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Counting atoms and molecules

When conducting a chemical reaction, it is often important to mix reactants in the correct proportions. This prevents contamination of the products by wasted reactants.

However, atoms are very small and impossible to count out. In order to estimate the number of atoms in a sample of an element, it is necessary to find their mass.

The mass of an atom is quantified in terms of relative atomic mass.


Relative atomic mass

The relative atomic mass (Ar) of an element is the mass of one of its atoms relative to 1/12 the mass of one atom of carbon-12.

Most elements have more than one isotope. The Ar of the element is the average mass of the isotopes, taking into account the abundance of each isotope. This is why the Ar of an element is frequently not a whole number.

average mass of an atom × 12mass of one atom of carbon-12

relative atomic mass(Ar)

=


Relative molecular mass

Example: what is the Mr of H2SO4?

(2 × H) + (1 × S) + (4 × O)1. Count number of atoms

(2 × 1.0) + (1 × 32.1) + (4 × 16.0)

2. Substitute the Ar values

2.0 + 32.1 + 64.0 = 98.13. Add the values together

The relative molecular mass (Mr) of a covalent substance is the mass of one molecule relative to 1/12 the mass of one atom of carbon-12.

Mr can be calculated by adding together the masses of each of the atoms in a molecule.


Relative formula mass

The equivalent of relative molecular mass for an ionic substance is the relative formula mass.

This is the mass of a formula unit relative to 1/12 the mass of one atom of carbon-12. It is calculated in the same way as relative molecular mass, and is represented by the same symbol, Mr.

Example: what is the Mr of CaCl2?

(1 × Ca) + (2 × Cl)1. Count number of atoms

(1 × 40.1) + (2 × 35.5)2. Substitute the Ar values

40.1 + 71.0 = 111.13. Add the values together


Calculating relative formula mass


Moles and Avogadro's number


Moles, mass and Ar / Mr


Moles, mass and Mr calculations


Avogadro’s law

In 1811 the Italian scientist Amedeo Avogadro developed a theory about the volume of gases.

Avogadro’s law:

Equal volumes of different gases at the same pressure and temperature will contain equal numbers of particles.

For example, if there are 2 moles of O2 in 50 cm3 of oxygen gas, then there will be 2 moles of N2 in 50 cm3 of nitrogen gas and 2 moles of CO2 in 50 cm3 of carbon dioxide gas at the same temperature and pressure.

Using this principle, the volume that a gas occupies will depend on the number of moles of the gas.


Molar volumes of gases

If the temperature and pressure are fixed at convenient standard values, the molar volume of a gas can be determined.

Standard temperature is 273 K and pressure is 100 kPa.

At standard temperature and pressure, 1 mole of any gas occupies a volume of 22.7 dm3. This is the molar volume.

Example: what volume does 5 moles of CO2 occupy?

volume occupied = no. moles × molar volume

= 5 × 22.7

= 113.5 dm3


Ideal gas equation

The ideal gas equation relates pressure, volume, number of moles and temperature for a gas.

pV = nRT

p = pressure in Pa

V = volume in m3

How is the number of moles in a gas at other temperatures and pressures calculated?

n = number of moles

R = gas constant: 8.31JK-1 mol-1

T = temperature in Kelvin

A gas that obeys this law under all conditions is called an ideal gas.


Ideal gas equation: converting units

It is very important when using the ideal gas equation that the values are in the correct units.

The units of pressure, volume or temperature often need to be converted before using the formula.

Pressure

to convert kPa to Pa: × 1000

Volume

to convert dm3 to m3:to convert cm3 to m3:

÷ 1000 (103)÷ 1 000 000 (106)

Temperature

to convert °C to Kelvin: + 273


Calculating the Mr of gases


Using the ideal gas equation


Ideal gas calculations


Types of formulae

The empirical formula of a compound shows the relative numbers of atoms of each element present,

using the smallest whole numbers of atoms.

The molecular formula of a compound gives the actual numbers of atoms of each element in a molecule.

For example, the empirical formula of hydrogen peroxide is HO – the ratio of hydrogen to oxygen is 1:1.

The molecular formula of hydrogen peroxide is H2O2 – there are two atoms of hydrogen and two atoms of oxygen in each molecule.


Determining empirical formulae


Percentage by mass

Elemental analysis is an analytical technique used to determine the percentage by mass of certain elements present in a compound.

To work out the empirical formula, the total mass of the compound is assumed to be 100 g, and each percentage is turned into a mass in grams.

If necessary, the mass of any elements not given by elemental analysis is calculated. The empirical formula of the compound can then be calculated as normal.


Calculating empirical formulae


Calculating molecular formulae

The molecular formula can be found by dividing the Mr by the relative mass of the empirical formula.

Example: What is the molecular formula of hydrogen peroxide given that its empirical formula is HO and the Mr is 34?

empirical formula mass = H + O = 1.0 + 16.0 = 17

1. Determine relative mass of empirical formula:

2. Divide Mr by mass of empirical formula to get a multiple:

HO × 2 = H2O2 3. Multiply empirical formula by multiple:

relative molecular mass 34mass of empirical formula 17multiple = = = 2


Formulae calculations


Glossary


What’s the keyword?


Multiple-choice quiz


Balancing equations

An important principle in chemical reactions is that matter cannot be created or destroyed. It is important that symbol equations are balanced.

A balanced equation has the same number of each type of atom on each side of the equation.

Na + Cl2 NaCl1 sodium 2 chlorine 1 sodium 1 chlorine

Balanced:

This shows that two moles of sodium react with one mole of chlorine to make two moles of sodium chloride.

2Na + Cl2 2NaCl2 sodium 2 chlorine 2 sodium 2 chlorine

Unbalanced:


Balancing unfamiliar equations


Balancing ionic equations

Equations containing ions should have the same overall charge on each side in order to be balanced.

This can be achieved by balancing the equation in the normal way:

Ca2+ + Cl- → CaCl22 calcium 1 chloride 2 calcium 2 chloride

+1 charge no charge

2 calcium 2 chloride 2 calcium 2 chloride

Ca2+ + 2Cl- → CaCl2

no charge no charge

Balanced:

Unbalanced:


Balancing ionic equations problems


State symbols

State symbols are letters that are added to a formula to indicate what state each reactant and product is in.

The four state symbols are:

These are added after the formula in brackets and subscript. For example:

2H2(g) + O2(g) 2H2O(g)

s

l

g

aq

solid

liquid

gas

aqueous


Adding state symbols


Reacting masses


Calculating reacting masses

To calculate the mass of a product given the mass of a reactant, use the following steps:

1. Calculate no. moles of reactant:

no. moles = mass / Mr

2. Determine mole ratio of reactant to product:

ensure the equation is balanced

3. Calculate no. moles of product:

use the mole ratio

4. Calculate mass of product:mass = moles × Mr


Reacting masses example

What mass of sodium chloride is produced if 2.30 g of sodium is burnt in excess chlorine?

3. Calculate no. moles of NaCl:

0.100 moles Na = 0.100 moles NaCl

4. Calculate mass of NaCl:

mass = moles × Mr

= 0.100 × 58.5= 5.85 g

1. Calculate no. moles of Na:

no. moles = mass / Mr

= 2.30 / 23.0= 0.100

2. Determine mole ratio of Na to NaCl:

2Na + Cl2 2NaClratio = 2:2

= 1:1


Reacting masses calculations


More reacting masses calculations


What is concentration?

The concentration of a solution is a measure of how much solute is dissolved per unit of solvent.

amount of solute is measured in moles

volume of solvent is measured in dm3

concentration is measured in mol dm-3.

concentration = amount of solute / volume of solvent

Volumes are often expressed in cm3, so a more useful equation includes a conversion from cm3 to dm3.

concentration = (no. moles × 1000) / volume

mol dm3 cm3


Concentration, moles and volume


Concentration calculations


Standard solutions

A standard solution is a solution of known concentration.

Standard solutions are made by dissolving an accurately weighed mass of solid in a known volume of solvent using a volumetric flask.

The volumetric flask has a thin neck, which is marked with a line so it can be filled accurately to the correct capacity.

The standard solution can then be used to find the concentration of a second solution with which it reacts. This is known as volumetric analysis or titration.


Preparing standard solutions


A titration is a procedure used to identify the concentration of a solution by reacting it with a solution of known concentration and measuring the volume required for a complete reaction.

Once the number of moles for the solution is known, the concentration can be easily calculated.

The number of moles in the standard solution is calculated. Using a balanced equation for the reaction, the number of moles in the solution of unknown concentration can also be calculated.

What is a titration?


Carrying out a titration


Titration calculations examples

What is the concentration of an NaOH solution if 25.0 cm3 is neutralized by 23.4 cm3 0.998 mol dm-3 HCl solution?

3. Calculate no. moles of NaOH:

0.0234 moles HCl = 0.0234 moles NaOH

4. Calculate conc. of NaOH: = (0.0234 × 1000) / 25.0

conc. = (moles × 1000) / volume

= 0.936 mol dm-3

1. Calculate no. moles HCl:

moles = (conc. × volume) / 1000

= 0.0234= (0.998 × 23.4) / 1000

2. Determine ratio of NaOH to HCl:

NaOH + HCl NaCl + H2O

ratio NaOH:NaCl = 1:1


Titration calculations


More titration calculations


What are the different types of yield?

The theoretical yield is the maximum mass of product expected from the reaction, calculated

using reacting masses.

To calculate the percentage yield, the theoretical yield and the actual yield must be calculated.

The actual yield is the mass of product that is actually obtained from the real chemical reaction.

The percentage yield of a chemical reaction shows how much product was actually made compared with

the amount of product that was expected.


Calculating yield

The percentage yield of a reaction can be calculated using the following equation:

percentage yield = (actual yield × 100) / theoretical yield

Example: What is the percentage yield of a reaction where the theoretical yield was 75 kg but the actual yield was 68 kg?

percentage yield = (actual yield × 100) / theoretical yield

= 90.7%

= (68 × 100) / 75


What is atom economy?

Atom economy is another measure of the efficiency of a chemical reaction. It is the mass of reactants that end up as the desired product – this is calculated as a percentage.

This concept is useful to chemical industry, because it takes into account the atoms that end up in unwanted waste products as well as the yield of the reaction.

This means a process that produces several worthlessby-products could have a high yield but a low atom economy.

Reactions with a high atom economy tend to be more environmentally friendly as they tend to produce less waste, use fewer raw materials and use less energy.


Calculating atom economy

Example: What is the atom economy of a reaction where the actual yield was 25 000 tonnes but the mass of the reactants was 30 000 tonnes?

= 83.3%

total mass of reactants

mass of desired products × 100atom economy =

mass of desired products × 100total mass of reactants

atom economy =

25 000 × 100

30 000=


Yield and atom economy calculations


Glossary


What’s the keyword?


Multiple-choice quiz

1 of 29© boardworks ltd 2009. 2 of 29© boardworks ltd 2009

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