MASS AND VOLUME RELATIONSHIP

TOPIC: MASS AND VOLUME RELATIONSHIPS IN REACTIONS

Mass and volume relationships in reactions describe how quantities of substances relate to one another during chemical changes. Chemical reactions obey the Law of Conservation of Mass, which states that matter is neither created nor destroyed. This means the total mass of reactants must equal the total mass of products. Because atoms combine in fixed ratios, chemists can use balanced chemical equations to calculate how much of each substance is required or produced in a reaction.

Mass relationships are based on the mole concept. In a balanced chemical equation, the numbers in front of each substance represent the ratio in which reactants combine and products form. These numbers give the mole ratio. Once a substance’s molar mass is known, it can be converted between mass and moles. For example, if a reaction shows that two moles of hydrogen combine with one mole of oxygen to form water, it becomes possible to calculate how many grams of water can be produced from a known mass of hydrogen or oxygen.

Volume relationships apply mainly to gases. According to Avogadro’s Law, equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. This means gases react in simple whole-number ratios by volume. For example, one volume of nitrogen reacts with three volumes of hydrogen to form two volumes of ammonia gas. These volume ratios correspond directly to the mole ratios in the balanced equation. At standard temperature and pressure (STP), one mole of any gas occupies 22.4 dm³, while at room temperature it occupies about 24 dm³. This molar volume helps convert between moles of gases and their actual volumes.

In many reactions, mass and volume relationships are used together. This happens when a solid reacts to produce a gas, or when a gas reacts with a liquid or solid. For example, when calcium carbonate decomposes to form calcium oxide and carbon dioxide, the solid calcium carbonate can be measured in grams while the carbon dioxide can be measured in volume. Converting the mass to moles allows the volume of gas formed to be calculated using the molar volume of a gas.

Understanding mass and volume relationships is very important in chemistry. It helps determine the exact amount of reactants needed in laboratory experiments and industrial processes, predict the amount of product formed, avoid waste of chemicals, and ensure reactions are efficient. These relationships form the foundation of stoichiometry and quantitative chemistry.

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 1023 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 Carbon - 12.

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 one - twelfth the mass of one atom of Carbon - 12. It indicates the mass of a n atom of  an element. For example; 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 1023 atoms; 1 mole of hydrogen having an atomic mass of 2.0g contains 6.02 x 1023 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 one - twelfth the mass of one atom of carbon - 12. 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 not only to the relative mass of a molecule but also that of an ion or radical.

Example: Calculate the Relative Molecular Mass of: 
1. Magnesium Chloride
2. Sodium Hydroxide
3. Calcium trioxocarbonate
[Mg=24, Cl=35.5, O=16, H=1, Ca=40, C=12]

Solution

(solve in your note)

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 dm3. Thus, 1 mole of oxygen gas of molar mass 32.0gmol-1 occupies a volume of 22.4 dm3 at s.t.p and 1 mole of helium gas of molar mass 4.0gmol-1 occupies a volume of 22.4 dm3 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 

Molar mass = mass (g)Amount (moles) i.e. M = mn gmol-1

Amount = number of moles

Molar Volume of gas = Volume (cm3 or dm3)Amount (mole) i.e. Vm = vn dm3mol-1

Amount = Reacting mass (g)Molar mass (gmol-1) Also, amount of substance = number of particlesAvogadro's constant)

But Avogadro’s constant = 6.02 x 1023

Combining the two equations: Reacting mass (g)Molar mass (gmol-1) = number of particles 6.02 x 1023

Example:

1. What is the mass of 2.7 mole of Aluminium (AL = 27)?

Solution

Amount = Reacting mass (g)Molar mass (gmol-1)

Reacting mass = Amount x Molar mass

= 2.7 mole x 27gmol-1 = 72.9g

1. What is the number of oxygen atoms in 32g of the gas? (O = 16, NA = 6.02 x 1023)

Solution

Reacting mass (g)Molar mass (gmol-1) = number of particles 6.02 x 1023

Number of atoms = Reacting mass x 6.02 x 1023Molar mass (gmol-1)

Molar mass of O2 = 16 x 2 = 32gmol-1

Number of atoms = 32g  x  6.02 x 102332gmol-1 =  6.02 x 1023

The number of O2 atoms is 6.02 x 1023

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 stoichiometry of 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

1. Calculate the mass of solid product obtained when 16.8g of NaHCO3 was heated strongly until there was no further change.

Solution

The equation for the reaction is:  2NaHCO3(aq)    ⇢  Na2CO3(s) + H2O(g) + CO2(g) 

Molar mass of NaHCO3(aq) = 23 + 12 + 16 x 3 = 84gmol-1
Molar mass of Na2CO3  = 23 x 2 + 12 + 16 + 3 = 106gmol-1

From the Equation:

2 moles NaHCO3 produces 1 mole of Na2CO3
2 x 84g NaHCO3 produces 106g Na2CO3
16.8g NaHCO3 will produce Xg Na2CO3
Xg Na2CO3 = 106g x 16.8g2 x 84g = 10.6g

Mass of solid product obtained = 10.6g

1. 20cm3 of CO was mixed and sparked with 200cm3 of air containing 215 of O2. If all the volumes are measured at s.t.p, calculate the total volume of the resulting gases.

Solution

In 200cm3 of air, Volume of O2 = 21 x 200cm3 100= 42cm3

Volume of N2 and rare gases = 200 - 42 = 158cm3

The equation for the reaction is:   2CO(g) + O2(g) ⇢ 2CO2(g)
Volume Ratio:       2       :    1       :     2
Before Sparking: 20cm3 42cm3
Reacting Volume: 20cm3 10cm3
After Sparking: 32cm3 20cm3
Volume of resulting Gases = 32 + 20+ 158 = 210cm3

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TOPIC: ACID - BASE REACTIONS

TITRATION 
There are two types of quantitative analysis namely: volumetric and gravimetric analysis.  Volumetric analysis is based on volume measurement while gravimetric analysis involves direct mass measurement. 
Volumetric analysis is carried out using Titration.  In titration, a standard solution (one of known concentration must be using be used to react with a solution of unknown of concentration) 

PREPARATION OF A STANDARD SOLUTION 
A standard solution is a solution of which the concentration is known .A standard solution is prepared by weighing a pure solute, for instance, and dissolving it in a suitable solvent, usually water, and making up the solution to a definite volume in a volumetric flask. 
For instance, a solution known to contain exactly 10.6g of anhydrous sodium trioxocarbonate (IV), Na2CO3, in 1 dm3 of solution is a standard solution. 

Preparation of 0.1mol/dm3NaOH          
40g NaOH dissolved in 1 dm3 of the water gives 1.0mol/dm3 solution  
XgNaOH will be dissolved in 1 dm3 of water to give 0.1mol/dm3  
Xg = 40g x 0.1mol/dm3  
1.0mol/dm3 = 4g 
Therefore, 4g of sodium hydroxide pellet is measured, dissolved in water and made up to 1dm3 mark to obtain 0.1mol/dm3NaOH

INDICATORS FOR ACID/BASE TITRATION 
Acid-base indicators are dyes that change colour according to the pH of the medium. The table below shows some titration and their suitable indicator: 

CONCENTRATION The concentration of a solution is the amount of solute in a given volume of the solution.  It can be expressed as mol/dm3 or g/dm3. 

Molar concentration: The molar concentration of a compound is one which contains one mole or the molar mass of the compound in 1dm3 of the solution. The unit of molar concentration is mol/dm3 

Mass concentration: The mass concentration of a compound is the mass of the compound contained in 1 dm3 of solution. The unit is g/dm3 

Relationship between Molar concentration and Mass concentration 

Concentration = number of molesvolume  = n/V………………(i) 
Number of moles, n = C x V ……………………..(ii) 
But, number of moles, n= m/M Where M = molar mass and m = mass 
Substituting n=m/M into …….(ii) We have m/M = C x V That is, m/V = C x M 
But m/V = mass concentration 

Therefore, mass concentration = molar concentration x molar mass = C x M

ASSIGNMENT (20marks)

Write out extensively the precautions to be taken when carrying out a titration experiment.