Welcome dear students! Today we are going to learn about Atoms and Molecules from Class 9 Science. Ancient Indian and Greek philosophers have always wondered about the unknown and unseen form of matter. The idea of divisibility of matter was considered long back in India, around 500 BC. An Indian philosopher Maharishi Kanad postulated that if we go on dividing matter, we shall get smaller and smaller particles. Ultimately, a stage will come when we shall come across the smallest particles beyond which further division will not be possible. He named these particles Parmanu. Another Indian philosopher, Pakudha Katyayama, elaborated this doctrine and said that these particles normally exist in a combined form which gives us various forms of matter. [CHECKPOINT]
Around the same era, ancient Greek philosophers Democritus and Leucippus suggested that if we go on dividing matter, a stage will come when particles obtained cannot be divided further. Democritus called these indivisible particles atoms, meaning indivisible. All this was based on philosophical considerations and not much experimental work to validate these ideas could be done till the eighteenth century. By the end of the eighteenth century, scientists recognised the difference between elements and compounds and naturally became interested in finding out how and why elements combine and what happens when they combine. [CHECKPOINT]
Antoine L. Lavoisier laid the foundation of chemical sciences by establishing two important laws of chemical combination. We will now study the Laws of Chemical Combination. The following two laws were established after much experimentations by Lavoisier and Joseph L. Proust. First is the Law of Conservation of Mass. Is there a change in mass when a chemical change takes place? Let us perform Activity three point one. Take one of the following sets, X and Y of chemicals. Set X contains copper sulphate, barium chloride, or lead nitrate. Set Y contains sodium carbonate, sodium sulphate, or sodium chloride. [CHECKPOINT]
Prepare separately a five percent solution of any one pair of substances listed under X and Y each in ten millilitres in water. Take a little amount of solution of Y in a conical flask and some solution of X in an ignition tube. Hang the ignition tube in the flask carefully so that the solutions do not get mixed. Put a cork on the flask. In this setup, as shown in Figure three point one, we see a conical flask with a cork, containing solution Y. Inside it hangs a small ignition tube containing solution X, suspended by a thread. Weigh the flask with its contents carefully. [CHECKPOINT]
Now tilt and swirl the flask, so that the solutions X and Y get mixed. Weigh again. Observe what happens in the reaction flask. You will notice a chemical reaction has taken place. We put a cork on the mouth of the flask to prevent any gas or vapour from escaping. You will find that the mass of the flask and its contents does not change. These observations are in agreement with the law of conservation of mass. The Law of conservation of mass states that mass can neither be created nor destroyed in a chemical reaction. [CHECKPOINT]
Let us solve the first in-text question. In a reaction, five point three grams of sodium carbonate reacted with six grams of acetic acid. The products were two point two grams of carbon dioxide, zero point nine grams water and eight point two grams of sodium acetate. Show that these observations are in agreement with the law of conservation of mass. The total mass of reactants is five point three grams plus six grams, which equals eleven point three grams. The total mass of products is two point two grams plus zero point nine grams plus eight point two grams, which also equals eleven point three grams. Since the mass of reactants equals the mass of products, the law of conservation of mass is verified. [CHECKPOINT]
Second question: Hydrogen and oxygen combine in the ratio of one to eight by mass to form water. What mass of oxygen gas would be required to react completely with three grams of hydrogen gas? Since the ratio is one to eight, for one gram of hydrogen we need eight grams of oxygen. Therefore, for three grams of hydrogen, we need three multiplied by eight, which equals twenty-four grams of oxygen. Third question: Which postulate of Dalton atomic theory is the result of the law of conservation of mass? The postulate that atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction, is the result of this law. [CHECKPOINT]
Fourth question: Which postulate of Dalton atomic theory can explain the law of definite proportions? The postulate that the relative number and kinds of atoms are constant in a given compound explains this law. Next, we study the Law of Constant Proportions. Lavoisier, along with other scientists, noted that many compounds were composed of two or more elements and each such compound had the same elements in the same proportions, irrespective of where the compound came from or who prepared it. In a compound such as water, the ratio of the mass of hydrogen to the mass of oxygen is always one to eight, whatever the source of water. [CHECKPOINT]
Thus, if nine grams of water is decomposed, one gram of hydrogen and eight grams of oxygen are always obtained. Similarly in ammonia, nitrogen and hydrogen are always present in the ratio fourteen to three by mass, whatever the method or the source from which it is obtained. This led to the law of constant proportions which is also known as the law of definite proportions. This law was stated by Proust as: In a chemical substance the elements are always present in definite proportions by mass. The next problem faced by scientists was to give appropriate explanations of these laws. [CHECKPOINT]
British chemist John Dalton provided the basic theory about the nature of matter. Dalton picked up the idea of divisibility of matter, which was till then just a philosophy. He took the name atoms as given by the Greeks and said that the smallest particles of matter are atoms. His theory was based on the laws of chemical combination. Dalton atomic theory provided an explanation for the law of conservation of mass and the law of definite proportions. John Dalton was born in a poor weaver family in seventeen sixty-six in England. He began his career as a teacher at the age of twelve. Seven years later he became a school principal. [CHECKPOINT]
In seventeen ninety-three, Dalton left for Manchester to teach mathematics, physics and chemistry in a college. He spent most of his life there teaching and researching. In eighteen oh eight, he presented his atomic theory which was a turning point in the study of matter. According to Dalton atomic theory, all matter, whether an element, a compound or a mixture is composed of small particles called atoms. The postulates of this theory may be stated as follows. First, all matter is made of very tiny particles called atoms, which participate in chemical reactions. Second, atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction. [CHECKPOINT]
Third, atoms of a given element are identical in mass and chemical properties. Fourth, atoms of different elements have different masses and chemical properties. Fifth, atoms combine in the ratio of small whole numbers to form compounds. Sixth, the relative number and kinds of atoms are constant in a given compound. You will study in the next chapter that all atoms are made up of still smaller particles. We might think that if atoms are so insignificant in size, why should we care about them? This is because our entire world is made up of atoms. We may not be able to see them, but they are there, and constantly affecting whatever we do. [CHECKPOINT]
Through modern techniques, we can now produce magnified images of surfaces of elements showing atoms. Figure three point two shows an image of the surface of silicon. Now let us move on to section three point two: What is an Atom? Have you ever observed a mason building walls, from these walls a room and then a collection of rooms to form a building? What is the building block of the huge building? What about the building block of an ant-hill? It is a small grain of sand. Similarly, the building blocks of all matter are atoms. How big are atoms? [CHECKPOINT]
Atoms are very small, they are smaller than anything that we can imagine or compare with. More than millions of atoms when stacked would make a layer barely as thick as this sheet of paper. Atomic radius is measured in nanometres. One divided by ten to the power of nine metres equals one nanometre. One metre equals ten to the power of nine nanometres. Let us look at the relative sizes of various objects. The radius of a hydrogen atom is ten to the power of minus ten metres. The radius of a water molecule is ten to the power of minus nine metres. [CHECKPOINT]
The radius of a haemoglobin molecule is ten to the power of minus eight metres. The radius of a grain of sand is ten to the power of minus four metres. The radius of an ant is ten to the power of minus three metres. The radius of an apple is ten to the power of minus one metres. Now, what are the modern day symbols of atoms of different elements? Dalton was the first scientist to use the symbols for elements in a very specific sense. When he used a symbol for an element he also meant a definite quantity of that element, that is, one atom of that element. [CHECKPOINT]
Figure three point three shows symbols for some elements as proposed by Dalton, using circles with internal markings to represent hydrogen, phosphorus, copper, gold, carbon, sulphur, lead, platina, oxygen, iron, silver, and mercury. Berzilius suggested that the symbols of elements be made from one or two letters of the name of the element. In the beginning, the names of elements were derived from the name of the place where they were found for the first time. For example, the name copper was taken from Cyprus. Some names were taken from specific colours. For example, gold was taken from the English word meaning yellow. [CHECKPOINT]
Now-a-days, IUPAC, the International Union of Pure and Applied Chemistry, is an international scientific organisation which approves names of elements, symbols and units. Many of the symbols are the first one or two letters of the element name in English. The first letter of a symbol is always written as a capital letter and the second letter as a small letter. For example, hydrogen is H, aluminium is Al and not AL, cobalt is Co and not CO. Symbols of some elements are formed from the first letter of the name and a letter, appearing later in the name. Examples are chlorine, Cl, and zinc, Zn. [CHECKPOINT]
Other symbols have been taken from the names of elements in Latin, German or Greek. For example, the symbol of iron is Fe from its Latin name ferrum, sodium is Na from natrium, potassium is K from kalium. Therefore, each element has a name and a unique chemical symbol. Let us review the symbols for some elements from Table three point one. Aluminium is Al, Argon is Ar, Barium is Ba, Boron is B, Bromine is Br, Calcium is Ca, Carbon is C, Chlorine is Cl, Cobalt is Co, Copper is Cu, Fluorine is F, Gold is Au, Hydrogen is H, Iodine is I, Iron is Fe, Lead is Pb, Magnesium is Mg, Neon is Ne, Nitrogen is N, Oxygen is O, Potassium is K, Silicon is Si, Silver is Ag, Sodium is Na, Sulphur is S, Uranium is U, and Zinc is Zn. [CHECKPOINT]
The above table is given for you to refer to whenever you study about elements. Do not bother to memorise all in one go. With the passage of time and repeated usage you will automatically be able to reproduce the symbols. Next, we discuss atomic mass. The most remarkable concept that Dalton atomic theory proposed was that of the atomic mass. According to him, each element had a characteristic atomic mass. The theory could explain the law of constant proportions so well that scientists were prompted to measure the atomic mass of an atom. [CHECKPOINT]
Since determining the mass of an individual atom was a relatively difficult task, relative atomic masses were determined using the laws of chemical combinations and the compounds formed. Let us take the example of a compound, carbon monoxide, CO, formed by carbon and oxygen. It was observed experimentally that three grams of carbon combines with four grams of oxygen to form CO. In other words, carbon combines with four-thirds times its mass of oxygen. Suppose we define the atomic mass unit, earlier abbreviated as amu, but according to the latest IUPAC recommendations, it is now written as u, unified mass, as equal to the mass of one carbon atom, then we would assign carbon an atomic mass of one point zero u and oxygen an atomic mass of one point three three u. [CHECKPOINT]
However, it is more convenient to have these numbers as whole numbers or as near to a whole numbers as possible. While searching for various atomic mass units, scientists initially took one-sixteenth of the mass of an atom of naturally occurring oxygen as the unit. This was considered relevant due to two reasons. First, oxygen reacted with a large number of elements and formed compounds. Second, this atomic mass unit gave masses of most of the elements as whole numbers. However, in nineteen sixty-one for a universally accepted atomic mass unit, carbon-twelve isotope was chosen as the standard reference for measuring atomic masses. [CHECKPOINT]
One atomic mass unit is a mass unit equal to exactly one-twelfth the mass of one atom of carbon-twelve. The relative atomic masses of all elements have been found with respect to an atom of carbon-twelve. Imagine a fruit seller selling fruits without any standard weight with him. He takes a watermelon and says, this has a mass equal to twelve units. He makes twelve equal pieces of the watermelon and finds the mass of each fruit he is selling, relative to the mass of one piece of the watermelon. Now he sells his fruits by relative fruit mass unit, as shown in Figure three point four. [CHECKPOINT]
The figure shows a whole watermelon, then it is cut into twelve equal pieces, representing one-twelfth of a watermelon, and finally the fruit seller uses these pieces to weigh other fruits. Similarly, the relative atomic mass of the atom of an element is defined as the average mass of the atom, as compared to one-twelfth the mass of one carbon-twelve atom. Table three point two gives the atomic masses of a few elements. Hydrogen is one u, Carbon is twelve u, Nitrogen is fourteen u, Oxygen is sixteen u, Sodium is twenty-three u, Magnesium is twenty-four u, Sulphur is thirty-two u, Chlorine is thirty-five point five u, and Calcium is forty u. [CHECKPOINT]
Now, how do atoms exist? Atoms of most elements are not able to exist independently. Atoms form molecules and ions. These molecules or ions aggregate in large numbers to form the matter that we can see, feel or touch. Let us answer the questions for this section. First, define the atomic mass unit. One atomic mass unit is a mass unit equal to exactly one-twelfth the mass of one atom of carbon-twelve. Second, why is it not possible to see an atom with naked eyes? Atoms are extremely small, smaller than anything we can imagine or compare with, and they do not exist independently in most cases, making them invisible to the naked eye. [CHECKPOINT]
Next, we learn about molecules. A molecule is in general a group of two or more atoms that are chemically bonded together, that is, tightly held together by attractive forces. A molecule can be defined as the smallest particle of an element or a compound that is capable of an independent existence and shows all the properties of that substance. Atoms of the same element or of different elements can join together to form molecules. Let us look at molecules of elements. The molecules of an element are constituted by the same type of atoms. Molecules of many elements, such as argon, Ar, and helium, He, are made up of only one atom of that element. [CHECKPOINT]
But this is not the case with most of the non-metals. For example, a molecule of oxygen consists of two atoms of oxygen and hence it is known as a diatomic molecule, O₂. If three atoms of oxygen unite into a molecule, instead of the usual two, we get ozone, O₃. The number of atoms constituting a molecule is known as its atomicity. Metals and some other elements, such as carbon, do not have a simple structure but consist of a very large and indefinite number of atoms bonded together. Let us look at the atomicity of some non-metals from Table three point three. Argon is monoatomic. Helium is monoatomic. Oxygen is diatomic. Hydrogen is diatomic. Nitrogen is diatomic. Chlorine is diatomic. Phosphorus is tetra-atomic. Sulphur is poly-atomic. [CHECKPOINT]
Now, molecules of compounds. Atoms of different elements join together in definite proportions to form molecules of compounds. Few examples are given in Table three point four. Water, H₂O, is formed by hydrogen and oxygen in a mass ratio of one to eight. Ammonia, NH₃, is formed by nitrogen and hydrogen in a mass ratio of fourteen to three. Carbon dioxide, CO₂, is formed by carbon and oxygen in a mass ratio of three to eight. Let us perform Activity three point two. Refer to Table three point four for ratio by mass of atoms present in molecules and Table three point two for atomic masses of elements. Find the ratio by number of the atoms of elements in the molecules of compounds given in Table three point four. [CHECKPOINT]
The ratio by number of atoms for a water molecule can be found as follows. For hydrogen, the ratio by mass is one, atomic mass is one u, so mass ratio divided by atomic mass is one divided by one, which equals one. For oxygen, the ratio by mass is eight, atomic mass is sixteen u, so mass ratio divided by atomic mass is eight divided by sixteen, which equals one half. To get the simplest ratio, we multiply both by two. Thus, the ratio by number of atoms for water is H to O equals two to one. Next, what is an ion? Compounds composed of metals and non-metals contain charged species. The charged species are known as ions. [CHECKPOINT]
Ions may consist of a single charged atom or a group of atoms that have a net charge on them. An ion can be negatively or positively charged. A negatively charged ion is called an anion and the positively charged ion, a cation. Take, for example, sodium chloride, NaCl. Its constituent particles are positively charged sodium ions, Na⁺, and negatively charged chloride ions, Cl⁻. A group of atoms carrying a charge is known as a polyatomic ion. We shall learn more about the formation of ions in Chapter four. Table three point five shows some ionic compounds. Calcium oxide is composed of calcium and oxygen with a mass ratio of five to two. Magnesium sulphide is composed of magnesium and sulphur with a mass ratio of three to four. Sodium chloride is composed of sodium and chlorine with a mass ratio of twenty-three to thirty-five point five. [CHECKPOINT]
Now we move to writing chemical formulae. The chemical formula of a compound is a symbolic representation of its composition. The chemical formulae of different compounds can be written easily. For this exercise, we need to learn the symbols and combining capacity of the elements. The combining power or capacity of an element is known as its valency. Valency can be used to find out how the atoms of an element will combine with the atom or atoms of another element to form a chemical compound. The valency of the atom of an element can be thought of as hands or arms of that atom. Human beings have two arms and an octopus has eight. [CHECKPOINT]
If one octopus has to catch hold of a few people in such a manner that all the eight arms of the octopus and both arms of all the humans are locked, how many humans do you think the octopus can hold? Represent the octopus with O and humans with H. Can you write a formula for this combination? Do you get OH₄ as the formula? The subscript four indicates the number of humans held by the octopus. The valencies of some common ions are given in Table three point six. We will learn more about valency in the next chapter. Table three point six lists names and symbols of some ions. For valency one, we have Sodium, Na⁺, Potassium, K⁺, Silver, Ag⁺, Copper one, Cu⁺, Hydrogen, H⁺, Hydride, H⁻, Chloride, Cl⁻, Bromide, Br⁻, Iodide, I⁻, Ammonium, NH₄⁺, Hydroxide, OH⁻, Nitrate, NO₃⁻, and Hydrogen carbonate, HCO₃⁻. [CHECKPOINT]
For valency two, we have Magnesium, Mg²⁺, Calcium, Ca²⁺, Zinc, Zn²⁺, Iron two, Fe²⁺, Copper two, Cu²⁺, Oxide, O²⁻, Sulphide, S²⁻, Carbonate, CO₃²⁻, Sulphite, SO₃²⁻, and Sulphate, SO₄²⁻. For valency three, we have Aluminium, Al³⁺, Iron three, Fe³⁺, Nitride, N³⁻, and Phosphate, PO₄³⁻. Some elements show more than one valency. A Roman numeral shows their valency in a bracket. The rules that you have to follow while writing a chemical formula are as follows. First, the valencies or charges on the ion must balance. Second, when a compound consists of a metal and a non-metal, the name or symbol of the metal is written first. [CHECKPOINT]
For example, calcium oxide, CaO, sodium chloride, NaCl, iron sulphide, FeS, copper oxide, CuO, where oxygen, chlorine, sulphur are non-metals and are written on the right, whereas calcium, sodium, iron and copper are metals, and are written on the left. Third, in compounds formed with polyatomic ions, the number of ions present in the compound is indicated by enclosing the formula of ion in a bracket and writing the number of ions outside the bracket. For example, Mg(OH)₂. In case the number of polyatomic ion is one, the bracket is not required. For example, NaOH. Now let us look at formulae of simple compounds. The simplest compounds, which are made up of two different elements are called binary compounds. [CHECKPOINT]
Valencies of some ions are given in Table three point six. You can use these to write formulae for compounds. While writing the chemical formulae for compounds, we write the constituent elements and their valencies as shown below. Then we must crossover the valencies of the combining atoms. Let us work through the examples. Example one: Formula of hydrogen chloride. Symbol H and Cl. Valency one and one. Formula of the compound would be HCl. Example two: Formula of hydrogen sulphide. Symbol H and S. Valency one and two. Formula is H₂S. Example three: Formula of carbon tetrachloride. Symbol C and Cl. Valency four and one. Formula is CCl₄. [CHECKPOINT]
For magnesium chloride, we write the symbol of cation, Mg²⁺, first followed by the symbol of anion, Cl⁻. Then their charges are criss-crossed to get the formula. Example four: Formula of magnesium chloride. Symbol Mg and Cl. Charge two plus and one minus. Formula is MgCl₂. Thus, in magnesium chloride, there are two chloride ions for each magnesium ion. The positive and negative charges must balance each other and the overall structure must be neutral. Note that in the formula, the charges on the ions are not indicated. Some more examples. Example a: Formula for aluminium oxide. Symbol Al and O. Charge three plus and two minus. Formula is Al₂O₃. [CHECKPOINT]
Example b: Formula for calcium oxide. Symbol Ca and O. Charge two plus and two minus. Here, the valencies of the two elements are the same. You may arrive at the formula Ca₂O₂. But we simplify the formula as CaO. Example c: Formula of sodium nitrate. Symbol Na and NO₃. Charge one plus and one minus. Formula is NaNO₃. Example d: Formula of calcium hydroxide. Symbol Ca and OH. Charge two plus and one minus. Formula is Ca(OH)₂. Note that the formula of calcium hydroxide is Ca(OH)₂ and not CaOH₂. We use brackets when we have two or more of the same ions in the formula. [CHECKPOINT]
Here, the bracket around OH with a subscript two indicates that there are two hydroxyl groups joined to one calcium atom. In other words, there are two atoms each of oxygen and hydrogen in calcium hydroxide. Example e: Formula of sodium carbonate. Symbol Na and CO₃. Charge one plus and two minus. Formula is Na₂CO₃. In the above example, brackets are not needed if there is only one ion present. Example f: Formula of ammonium sulphate. Symbol NH₄ and SO₄. Charge one plus and two minus. Formula is (NH₄)₂SO₄. Let us answer the questions for this section. First, write down the formulae of sodium oxide, aluminium chloride, sodium sulphide, and magnesium hydroxide. Sodium oxide is Na₂O. Aluminium chloride is AlCl₃. Sodium sulphide is Na₂S. Magnesium hydroxide is Mg(OH)₂. [CHECKPOINT]
Second, write down the names of compounds represented by the following formulae. Al₂(SO₄)₃ is aluminium sulphate. CaCl₂ is calcium chloride. K₂SO₄ is potassium sulphate. KNO₃ is potassium nitrate. CaCO₃ is calcium carbonate. Third, what is meant by the term chemical formula? The chemical formula of a compound is a symbolic representation of its composition. Fourth, how many atoms are present in a H₂S molecule and a PO₄³⁻ ion? In an H₂S molecule, there are two hydrogen atoms and one sulphur atom, making three atoms in total. In a PO₄³⁻ ion, there is one phosphorus atom and four oxygen atoms, making five atoms in total. [CHECKPOINT]
Now we study molecular mass. In section three point two point two we discussed the concept of atomic mass. This concept can be extended to calculate molecular masses. The molecular mass of a substance is the sum of the atomic masses of all the atoms in a molecule of the substance. It is therefore the relative mass of a molecule expressed in atomic mass units, u. Let us solve Example three point one. Part a: Calculate the relative molecular mass of water, H₂O. Solution: Atomic mass of hydrogen is one u, oxygen is sixteen u. So the molecular mass of water, which contains two atoms of hydrogen and one atom of oxygen is two multiplied by one plus one multiplied by sixteen, which equals eighteen u. [CHECKPOINT]
Part b: Calculate the molecular mass of HNO₃. Solution: The molecular mass of HNO₃ equals the atomic mass of H plus the atomic mass of N plus three times the atomic mass of O. This is one plus fourteen plus forty-eight, which equals sixty-three u. Next, formula unit mass. The formula unit mass of a substance is a sum of the atomic masses of all atoms in a formula unit of a compound. Formula unit mass is calculated in the same manner as we calculate the molecular mass. The only difference is that we use the word formula unit for those substances whose constituent particles are ions. [CHECKPOINT]
For example, sodium chloride as discussed above, has a formula unit NaCl. Its formula unit mass can be calculated as one multiplied by twenty-three plus one multiplied by thirty-five point five, which equals fifty-eight point five u. Let us solve Example three point two. Calculate the formula unit mass of CaCl₂. Solution: Atomic mass of Ca plus two multiplied by atomic mass of Cl equals forty plus two multiplied by thirty-five point five, which equals forty plus seventy-one, which equals one hundred and eleven u. Now let us answer the questions for this section. First, calculate the molecular masses of H₂, O₂, Cl₂, CO₂, CH₄, C₂H₆, C₂H₄, NH₃, CH₃OH. H₂ is two multiplied by one equals two u. O₂ is two multiplied by sixteen equals thirty-two u. Cl₂ is two multiplied by thirty-five point five equals seventy-one u. CO₂ is twelve plus two multiplied by sixteen equals forty-four u. CH₄ is twelve plus four multiplied by one equals sixteen u. [CHECKPOINT]
C₂H₆ is two multiplied by twelve plus six multiplied by one equals thirty u. C₂H₄ is two multiplied by twelve plus four multiplied by one equals twenty-eight u. NH₃ is fourteen plus three multiplied by one equals seventeen u. CH₃OH is twelve plus three multiplied by one plus sixteen plus one equals thirty-two u. Second, calculate the formula unit masses of ZnO, Na₂O, K₂CO₃, given atomic masses of Zn equals sixty-five u, Na equals twenty-three u, K equals thirty-nine u, C equals twelve u, and O equals sixteen u. ZnO is sixty-five plus sixteen equals eighty-one u. Na₂O is two multiplied by twenty-three plus sixteen equals sixty-two u. K₂CO₃ is two multiplied by thirty-nine plus twelve plus three multiplied by sixteen equals one hundred and thirty-eight u. [CHECKPOINT]
Let us review what you have learnt. During a chemical reaction, the sum of the masses of the reactants and products remains unchanged. This is known as the Law of Conservation of Mass. In a pure chemical compound, elements are always present in a definite proportion by mass. This is known as the Law of Definite Proportions. An atom is the smallest particle of the element that cannot usually exist independently and retain all its chemical properties. A molecule is the smallest particle of an element or a compound capable of independent existence under ordinary conditions. It shows all the properties of the substance. A chemical formula of a compound shows its constituent elements and the number of atoms of each combining element. [CHECKPOINT]
Clusters of atoms that act as an ion are called polyatomic ions. They carry a fixed charge on them. The chemical formula of a molecular compound is determined by the valency of each element. In ionic compounds, the charge on each ion is used to determine the chemical formula of the compound. Now we will solve the chapter exercises. Exercise one: A zero point two four gram sample of compound of oxygen and boron was found by analysis to contain zero point zero nine six gram of boron and zero point one four four gram of oxygen. Calculate the percentage composition of the compound by weight. Total mass is zero point two four gram. Percentage of boron is zero point zero nine six divided by zero point two four multiplied by one hundred, which equals forty percent. [CHECKPOINT]
Percentage of oxygen is zero point one four four divided by zero point two four multiplied by one hundred, which equals sixty percent. Exercise two: When three point zero grams of carbon is burnt in eight point zero zero grams oxygen, eleven point zero zero grams of carbon dioxide is produced. What mass of carbon dioxide will be formed when three point zero zero grams of carbon is burnt in fifty point zero zero grams of oxygen? Which law of chemical combination will govern your answer? According to the law of constant proportions, carbon and oxygen combine in a fixed mass ratio of three to eight to produce carbon dioxide. When three grams of carbon reacts with fifty grams of oxygen, only eight grams of oxygen will be used, and the remaining forty-two grams will be left unreacted. [CHECKPOINT]
The mass of carbon dioxide produced will still be eleven grams. This is governed by the law of constant proportions. Exercise three: What are polyatomic ions? Give examples. Polyatomic ions are groups of atoms that carry a net charge and act as a single unit in chemical reactions. Examples include ammonium, NH₄⁺, hydroxide, OH⁻, nitrate, NO₃⁻, carbonate, CO₃²⁻, sulphate, SO₄²⁻, and phosphate, PO₄³⁻. Exercise four: Write the chemical formulae of the following. Magnesium chloride is MgCl₂. Calcium oxide is CaO. Copper nitrate is Cu(NO₃)₂. Aluminium chloride is AlCl₃. Calcium carbonate is CaCO₃. [CHECKPOINT]
Exercise five: Give the names of the elements present in the following compounds. Quick lime is calcium oxide, containing calcium and oxygen. Hydrogen bromide contains hydrogen and bromine. Baking powder is sodium hydrogen carbonate, containing sodium, hydrogen, carbon, and oxygen. Potassium sulphate contains potassium, sulphur, and oxygen. Finally, let us engage in the Group Activity. Play a game for writing formulae. Example one: Make placards with symbols and valencies of the elements separately. Each student should hold two placards, one with the symbol in the right hand and the other with the valency in the left hand. Keeping the symbols in place, students should criss-cross their valencies to form the formula of a compound. [CHECKPOINT]
Example two: A low cost model for writing formulae. Take empty blister packs of medicines. Cut them in groups, according to the valency of the element, as shown in the figure. Now, you can make formulae by fixing one type of ion into other. For example, for sodium ion and sulphate ion, two sodium ions can be fixed on one sulphate ion. Hence, the formula will be Na₂SO₄. Do it yourself: Now, write the formula of sodium phosphate. Since sodium has a valency of one and phosphate has a valency of three, you will need three sodium ions for one phosphate ion, giving the formula Na₃PO₄. Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]