# Mole Concept and Stoichiometry

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1 1 Mole Concept and Stoichiometry Concept Mole Concept Introduction This is our common experience that when we go to market to buy something, a few things we always get in definite numbers. For example, eggs we get in a number of 12 which is called a dozen, playing cards we get in a number of 52 which is called a pack, papers we get in a number of 480 sheets which is called a ream. Dozen, pack and ream are units to describe those items. Following chart shows certain items which we get in definite numbers and the units in which these items are described. Substance Unit Number Drawing pins Gross 144 Football boots Pair 2 Eggs Dozen 12 Playing cards Pack 52 Paper Ream 480 sheets Carbon Mole 6.02 x particles What is a mole? In Latin, mole means massive heap of material. In chemistry, it is a unit which is used to describe an amount of atoms, ions and molecules. It enables chemists to count these particles by weighing. According to 14 th conference of National Institutes of Standards and Technology ( NIST ) held in 1971, the definition of mole is given as follows. Mole is the amount of substance of a system which contains as many elementary entities as there are atoms in kilogram of carbon 12. Its symbol is mol. When the mole is used, the elementary entities must be specified and they may be atoms, molecules, ions, electrons, other particles or specified group of such particles. Visualizing a mole as a pile of particles, however, is just one way to understand this concept. A sample of a substance has mass, volume ( generally used with gases ) and number of particles that is proportional to the chemical amount ( measured in moles ) of the sample. For example, one mole of oxygen ( O 2 ) occupies a volume of 22.4 liters at

2 2 standard temperature and pressure ( STP = O 0 C and 1 atm ), has a mass of grams and contains about x molecules of oxygen. Measuring one of these quantities allows the calculation of the others and this is frequently done in stoichiometry. One interpretation : A specific number of particles When a quantity of particles is to be described, mole is a grouping unit analogous to groupings such as pair, dozen or gross, in that all of these words represent specific numbers of objects. The main difference between the mole and the other grouping units is the magnitude of the number represented and how that number is obtained. One mole is an amount of substance containing Avogadro s number of particles. Avogadro s number is equal to x Unlike pair, dozen and gross, the exact number of particles in a mole can not be counted. There are several reasons for this. First, the particles are too small and can not be seen even with a microscope. Second, as naturally occurring carbon contains approximately % carbon 12, the sample would need to be purified to remove every atom of carbon 13 and carbon 14. Third, as the number of particles in a mole is tied to the mass of exactly 12 grams of carbon 12, a balance would need to be constructed that could determine if the sample was one atom over or under exactly 12 grams. If the first two requirements were met, it would take one million machines counting one million atoms each second more than 19,000 years to complete the task. So, practically it can be treated as impossible though is theoretically possible. Obviously, if the number of particles in a mole can not be counted, the value must be measured indirectly and with every measurement there is some degree of uncertainty. Therefore, the number of particles in a mole, Avogadro s constant ( N A ) can only be approximately found through experimentation and thus its reported values will vary slightly ( at the tenth decimal place ) based on the method used for measurement.. Most methods agree to four significant figures, so N A is generally said to equal x particles per mole and this value is usually sufficient for solving common problems. Another key point is that the formal definition of a mole does not include a value for Avogadro s constant and this is probably due to the inherent uncertainty in its measurement. As for the difference between Avogadro s constant and Avogadro s number, they are numerically equivalent but the former has the unit of mol -1 whereas the latter is a pure number with no unit. Avogadro s constant is related to the number of atoms or molecules present in the volume occupied by one mole of a gas at S.T.P. conditions or it is the number of atoms or molecules present in one mole of a substance. Avogadro s number is a pure number just as dozen or gross and it is dimensionless. It equals the number of particles ( electrons, atoms, ions, molecules etc. ) of anything which is numerically equivalent to the number of atoms present in kg of carbon 12. Activity 1 - In a boys hostel, 1206 eggs were brought for the breakfast of students. What is the number of moles of eggs brought?

3 3 A second interpretation : A specific mass Atoms and molecules are incredibly small and even a tiny chemical sample contains an unimaginable number of them. Therefore, counting the number of atoms or molecules in a sample is impossible. The multiple interpretations of the mole allow us to bridge the gap between the submicroscopic world of atoms and molecules and the macroscopic world that we can observe. To determine the chemical amount of a sample, we use the substance s molar mass, the mass per mole of particles. We will use carbon 12 as an example because it is the standard for the formal definition of the mole. According to definition, one mole of carbon 12 is 12 g/mol. However the molar mass for the element carbon is g/mol. Why are they different? To answer that question, few terms need to be clarified. In the periodic table, we observe that most of the atomic weights are not integers. The atomic weight is a weighed average of the atomic masses of the natural isotopes of the element. For example, bromine has two natural isotopes with atomic masses of 79 u and 81 u. The unit u represents the atomic mass unit and is used in place of grams because the value would be inconveniently small. These two isotopes of bromine are present in nature in almost equal amounts, so the atomic weight of the element bromine is ( i.e. nearly 80, the arithmetic mean of 79 and 81 ). A similar situation exists for chlorine but chlorine 35 is almost three times as abundant as chlorine 37, so the atomic weight of chlorine is Technically, atomic weights are ratios of the average atomic mass to the unit u and that is why they do not have units. Sometimes, atomic weights are given the unit u but this is not quite correct according to the International Union of Pure and Applied Chemistry ( IUPAC ). To find the molar mass of an element or compound, determine the atomic, molecular or formula weight and express that value as g/mol. For bromine and chlorine, the molar masses are g/mol and g/mol respectively. Sodium chloride ( NaCl ) has a formula weight of ( atomic weight of Na + atomic weight of Cl ) and a molar mass of g / mol. Formaldehyde ( CH 2 O ) has a molecular weight ( atomic weight of C + 2 [ atomic weight of H ] + atomic weight of O ) and a molar mass of g/mol. The concept of molar mass enables chemist to measure the number of submicroscopic particles in a sample without counting them directly simply by determining the chemical amount ( in grams )of a sample. To find the chemical amount of a sample, chemists measure its mass and divide by its molar mass. Multiplying the chemical amount ( in moles ) by Avogadro s constant ( N A ) yields the number of particles present in the sample.

6 6 Concept Applications of Avogadro s law Avogadro s law has been used at number of places. Some of its applications are described below. 1) To determine atomicity of gases The number of atoms present in one molecule of an element is called its atomicity. Consider the following reaction: Hydrogen + Chlorine Hydrogen chloride 1 volume 1 volume 2 volumes ( According to Gay-Lussac s law ) n molecules n molecules 2n molecules ( According to Avogadro s law ) 1 atom 1 atom 2 compound - atoms (molecules) ½ atom ½ atom 1 molecule 1 molecule of hydrogen chloride = ½ molecule of hydrogen + ½ molecule of chlorine But ½ molecule of hydrogen = 1 atoms of hydrogen. So 1 molecule of hydrogen = 2 atoms of hydrogen Hence atomicity of hydrogen molecule is 2. Activity 1 - One volume of nitrogen combines with three volumes of hydrogen to form two volumes of ammonia gas. Determine the atomicity of nitrogen gas. 2) Explanation of Gay Lussac s law of combining volumes Gay Lussac suggested that one volume of hydrogen and one volume of chlorine react to produce two volumes of hydrogen chloride gas. According to Avogadro s law if, 1 volume of hydrogen contains n molecules of the gas, then 1 volume of chlorine also contains n molecules of the gas 2 volumes of hydrogen chloride contain 2n molecules of the gas Hydrogen + Chlorine Hydrogen chloride 1 volume 1 volume 2 volumes ( According to Gay-Lussac s law ) n molecules n molecules 2n molecules ( According to Avogadro s law ) But hydrogen and chlorine are diatomic. 2 atoms 2 atoms 2 molecules 1 atom 1 atom 1 molecule

7 7 It is seen that 1 molecule of hydrogen chloride is formed when 1 atom of hydrogen combines with 1 atom of chlorine. Thus, Avogadro s law explains Gay Lussac s law of combining volumes. 3) Modification of Dalton s atomic theory Avogadro suggested the presence of two particles an atom and a molecule. An atom is the smallest particle of an element that can take part in a chemical reaction. It does not have independent existence and it can not be split into similar particles. A molecule is the smallest particle of matter that exists in free state. It is made up of two or more atoms of the same or different elements. A molecule can be divided to give the respective atoms. Let us take the example of reaction between hydrogen and chlorine. Hydrogen Chlorine Hydrogen chloride 1 volume 1 volume 2 volumes n molecules n molecules 2n molecules ( According to Avogadro s hypothesis ) 1 molecule 1 molecule 2 molecules ½ molecule ½ molecule 1 molecule This shows that 1 molecule of hydrogen chloride is formed from ½ molecule of hydrogen and ½ molecule of chlorine. Since molecules can be divided, Avogadro s hypothesis is true. 4) To derive the relation between molecular weight and vapour density According to Avogadro s law, Mass of n molecules of a gas or vapour Vapour density at STP = Mass of n molecules of hydrogen Mass of 1 molecule of a gas or vapour Vapour density = Mass of 1 molecule of hydrogen Mass of 1 molecule of a gas or vapour Vapour density = Mass of 2 atoms of hydrogen Mass of 1 molecule of a gas or vapour Vapour density = x ( Mass of 1 atoms of hydrogen )

8 8 1 Mass of 1 molecule of a gas or vapour Vapour density = --- x Mass of 1 atom of hydrogen 1 Vapour density ( VD ) = -- x Molecular weight 2 2 x Vapour density ( 2 V.D ) = Molecular weight Activity 2 - Find out the molecular weight of NO 2 gas and determine its vapour density. 5) To derive the relation between gram molecular weight ( GMW) and gram molecular volume (GMV) Density of any gas = Weight of 1 liter of that gas at STP Let us determine the molar volume of hydrogen ( any gas for that matter ) Density of hydrogen = 0.09 g/liter Molecular weight of hydrogen = Molecular weight in grams Molar volume = = = 22.4 liters Density of hydrogen 0.09 Thus gram molecular masses ( one gram mole ) of all gases at STP occupy a volume of 22.4 liters. Activity 3 - Derive the relation between gram molecular volume and vapour density of a volatile compound. 6) To determine the formula of a gaseous molecule Let us take an example of ammonia gas. Experimentally it is proved that two volumes of ammonia on decomposition give one volume of nitrogen and three volumes of hydrogen. Ammonia Nitrogen + Hydrogen 2 volumes 1 volume 3 volumes ( According to Gay-Lussac s law ) 2 molecules 1 molecule 3 molecules ( According to Avogadro s law ) 2 molecules 2 atoms 6 atoms ( Nitrrogen and hydrogen are diatomic ) 1 molecule 1 atom 3 atoms

9 9 1 molecule of ammonia = 1 atom of nitrogen + 3 atoms of hydrogen Hence, molecular formula of ammonia is NH 3. Activity 4 - Two volumes of nitrogen dioxide, on decomposition, give two volumes of nitric oxide and one volume of oxygen. Prove that molecular formula of nitrogen dioxide is NO 2. Check your understanding : 1) 500 ml of nitric oxide react with 300 ml of oxygen to form nitrogen dioxide. What is the composition of the resulting mixture? ( Answer 500 ml NO ml O 2 ) 2) 6 grams of a gas occupies a volume of 700 ml at 70 0 C and 70 cm pressure. If 1 liter of hydrogen weighs 0.9 gram at STP, what is the vapour density of the gas? ( Answer 13) 3) The molecular weight of a gas is 44. What is the weight of the gas which occupies a volume of 20 liters at 27 0 C and 70 cm pressure? ( Answer gram )

10 10 Concept Stoichiometry Introduction Stoichiometry is simply the mathematics behind chemistry. Given enough information, one can use stoichiometry to calculate masses, moles and percents within a chemical equation. It is a mathematical method to calculate quantitative information from chemical formulae or equation of chemical reaction. Every chemical reaction has its characteristic proportions. Chemical formulae, equations, atomic weights and molecular weights are used to find out these proportions. Determination of what and how much is used and produced in chemical processes is the major concern of stoichiometry.. What is stoichiometry? Stoichiometry is a branch of chemistry that deals with the quantitative relationships that exist among the reactants and products in chemical reactions. In a balanced chemical reaction, the relations among quantities of reactants and products typically form a proportion of whole numbers. For example, in a reaction that forms ammonia (NH 3 ), exactly one molecule of nitrogen (N 2 ) reacts with three molecules of hydrogen (H 2 ) to produce two molecules of NH 3 : N 2 + 3H 2 2NH 3 Stoichiometry can be used to derive various kinds of information from the balanced chemical equation. Depending upon its use, it can be broadly classifies as reaction stoichiometry, composition stoichiometry, gas stoichiometry etc. 1) Reaction stoichiometry describes the quantitative relationships among substances as they participate in chemical reactions. In the example above, reaction stoichiometry describes the 1:3:2 proportion s of molecules of nitrogen, hydrogen, and ammonia. 2) Composition stoichiometry describes the quantitative (mass) relationships among elements in compounds. For example, composition stoichiometry describes the nitrogen to hydrogen (mass) relationship in the compound ammonia i.e. 28 grams of nitrogen react with 6 grams of hydrogen to form 34 grams of ammonia.

11 11 Stoichiometry can be used to calculate quantities such as the amount of products that can be produced with given reactants and percent yield (the percentage of the given reactant that is made into the product). A stoichiometric amount or stoichiometric ratio of a reagent is the amount or ratio where, it is assumed that the reaction proceeds to completion and : 1. all reagent is consumed, 2. there is no shortfall of reagent, and 3. no residues remain. 3) Gas stoichiometry deals with reactions involving gases, where the gases are at a known temperature, pressure, and volume and can be assumed to be ideal gases. For gases, the volume ratio is ideally the same by the ideal gas law, but the mass ratio of a single reaction has to be calculated from the molecular masses of the reactants and products. In practice, due to the existence of isotopes, molar masses are used instead when calculating the mass ratio. Stoichiometry is based on the law of conservation of mass. The mass of the reactants equals the mass of the products. Illustrations 1) Reaction stoichiometry - How many moles of magnesium will react with one mole of oxygen? A balanced chemical equation for the reaction between magnesium and oxygen can be written as follows. 2 Mg + O 2 2 MgO The reaction stoichiometry shows that Mg, O 2 and MgO take part in the reaction in the proportion 2:1:2. So 2 moles of Mg will react with one mole of oxygen. Activity 1 Describe the reaction stoichiometry in the following reaction. 4 NH 3 (g) + 5 O 2 (g) 4 NO (g) + 6 H 2 O ( l ) 2) Composition stoichiometry - How many grams of oxygen do you need to burn 1.00 gram of Mg? (Note: The balanced equation for this reaction was obtained in the previous question.) The balanced equation shows that 24 grams ( 2 moles ) of Mg react with 32 grams ( 1 mole ) of oxygen to produce 56 grams ( 2 moles ) of MgO. So 1.33 gram of O 2 will be needed to burn 1 gram of Mg metal.

12 12 Activity 2 - Describe the composition stoichiometry in the following reaction. (NH 4 ) 2 Cr 2 O 7 (s) Cr 2 O 3 (s) + N 2 (g) + 4 H 2 O (g) 3) Gas Stoichiometry Calculate the volume of gaseous NO 2 produced at STP from the combustion of 100 grams of NH 3 by the reaction 4NH 3 (g) + 7O 2 (g) 4NO 2 (g) + 6H 2 O (g ) First we find out the moles of NH 3. Moles of NH 3 = 100 / 17 = 5.88 moles There is a 1:1 molar ratio of NH 3 to NO 2 in the above balanced combustion reaction, so 5.88 mol of NO 2 will be formed. We will employ the ideal gas law to solve for the volume at 0 C (273 K) and 1 atmosphere using the gas law constant of R = L atm K 1 mol 1 : PV = n R T or V = n R T / P. Then, V = 5.88 x x 273 = liters Activity 3 - Describe the gas stoichiometry in the following reaction. Limiting reagents C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (g) Sometimes when reactions occur between two or more substances, one reactant runs out before the other. That is called the "limiting reagent". Often, it is necessary to identify the limiting reagent in a problem. Illustration: A chemist only has grams of C 2 H 2 and an unlimited supply of oxygen and he desires to produce as much CO 2 as possible. If he uses the equation below, how much oxygen should he add to the reaction? 2C 2 H 2 (g ) + 5O 2 (g) ---> 4CO 2 (g) + 2 H 2 O(l) Here acetylene, C 2 H 2 is the limiting reagent. To solve this problem, it is necessary to determine how much oxygen should be added if all of the reactants were used up (this is the way to produce the maximum amount of CO 2 ). First, we calculate the number of moles in 6.50 g of C 2 H 2. We know that molecular weight of CO 2 is 26. Therefore 1 mole of C 2 H 2 weighs 26 g.then 6.5 gram of C 2 H 2 is equivalent to 6.5 / 26 = 0.25 moles Then, because there are five (5) molecules of oxygen to every two (2) molecules of C 2 H 2, we need to multiply the result by 5/2 to get the total molecules of oxygen. Then we convert to grams to find the amount of oxygen that needs to be added:

13 x 5/2 = moles of O 2. Since 1 mole of O 2 weighs 32 grams, moles of O 2 will weigh x 32 = 20 grams Activity 4 :If 20.0 g of iron (III) oxide (Fe 2 O 3 ) are reacted with 8.00 g aluminium (Al) in the thermite reaction, which reactant is limiting? Fe 2 O 3 (s) + 2 Al (s) Al 2 O 3 (s) + 2 Fe ( l ) Check your understanding : 1) A reaction takes place as Na 3 PO 4 + Al(NO 3 ) 3 AlPO NaNO 3 How many moles of sodium nitrate are produced when two moles of sodium phosphate react with two moles of aluminium nitrate? ( Answer 6 moles ) 2) What weight of sulphuric acid will be required to dissolve 6 grams of magnesium carbonate according to the equation MgCO 3 + H 2 SO 4 MgSO 4 + H 2 O + CO 2? 3) Potassium chlorate decomposes as follows. ( Answer 7 grams ) 2 KClO 3 2 KCl + 3 O 2 4) If 4.9 grams of KClO 3 is decomposed, what volume of O 2 is produced at STP? ( Answer liters )

14 14 Concept Percentage composition of a compound by mass Percentage composition of a compound by mass can be defined as follows. Percentage composition by mass is the mass of solute ( or one component ) divided by the mass of solute + solvent ( mass of all the components ) multiplied by 100. Or The percentage composition of a compound is a relative measure of the mass of each different element present in the compound. How to calculate the percentage composition of a compound? (i) Calculate the molecular mass, ( molecular weight, formula weight or formula mass ) MM of the compound. (ii) Calculate the total mass of each element present in the formula of the compound (iii) % composition = ( total mass of element present / molecular mass ) x 100 Illustration 1) Calculate the percent by weight of each element present in sodium sulfate (Na 2 SO 4 ). Calculate the molecular mass MM of the compound Na 2 SO 4. MM = ( 2 x 23 ) + ( 1 x 32 ) + ( 4 x 16 ) = 142 Total mass of Na present = 2 x 23 = 46 Total mass of S present = 1 x 32 = 32 Total mass of O present = 4 x 16 = 64 % of Na = ( 46 / 142 ) x 100 = 32.4 % % of S = ( 32 / 142 ) x 100 = % % of O = ( 64 / 142 ) x 100 = % Total percentage = = 100 2) Calculate the percent by weight of each element in sodium chloride (NaCl) Calculate the molecular mass MM of the compound Na Cl MM = ( 1 x 23 ) + ( 1 x 35.5 ) = Total mass of Na present = 1 x 23 = 23 Total mass of Cl present = 1 x 35.5 = 35.5 % of Na = 23 / 58.5 = ; % of Cl = 35.5 / 58.5 = Total percentage = = 100

15 15 Activity 1 Calculate the percent by weight of each element present in calcium phosphate Ca 3 (PO 4 ) 2.Calculate the molecular mass MM of the compound Ca 3 (PO 4 ) 2. Activity 2 A compound with molecular mass 74 contains elements Ca, O and H. If the % of Ca is and that of H is 2.73, find the atomic mass of Ca in the compound. Check your understanding : 1) Ammonium dichromate (NH 4 ) 2 Cr 2 O 7 has a molecular mass 252. What is the % of N, H, Cr and O in the compound ( Answer N = %, H = 3.18 %, Cr = %, O = % ) 2) What is the atomic mass of potassium, if the % of nitrogen is 13.9 and that of oxygen is 47.5 in a compound containing K,N and O with molecular mass 101? ( Answer 39 ) 3) A compound contains the elements Mg, N and O. The percentage of these elements in a compound is 16.2, 18.9 and 64 respectively. What is the molecular weight of this compound? ( Answer 148 )

16 16 Concept Empirical and Molecular Formula Introduction In chemistry, generally we come across three types of formulae of compounds viz. Empirical formula, Molecular formula and Structural formula. Empirical formula of a compound shows the ratio of elements present in that compound. Molecular formula of a compound shows how many atoms of each element are present in a molecule of that compound. Structural formula of a compound shows the covalent bonds ( single, double or triple ) between every two atoms and placement of the atoms in the molecule of that compound. Empirical and Molecular formula Let us explain empirical and molecular formula in some more details. Empirical formula represents the simplest whole number ratio of atoms of each element present in a compound. The empirical formula is the molecular formula reduced to its lowest common denominator. The empirical formula makes no reference to isomerism, structure or absolute number of atoms. It does not indicate the actual number of atoms of the elements present in the molecule. For example, formaldehyde, acetic acid and glucose have the same empirical formula, CH 2 O. The elements C, H and O are present in the proportion 1:2:1 in all these compounds. Empirical formula represents the simplest whole number ratio of atoms of each element present in a compound. However, the molecular formulae of formaldehyde, acetic acid and glucose are HCHO, CH 3 COOH and C 6 H 12 O 6 respectively. Empirical formula and molecular formula are same for formaldehyde. But they are different for acetic acid and glucose. Acetic acid has double the number of atoms and glucose has six times the number of atoms of formaldehyde. Molecular formula is a group of elemental symbols and possibly subscript numbers which represent the composition of a molecule. Molecular formula shows the exact number of atoms of each element in the molecule or how many atoms of each element are present in the molecule of a compound. In the above examples, all the compounds have the same proportion of C, H and O. But the actual number of atoms of each element are different in formaldehyde, acetic acid and glucose.

17 17 Another example is that ethylene ( C 2 H 4 ), propene ( C 3 H 6 ) and butane ( C 4 H 8 ) all have the same empirical formula CH 2 but their molecular formulae are different. They have the same proportion of C to H as 1:2 but actual number of C and H atoms are 1.5 times in propene and 2 times in butene as compared to those in ethylene. Activity 1 - Write the molecular formula of two cycloalkanes which have the empirical formula CH 2. Activity 2 Two organic compounds have the same empirical formula and molecular formula C 2 H 6 O but different structural formula. Suggest the names of these compounds. The relation between empirical formula and molecular formula can be given as follows. It follows then that : Molecular formula = Integer x Empirical formula Molecular formula weight = Integer x Empirical formula weight Check your understanding : 1) Find the empirical formula of a compound of carbon and hydrogen which contains 80 % carbon. ( Answer CH 3 ) 2 ) A compound has an empirical formula CH 2 O and a vapour density 30. What is its molecular formula? ( Answer C 2 H 4 O 2 ) 3) An organic compound contains the elements C, H and Br. Their % in the compound is as follows. C = %, H = 2.13 % and Br = % The vapour density of the compound is 94. What is the molecular formula of the compound? ( Answer C 2 H 4 Br 2 )

18 18 Concept Introduction How to calculate Empirical Formula from Percentage Composition To calculate empirical formula from percentage composition, we follow the steps : 1) Assume 100g of sample 2) Convert all percentages to a mass in grams, eg, 21% = 21g, 9% = 9g 3) Find the relative atomic mass (r.a.m) of each element present using the Periodic Table 4) Calculate the moles of each element present: n = mass r.a.m 5) Divide the moles of each element by the smallest of these to get a mole ratio 6) If the numbers in the mole ratio are all whole numbers (integers) convert this to an empirical formula 7) If the numbers in the mole ratio are NOT whole numbers, you will need to further manipulate these until the mole ratio is a ratio of whole numbers (integers). Recognizing the decimal equivalent of common fractions is very helpful. Common Decimal Equivalent Fraction Mole Ratio Example /8 1 : converts to 1 : 9/8 multiply throughout by 8 to give 8 : /4 1 : 0.25 converts to 1 : 1/4 multiply throughout by 4 to give 4 : /3 1 : 1.33 converts to 1 : 4/3 multiple throughout by 3 to give 3 : /8 1 : converts to 1 : 11/8 multiply throughout by 8 to give 8 : /2 2 : 1.5 converts to 2 : 3/2 multiply throughout by 2 to give 4 : /8 1 : converts to 1 : 13/8 multiply throughout by 8 to give 8 : /3 2 : 1.66 converts to 2 : 5/3 multiply throughout by 3 to give 6 : /8 1 : converts to 1 : 7/8 multiply throughout by 8 to give 8 : 7

19 19 Illustration 1) A compound is found to contain 47.25% copper and 52.75% chlorine. Find the empirical formula for this compound. Element Cu Cl mass in grams r.a.m moles = mass r.a.m = = 1.49 divide throughout by lowest number = = 2.01 = 2 Empirical formula for this compound is CuCl 2 2) A compound with a molecular mass of 34 g/mol is known to contain 5.88 percent hydrogen and percent oxygen. Find the molecular formula for this compound. First find the empirical formula for this compound. Element H O mass in grams r.a.m moles = mass r.a.m = = 5.88 divide throughout by the smallest number = = 1 Empirical formula is HO Calculate the empirical formula mass: = 17.0 g/mol Molecular formula weight = Integer x Empirical formula weight 34 = n x 17 So, n = 2 Molecular Formula is 2 x (HO) which is H 2 O 2 3) A substance on analysis gave Na = 43.4 %, C = 11.3 % and O = 45.3 %. Find out its empirical and molecular formula. Element Na C O % Atomic mass Mole = mass / r.a.m / 23 = / 12 = / 16 = 2.83 Simplest ratio 1.88 / 0.94 = / 0.94 = / 0.94 = 3

20 20 Empirical formula is Na 2 CO 3. No information is given about the molecular formula weight. A compound with molecular formula Na 2 CO 3 does exist. So molecular formula and empirical formula of this compound is same i.e. Na 2 CO 3. Activity 1 - A compound has three elements C, H and O. Their mole % in the compound is 3.331, and respectively. The molecular mass of the compound is 180. What is the molecular formula of the compound? Activity 2 - Given that the empirical formula of a compound is CH and the molar mass is 104 g/mol, calculate the molecular formula. Check your understanding : 1) A compound contains % carbon, 4.92 % hydrogen and % oxygen. Find the empirical formula of the compound. ( Answer C 7 H 6 O 2 ) 2) Find the empirical formula for a compound consisting of 63% Mn and 37% O. ( Answer - MnO 2 ) 3) The empirical formula for vitamin C is C 3 H 4 O 3. Experimental data indicates that the molecular mass of vitamin C is 176. What is the molecular formula of vitamin C? ( Answer C 6 H 8 O 6 )

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