Chemistry 11. Unit 5 The Mole Concept

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1 1 Chemistry 11 Unit 5 The Mole Concept

2 1. Atomic mass and Avodagro s hypothesis It has been studied that during a chemical reaction, atoms that make up the starting material rearrange to form new and different

2 2 1. Atomic mass and Avodagro s hypothesis It has been studied that during a chemical reaction, atoms that make up the starting material rearrange to form new and different molecules which are called compounds. There is a question: how do we know how many atoms/particles of each type of starting materials will be involved in a chemical reaction?

3 3 John Dalton was the first who tried to determine how much of one element could combine with a given amount of another element. He made an assumption: all elements react to form compound in 1:1 ratio. Based on this assumption, he assigned an arbitrary masses to each element. He gave the value of 1 to hydrogen as it is the lightest. Other elements were given the arbitrary masses with respect to hydrogen. English chemist John Dalton FRS ( )

4 4 Example: The reaction between 2.74 g of hydrogen gas and g of chlorine gas makes 100 g of hydrogen chloride. What is the mass of chlorine? Solution: Assuming hydrogen chloride contains one atom each of hydrogen and chlorine, the chlorine atom is therefore times heavier than the hydrogen. Since hydrogen is assigned a mass of 1, chlorine has a mass of g 2.74g = 35.5

5 5 By investigating some simple compounds between elements, Dalton developed a list of relative masses for some elements, in which all mass values were determined assuming that hydrogen has the mass of 1. This mass scaling worked for some elements only. The source of error is the assumption that compounds contain one atom of each element. Obviously, it is not correct! (We can find many counter examples from Chapter 4!)

6 Joseph Gay-Lussac, on the other hand, explored the reactions between gases at same temperature and pressure. He observed that many gases combined in simple whole number ratios.

6 6 Joseph Gay-Lussac, on the other hand, explored the reactions between gases at same temperature and pressure. He observed that many gases combined in simple whole number ratios. Example: 1L of hydrogen gas reacts with 1L of chlorine gas to produce 2L of hydrogen chloride gas. This leads to the Gay-Lussac law of combining volume. French chemist Joseph Gay-Lussac ( )

7 7 Observations by Gay-Lussac seemed to be unrelated to masses of elements. However, Amedeo Avogadro interpreted his observations from a different angle, and proposed the following hypothesis to explain the whole number ratios. Avogadro hypothesis: Equal volumes of different gases, at the same temperature and pressure, contain the same number of particles. Italian chemist Amedeo Avogadro ( )

8 8 That means if 1L of gas A reacts with 1L of gas B, then the compound formed will have the formula AB. Similarly, if 2L of gas A reacts with 1L of gas B, then the molecule formed will have the formula of A 2 B. Despite being neglected by his contemporaries for many years, Avogadro s hypothesis was finally proven correct, and it is known nowadays as the Avogadro s law! V Mathematically it looks like: n = k

9 9 2. The Mole The Avogadro s hypothesis allows us to determine the formula of a compound by finding the ratio of the volumes of gases needed. For example, 1L of N 2 gas reacts completely with 3L of H 2 gas to produce 2L of ammonia gas. Therefore, the formula of ammonia should be NH 3. This hypothesis, however, only works for gas compounds. How do we determine how much of one element reacts in a reaction if it is a solid?

10 10 We can measure mass instead Mass of a solid is related to how many atoms it has. But how are they related to one another? We need a scale: Unified atomic mass unit (amu)! It is something similar to Dalton s scale of masses except that the reference is no longer hydrogen but carbon-12 (a specific isotope of carbon which has 6 protons and 6 neutrons). One amu is defined as 1 atom. 12 the mass of a carbon-12

11 11 The amu scale only shows the relative masses of the elements. How about their actual masses? Two useful definitions: (1) A mole is the number of carbon atoms in exactly 12 g of carbon. (2) Molar mass is the mass of one mole of particles.

12 12 Molar mass of an element can be read from the periodic table. Numerically, molar mass is equivalent to the relative atomic mass.

13 13 Example: What is the molar mass of iron (III) sulphate? Example: Find the molar mass of Ag 2 SO 4.

14 14 3. Relating mole, mass, volume and number of particles The molar mass of a compound is the mass for 1 mole of such substance. Given the molar mass of a substance is X g/mol. There are two possible conversion factors: 1 mol X g The number of particles in X g of the substance is 1 mole. X g 1 mol Every mole of the substance has the mass of X g.

15 15 Avogadro s hypothesis says that gases of equal volume at the same temperature and pressure have the same number of particles. Standard conditions (STP): 0 C and kpa At standard conditions, one mole of gas occupies 22.4 L of space. Two possible conversion factors: 1 mol 22.4 L The number of particles in 22.4 L of gas at STP is 1 mole L 1 mol Each mole of gas has the volume of 22.4 L at STP

16 16 Mole is a fundamental unit of counting in chemistry. Experimentally, its value is This value is called Avogadro number, and has no units. Two conversion factors: 1 mol particles Every particles are called 1 mole particles 1 mol 1 mole of substance contains particles

17 17 Mole is the central quantity that connects mass, volume and number of molecules. It serves as a bridge to enable the conversion between these values.

18 18 Example: What is the volume occupied by 50.0 g of NH 3 at STP? [65.9L] Example: What is the mass of atoms of Cl? [ g] Example: How many oxygen atoms are contained in 75 L of SO 3 at STP? [ ]

19 19 Sometimes, the volume of a gas is not measured at STP. Rather, its density is given. Recall the definition of density: d = m V Depending upon which quantity is missing, you can use either of the following forms: If d is unknown If m is unknown If V is unknown d = m V m = V d V = m d Note that V = 22.4 L at STP cannot be used for solids or liquids.

20 20 Example: What is the volume occupied by 3.00 mol of ethanol given its density is g/ml? [175 ml] Example: How many moles of Hg(l) are contained in 100 ml of Hg(l) given its density is 13.6 g/ml? [6.78 mol]

21 21 If the density of a gas is unknown, then the following equation can be used: d = molar mass of gas molar volume ofgas Example: What is the density of O 2 at STP? [1.43 g/l]

22 22 Example: A 2.50 L bulb contains 4.91 g of a gas at STP. What is the molar mass of the gas? [1.96 g/l] Example: Al 2 O 3 (s) has a density of 3.97 g/ml. How many Al atoms are in 100 ml of Al 2 O 3? [ atoms]

23 23 4. Percentage composition By definition, percentage composition is the percentage by mass of each element in the chemical formula. Example: What is the percentage composition of H 2 SO 4? Molar mass of H 2 SO 4 = 2x x x4 = 98.1 Percentage of H: 2 100% = 2% 98.1 Percentage of S: % = 32.7% 98.1 Percentage of O: % = 65.2% 98.1

24 24 Example: Calculate the percentage composition of C 2 H 4 O 2. [C: 40%, H: 6.7%, O: 53.3%] Example: Calculate the percentage of the bold species in Fe 2 (SO 4 ) 3 9H 2 O. [51.3%]

25 25 5. Empirical and molecular formulas The composition of a substance can be expressed in either empirical formula or molecular formula. Empirical formula: The smallest whole-number ratio of atoms that represents the composition of a substance. Molecular formula: The whole-number ratio that shows the actual composition of a substance. These two formulas are related by an integer multiple.

26 26 For example, alkenes (such as ethene C 2 H 4, propene C 3 H 6, butane C 4 H 8 and pentene C 5 H 10 ) have the same empirical formula CH 2 but different molecular formulas. (That s why they are different compounds!)

27 27 Example: What is the empirical formula of a compound consisting of 80.0% C and 20.0% H? Assume 100 g of the compound. Mass of C = 80.0% x 100 g = 80.0 g Mass of H = 20.0% x 100 g = 20.0 g Use the mass to determine the moles of each element. Mole C = 80.0 g 12.0 g/mol = 6.67 mol Mole H = 20.0 g 1.0 g/mol = 20 mol Determine the smallest ratio by dividing by the smallest number of moles. # C = 6.67 mol 6.67 mol = 1 # H = 20 mol 6.67 mol = 3 Empirical formula is CH 3.

28 28 Practice: A compound contains 58.5% C, 7.3% H, and 34.1% N. What is the empirical formula of the compound? [C 2 H 3 N]

29 29 Challenging example: What is the empirical formula of a compound consisting of 81.8% C and 18.2% H? [C 3 H 8 ]

30 30 The molecular formula can be found by finding the integer multiple which is the ratio of the molar mass of the compound and the mass of the empirical formula which is called empirical mass. N = molar mass empirical mass Molecular formula is just the product of this multiple and the empirical formula. Molecular formula = N empirical formula

31 31 Example: The empirical formula of a compound is SiH 3. If mol of the compound has a mass of 1.71 g, what is the molecular formula of the compound? The empirical mass = (1.0) = 31.1 g/mol The molar mass = The integer multiple = 1.71 g mol = 62.2 g/mol 62.2 g/mol 31.1 g/mol = 2 The molecular formula = 2(SiH 3 ) = Si 2 H 6

32 32 Practice: A certain sugar has a chemical composition of 40% C, 6.6% H and 53.3% O. The molar mass is 180 g/mol. What is the molecular formula? [C 6 H 12 O 6 ]

33 33 6. Molar concentration Solutions are homogeneous mixtures of substances in which there is no boundary between them. Solutions can be of different forms. (Review Chapter 3)

34 34 The substances in a mixture can be divided into solute (in small quantity) or solvent (in large quantity). To determine how much solute is dissolved into the solvent, a value called concentration is calculated. This quantity is defined as: c = amount of solute volume ofsolvent The number of moles of solute contained in 1 L of solution is called molar concentration or molarity (M).

35 35 Molar concentration can be calculated using the following equation: M = mol V Symbolically, the relationship between molarity, mole and volume can be shown by the following triangles:

36 36 Example: What is the [NaCl] in a solution containing 5.12 g of NaCl in ml of solution? To solve this problem, you need to find the number of moles of NaCl dissolved. Mole of NaCl = 5.12 g 58.5 g/mol Concentration of NaCl = = mol mol L = M

37 37 Example: What mass of NaOH is contained in 3.50 L of M NaOH solution? The number of moles of NaOH dissolved is given by # mol NaOH = M 3.50 L = mol The molar mass of NaOH is 40.0 g The mass of NaOH = mol 40.0 g/mol = 28.0 g

38 38 Example: What is the molarity of pure sulphuric acid having a density of g/ml? Note that molarity = mole volume = mass molar mass 1 volume = density molar mass Hence, the molarity of pure sulphuric acid is [H 2 SO 4 ] = g L 1 mol 98.1 g = 18.7 M

39 39 7. Dilution When two solutions are mixed, the resulting solution has a volume and total number of moles equal to the sum of the individual volume and individual number of moles of chemical found in the separate solutions. The molarity of the resulting solution is therefore given by Molarity = total number of moles of chemical totalvolume ofthe mixture

40 40 Consider the following: Original stock solution Diluted solution The number of moles of solute in the initial solution is equal to the number of moles of solute in the diluted solution. Therefore M 1 V 1 = M 2 V 2

41 41 Example: If ml of M NaCl is added to ml of water, what is the resulting [NaCl] in the mixture? [0.200 M] Example: What volume of 6.00 M HCl is needed to make 2.00 L of M HCl? [ L]

42 42 Example: If ml of M NaCl is added to ml of M NaCl, what is the resulting [NaCl] in the mixture? Calculate the total number of moles of NaCl: Moles = = mol Molarity = mol L = M

43 43 Challenging question: What volume of 4.50 M HCl can be made by mixing 5.65 M HCl with ml of 3.55 M HCl? [456.5 ml]

Vijaykumar N. Nazare

Std-XI science Unit 1: Some Basic Concepts of Chemistry Vijaykumar N. Nazare Grade I Teacher in Chemistry (Senior Scale) vnn001@ chowgules.ac.in 1.1 IMPORTANCE OF CHEMISTRY Chemistry is the branch of science