Atoms and reactions

Revision for resit F321

Aims

• Atomic structure• Isotopes and relative masses• The mole• Calculations using the mole• Acids and bases • Reactions of acids and bases

THE STRUCTURE OF ATOMS

Atoms consist of a number of fundamental particles,the most important are ...

Mass / kg Charge / C Relative mass

Relative charge

PROTON

NEUTRON

ELECTRON

0

-1

+111

1836

1

Calculate the mass of a carbon-12 atom; it has 6 protons, 6 neutrons and 6 electrons

9.109 x 10-31 1.602 x 10-19

1.672 x 10-27 1.602 x 10-19

1.675 x 10-27 0

6 x 1.672 x 10-27 + 6 x 1.675 x 10-27 + 6 x 9.109 x 10-31 = 2.0089 x 10-26 kg

MASS NUMBER AND ATOMIC NUMBER

Atomic Number (Z) Number of protons in the nucleus of an atom

Mass Number (A) Sum of the protons and neutrons in the nucleus

Na23

11

Mass Number (A)PROTONS + NEUTRONS

Atomic Number (Z)PROTONS

RELATIVE MASSES

Relative Atomic Mass (Ar)The mass of an atom relative to the 12C isotope having a value of 12.000

Ar = average mass per atom of an element x 12 mass of one atom of carbon-12

Relative Isotopic MassSimilar, but uses the mass of an isotope 238U

Relative Molecular Mass (Mr)Similar, but uses the mass of a molecule CO2, N2

Relative Formula MassUsed for any formula of a species or ion NaCl, OH¯

Atomic structure summary

• The nucleus contains protons (positively charged) and neutrons (neutrally charged i.e. no charge).

• The atomic number (proton number) is equal to the number of protons in the atom’s nucleus.

• The mass number is the total number of protons and neutrons in the nucleus.

• Ions do not have the same number of electrons as protons, and so have an overall charge.

Isotopes and relative masses

• Isotopes are atoms having the same number of protons but different numbers of neutrons.

• The relative atomic mass is the weighted mean mass of an atom relative to 12C, so that carbon is exactly 12 on this scale.

• The average relative atomic mass is equal to the sum of each isotope’s mass for an element x its relative abundance.

• The relative formula mass of a compound is equal to the sum of the individual relative atomic masses.

Definitions so far:

• Isotopes• Atomic number• Mass number• Ion• Relative isotopic mass• Relative atomic mass• Relative molecular mass• Relative formula mass

ion A positively or negatively charge atom or (covalently bonded) group of atoms (a molecular ion).

isotopes Atoms of the same element with different numbers of neutrons and different masses.

atomic (proton) number

The number of protons in the nucleus of an atom.

mass (nucleon) number

The number of particles (protons and neutrons) in the nucleus.

relative atomic mass, Ar

The weighted mean mass of an atom of an element compared with one-twelfth of the mass of an atom of carbon-12.

relative formula mass

The weighted mean mass of a formula unit compared with one-twelfth of the mass of an atom of carbon-12.

relative isotopic mass

The mass of an atom of an isotope compared with one-twelfth of the mass of an atom of carbon-12.

relative molecular mass, Mr

The weighted mean mass of a molecule compared with one-twelfth of the mass of an atom of carbon-12.

WHAT IS IT? The standard unit of amount of a substance - just as the standard unit of length is a METREIt is just a number, a very big numberIt is also a way of saying a number in wordslike DOZEN for 12

GROSS for 144

HOW BIG IS IT ? 602200000000000000000000 (approx) - THAT’S BIG !!!It is a lot easier to write it as 6.022 x 1023

And anyway it doesn’t matter what the number isas long as everybody sticks to the same value !

WHY USE IT ? Atoms and molecules don’t weigh much so it iseasier to count large numbers of them.In fact it is easier to weigh substances.

Using moles tells you :- how many particles you get in a certain massthe mass of a certain number of particles

THE MOLE – AN OVERVIEW

Empirical formula

Analysis showed that 0.6075g of Mg combines with 3.995g of bromine to form a compound. Find the empirical formula:Ar Mg 24.3 Br 79.9

Mg: Br :

0.025: 0.0501:2

MgBr2

Molecular formula

A compound has an empirical formula of CH2 and a relative molecular mass, Mr of 56.0. What is its molecular formula?• Empirical formula mass of CH2: = 12 + (1x2) = 14.0

• Number of CH2 units in the molecule:

• Molecular formula: (4xCH2) = C4H8

The mole summary

• A mole is the S.I. unit for amount of substance and has units of mol.

• One mole of a substance is simply the relative formula mass for a compound, or relative atomic mass for an element in grams.

• The empirical formula is the simplest whole-number ratio of atoms of each element present in a compound.

• The molecular formula is the actual number of atoms of each element in a molecule.

CALCULATING THE NUMBER OF MOLES OF A SINGLE SUBSTANCE

moles = mass / molar mass mass = moles x molar mass

molar mass = mass / moles

UNITS

mass g or kgmolar mass g mol-1 or kg mol-1

THE MOLE

MOLES = MASS MOLAR MASS

MASS

MOLES x MOLAR MASS

COVER UP THE VALUE YOU WANT AND THE METHOD

OF CALCULATION IS REVEALED

1. Calculate the number of moles of oxygen molecules in 4g oxygen molecules have the formula O2

relative mass will be 2 x 16 = 32 ; molar mass will be 32g mol-1

moles = mass = 4g = 0.125 mol molar mass 32g mol -1

2. What is the mass of 0.25 mol of Na2CO3 ?

Relative Molecular Mass of Na2CO3 = (2x23) + 12 + (3x16) = 106 Molar mass of Na2CO3 = 106g mol-1

mass = moles x molar mass = 0.25 x 106 = 26.5g

MOLES OF A SINGLE SUBSTANCE

CaCO3 + 2HCl ———> CaCl2 + CO2 + H2O

1. What is the relative formula mass of CaCO3? 40 + 12 + (3 x 16) = 100

2. What is the mass of 1 mole of CaCO3 100 g

3. How many moles of HCl react with 1 mole of CaCO3? 2 moles

4. What is the relative formula mass of HCl? 35.5 + 1 = 36.5

5. What is the mass of 1 mole of HCl? 36.5 g

6. What mass of HCl will react with 1 mole of CaCO3 ? 2 x 36.5g = 73g

7. What mass of CO2 is produced ? moles of CO2 = moles of CaCO3

moles of CO2 = 0.001 molesmass of CO2 = 0.001 x 44 = 0.044g

REACTING MASS CALCULATIONS

CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION

UNITS concentration mol dm-3

volume dm3

BUT IF... concentration mol dm-3

volume cm3

THE MOLE

MOLES

CONC x VOLUME

COVER UP THE VALUE YOU WANT AND THE

METHOD OF CALCULATION IS

REVEALED

MOLES = CONCENTRATION x VOLUME

MOLES = CONCENTRATION (mol dm-3) x VOLUME (dm3)

MOLES = CONCENTRATION (mol dm-3) x VOLUME (cm3) 1000

The original solution has a concentration of 0.100 mol dm-3

This means that there are 0.100 mols of solute in every 1 dm3 (1000 cm3) of solutionTake out 25.00 cm3 and you will take a fraction 25/1000 or 1/40 of the number of moles

moles in 1dm3 (1000cm3) = 0.100moles in 1cm3 = 0.100/1000moles in 25cm3 = 25 x 0.100/1000 = 2.5 x 10-3 mol

250cm3

25cm3

250cm3

CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION

concentration of solution in the graduated flask = 0.100 mol dm-3

volume pipetted out into the conical flask = 25.00 cm3

THE MOLE

1 Calculate the moles of sodium hydroxide in 25cm3 of 2M NaOH

moles = conc x volume in cm3

1000

= 2 mol dm-3 x 25cm3 = 0.05 moles 1000

2 What volume of 0.1M H2SO4 contains 0.002 moles ?

volume = 1000 x moles (re-arrangement of above)(in cm3) conc

= 1000 x 0.002 = 20 cm3

0.1 mol dm-3

MOLE OF SOLUTE IN A SOLUTION

MOLES = CONCENTRATION x VOLUME

STANDARD SOLUTION

‘ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY’

4.240g of Na2CO3 was placed in a clean The solution was transferredbeaker and dissolved in de-ionised water quantitatively to a 250 cm3

graduated flask and made upto the mark with de-ionised(or distilled) water.

What is the concentration of the solution in mol dm-3 ?

mass of Na2CO3 in a 250cm3 solution = 4.240gmolar mass of Na2CO3 = 106g mol -1

no. of moles in a 250cm3 solution = 4.240g / 106g mol -1 = 0.04 mol

Concentration is normally expressed as moles per dm3 of solutionTherefore, as it is in 250cm3, the value is scaled up by a factor of 4

no. of moles in 1000cm3 (1dm3) = 4 x 0.04 = 0.16 mol ANS. 0.16 mol dm-3

STANDARD SOLUTION

How to work out how much to weigh out

A chemist needs to make up a 250cm3 standard solution of 0.100M sodium carbonate from anhydrous sodium carbonate. How much will they need to weigh out?

What concentration is the solution to be? = 0.100 mol dm-3

How many moles will be in 1 dm3 ? = 0.100 molHow many moles will be in 250cm3 ? = 0.100/4 = 0.025 mol

What is the formula of anhydrous sodium carbonate? = Na2CO3

What is the relative formula mass? = 106What is the molar mass? = 106g mol -1

What mass of Na2CO3 is in 0.025 moles = 0.025 x 106 = 2.650gof Na2CO3 ? (mass = moles x molar mass)

ANS. The chemist will have to weigh out 2.650g, dissolve it in waterand then make the solution up to 250cm3 in a graduated flask.

Care must be taken when dealing with reactions that do not have a 1:1 molar ratio. If you don’t understand what an equation tells you, it is easy to make a mistake.

2NaOH + H2SO4 ——> Na2SO4 + 2H2O

you need 2 moles of NaOH to react with every 1 mole of H2SO4

i.e moles of NaOH = 2 x moles of H2SO4

or moles of H2SO4 = moles of NaOH 2

VOLUMETRIC CALCULATIONS

REMEMBER... IT IS NOT A MATHEMATICAL EQUATION

2 moles of NaOH DO NOT EQUAL 1 mole of H2SO4

More examples follow

Calculate the volume of sodium hydroxide (concentration 0.100 mol dm-3)required to neutralise 20cm3 of sulphuric acid of concentration 0.120 mol dm-3.

2NaOH + H2SO4 ——> Na2SO4 + 2H2O

you need 2 moles of NaOH to react with every 1 mole of H2SO4

therefore moles of NaOH = 2 x moles of H2SO4

moles of H2SO4 = 0.120 x 20/1000 (i)

moles of NaOH = 0.100 x V/1000 (ii)where V is the volume of alkali in cm3

substitute numbers moles of NaOH = 2 x moles of H2SO4

0.100 x V/1000 = 2 x 0.120 x 20/1000

cancel the 1000’s 0.100 x V = 2 x 0.120 x 20

re-arrange Volume of NaOH (V) = 2 x 0.120 x 20 = 48.00 cm3

0.100

VOLUMETRIC CALCULATIONS

ONE MOLE OF ANY GAS OR VAPOUR OCCUPIES 24dm3 at stp

1. Calculate the volume occupied by 0.25 mols of carbon dioxide at stp

1 mol of carbon dioxide will occupy a volume of 24 dm3 at stp0.25 mol of carbon dioxide will occupy a volume of 24 x 0.25 dm3 at stp0.25 mol of carbon dioxide will occupy a volume of 6 dm3 at stp

2. Calculate the volume occupied by 0.08g of methane (CH4) at stp

Relative Molecular Mass of CH4 = 12 + (4x1) = 16

Molar Mass of CH4 = 16g mol-1

Moles = mass/molar mass 0.08g / 16g mol-1 = 0.005 mols

1 mol of methane will occupy a volume of 24 dm3 at stp0.005 mol of carbon dioxide will occupy a volume of 24 x 0.005 dm3 at stp0.005 mol of carbon dioxide will occupy a volume of 0.12 dm3 at stp

MOLAR VOLUME

stp = standard temperature and pressure (298K and 105 Pa)ONE MOLE OF ANY GAS OR VAPOUR OCCUPIES 24dm3 at room temperature and pressure

Calculations using the mole summary

• Mass calculations: calculate the amount of substance and then use the chemical equation to deduce the moles of required substance.

• Gas calculations: 1 mol of any gas occupies 24  000 cm3 or 24 dm3 at room temperature.

• Solution calculations: the amount of substance dissolved is equal to the concentration x the volume of solution (in dm3).

• A dilute solution consists of a small amount of dissolved solute. A concentrated solution consists of a large amount of solute

Some more definitions• Amount of substance• Avogadro constant• The mole• Molar mass• Empirical formula• Molecule• Molecular formula• Molar volume• Concentration• Standard solution• Stoichiometry

Avogadro constant, NA

The number of atoms per mole of the carbon-12 isotope (6.02 × 1023 mol–1).

amount of substance

The quantity whose unit of the mole. Chemists use ‘amount of substance’ as a means of counting atoms.

mole The amount of any substance containing as many particles as there are carbon atoms in exactly 12 g of the carbon-12 isotope.

molar mass, M The mass mole of a substance. The units of molar mass are g mol–1.

molar volume The volume per mole of a gas. The units of molar volume are dm3 mol–1. At room temperature and pressure the molar volume is approximately 24.0 dm3 mol–1.

empirical formula

The simplest whole-number ratio of atoms of each element present in a compound.

molecular formula

The number of atoms of each element in a molecule.

molecule A small group of atoms held together by covalent bonds.

concentration The amount of solute, in mol, per 1 dm3 (1000 cm3) of solution.

standard solution

A solution of known concentration. Standard solutions are normally used in titrations to determine unknown information about another substance.

stoichiometry The molar relationship between the relative quantities of substances taking part in a reaction.

Acids and Bases3 common acids:1. Sulfuric acid H2SO4

2. Hydrochloric acid HCl3. Nitric acid HNO3

Acids are proton donors

Weaker acids:1. Ethanoic acid CH3COOH2. Methanoic acid HCOOH3. Citric acid C6H8O7

All acids contain H+ ions this is the active ingredient in acids and it is responsible for acid reactions. They have a pH of less than 7

Bases and AlkalisA base is a proton acceptor. An alkali is a base that dissolves is water forming OH- ions

Common bases:Metal oxides: MgO, CuOMetal hydroxides: NaOH, Mg(OH)2

Ammonia: NH3

Common Alkalis1. Sodium hydroxide NaOH2. Potassium hydroxide KOH3. Ammonia NH3

H+(aq) + OH-

(aq) H2O(l)

Acid base summary

An acid is a hydrogen ion (H+) or proton donor in solution, whereas a base is a hydrogen ion or proton acceptor in solution.

Hydrochloric acid (HCl), sulfuric acid (H2SO4) and nitric acid (HNO3) are common acids.

Bases include metal oxides (e.g. MgO), metal hydroxides (e.g. NaOH) and ammonia (NH3).

Alkalis are soluble bases and form hydroxide ions, OH-, in solution.

Acid base reactions

1. Acid + carbonate salt + CO2 + H2O2. Acid + base salt + water3. Acid + alkali salt + water4. Metal + acid salt + hydrogen

You must be able to write balanced equations for all of these reactions.

Water of crystallisation

Or how much water is in a compoundHow would you calculate the water of crystallisation?Mass of hydrated saltMass of anhydrous saltMass of water that was in theHydrated salt

Determine the formula of hydrated magnesium sulfate

Mass of hydrated salt: 4.312gMass of anhydrous salt: 2.107gSo mass of H2O in MgSO4.xH2O = 2.205Calculate the amount in mol of anhydrous MgSO4

MgSO4 = 24.3 + 32.1+ (16.0 x 4) = 120.4gmol-1

n(MgSO4) = = 0.0175 molCalculate the amount in mol of watern(H2O) = MgSO4 : H2O0.0175: 0.1225 (divide by the smallest number)1: 7 MgSO4. 7 H2O

Acid base reactions summary

• Salts are formed when a hydrogen ion from the acid is replaced by a metal ion, or an ammonium ion.

• Acids react with bases to form a salt and water only; they react with metal carbonates to form a salt, water and carbon dioxide gas.

• Metals react with acids to form a salt and hydrogen gas.

• Salts may chemically combine with water as water of crystallisation in hydrated salts. (Without water in anhydrous salts.)

Final definitions

• Salt• Hydrated• Anhydrous• Water of crystallisation

Salt A chemical compound formed from an acid, when a H+ ion from the acid has been replaced by a metal ion or another positive ion, such as the ammonium ion, NH4

+.

hydrated Crystalline and containing water molecules.

anhydrous A substance that contains no water molecules.

water of crystallisation

Water molecules that form an essential part of the crystalline structure of a compound.