02 1023
Mole and MassExample:
Relative atomic mass for carbon, C = 12.01
Mass of 1 C atom = 12.01 amu
Mass of 1 mol C atoms = 12.01 g
Mass of 1 mol C atoms consists of 6.02 x 1023 C atoms
= 12.01 g
Mass of 1 C atom =
= 1.995 x 10-23 g
2310 x 6.02 g 01.12
12.01 amu = 1.995 x 10-23 g
1 amu =
= 1.66 x 10-23 g
amu 12.01g 10 x .9951 23
Example:
From the periodic table, Ar of nitrogen, N is
� The mass of 1 N atom =
� The mass of 1 mol of N atoms =
� The molar mass of N atom =
� The molar mass of nitrogen gas =
The nucleon number of N =
14.0114.01
14.01 amu14.01 amu
14.01 g14.01 g
14.01 g mol14.01 g mol11
28.02 g mol28.02 g mol11
1414
Mr of CH4 is
� The mass of 1 CH4 molecule =
� The mass of 1 mol of CH4 molecules =
� The molar mass of CH4 molecule =
16.0516.05
16.05 amu16.05 amu
16.05 g16.05 g
16.05 g 16.05 g molmol11
Learning Outcome
At the end of this topic, students should be able to:(a) Interconvert between moles, mass,
number of particles, molar volume of gas at STP and room temperature.
(b) Define the terms empirical & molecular formulae
(c) Determine empirical and molecular formulae from mass composition or from mass composition or combustion data.combustion data.
Example 1:Calculate the number of moles of molecules for
3.011 x 1023 molecules of oxygen gas.
Solution:
6.02 x 1023 molecules of O2
3.011 x 1023 molecules of O2
= 0.5000 mol of O2 molecules
1 mol of O1 mol of O22 molecules molecules
molecules 10 6.02mol 1 molecules
23 2310011.3
Example 2:Calculate the number of moles of atoms for
1.204 x 1023 molecules of nitrogen gas.
Solution:
6.02 x 1023 molecules of N2
2 mol of N atoms
1.204 x 1023 molecules of N2
= 0.4000 mol of N atoms
1 mol of N1 mol of N22 molecules molecules
molecules mol 2 molecules
23
23
1002.610204.1
Example 3:Calculate the mass of 0.25 mol of chlorine gas.
Solution:
1 mol Cl2
0.25 mol Cl2
18 g
or
mass = mol x molar mass
= 0.25 mol x (2 x 35.45 g mol-1)
= 18 g
2 2 35.45 g 35.45 g
mol 1mol 0.25 g 35.45 2
Example 4:Calculate the mass of 7.528 x 1023 molecules of
methane, CH4Solution:
6.02 x 1023 CH4 molecules (12.01 + 4(1.01)) g
7.528 x 1023 CH4 molecules
= 20.06 g
23
23
1002.6107.528 g 05.16
Molar Volume of Gases Avogadro (1811) stated that equal volumes of gases
at the same temperature and pressure contain equal number of molecules
Molar volume is a volume occupied by 1 mol of gas
At standard temperature and pressure (STP), the molar volume of an ideal gas is 22.4 L mol 1
Standard Temperature and Pressure
273.15 K 1 atm 760 mmHg
0 C 101325 N m-2
101325 Pa
Standard Molar Volume
At room conditions (1 atm, 25 C), the molar volume of a gas = 24 L mol-1
Example 1:
Calculate the volume occupied by 1.60 mol of Calculate the volume occupied by 1.60 mol of ClCl22 gas at STP. gas at STP.Solution:At STP,At STP,1 mol Cl1 mol Cl22 occupiesoccupies
1.60 mol Cl1.60 mol Cl22occupies occupies
= 35.8 L= 35.8 L
22.4 L
mol 1L 4.22mol 60.1
Example 2:
Calculate the volume occupied by 19.61 g of Calculate the volume occupied by 19.61 g of NN22 at STP at STPSolution:1 mol of N1 mol of N22 occupiesoccupies 22.4 L22.4 L
of N of N22 occupies occupies
= 15.7 L= 15.7 L
mol 1
L 22.4mol )01.14(2
61.19
1mol g 2(14.01)g 61.19
Example 3:0.50 mol methane, CH0.50 mol methane, CH44 gas is kept in a cylinder at gas is kept in a cylinder at STP. Calculate:STP. Calculate:(a)(a) The mass of the gasThe mass of the gas(b)(b) The volume of the cylinderThe volume of the cylinder(c)(c) The number of hydrogen atoms in the cylinderThe number of hydrogen atoms in the cylinderSolution:(a)(a) Mass of 1 mol CHMass of 1 mol CH44 ==
Mass of 0.50 mol CHMass of 0.50 mol CH44 ==
= 8.0 g= 8.0 g
16.05 g
mol 1mol 0.50g 05.16
(b)(b)At STP;At STP; 1 mol CH1 mol CH44 gas gas occupiesoccupies
0.50 mol CH0.50 mol CH44 gas gas occupies occupies
= 11 L = 11 L
(c)(c) 1 mol of CH1 mol of CH44 molecules molecules 4 mol of H atoms4 mol of H atoms
0.50 mol of CH0.50 mol of CH44 molecules molecules 2 mol of H atoms2 mol of H atoms
1 mol of H atoms1 mol of H atoms
2 mol of H atoms2 mol of H atoms
1.2 x 10 1.2 x 102424 atoms atoms
22.4 L
6.02 x 1023 atoms
2 x 6.02 x 1023 atoms
mol 1mol 0.50 L 4.22
Exercise A sample of CO2 has a volume of 56 cm3 at STP.
Calculate:
a) The number of moles of gas molecules (0.0025 mol)
b) The number of CO2 molecules (1.506 x 1021 molecules)
c) The number of oxygen atoms in the sample (3.011x1021atoms)
Notes: 1 dm3 = 1000 cm3
1 dm3 = 1 L
Empirical And Molecular Formulae
Empirical formula => => chemical formula that shows the simplest ratio of all elements in a molecule.
Molecular formula => => formula that show the actual number of atoms of each element in a molecule.
The relationship between empirical formula and molecular formula is :
Molecular formula = n ( empirical formula )
Example:
A sample of hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Its molar mass is 56. Determine the empirical formula and molecular formula of the compound.
== = = == = =
Mass
Number of moles
Simplest ratio
85.7
21.984
14.1584
1
7.1357
14.3
01.127.85
01.13.14
C H
Empirical formula = Empirical formula = CH 2
14.0356 n
4 3.99
84
2
HC formula Molecular )n(CH formula Molecular
Exercise: A combustion of 0.202 g of an organic sample that contains carbon, hydrogen and oxygen produce 0.361g carbon dioxide and 0.147 g water. If the relative molecular mass of the sample is 148, what is the molecular formula of the sample?
Answer : C6H12O4
At the end of this topic, students should be At the end of this topic, students should be able to:(a) (a) Define and perform calculation for each of for each of
the the following concentration measurements :following concentration measurements : i) molarity (M) ii) molality(m) iii) mole fraction, X iv) percentage by mass, % w/w v) percentage by volume, %v/v
Learning Outcome
Concentration of Solutions
A solution is a homogeneous mixture of two or more substances:
solvent + solute(s)
e.g: sugar + water – solution
sugar – solute
water – solvent
Concentration of a solution can be expressed in various ways :
a) molarityb) molalityc) mole fractiond) percentage by mass e) percentage by volume
a) Molarity
Molarity is the number of moles of solute in 1 litre of solution
Units of molarity: mol L-1
mol dm-3
M
(L) solution of volume(mol) solute of molesM molarity,
Example 1: Determine the molarity of a solution
containing 29.22 g of sodium chloride, NaCl in a 2.00 L solution.
Solution:
solution
NaClNaCl V
n M
L 00.2
mol )45.3599.22(
22.29
= 0.250 mol L-1
Example 2:How many grams of calcium chloride, CaCl2 should be used to prepare 250.00 mL solution with a concentration of 0.500 M
Solution:
solutionCaClCaCl x VM n22
= 0.500 mol L 1 250.00 10 3 L
massmolar x n CaClof mass2CaCl2
= (0.500 250.00 10 3 ) mol (40.08 + 2(35.45)) g mol 1
= 13.9 g
b) Molality
Molality is the number of moles of solute dissolved in 1 kg of solvent
Units of molality:mol kg-1
molal m
(kg) solvent of mass(mol) solute of moles m molality,
Example:What is the molality of a solution prepared by dissolving 32.0 g of CaCl2 in 271 g of water?
Solution:
1-CaCl mol g 2(35.45) 08.40g 0.32
n2
kg 10271
mol 98.1100.32
CaClof Molality 32
1kg mol 1.06
Exercise:Calculate the molality of a solution prepared by dissolving 24.52 g of sulphuric acid in 200.00 mL of distilled water. (Density of water = 1 g mL-1)
Ans = 1.250 mol kg-1
c) Mole Fraction (X)
Mole fraction is the ratio of number of moles of one component to the total number of moles of all component present.
For a solution containing A, B and C:
T
A
CBA
AA
nn
n n nn
Xof A, fraction Mol
Mol fraction is always smaller than 1
The total mol fraction in a mixture (solution) is equal to one.
XA + XB + XC + X….. = 1
Mole fraction has Mole fraction has no unit (dimensionless) since it is a ratio of two similar quantities.
Example:A sample of ethanol, C2H5OH contains 200.0 g of ethanol and 150.0 g of water. Calculate the mole fraction of(a) ethanol(b) waterin the solution.
Solution:
n ethanol = 1mol g 16.00)5(1.01) (2(12.01)g 0.200
n water = 1mol g 16.00)(2(1.01)g 0.150
X ethanol =
mol 02.18
0.150mol
07.450.200
mol 07.45
0.200
= 0.3477
X water = 1 0.3477= 0.6523
d) Percentage by Mass (%w/w)
Percentage by mass is defined as the percentage of the mass of solute per mass of solution.
Note: Mass of solution = mass of solute + mass of solvent
100xsolutionof masssoluteof mass
ww%
Example:A sample of 0.892 g of potassium chloride, KCl is dissolved in 54.362 g of water. What is the percent by mass of KCl in the solution?
Solution:
%100g 54.362g 0.892
g 892.0 mass %
= 1.61%
Exercise:A solution is made by dissolving 4.2 g of sodium chloride, NaCl in 100.00 mL of water. Calculate the mass percent of sodium chloride in the solution.
Answer = 4.0%
e) Percentage by Volume (%V / V)
Percentage by volume is defined as the percentage of volume of solute in milliliter per volume of solution in milliliter.
Note:
solutionof volume solutionof mass
solutionof Density
100 x solution of volumesolute of volume v
v % 100 x solution of volumesolute of volume v
v %
Example 1: 25 mL of benzene is mixed with 125 mL
of acetone. Calculate the volume percent of benzene solution.
Solution:
100% mL 125 mL 25
mL 25 volume%
= 17%
Example 2:A sample of 250.00 mL ethanol is labeled as 35.5% (v/v) ethanol. How many milliliters of ethanol does the solution contain?
Solution:
100%mL 00.250%5.35
Vethanol
%100VV
ethanolof volume%solution
ethanol
= 88.8 mL
Example 3:A 6.25 m of sodium hydroxide, NaOH solution has has a density of 1.33 g mL-1 at 20 ºC. Calculate the concentration NaOH in:(a) molarity(b) mole fraction(c) percent by mass
(a) M = solution
NaOHVn
6.25 m of NaOH there is 6.25 mol of NaOH in 1 kg of water
for a solution consists of 6.25 mol of NaOH and 1 kg of water;
V solution = solution
solutionmass
mass solution = mass NaOH + mass water
mass NaOH = n NaOH molar mass of NaOH = 6.25 mol (22.99 + 16.00 + 1.01) g mol 1
= 250 g
mass solution = 250 g + 1000 g = 1250 g
V solution = 1mL g 1.33g 1250
M NaOH =
L 10
33.11250
mol 25.63
= 6.65 mol L 1
(b) X NaOH = waterNaOH
NaOHnn
n
1 kg of water contains 6.25 mol of NaOH
n water = waterof mass molar
mass water
= 1mol g 16.00) (2(1.01)g 1000
X NaOH =
mol
02.181000 mol 25.6
mol 25.6
= 0.101
(c) %(w/w) of NaOH = waterNaOH
NaOHmassmass
mass
100%
= g 1000 g 250
g 250
100%
= 20.0%
Exercise:An 8.00%(w/w) aqueous solution of ammonia has a density of 0.9651 g mL-1. Calculate the(a) molality (b) molarity(c) mole fraction
of the NH solution
Answer: a) 5.10 mol kg-1
b) 4.53 mol L-1
c) 0.0842
MATTER1.3 Stoichiometry
Learning Outcome
At the end of the lesson, students should be able to:
a) Determine the oxidation number of an element in a chemical formula.
b) Write and balance :i) Chemical equation by inspection methodii) redox equation by ion-electron method
Balancing Chemical Equation
A chemical equation shows a chemical reaction using symbols for the reactants and products.
The formulae of the reactants are written on the left side of the equation while the products are on the right.
Example:
x A + y B z C + w D
Reactants Products
A chemical equation must have an equal number of atoms of each element on each side of the arrow
The number x, y, z and w, showing the relative number of molecules reacting, are called the stoichiometric coefficients.
A balanced equation should contain the smallest possible whole-number coefficients
The methods to balance an equation: a) Inspection method b) Ion-electron method
Inspection Method
1. Write down the unbalanced equation. Write the correct formulae for the reactants and products.
2. Balance the metallic atom, followed by non-metallic atoms.
3. Balance the hydrogen and oxygen atoms.
4. Check to ensure that the total number of atoms of each element is the same on both sides of equation.
Example: Balance the chemical equation by applying the inspection method.
NH3 + CuO → Cu + N2 + H2O
Exercise
Balance the chemical equation below by applying inspection method.
1. Fe(OH)3 + H2SO4 → Fe2(SO4)3 + H2O
2. C6H6 + O2 → CO2 + H2O
3. N2H4 + H2O2 → HNO3 + H2O
4. ClO2 + H2O → HClO3 + HCl
Redox Reaction
Mainly for redox (reduction-oxidation) reaction
Oxidation is defined as a process of electron loss. The substance undergoes oxidation
loses one or more electrons.
increase in oxidation number
act as an reducing agent (electron donor)
Half equation representing oxidation: Mg Mg2+ 2e Fe2+ Fe3+ + e 2Cl- Cl2 + 2e
Reduction is defined as a process of electron gain. The substance undergoes reduction
gains one or more electrons.
decrease in oxidation number
act as an oxidizing agent (electron acceptor)
Half equation representing reduction: Br2 + 2e → Br-
Sn4+ + 2e → Sn2+
Al3+ + 3e → Al
Oxidation numbers of any atoms can be determined by applying the following rules:
1. For monoatomic ions, oxidation number = the charge on the ione.g: ion oxidation number
Na+ +1Cl- -1Al3+ +3S2- -2
2. For free elements, e.g: Na, Fe, O2, Br2, P4, S8 oxidation number on each atom = 0
1. For most cases, oxidation number for O = -2 H = +1
Halogens = -1
Exception:
1. H bonded to metal (e.g: NaH, MgH2) oxidation number for H = -1
2. Halogen bonded to oxygen (e.g: Cl2O7) oxidation number for halogen = +ve
3. In a neutral compound (e.g: H2O, KMnO4) the total of oxidation number of every atoms that made up the molecule = 0
4. In a polyatomic ion (e.g: MnO4-, NO3
-) the total oxidation number of every atoms that made up the molecule = net charge on the ion
Exercise
1. Assign the oxidation number of Mn in the following chemical compounds.i. MnO2 ii. MnO4
-
1. Assign the oxidation number of Cl in the following chemical compounds.i. KClO3 ii. Cl2O7
2-
1. Assign the oxidation number of following:i. Cr in K2Cr2O7ii. U in UO2
2+
iii. C in C2O42-
Balancing Redox Reaction
Redox reaction may occur in acidic and basic solutions.
Follow the steps systematically so that equations become easier to balance.
Balancing Redox Reaction In Acidic Solution
Fe2+ + MnO4- → Fe3+ + Mn2+
1. Separate the equation into two half-reactions: reduction reaction and oxidation reactioni. Fe2+ → Fe3+ ii. MnO4
- → Mn2+
1. Balance atoms other than O and H in each half-reaction separately
i. Fe2+ → Fe3+
ii. MnO4- → Mn2+
3. Add H2O to balance the O atomsAdd H+ to balance the H atoms
i. Fe2+ → Fe3+ ii. MnO4
- + → Mn2+ +
4. Add electrons to balance the charges
i. Fe2+ → Fe3+ + ii. MnO4
- + 8H+ + → Mn2+ + 4H2O
4H2O8H+
1 e
5 e
3.Multiply each half-reaction by an integer, so that number of electron lost in one half-reaction equals the number gained in the other.
i. 5 x (Fe2+ → Fe3+ + 1e)5Fe2+ → 5Fe3+ + 5e
ii. MnO4- + 8H+ + 5e → Mn2+ + 4H2O
1. Add the two half-reactions and simplify where possible by canceling the species appearing on both sides of the equation.
i. 5Fe2+ → 5Fe3+ + 5eii. MnO4
- + 8H+ + 5e → Mn2+ + 4H2O___________________________________5Fe2+ + MnO4
- + 8H+ → 5Fe3+ + Mn2+ + 4H2O
___________________________________
5. Check the equation to make sure that there are the same number of atoms of each kind and the same total charge on both sides.
5Fe 2+ + MnO 4- + 8H + → 5Fe 3+ + Mn 2+ + 4H 2O
Total charge reactant= 5(+2) + ( -1) + 8(+1)= + 10 - 1 + 8= +17
Total charge product= 5(+3) + (+2) + 4(0)= + 15 + (+2)= +17
5Fe 2+ + MnO 4- + 8H + → 5Fe 3+ + Mn 2+ + 4H 2O
Total charge reactant= 5(+2) + ( -1) + 8(+1)= + 10 - 1 + 8= +17
Total charge product= 5(+3) + (+2) + 4(0)= + 15 + (+2)= +17
Exercise: In Acidic Solution
C2O42- + MnO4
- + H+ → CO2 + Mn2+ + H2O
Solution :
Balancing Redox Reaction In Basic Solution
1. Firstly balance the equation as in acidic solution.
2. Then, add OH- to both sides of the equation so that it can be combined with H+ to form H2O.
3. The number of OH- added is equal to the number of H+ in the equation.
Example: In Basic Solution
Cr(OH)3 + IO3- + OH- → CrO3
2- + I- + H2O
Exercise:
1. H2O2 + MnO4- + H+ O2 + Mn2+ + H2O
(acidic medium)2. Zn + SO4
2- + H2O Zn2+ + SO2 + 4OH-
(basic medium)3. MnO4
- + C2O42- + H+ Mn2+ + CO2 + H2O
(acidic medium)4. Cl2 ClO3
- + Cl- (basic medium)
Stoichiometry
Stoichiometry is the quantitative study of reactants and products in a chemical reaction.
A chemical equation can be interpreted in terms of molecules, moles, mass or even volume.
C3H8 + 5O2 3CO2 + 4H2O
1 molecule of C3H8 reacts with 5 molecules of O2 to produce 3 molecules of CO2 and 4 molecules of H2O
6.02 x 1023 molecules of C3H8 reacts with 5(6.02 x 1023) molecules of O2 to produce 3(6.02 x 1023) molecules of CO2 and 4(6.02 x 1023) molecules of H2O
C3H8 + 5O2 3CO2 + 4H2O
1 mol of C3H8 reacts with 5 moles of O2 to produce 3 moles of CO2 and 4 moles of H2O
44.09 g of C3H8 reacts with 160.00 g of O2 to produce 132.03 g of CO2 and 72.06 g of H2O
5 moles of C3H8 reacts with 25 moles of O2 to produce 15 moles of CO2 and 20 moles of H2O
At room condition, 25 ºC and 1 atm pressure;22.4 dm3 of C3H8 reacts with 5(22.4 dm3) of O2 to produce 3(22.4 dm3) of CO2
Example 1:How many grams of water are produced in the oxidation of 0.125 mol of glucose?
C6H12O6(s) + O2(g) CO2(g) + H2O(l)
Solution:From the balanced equation;1 mol C6H12O6 produce 6 mol H2O
0.125 mol C6H12O6 produce H2O
mass of H2O = (0.125 x 6) mol x (2.02 + 16.00) g mol-1
= 13.5 g
mol 1mol 6mol 125.0
Example 2:Ethene, C2H4 burns in excess oxygen to form carbon dioxide gas and water vapour.(a) Write a balance equation of the
reaction(b) If 20.0 dm3 of carbon dioxide gas is
produced in the reaction at STP, how many grams of ethene are used?
Solution:(a) C2H4 + O2 CO2 + H2O
(b) 22.4 dm3 is the volume of 1 mol CO2
20.0 dm3 is the volume of CO2
2 mol CO2 produced by 1 mol C2H4
mol CO2 produced by C2H4
3
3
dm 4.22mol 1dm 0.20
4.220.20
mol 2
mol 1mol 4.220.20
mass ethane1-mol g 4(1.01)] [2(12.01) x mol
24.220.20
= 12.5 g
Learning Outcome
At the end of this topic, students should be able to:
a) Define the limiting reactant and percentage yield
b) Perfome stoichiometric calculations using mole concept including limiting reactant and percentage yield.
Limiting Reactant/Reagent
Limiting reactant is the reactant that is completely consumed in a reaction and limits the amount of product formed
Excess reactant is the reactant present in quantity greater than necessary to react with the quantity of limiting reactant
Example:3H2 + N2 2NH3If 6 moles of hydrogen is mixed with 6 moles of nitrogen,
how many moles of ammonia will be produced?
Solution:3 mol H2 reacts with 1 mol N2
6 mol H2 reacts with
mol 3
mol 1 mol 6
= 2 mol N 2
N2 is the excess reactantH2 is the limiting reactant limits the amount of products formed
3 mol H2 produce 2 mol NH3
6 mol H2 produce
mol 3mol 2 mol 6
= 4 mol NH 3
or1 mol N2 react with 3 mol H2
6 mol N2 react with mol NH3
mol 1mol 3 mol 6
= 18 mol H 2
H 2 is not enough
H 2 limits the amount of products formed
limiting reactant
3 mol H2 produce 2 mol NH3
6 mol N2 produce mol NH3
= 4 mol NH3
mol 3mol 2 mol 6
Exercise:Consider the reaction:
2 Al(s) + 3Cl2(g) 2 AlCl3(s)
A mixture of 2.75 moles of Al and 5.00 moles of Cl2 are allowed to react.
(a) What is the limiting reactant?(b) How many moles of AlCl3 are formed?(c) How many moles of the reactant remain at the end of the reaction?
PERCENTAGE YIELD
The amount of product predicted by a balanced equation is the theoretical yield
The theoretical yield is never obtain because: 1. The reaction may undergo side reaction 2. Many reaction are reversible 3. There may be impurities in the reactants
4. The product formed may react further to form other product
5. It may be difficult to recover all of the product from the reaction medium
The amount product actually obtained in a reaction is the actual yield
Percentage yield is the percent of the actual yield of a product to its theoretical yield
100 x yield
yield yield %ltheoretica
actual
Example 1:Benzene, C6H6 and bromine undergo reaction as follows:C6H6 + Br2 C6H5Br + HBr
In an experiment, 15.0 g of benzene are mixed with excess bromine
(a) Calculate the mass of bromobenzene, C6H5Br that would be produced in the reaction.
(b) What is the percent yield if only 28.5 g of bromobenzene obtain from the experiment?