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In today’s Chemistry class, We will be learning about Mass / Volume Relationship. We hope you enjoy the class!
CONTENT
 Mole and Molar Quantities
 Relative Atomic Mass and Relative Molecular Mass.
 Calculations involving Mass and Volume.
MOLE AND MOLAR QUANTITIES
THE MOLE
A mole is a number of particles of a substance which may be atoms, ions, molecules or electrons. This number of particles is approximately 6.02 x 10^{23} in magnitude and is known as Avogadro’s number of particles.
The mole is defined as the amount of a substance which contains as many elementary units as there are atoms in 12g of Carbon12.
RELATIVE ATOMIC MASS
The relative atomic mass of an element is the number of time the average mass of one atom of that element is heavier than onetwelfth the mass of one atom of Carbon12. It indicates the mass of an atom of an element. For e.g, the relative atomic mass of hydrogen, oxygen, carbon, sodium and calcium are 1, 16, 12, 23, and 40 respectively.
The atomic mass of an element contains the same number of atoms which is 6.02 x 10^{23}atoms; 1 mole of hydrogen having an atomic mass of 2.0g contains 6.02 x 10^{23} atoms.
RELATIVE MOLECULAR MASS
The relative molecular mass of an element or compound is the number of times the average mass of one molecule of it is heavier than onetwelfth the mass of one atom of Carbon12
It is the sum of the relative atomic masses of all atoms in one molecule of that substance. It is also called the formula mass. The formula mass refers not only to the relative mass of a molecule but also that of an ion or radical.
CALCULATION
Calculate the relative molecular mass of:
 Magnesium chloride
 Sodium hydroxide
 Calcium trioxocarbonate
[Mg=24, Cl=35.5, Na=23, O=16, H=1, Ca=40,C=12]
Solution:
 MgCl_{ 2} = 24 + 35.5×2 = 24 + 71 = 95gmol^{1}
 NaOH = 23 + 16 + 1 = 40gmol^{1}
 CaCO_{3} = 40 + 12 +16×3 = 100gmol^{1}
EVALUATION
 What is the relative molecular mass of a compound?
 Calculate the relative molecular mass of (a) NaNO_{3} (b) CuSO_{4}.5H_{2}O
MOLAR VOLUME OF GASES
The volume occupied by 1 mole of a gas at standard conditions of temperature and pressure (s.t.p) is 22.4 dm^{3}. Thus 1 mole of oxygen gas of molar mass 32.0gmol^{1} occupies a volume of 22.4dm^{3} at s.t.p and 1 mole of helium gas of molar mass 4.0gmol^{1} occupies a volume of 22.4 dm^{3} at s.t.p.
Note: When the conditions of temperature and pressure are altered, the molar volume will also change. Also, standard temperature = 273K and standard pressure = 760mmHg.
RELATIONSHIP BETWEEN QUANTITIES
$Molarmass=\frac{mass\left(g\right)}{Amount\left(moles\right)}i.eM=\frac{m}{n}gmol\u20131$
Note: Amount = Number of moles
$Molarvolumeofgas=\frac{volume(c{m}^{3}ord{m}^{3})}{Amount\left(mole\right)}i.e{V}_{m}=\frac{v}{n}d{m}^{3}mo{l}^{\u20131}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Amount=\frac{Reactingmass\left(g\right)}{Molarmass\left(gmo{l}^{\u20131}\right)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Also,Amountofsubs\mathrm{tan}ce=\frac{Numberofparticles}{Avogadro\u2018scons\mathrm{tan}t}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}But,Avogadro\u2018scons\mathrm{tan}t=6.02\times {10}^{23}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathit{C}\mathit{o}\mathit{m}\mathit{b}\mathit{i}\mathit{n}\mathit{i}\mathit{n}\mathit{g}\mathbf{}\mathit{t}\mathit{h}\mathit{e}\mathbf{}\mathit{t}\mathit{w}\mathit{o}\mathbf{}\mathit{e}\mathit{x}\mathit{p}\mathit{r}\mathit{e}\mathit{s}\mathit{s}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathbf{:}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{\mathbf{R}\mathbf{e}\mathbf{a}\mathbf{c}\mathbf{t}\mathbf{i}\mathbf{n}\mathbf{g}\mathbf{}\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}}{\mathbf{M}\mathbf{o}\mathbf{l}\mathbf{a}\mathbf{r}\mathbf{}\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}\mathbf{}}\mathbf{}=\frac{Numberofparticles}{6.02\times {10}^{23}}$
CALCULATIONS
 What is the mass of 2.7 mole of Aluminium (Al=27)?
Solution:
$Amount=\frac{Reactingmass}{MolarMass}$
Reacting mass = Amount x Molar mass
= 2.7mole x 27 gmol^{1} = 72.9g.
 What is the number of oxygen atoms in 32g of the gas? (O=16, N_{A} = 6.02 x 10^{23})
Solution:
$\frac{Reactingmass}{Molarmass}=\frac{Numberofatoms}{6.02x{10}^{23}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Numberofatoms=\frac{Reactingmassx6.02x{10}^{23}}{Molarmass}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Molarmassof{O}_{2}=16\times 2=32gmo{l}^{\u20131}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Numberofatoms=\frac{32gx6.02x{10}^{23}}{32gmo{l}^{\u20131}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=6.02x{10}^{23}\phantom{\rule{0ex}{0ex}}Thenumberofoxygenatomsis6.02x{10}^{23}$
EVALUATION
 Define the molar volume of a gas
 How many molecules are contained in 1.12dm^{3} of hydrogen gas at s.t.p.?
STOICHIOMETRY OF REACTION
The calculation of the amounts (generally measured in moles or grams) of reactants and products involved in a chemical reaction is known as the stoichiometry of the reaction. In other words, the mole ratio in which reactants combine and products are formed gives the stoichiometry of the reactions.
From the stoichiometry of a given balanced chemical equation, the mass or volume of the reactant needed for the reaction or products formed can be calculated.
CALCULATION OF MASSES OF REACTANTS AND PRODUCTS
 Calculate the mass of solid product obtained when 16.8g of NaHCO_{3}was heated strongly until there was no further change.
Solution:
The equation for the reaction is:
2NaHCO_{3(s)} → Na_{2}CO_{3(s)} + H_{2}O_{(g)} CO_{2(g)}
Molar mass of NaHCO_{3} = 23 + 12 + 16×3 = 84gmol^{1}
Molar mass of Na_{2}CO_{3} = 23×2 +12+16×3 = 106gmol^{1}
From the equation:
2 moles NaHCO_{3} produces 1 mole Na_{2}CO_{3}
2x84g NaHCO_{3} produces 106g Na_{2}CO_{3}
16.8g NaHCO_{3} will produce Xg Na_{2}CO_{3}
$XgNa2CO3=\frac{106gx16.8g}{2\times 84g}=10.6g$
Mass of solid product obtained = 10.6g
 Calculate the number of moles of CaCl_{2} that can be obtained from 25g of limestone [CaCO_{3}] in the presence of excess acid.
Solution:
The equation for the reaction is:
CaCO_{3(s) }+ 2HCl → CaCl_{2(s)} + H_{2}0_{(l)} + CO_{2(g)}
$Numberofmoles=\frac{Reactingmass}{MolarMass}$
Molar mass of CaCO_{3} = 40 + 12 + 16×3 = 100gmol^{1}
$NumberofmolesofCaCO3=\frac{25g}{100gmo{l}^{\u20131}}=0.25mole$
From the equation of reaction,
1 mole CaCO_{3} yields 1 mole CaCl_{2}
Therefore, 0.25 mole CaCO_{3} yielded 0.25 mole CaCl_{2.}
EVALUATION
 What does the term ‘Stoichiometry of reaction’ mean?
Ethane [C_{2}H_{6}] burns completely in oxygen. What amount in moles of CO_{2}will be produced when 6.0g of ethane is completely burnt in oxygen?
CALCULATION OF VOLUME OF REACTING GASES
 In an experiment, 10cm^{3} of ethene [C_{2}H_{4}] was burnt in 50cm^{3} of oxygen.

 A. Which gas was supplied in excess? Calculate the volume of the excess gas remaining at the end of the reaction.
 B. Calculate the volume of CO_{2 }gas produced

Solution:
The equation for the reaction is:
C_{2}H_{4(g) }+ 3O_{2(g)} → 2CO_{2(g)} + 2H_{2}O_{(g)}
A. From the equation,
1 mole of ethene reacts with 3mole of oxygen
1 volume of ethene reacts with 3 volumes of oxygen
10cm^{3} of ethene will react with 30cm^{3} of oxygen
Since 50cm^{3} of oxygen was supplied, oxygen was in excess
Hence volume of the excess gas = initial volume – volume used up = 5030 = 20cm^{3}
B. 1 volume of ethene produces 2 volumes of CO_{2}
10 cm3 of ethene will produce 20cm3 of CO_{2 }
Therefore, 20cm^{3} of CO_{2} was produced
 20cm^{3} of CO was mixed and sparked with 200cm^{3} of air containing 21% of O_{2}. If all the volumes are measured at s.t.p, calculate the total volume of the resulting gases.
Solution:
In 200cm^{3} of air,
Volume of O_{2} = 21 x 200cm^{3} = 42cm^{3}
100
Volume of N_{2} and rare gases = 20042 = 158cm^{3}
The equation for the reaction is:
2CO_{(g)} + O_{2(g)} → 2CO_{2(g)}
Volume ratio 2 : 1 : 2
Before sparking 20cm^{3} 42cm^{3}
Reacting volume 20cm^{3} 10cm^{3}
After sparking 32cm^{3} 20cm^{3}
Volume of resulting gases = 32 + 20 + 158 = 210cm^{3}
^{ }
GENERAL EVALUATION/REVISION
 Find the volume of oxygen produced by 1 mole of KClO_{3} at s.t.p. in the following reaction: 2KClO_{3(s) }→ 2KCl_{(s)} + 30_{2(g)}
 Define the term ‘Relative atomic mass’
 Balance the following redox equations I^{–} + MnO_{4}^{– } IO_{3}^{–} + MnO_{2 }in basic medium
 Write the symbols of the following elements: mercury, silver, gold, lead, tin, antimony.
 Define valency.
READING ASSIGNMENT
New School Chemistry for Senior Secondary School by O. Y. Ababio, Pg 156164
WEEKEND ASSIGNMENT
SECTION A: Write the correct option ONLY
 Amount of a substance is expressed in a. mole b. grams c. kilograms d. mass
 Determine the mass of CO_{2} produced by burning 104g of ethyne [C_{2}H_{2}] a. 256g b.352g c. 416g d. 512g
 The mole ratio in which reactants combine and products are formed is known as a. rate of reaction b. stoichiometry of reaction C. equation of reaction d. chemical reaction
 The unit for relative molecular mass is A. mole B. gmol^{1} C. grams D. mass
 What mass of Pb(NO_{3})_{2} would be required to 9g of PbCl_{2} on the addition of excess NaCl solution? [Pb=207, Na=23, O=16, N=14] A. 10.7g B. 1.2g C. 6.4g D. 5.2g
SECTION B
 Calculate the number of molecules of CO_{2} produced when 10g of CaCO_{3} is treated with 100cm^{3} of 0.20moldm^{3}HCl
 Calculate the volume of nitrogen that will be produced at s.t.p. from the decomposition of 9.60g ammonium dioxonitrate (III), NH_{4}NO_{2}.
We have come to the end of this class. We do hope you enjoyed the class?
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In our next class, we will be learning about AcidBase Reactions. We are very much eager to meet you there.
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