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MATTER 1.1ATOMS AND MOLECULES MATTER CONTENTS Define relative atomic mass and relative molecular mass based on the C-12 scale Analyze mass spectra in.

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Presentation on theme: "MATTER 1.1ATOMS AND MOLECULES MATTER CONTENTS Define relative atomic mass and relative molecular mass based on the C-12 scale Analyze mass spectra in."— Presentation transcript:

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2 MATTER

3 1.1ATOMS AND MOLECULES MATTER CONTENTS Define relative atomic mass and relative molecular mass based on the C-12 scale Analyze mass spectra in terms of isotopic abundances and calculate A r Navigation button

4 1.1ATOMS AND MOLECULES MATTER CONTENTS Define relative atomic mass and relative molecular mass based on the C-12 scale Analyze mass spectra in terms of isotopic abundances and calculate A r Navigation button Definition of relative atomic mass and molecular mass based on the C-12 Define relative atomic mass and relative molecular mass based on the C-12 scale

5 1.1ATOMS AND MOLECULES MATTER CONTENTS Define relative atomic mass and relative molecular mass based on the C-12 scale Analyze mass spectra in terms of isotopic abundances and calculate A r Navigation button Animation of mass spectrometer and how to calculate the atomic mass from its isotopic abundances Analyze mass spectra in terms of isotopic abundances and calculate A r

6 Relative atomic mass, A r of an element: A r = average mass of one atom of the element mass of one atom C-12 1 12 x TOPIC 1: MATTER SUB TOPIC: Atoms and Molecules Objective: At the end of the lesson, you should be able to: 1) Define relative atomic mass and relative molecular mass based on C-12 scale Back to content

7 Relative molecular mass, M r of a molecular substance: M r = average mass of one molecule of the substance mass of one atom C-12 1 12 x The mass of an atom depends on the number of electrons, protons and neutrons it contains. But atoms are extremely small particles. Clearly we cannot weigh a single atom, but it is possible to determine the mass of one atom relative to another experimentally. By international agreement, atomic mass unit is define as a mass of the atom exactly equal to one-twelfth the mass of one carbon-12 atom.

8 (1) Vaporisation chamber Sample of the element is vaporised into gaseous atom. MASS SPECTROMETER (Five part of mass spectrometer and its functions) To vacuum pump

9 (2) Ionisation chamber A hot filament emits high-energy electrons. When the electrons collide with the gaseous sample (atom or molecule), positive ions are produced by dislodging an electron from each atom or molecule MASS SPECTROMETER (Five part of mass spectrometer and its functions)

10 To vacuum pump MASS SPECTROMETER (Five part of mass spectrometer and its functions)

11 To vacuum pump (3) Acceleration chamber The positive ions are accelerated by an electric field towards the two oppositely charged plates. The electric field is produced by a high voltage between the two plates. The emerging ions are of high and constant velocity. MASS SPECTROMETER (Five part of mass spectrometer and its functions)

12 To vacuum pump (4) Magnetic Field The positive ions are separated and deflected into a circular path by a magnet according to its m/e ratio. MASS SPECTROMETER (Five part of mass spectrometer and its functions)

13 To vacuum pump 37 Cl + 35 Cl + Magnetic field 35 Cl + MASS SPECTROMETER (Five part of mass spectrometer and its functions) 37 Cl + has a higher ratio of m/e than 35 Cl +, hence it is deflected less

14 To vacuum pump (5) Ion Detector The numbers of ions and types of isotopes are recorded as a mass spectrum. MASS SPECTROMETER (Five part of mass spectrometer and its functions) 19 20 21 22 23

15 TOPIC 1: MATTER SUB TOPIC: Atoms and Molecules Objective: At the end of the lesson, you should be able to: 2) Analyze mass spectra in terms of isotopic abundances and calculate A r 19 20 21 22 23 20 Ne 90.92% 21 Ne 0.26% 22 Ne 8.82% This is the diagram of mass spectra obtained from mass spectrometer with the element Neon with three different isotopes at its own abundances.

16 TOPIC 1: MATTER SUB TOPIC: Atoms and Molecules Objective: At the end of the lesson, you should be able to: 2) Analyze mass spectra in terms of isotopic abundances and calculate A r Relative abundance m/e 63 8.1 9.1 Relative abundance Isotopes of Mg 24 25 26 The height is proportional to the amount of isotope present m/e ratio or nucleon no. or relative atomic mass Typical mass spectrum of Magnesium At this stage, you need to know that the mass spectrum diagram consist of four important element. Let’s takes mass spectrum diagram of Magnesium as an example.

17 TOPIC 1: MATTER SUB TOPIC: Atoms and Molecules Objective: At the end of the lesson, you should be able to: 2) Analyze mass spectra in terms of isotopic abundances and calculate A r Since relative abundance can be presented in 2 ways, ratio & percentage, make sure you identify them before do the next step. Relative ratio 8.0 3.5 85 87m/e Mass spectrum in ratio Percentage % 70 30 85 87m/e Mass spectrum in percentage Sum of abundances = 100 i.e. 70 + 30 = 100 Sum of abundances is not 100 i.e. 8.0 + 3.5 = 11.5

18 TOPIC 1: MATTER SUB TOPIC: Atoms and Molecules Objective: At the end of the lesson, you should be able to: 2) Analyze mass spectra in terms of isotopic abundances and calculate A r Why mass spectrum formula is very important? This formula is useful to calculate the average atomic mass of element!! This is general mass spectrum formula could be used and you have to memorize it! A r =  (m i x Q i )  Q i Relative atomic mass Isotopes masses Relative abundances Sum of relative abundances ExampleExercise Past year Question

19 TOPIC 1: MATTER SUB TOPIC: Atoms and Molecules Objective: At the end of the lesson, you should be able to: 2) Analyze mass spectra in terms of isotopic abundances and calculate A r Relative abundance 63 9.1 8.1 24 Calculate the relative molecular mass from the above mass spectrum. Example 1 From this diagram, we can deduce that: 1)The element has 3 isotopes: m 1, m 2 & m 3 with value 24, 25 & 26 respectively 2)Type of abundance:  Q i = Q 1 + Q 2 + Q 3 ; 63 + 9.1 + 8.1 = 80.2. Not equal to 100. Thus, ratio type. 3)Write the equation: A r =  (m i x Q i )  (Q i ) A r = (m 1 x Q 1 ) + (m 2 x Q 2 ) + (m 3 x Q 3 ) Q 1 + Q 2 + Q 3 4)Substitute the data into the equation to get the answer. 5)A r = ( x ) + ( x ) + ( x ) m1m1 m2m2 m3m3 Q3Q3 Q1Q1 Q1Q1 Q2Q2 Q2Q2 + +Q3Q3 2526m/e242526 63 9.1 8.1

20 TOPIC 1: MATTER SUB TOPIC: Atoms and Molecules Objective: At the end of the lesson, you should be able to: 2) Analyze mass spectra in terms of isotopic abundances and calculate A r Relative abundance 63 9.1 8.1 24 Calculate the relative molecular mass from the above mass spectrum. Example 1 From this diagram, we can deduce that: 1)The element has 3 isotopes: m 1, m 2 & m 3 with value 24, 25 & 26 respectively 2)Type of abundance:  Q i = Q 1 + Q 2 + Q 3 ; 63 + 9.1 + 8.1 = 80.2. Not equal to 100. Thus, ratio type. 3)Write the equation: A r =  (m i x Q i )  (Q i ) A r = (m 1 x Q 1 ) + (m 2 x Q 2 ) + (m 3 x Q 3 ) Q 1 + Q 2 + Q 3 4)Substitute the data into the equation to get the answer. 5)A r = ( x ) + ( x ) + ( x ) = 24 2425268.1 Q1Q1 63 Q2Q2 9.1 + +Q3Q3 2526m/e 9.1 8.1 63

21 TOPIC 1: MATTER SUB TOPIC: Atoms and Molecules Objective: At the end of the lesson, you should be able to: 2) Analyze mass spectra in terms of isotopic abundances and calculate A r The ratio of relative abundance of naturally occurring of copper is as follow: Based on the carbon-12 scale, the relative atomic mass of 63 Cu = 62.9396 and 65 Cu = 64.9278. Calculate the A r of copper. Example 2 From question, we can deduce that: 1)Copper has 2 isotopes: 63 Cu and 65 Cu with value 62.9396 and 64.9278 respectively 2) means 3)Write the equation: A r Cu =  (m i x Q i ) Q i A r Cu = (m 1 x Q 1 ) + (m 2 x Q 2 ) Q 1 + Q 2 4)Substitute the data into the equation A r Cu = (62.9396 x 2.333) + (64.9278 x 1) 2.333 + 1 A r Cu = 63.5 63 Cu 65 Cu = 2.333 63 Cu 65 Cu = 2.333 63 Cu 65 Cu = 2.333 1

22 TOPIC 1: MATTER SUB TOPIC: Atoms and Molecules Objective: At the end of the lesson, you should be able to: 2) Analyze mass spectra in terms of isotopic abundances and calculate A r 1.Naturally occurring potassium consists of potassium-39 and potassium-41. Calculate the percentage of each isotope present if the average is 39.1. (K-39 = 95%, K-41 = 5%) 2. Relative abundance of Rb is given below: (a) What is the percentage abundance of each of the isotopes? (Rb-85 = 72%, Rb-87 = 28%) (b) Calculate the relative atomic mass of Rubidium. (A r Rb = 85.56) Exercises 85 87 m/e 18 7 solution

23 TOPIC 1: MATTER SUB TOPIC: Atoms and Molecules Objective: At the end of the lesson, you should be able to: 2) Analyze mass spectra in terms of isotopic abundances and calculate A r SOLUTION STEP1: Write the equation A r K =  (m i x Q i )  Q i = (m 1 x Q 1 ) + (m 2 x Q 2 ) Q 1 + Q 2 STEP2: List the data given A r K = 39.1, m 1 = 39, m 2 =41, Q 1 =?, Q 2 = ? STEP3: Substitute the data into the equation A r K = (m 1 x Q 1 ) + (m 2 x Q 2 ) Q 1 + Q 2 39.1 = (39 x Q 1 ) + (41 x (100 – Q 1 )) 100 Q 1 = 95%; Q 2 = 100 – 95 = 5% Abundance of K-39 = 95% and abundance of K-41 = 5% K only has 2 isotopes; K-39 & K-41 Assume that the abundance of potassium in percentage. Hence, Q 1 + Q 2 = 100 & Q 2 = 100 – Q 1 back

24 TOPIC 1: MATTER SUB TOPIC: Atoms and Molecules Objective: At the end of the lesson, you should be able to: 2) Analyze mass spectra in terms of isotopic abundances and calculate A r SOLUTION (a) STEP1: List down the abundances of isotopes Rb Abundance of Rb-85 = 18 Abundance of Rb-87 =7 STEP2: Convert it to percentage Abundance of Rb-85 in %: 18 / (18 + 7) x 100% = 72% Abundance of Rb-87 in %: 100 – 72 = 28% (b) STEP1: Write the equation A r Rb = (m 1 x Q 1 ) + (m 2 x Q 2 ) Q 1 + Q 2 STEP2: List down the data given m 1 = 85, Q 1 = 72, m 2 = 87, Q 2 = 28, A r Rb =? STEP3: Substitute the data into the equation A r Rb = (85 x 72 ) + (87 x 28) 100 = 85.56 Rb has 2 isotopes; Rb-85 & Rb-87 back

25 TOPIC 1: MATTER SUB TOPIC: Atoms and Molecules Objective: At the end of the lesson, you should be able to: 2) Analyze mass spectra in terms of isotopic abundances and calculate A r 1.(i) Define isotope. (ii) A sample of chlorine consists of 76% 35 Cl and 24% 37 Cl. Calculate the relative atomic mass of chlorine if the masses of isotopes are 34.96 amu and 36.97 amu respectively. (Mac 01) 2. Mass spectrometer is used to determine the relative atomic mass. State the main function for the component or part in the mass spectrometer: (a) Vacuum pump (b) Vapourisation chamber (c) Electric field (d) Magnetic field. (Jan 00)

26 TOPIC 1: MATTER SUB TOPIC: Atoms and Molecules WELL DONE! Try some past paper questions Objective: At the end of the lesson, you should be able to: 2) Analyze mass spectra in terms of isotopic abundances and calculate Ar


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