However it is possible to compare the masses of atoms, they are as well small to be weighed. A chemical reaction always occurs between large numbers of reactant particles. The products which are formed as well include a large number of product particles. Chemists thus employ a large number of particles as a base unit whenever comparing amounts of different substances reacting or are formed in chemical reactions. This fundamental unit is the mole and the mole concept is one of the most significant concepts in Chemistry. The mole concept is applicable to each and every chemical processes.
The Mole Defined:
The mole is the amount of any substance which includes as many elementary particles as there are atoms in 0.012 kg of carbon-12. Now 0.012 kg of carbon - 12 includes the Avogadro number of atoms, therefore one mole of a substance is that amount that includes Avogadro number of particles. The elementary particles might be molecules, atoms, ions, electron and so on and should be specified. Avogadro number has been found out experimentally and is 6.02 x 1023, a very large number certainly. A mole of a substance thus gives a quantity of material which can be computed for use in the laboratory.
1 mole = 6 02 x 1023 particles
The Mole and the Molar Mass:
The atomic mass of an element is the mass of one atom of the element relative to the mass of one atom of Carbon-12 that by convention is represented a mass of 12.00 amu.
6.02 x 1023 atoms of C-12 = 12g
1 atom of C-12 = 12g/6.02 x 1023
Take helium (that is, relative atomic mass 4) as our first illustration.
Mass of 1 atom of He = (4/12) x (12/6.02 x 1023) g
Mass of 1 mole He = [(4x12)/(12)] x [(6.02 x 1023)/(6.02 x 1023)]
= 4 g
Take oxygen (having relative atomic mass 15.99)
Mass of 1 atom of O = [(15.999 x 12)/(12 x 6.02 x 1023)] x 1023 x 12 g
= 15.999 g
From the above computation it is evident that the mole can as well be stated as the quantity of a substance whose mass in grams is numerically equivalent to the atomic, molecular or formular mass of the substance. The molar mass of a compound is the number of grams of the compound required to make up one mole of the compound that is, includes 6.02 x 1023 molecules of the compound.
By this new definition of the mole we can compute the number of:
Table: A few substances and their molar masses
Substance Formular Formular mass Molar mass (g)
Hydrogen gas H2 2.0 2.0 g
Magnesium Mg 24.0 24.0g
Methane CH4 18.0 18.0g
Water H20 18.0 18.0g
Common salt NaCl 58.5 58.5g
Finding the Formula of a Compound:
This comprises the computation of the quantities of the elements comprised in the formation of the compounds. The theory of a mole is central in this kind of computations.
0.24g Magnesium is burnt in oxygen. The mass of magnesium oxide is found to be 0.40 g. Find out (a) the formula of magnesium oxide (b) the molar mass of magnesium oxide.
Solution: Mass of oxygen in magnesium oxide
= 0.40 - 0.24 = 0.16 g
0.24g Mg merges with 0.16g O
OR (multiply x 100 to take away the decimal point)
24g Mg merges with 16g O to form 40g of magnesium oxide.
Mole of magnesium = 24/24 = 1
Mole of Oxygen = 16/16 = 1
1 mole of magnesium merges with 1 mole of oxygen in magnesium oxide
Formula = MgO
Molar mass = 24 + 16 = 40g.
Computing the percentages of elements in a compound:
Illustration: Compute the percentage of oxygen in water.
Solution: The formula of water is H2O
Formula mass = 2 + 16 = 18g
Mass of 1 mole = 18g.
Percentage of oxygen = (16/18) x 100 = 88.9
Employing Empirical Formula to compute unknown Atomic Mass of an Element:
The empirical formula of non-metallic oxide is XO2, here 'X' stands for the chemical symbol of the element having unknown atomic mass. 0.80g of X is burnt in oxygen and the mass of the metal oxide found to be 1.60g.
Given that the atomic mass of oxygen is 16, find out the atomic mass of X.
The mass of oxygen in compound = 1.60 - 0.80 = 0.80g
Mole of oxygen = 0.80/16 = 0.05
From the formula XO2,
Ratio X: O is 1:2
2 mol O merge with 1 mole x
0.05 mol O will merge with 0.05 x 1/2 = 0.025 mol X
0.025 mol X = 0.8g
1 mol X = 0.8/0.025 = 32 g
Molar mass is 32
Mole in Chemical Reactions (Yield and percentage yield):
The above is much significant in chemistry as it allows for the computation of yields expected even before the experiment is taken out. A balanced chemical equation of the reaction is all that is needed.
FeO (s) + CO (g) → Fe (s) + CO2 (g)
Mole ratio of reactant of product
1: 1 1: 1
1 mole FeO + 1 mole CO → 1 mole Fe(s) + 1 mole CO2 (g) or in terms of masses.
72g FeO + 28g CO → 56g Fe + 44g CO2
Assume that we need to produce 2.8g Fe(s)
Mole Fe(s) needed = 2.8/5.6 = 0.05
Mole FeO required as reactant = 0.05 x 72 = 3.6 g
Mole CO = .05 x 28 = 1.4 g
Computation of percentage yield:
In most of the reactions the conversion of reactant to product is not frequently complete. The percentage yield provides the ratio of the experimental yield to a theoretical yield supposing complete reaction.
Taking the last illustration:
Assume that 3.6g of FeO reacts with adequate CO (g) and only 2.2g of Fe(s) is made up of.
% yield = (Yield from experiment/yield expected) x 100
= (2.2/2.8) x 100
= 78.571 %
Molar volume for gases:
At a temperature of 273 and pressure of 1 atmosphere generally known as (STP), a mole of any gas occupies a volume of 22.4 dm3. This is known as the molar volume.
1 mole of a gas occupies a volume of 22.4 dm3 at S.T.P. Now taking the reaction:
FeO (s) + CO (g) → Fe(s) + CO2 (g)
22.4 dm3 of CO (g) = 1 mole CO (g)
22.4 dm3 of CO2 (g) = 1 mole CO2 (g)
Illustration: Compute the volume of CO needed at S.T.P. to react fully with 3.6g FeO.
Mole ratio FeO: CO = 1:1
Mole FeO in 3.6g FeO = 3.6/72 = 0.05
Mole CO required = 0.05
Volume CO required at S.T.P.
= 0.05 x 22.4 dm3
= 0.112 dm3
Tutorsglobe: A way to secure high grade in your curriculum (Online Tutoring)
Expand your confidence, grow study skills and improve your grades.
Since 2009, Tutorsglobe has proactively helped millions of students to get better grades in school, college or university and score well in competitive tests with live, one-on-one online tutoring.
Using an advanced developed tutoring system providing little or no wait time, the students are connected on-demand with a tutor at www.tutorsglobe.com. Students work one-on-one, in real-time with a tutor, communicating and studying using a virtual whiteboard technology. Scientific and mathematical notation, symbols, geometric figures, graphing and freehand drawing can be rendered quickly and easily in the advanced whiteboard.
Free to know our price and packages for online chemistry tutoring. Chat with us or submit request at firstname.lastname@example.org
www.tutorsglobe.com offers Operations or Methods homework help, assignment help, case study, writing homework help, online tutoring assistance by computer science tutors.
Theory of Molecular Spectroscopy tutorial all along with the key concepts of symmetric stretch mode, Spectroscopic techniques, Asymmetric modes
theory and lecture notes of bipolar junction transistor all along with the key concepts of npn transistor, pnp transistor and base–emitter junction. tutorsglobe offers homework help, assignment help and tutor’s assistance on theory of bipolar junction transistor.
To study the functional block diagram of F.M radio receiver and measure its potential at several stages
Crystal geometry tutorial all along with the key concepts of Definition of crystal, Translational Symmetry, Lattice and Unit cell, Primitive and Non-Primitive cells, Bravais Lattice and Crystal structure
tutorsglobe.com anatomical adaptations assignment help-homework help by online hydrophytes tutors
tutorsglobe.com criticism of adam smiths definition assignment help-homework help by online adam smiths definition tutors
tutorsglobe.com alfred marshall definition assignment help-homework help by online theory of demand tutors
The remuneration committee is the cornerstone of the UK Code’s attempt to make sure that directors’ rewards are suitable. The UK Code says that this committee should be accountable for setting remuneration for executive directors and the chairman.
tutorsglobe.com phyllode assignment help-homework help by online leaf modification tutors
tutorsglobe.com animal behavior assignment help-homework help by online humanities tutors
Theory and lecture notes of Simplified Database System all along with the key concepts of Simplified Database System, Illustration of Database with Conceptual Data Model, Example of a simple Database. Tutorsglobe offers homework help, assignment help and tutor’s assistance on Simplified Database System.
Monosaccharides tutorial all along with the key concepts of Structure of glucose, Projection and perspective formulas, Fischer's projection, Measurement of optical activity and Haworth's projection formula
www.tutorsglobe.com offers fischer projection formulas homework help, fischer projection formulas assignment help, online tutoring assistance, organic chemistry solutions by online qualified tutor's help.
Theory and lecture notes of Post machines or tag machines all along with the key concepts of post machines or tag machines, Equivalence of TMs, PMs and Markov Algorithms, Morse code, FSM controller. Tutorsglobe offers homework help, assignment help and tutor’s assistance on Post machines or tag machines.
Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!