1.3 Equations Moles Notes for Students

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Unit 1: The Core Principles of Chemistryy

Formulae, equations and amounts of substance

An atom is the smallest unit of a chemical element/compound. (The smallest particle of an element)At t l ( h )

ATOM

Atoms are neutral (no charge) ‐+ve charge of proton + ‐ve of electron = 0

In an atom, there are three subatomic particles:‐ Proton (p)‐ Neutron (n)

Electron (e)

Packed in a small nucleus

Move rapidly around the nucleus of an atom

Modern Model of the Atom

‐

electron

nucleus

Proton

( positive charge)+

+

• Electrons move around the region of the atom.

‐

+

Neutron (neutral)

Elements• An element is a substance that cannot be broken down into two or more different substances.

Compounds

A compound is made of two or more elements chemically joined together

Compounds

• Gas– CO2, CH4 (methane)

• LiquidH O C H OH ( h l)– H2O, C2H5OH (ethanol)

• Solid– NaCl (sodium chloride)

– Mg(OH2) (magnesium hydroxide)

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Mass number (p + n )

XA

ZX Symbol of element

Atomic number (p)

Example

• Sodium 23Na

• 11

» 11 proton

» 12 neutron (A – Z = 23 – 11 = 12)( )

» 11 electron

• Try writing symbols for – H, O, N, F, S, Mg, Ne, Al, Si

Ions

• An ion is a particle that is electrically charged (positive or negative);

• an atom or molecule or group that has lost or gained one or more electrons

• When particle

loses e‐ → +ve ion (cation)

accept e‐ → –ve ion (anion)

Eg:

• 23Na ‐ e‐→ Na+ (11p, 12n, 10e‐)11

• 19F + e‐ → F‐ (9p, 10n, 10e‐)9

Molecule

• A molecule is an aggregate of at least two atoms in a definite arrangement held together

by chemical forces

• A particle containing two or more atoms• A particle containing two or more atoms joined together chemically.

• Cl2, H2O, CO2

DEFINITION

• An element is a substance that cannot be broken down into two or more different substances.

• A compound is made of two or more elements chemically joined together

• An ion is a particle that is electrically charged (positive or negative);

• an atom or molecule or group that has lost or gained one or more electrons

• A molecule is an aggregate of at least two atoms in a definite arrangement held together by chemical forces

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Chemical Formulas

• Empirical Formula – Shows the relativenumber of atoms of each element in the compound. It is the simplest formula, and is derived from masses of the elementsderived from masses of the elements.

• Molecular Formula – Shows the actual number of atoms of each element in the molecule of the compound.

What is Molecular Formula

Molecular formula

= (Empirical formula)n

Determining the Molecular Formula from Elemental Composition and Molar Mass ‐ IProblem:

The sugar burned for energy in cells of the body is Glucose(M=180.16% g/mol), elemental analysis shows that it contains 40.00 mass % C, 6.719 mass % H and 53.27% mass % O.

(a) Determine the empirical formula of glucose.

(b) h l l f l(b) Determine the molecular formula.

Solution:

Assume 100g

Mass Carbon = 40.00% = 40.00g C

Mass Hydrogen = 6.719% = 6.719g H

Mass Oxygen = 53.27% = 53.27g O

Determining the Molecular Formula from Elemental Composition and Molar Mass ‐ II

Element C H O

Mass 40.00 6.719 53.27

No of Moles 40.00g12.01= 3.3060 mol

6.719g1.008= 6.6657 mol

53.2716.00= 3.3294 mol

Simplest Ratio 3.30603.3294= 1

6.66571.008= 2

3.32943.294= 1

Empirical formula = CH2O

Determining the Molecular Formula from Elemental Composition and Molar Mass ‐ III

(b) Determining the Molecular Formula:

The formula weight of the empirical formula:

1 X 12.01 + 2 X 1.008 + 1 X 16.00 = 30.03

Whole‐number multiple = M of Glucose

empirical formula mass

= 180.16 = 6.00 = 6

30.03

Therefore the Molecular Formula is:

C 1 X 6 H 2 x 6 O 1 x 6 = C6H12O6

(a) Find the empirical formula of vitamin C if it consists of 40.9% carbon, 54.5% oxygen and 4.6% hydrogen by mass. (RAM: C = 12.0, H = 1.0, O = 16.0)

Let the mass of vitamin C analyzed be 100g.

Carbon Hydrogen Oxygen

Mass (g) 40.9 4.6 54.5

No of moles (mol) 40.9 = 3.4112 0

4.6 = 4.601 0

54.5 = 3.4116 0

The empirical formula of vitamin C is C3H4O3.

12.0 1.0 16.0

Relative number of moles

3.41 = 13.41

4.61 = 1.353.41

3.41 = 13.41

Simplest mole ratio

3 4 3

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~a process in which one set of substances called reactants is converted to a new set of substances called products.

CHEMICAL EQUATION

of substances called products.

• Chemical equation;xA + yB ⎯⎯→ zC + wD

( ) ~ an irreversible reaction( ) ~ reversible reactions.

Balancing a Chemical Equation

2 METHOD

INSPECTIONION‐ELECTRON

METHOD

~Mainly for balancing the oxidation – reduction

equations

INSPECTION METHODINSPECTION METHOD

HCl + Ca CaCl2 + H2

Step 1:Count the no of moles of atoms of eachelement on both product and reactant sides

Reactants Products

1 mol H atom 2 mol H atoms

1 mol Cl atoms 2 mol Cl atoms

1 mol Ca atoms 1 mol Ca atoms

Step 2:

Determine which elements are not balanced

• The no of moles of H and Cl are not balanced

Balance one element at a time using coefficients

Step 3:

~ Insertion of a 2 before HCl on the reactant side should be balance the equation:

2HCl + Ca CaCl2 + H2

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Step 4:

Reactants Products

Check the mass balance

Reactants Products

2 mol H atoms 2 mol H atoms

2 mol Cl atoms 2 mol Cl atoms

1 mol Ca atoms 1 mol Ca atoms

The equation is balanced!!

• The reactant that is completely used up in a

Limiting reactants

p y pchemical reaction and limits the quantities of products formed

Excess reactant

• A reactant that remains after a reaction is overis over

or• Reactants not completely used up

Limiting Reagents

6 green used up6 red left over

1.2

3 slices of cheeseExcess Reactant:bread reactants

Limiting reactant:cheese

The Cheese Sandwich Analogy

9 slices of Bread

product

Objective:At the end of the lesson, you should be able to:1) Perform stoichiometric calculations using the mole concept including limiting reactant & percentage yield.

Consider the following reaction;

Zn (s) + 2HCl (aq) → ZnCl2 (aq) + H2 (g)

Suppose 5 moles of HCl and 2 moles of Zn are mixed and allowed to react, what will be the moles of ZnCl2 produced?

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Solution

According to the equation, 1 mole Zn will react with 2 moles of HCl

1 mol Zn ≡ 2 mol HCl

∴2 mol Zn need 4 mol HCl

Thus, HCl in excess since the moles of HCl is 5 moles while Zn is limiting reactant

We can anticipate the moles of ZnCl2 produced by performing calculation based on moles of the limiting reactant, Zn.

Since 1 mol of Zn gives 1 mol of ZnCl2, therefore

2 mol of Zn produce 2 mol of ZnCl2.


Percentage yield

PERCENTAGE YIELD

The percentage yield is the ratio of the actual yield (obtained from experiment) to the theoretical yield multiplied by 100%

Percentage yield = actual yield x 100%

theoretical yield


Atom economy

ATOM ECONOMY

A reaction with good atom economy is very efficient at turning reactants into desired products with little waste.

Atom economy = mass of atoms in desired product x 100%

mass of atoms in reactants


QUESTIONS:

1. In one process, 637.2 g NH3 are allowed to react with 1142 g of CO2 as equation below:

2NH3 (g) + CO2 (g)→ (NH2)2CO (aq) + H2O (l)

(a) Which one is limiting reactant?

(b) How much of the excess reagent (in grams) is left at the end of the reaction?


2. Titanium is prepared by the reaction of titanium(IV)chloride with molten magnesium between 950oC and 1150oC as equation below:

TiCl4 (g) + 2Mg (l)→ Ti (s) + 2MgCl2 (l)

(a) If 3.54 x 107 g of TiCl4 are reacted with 1.13 x 107 g of Mg, calculate the theoretical yield of Ti in grams?

(b) Calculate the percent yield if 7.91 x 106 g of Ti are actually obtained.


Extra exercises1. 10.00 g of Zn are added into a 50.0 mL hydrochloric acid

solution with a molarity of 0.20 M to form H2 and ZnCl2. Calculate:

(a) The amount (in mol) of H2 produced

(b) The amount (in mol) of excess reactant remains after the reaction is complete.

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2. Carbon tetrachloride was prepared by reacting 100 g of carbon disulfide with 100 g of chlorine according to the reaction:

CS2 + 3Cl2 → CCl4 + S2Cl2

(Mr: CS2 = 76.2, Cl2 = 71.0, CCl4 = 154 )

Calculate the percentage yield if 65.0 g of CCl4 was obtained from the experiment.


3. 10.00 g of Zn are added into a 50.0 mL hydrochloric acid solution with a molarity of 0.20 M to form H2 and ZnCl2. Calculate:

(a) The amount (in mol) of H2 produced

(b) The amount (in mol) of excess reactant remains after the reaction is complete.


4. Carbon tetrachloride was prepared by reacting 100 g of carbon disulfide with 100 g of chlorine according to the reaction:

CS2 + 3Cl2 → CCl4 + S2Cl2

(Mr: CS2 = 76.2, Cl2 = 71.0, CCl4 = 154 )

Calculate the percentage yield if 65.0 g of CCl4 was obtained from the experiment.

~SALTS~• SALTS are ionic compounds that can be results from neutralization of acids and base• It is composed of cations and anions so that the product is electrically neutral

Acidic salt salt that hydrolyze to produce hydronium ions

Basic salt salt that hydrolyze to produce hydroxides ions

DOUBLE SALTSDouble salts are salts containing more than one cation or

anion

eg: Ammonium cobalt(II) sulfate, (NH4)2Co(SO4)2.6H2O and ammonium nickel sulfate, (NH4)2Ni(SO4)2.6H2O

"A d bl lt i b t bt i d b th"A double salt is a substance obtained by the combination of two different salts which crystallise together as a single substance but ionize as two distinct salts when dissolved in water.“

Double salts when dissolved in water dissociate into simple ions completely

SOLUBILITY OF IONIC SOLIDSSOLIDS

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SOLUBLE INSOLUBLE SLIGHTLY SOLUBLE

All group 1 metal compounds

All carbonates(apart from group 1 carbonate and ammonium carbonate

Calcium sulfate

All ammonium compounds

All hydroxides(except: Group 1 hydroxides, ammonium hydroxides, and barium

Calcium and strontium hydroxidescompounds hydroxides, and barium

hydroxide)hydroxides

All nitrates

All chlorides (except:AgCl and PbCl2)

All sulfates(except: strontium sulfate,barium sulfate

IONIC EQUATIONS

Write full equation and balance itFor dissolves ionic substances, write the ions

separatelyFor all solids (whether ionic or not) liquidsFor all solids (whether ionic or not), liquids

and gases, write full formulaCross out all the spectators ions

Note: Spectator ions are ions which take no part in reaction

Exercise

1. Predict the products of the following reaction and write a net ionic equation for the reaction

K PO ( ) C (NO ) ( ) ??K3PO4 (aq) + Ca(NO3)2 (aq) ??

2. Predict the precipitate produced by the following reaction, and write a net equation for the reaction

Al(NO3)3 (aq) + NaOH(aq) ??

Ionic Equation

• In the ionic equation, all strong electrolytes (strong acids, strong bases, and soluble ionic salts) are dissociated into their ions.

• This more accurately reflects the species that are found in the reaction mixture.

Formula EquationAgNO3(aq) + KCl(aq) → AgCl(s) + KNO3(aq)

Total Ionic EquationAg+(aq) + NO3

‐ (aq) + K+ (aq) + Cl‐ (aq) → AgCl (s) + K+ (aq) + NO3

‐ (aq)

Net Ionic Equation• To form the net ionic equation, cross out anything that does not change from the left side of the equation to the right.

Ag+ (aq) + NO3‐ (aq) + K+ (aq) + Cl‐ (aq) →

A Cl ( ) K+ ( ) NO ( )AgCl (s) + K+ (aq) + NO3‐ (aq)

• The only things left in the equation are those things that change (i.e., react) during the course of the reaction.

Ag+ (aq) + Cl‐ (aq) → AgCl (s)

Chemical Equations

a A + b B→ c C + d D

Mole ratios (can also be volume ratios for

Stoichiometry

= relative no. of moles of substances involved in a chemical reaction

gases)

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Molarity

A way of expressing concentrations

Molarity is the number of moles of solute dissolved in 1 dm3 (1000 cm3) of solution1 dm (1000 cm ) of solution.

Molarity = number of moles of solute

volume of solution (in dm3)

Unit: moles/dm3 (M)

Example 1.3 ‐ 5

25.0 cm3 of sodium hydroxide solution was titrateedagainst 0.067 M of sulphuric acid using methyl orange as indicator. The indicator changed colour from yellow to red when 22.5 cm3 of sulphuric acid had been added. Calculate the molarity of the sodium hydroxide solution.

Solution:

2NaOH(aq) + H2SO4 → Na2SO4 + 2H2O

Mole ratio of NaOH(aq) = 2 X Number of moles of H2SO4

Number of moles of H2SO4(aq) = 0.067 moldm‐3 X 22.5 X 10‐3 dm‐3

= 1.508 X 10‐3mol

Number of moles of NaOH (aq) = 2 X 1.508 X 10‐3 mol= 3.016 X 10‐3 mol

Molarity of NaOH(aq)= 3.012 X 10‐3 mol / 25.0 X 10‐3 mol= 0.1221 mol dm‐3

The molarity of NaOH is 0.121 M

MOLE CONCEPT

CONTENTS

Define mole in terms of mass & relate it with Avogadro constant

Definition of mole and its relationships with Avogadro number and molar

mass of element

Relate mole with molar volume of gases

CONTENTS

Define mole in terms of mass & relate it with Avogadro constant

Relate mole with molar volume of gases

Relationships between number of mole and molar volume of gases

Objective:At the end of the lesson, you should be able to:1) Define mole in in terms of mass & relate it with Avogadro constant

Definition: Mole (mol) is the amount of a substance that contains as many elementary entities (atoms, molecules or particles) as there are atoms in exactly 12.00g of the carbon‐12 isotope.

This number is called Avogadro’s number (NA)

NA = 6.02 x 1023

The idea of a unit to denote a particular number of objects is not new. For example, the pair ( 2 items ) and the dozen ( 12 items ) are all familiar units. Chemists measure atoms & molecules in moles!!

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Just as one dozen books contains 12 books, 1 mole of hydrogen atoms contains 6.02 x 1023 H atoms

1 dozen = 12


1 mole = 6.02 x 1023

1 mole of carbon‐12 atoms has a mass of exactly 12 g and contains 6.02 x 1023 atoms. This mass of carbon‐12 is called molar mass.

Definition: The molar mass is the mass of one mole of the substance.


For all substances, the molar mass in gram per mole is numerically equal to the formula weight in atomic mass units.

Molar mass = M (gmol‐1)


The mole concept and Avogadro number

1 mole of an element = 6.02 x 1023 atoms of that element

In term of molecule and ions

1 mole of molecules = 6.02 x 1023 molecules

1 mole of ions = 6.02 x 1023 ions

C12H22O11

Carbon12 atoms C12 moles C

Hydrogen22 atoms H22 moles H

Oxygen11 atoms O11 moles O

How to find the number of moles?

Number of moles = Number of moles

Avogadro constant

or

Number of particles

= Number of moles X (6.02 X 1023)

Number of moles = mass(g)

molar mass (g/mol)

or

Mass

= number of moles X molar mass

Example 1.3– 1What is the mass of 0.2 mole of calcium carbonate?(RAM: C = 12.0, O = 16.0, Ca = 40.1)

Solution:

The chemical formula of calcium carbonate is CaCO3.

The molar mass of calcium carbonate

= (40 1 + 12 0 + 3 X 16 0) gmol‐1= (40.1 + 12.0 + 3 X 16.0) gmol

= 100.1 Mass of calcium carbonate

= Number of moles X Molar mass

= 0.2mol X 100.1 gmol‐1

= 20.02g

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Example 1.3 – 2Calculate the number of gold atoms in 20g of gold atom. (RAM: Au = 197.0)

Solution:

Number of gold atoms in 20g of gold coin

= 20g X 6.02 X 1023mol‐1

197.0 gmol‐1

= 6.11 X 1022

Go to examples from edexcel text‐book

What is Molar Volume of Gases?• Go to examples from edexcel text‐book

1 mole of He 1 mole of O2 1 mole of CO2

At 0 °C and 1 atm (stp)

Objective:At the end of the lesson, you should be able to:2) Relate mole with molar volume of gases

Molar Volume of Gases

The volume occupied by one mol of any gas is called Molar Volume.

Standard Temperature and Pressure (STP) is a condition inStandard Temperature and Pressure (STP) is a condition in which gases are measured at 0oC and 1 atm pressure.

At STP, volume of 1 mol gases = 22.4 Liter or dm3.

The molar volume of gases at room temperature (25oC) and 1 atm pressure.

At RT, volume of 1 mol gases= 24.0 Liter or dm3.

Objective:At the end of the lesson, you should be able to:1) Define mole in in terms of mass & relate it with Avogadro constant2) Relate mole with molar volume of gases

Molar vol (STP) (L or dm3)

2.4)

Volume

Mass of element (m)

No. of moles (n)

No. of atoms (N)

m / M

n M

n NA

N / NA

n (2

e / n

Figure 1: The relationships between mass (in grams) of an element, number of moles, number of atoms and molar mass at STP. M is molar mass (g/mol) and NA is Avogadro’s number.


EXAMPLE 1

Helium (He) is a valuable gas used in industry. How many moles of He atoms are in 6.46 g of He?

Solution.According to Figure 1, to convert grams to moles we need the molar mass. In exam, the molar mass of element is given. We find that the molar mass of He is 4.0 g. This can be expressed as

4.0 g He ≡ 1 mol He

Thus, 6.46 g He = 6.46 g x 1mol

4.0 g

= 1.61 mol

Example 1.3 – 3Find the volume occupied by 3.55g of chloride gas at room temperature and pressure (Molar volume of gas at R.T.P. = 24.0 dm3 mol‐1, RAM: Cl = 35.5)

Solution:

Molar mass of chloride gas (CL2) = 35.5 X 2g mol‐1

= 71.0 gmol‐1

Number of moles of Cl2 = 3.55g

71.0g mol‐1

= 0.05 mol

Volume of Cl2= Number of moles of Cl2 X Molar volume

= 0.05 mol X 24.0 dm3mol‐1

= 1.2dm3

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Example 1.3 – 4a) Find the volume of hydrogen produced at R.T.P. when 2.43 g of magnesium reacts with excess hydrochloric acid. (RAM: Mg = 24.3; molar volume of gas at R.T.P. = 24.0 dm3mol‐1

Answer

Mg + 2HCl → MgCl2 + H2

From the equation, 1 mole Mg will form 1 mole of H2 (g).

No. of moles of H2 = No. of moles of Mg

= 2.43g / 24.3 g mol‐1

Volume of H2 = 0.1 mol X 24.0 dm3 mol‐1

Volume of H2 = 2.4 dm3


EXAMPLE 2

How many grams of CH2Cl2 are obtained in 2.88 moles of CH2Cl2? ( Ar

of H = 1.0, C = 12.0, Cl = 35.5)

SolutionSolution.

Firstly we have to calculate the molar mass of CH2Cl2 = C + 2H + 2Cl

= 12 + 2(1) + 2(35.5) = 85.0 g

1 mol CH2Cl2 ≡ 85.0 g CH2Cl2

Thus, 2.88 mol CH2Cl2 = 2.88 mol x 85.0 g

1 mol

= 244.8 g


EXAMPLE 3

(a) Determining the number of atoms in 12.3 g Li

(b) Calculate the number of Br atoms in 32 g Br2

(Ar Li = 7, H = 1, S = 32, O = 16)


EXAMPLE 4

How many grams are there in 6.5 L oxygen gas at STP?

(A O = 16 0)(Ar O = 16.0)


EXERCISES

1. How many atoms are present in 3.14 g of copper (Cu)? (Ar

Cu = 63.5)

2. Calculate the molar mass of a compound if 0.372 mole of it has a mass of 152 g.


3. Pheromones are a special type of compound secreted by the females of many insect species to attract the males for mating. One pheromone has the molecular formula C19H38O. Normally, the amount of this pheromone secreted by a female insect is about 1 0 x 10‐12 g Howsecreted by a female insect is about 1.0 x 10 g. How many molecules are there in this quantity? (Ar C = 12, H = 1, O =16)

4. The density of water is 1.00 g/mL at 4oC. How many water molecules are present in 2.56 mL of water at this temperature? (Ar H = 1, O = 16)

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EXERCISES

1. How many atoms are present in 3.14 g of copper (Cu)? (Ar

Cu = 63.5)

(2.95 x 1022 Cu atoms)

2. Calculate the molar mass of a compound if 0.372 mole of it has a mass of 152 g.

(408.6 gmol‐1)


3. Pheromones are a special type of compound secreted by the females of many insect species to attract the males for mating. One pheromone has the molecular formula C19H38O. Normally, the amount of this pheromone secreted by a female insect is about 1 0 x 10‐12 g How manyby a female insect is about 1.0 x 10 g. How many molecules are there in this quantity? (Ar C = 12, H = 1, O =16)

(2.13 x 109 pheromone molecules)

4. The density of water is 1.00 g/mL at 4oC. How many water molecules are present in 2.56 mL of water at this temperature? (Ar H = 1, O = 16)

(8.55 x 1022 H2O molecules)


ADVANCED QUESTION:

1. 10.64g of metal oxide M2O3 are reacted with excess hydrogen gas and produces 3.78 g of water and metal M as shown by the equation:

M2O3(s) + 3H2(g) → 2M(s) + 3H2O(g)

Calculate the relative molecular mass of M2O3, relative atomic mass of metal M and the mass of metal M produced in the above reaction. (Ar O = 16).

(152, 52, 7.28 g)

Concentration Units

The concentration of solutions is the quantity of dissolved substance per unit quantity of solvent in a solution. Concentration is measured in various ways:

Objective:At the end of the lesson, you should be able to:1) Perform calculations involving concentration of solution in terms of molarity, molality, mole fraction, percent by mass, percent by volume and parts per million and to include their respective units

Concentration is measured in various ways:

a) Concentration (formerly molarity)

b) Parts per million (ppm)

Objective:At the end of the lesson, you should be able to:1) Perform calculations involving concentration of solution in terms of molarity, molality, mole fraction, percent by mass, percent by volume and parts per million and to include their respective unit

(A)Concentration ( or molarity )

Concentration is defined as the number of moles of solute per cubic decimeter ( or liter ) of solutionsolute per cubic decimeter ( or liter ) of solution.

Symbol: c

Unit: moldm‐3 or molL‐1 or M

Molarity, c =Moles of solute (mol)

Volume of solution (dm3)

Example

l d d l b d l


An A‐Level student prepared a solution by dissolving 0.586 g of sodium carbonate, Na2CO3 in 250.0 cm3 of water. Calculate its concentration.

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Solution

Step1: Write the equation

Molar = mol of solute (mol) .

Volume of solution (dm3)

Step2: Calculate the no. of mol of Na2CO3 and convert volume to liter

Mol of Na2CO3 = mass / Mr

= 0.586 g / 23(2) + 12 + 16(3)

= 5.53 x 10‐3 mol

Volume of H2O in dm3 = 250 x 10‐3 dm3

Step3: Substitute into equation

Molar = 5.53 x 10‐3 mol / 250 x 10‐3 dm3

= 0.022 mol dm‐3


(B)Part per million, ppm

This unit is used when one component in a solution is present in a relatively very small amount.

Unit: ppmUnit: ppm

ppm =

Mass of solute

Mass of solutionx 106

Example

The concentration of calcium ions in blood is 100.0 ppm. Calculate the mass of calcium ions in 500 0 g of blood


Calculate the mass of calcium ions in 500.0 g of blood.


Solution

Step1: Write the equation

ppm = mass of calcium ion x 106

mass of solution

Step2: Substitute into equation

100 = mass Ca2+ x 106

500 g

Mass Ca2+ = 100 x 500 x 10‐6

= 0.05 g

EXERCISES

1. Calculate the molarity of the following solutions:

(a)0.10 mol of solute in 200 mL of solution

(b)1.5 mol of NaCl in 0.40 L solution

(c)0.5 mol of CuSO4.5H2O in 500 mL of solution

2. What will be the final molarity of the resulting solution after dilution?

(a) 300 mL of 4 M HCl + 200 mL H2O

(b) 10.0 mL of 5 M HNO3 is diluted to 500 mL

(c) 100 mL of 1.5 M NaOH + 400 mL 2.0 M NaOH

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Definition of Terms

• 1.3 (a) demonstrate an understanding of the terms atom, element, ion, molecule, compound empirical and molecular formulaecompound, empirical and molecular formulae

• 1.3 b. write balanced equations (full and ionic) for simple reactions, including the use of state symbols

Spec

• c. demonstrate an understanding of the terms relative atomic mass, amount of substance, molar mass and parts per million (ppm), eg : gases in the atmosphere, exhausts, water pollution

• d. calculate the amount of substance in a solution of known concentration (excluding titration calculations at this stage), eg use data from the concentrations of the various species in blood samples to perform calculations in mol dm‐3

Specs

• e. use chemical equations to calculate reacting masses and vice versa using the concepts of amount of substance and molar mass

• f. use chemical equations to calculate volumes of gases and vice versa using the concepts of amount of substance and molar volume of gases, eg calculation of the mass or volume of CO2 produced by combustion of a hydrocarbon (given a molar volume for the gas) writing full and ionic equations.

• g. use chemical equations and experimental results to deduce percentage yields and atom economies in laboratory and industrial processes and understand why they areprocesses and understand why they are important

Spec

• h. demonstrate an understanding of, and be able to perform, calculations using the Avogadro constant

• i. analyse and evaluate the results obtained from finding a formula or confirming an equation by experiment, eg the reaction of lithium with water and deducing the equationlithium with water and deducing the equation from the amounts in moles of lithium and hydrogen

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• j. make a salt and calculate the percentage yield of product, eg preparation of a double salt (ammonium iron(II) sulfate from iron, ammonia and sulfuric acid)ammonia and sulfuric acid)

• k. carry out and interpret the results of simple test tube reactions, such as displacements, reactions of acids, precipitations to relate the observations toprecipitations, to relate the observations to the state symbols used in equations and to practise

1.3 Equations Moles Notes for Students

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