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Www.soran.edu.iq Inorganic chemistry Assistance Lecturer Amjad Ahmed Jumaa  Calculating the number of atoms in a given amount of a compound.  Calculating.

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Presentation on theme: "Www.soran.edu.iq Inorganic chemistry Assistance Lecturer Amjad Ahmed Jumaa  Calculating the number of atoms in a given amount of a compound.  Calculating."— Presentation transcript:

1 www.soran.edu.iq Inorganic chemistry Assistance Lecturer Amjad Ahmed Jumaa  Calculating the number of atoms in a given amount of a compound.  Calculating the Frequency and Wavelength of an Electromagnetic Wave,.  Calculating the energy of a photon. 1

2 www.soran.edu.iq  Calculating the number of atoms in a given amount of a compound:  convert from grams of compound to moles of a particular atom to number of atoms. Let's try an example. Example: How many carbon atoms are present in (50.3 g) of (C 2 H 6 ). Solution: We started this problem in the above example, when we calculated the moles of ethane in (50.3 g ethane). To continuo, we need two additional conversion factors. One should represent the mole ratio between moles of( C ) atoms and moles of ethane molecules. The other conversion factor needed is Avogadro's number.

3 www.soran.edu.iq Step(1): the two conversion factors needed are:  You should come up with the following strategy: Grams of C 2 H 6 → moles of C 2 H 6 → moles of C → atoms of C. Step (2): = 2.01 x 10 24 C atom.

4 www.soran.edu.iq  Calculating the frequency and wavelength of an electromagnetic wave: All types of electromagnetic radiation move through a vacuum at a speed of about (3.00 x 10 8 m/s), which is called the speed of light( c), speed is an important property of a wave traveling through space and is equal to the product of the wavelength and the frequency of the wave. For electromagnetic waves : c = λv …………………….. (1). Equation (1) can be rearranged as necessary to solve for either the wavelength (λ) or the frequency (v). Example: A certain (AM) radio station broadcasts at a frequency (6.00 x 10 2 kHz). What is the wavelength of these radio wave in meters.

5 www.soran.edu.iq Solution: Step(1): solve the equation (1) algebraically for the wavelength (λ). c = λv Step (2): since the speed of light has units of (m/s). We must convert the frequency from units of (kHz) to (Hz)(s -1 ). 6.00 x 10 2 kHz x = 6.00 x 10 5 Hz = 6.00 x 10 5 s -1. Step (3): calculate the value of (λ) by substituting the known quantities into equation (2). λ = = 5.00 x 10 2 m. ……………………(2)

6 www.soran.edu.iq Example (2): What is the frequency of light that has a wavelength of (665) (nm). Solution: Step (1): solve equation (1) algebraically for the frequency (v). c = λv …………………..(3) Step (2): since the speed of light has units of (m/s). we must convert the wavelength from units of (nm) to (m). 665 nm x = 6.65 x 10 -7 m. Step (3): calculate the value of (v) by substituting the known quantities into equation (3).

7 www.soran.edu.iq = 4.51 x 10 14 s -1 = 4.51 x 10 14 Hz  Calculating the energy of a photon:  atoms and molecules could emit( or absorb) energy only in discrete quantities.  Planck gave the name (quantum) to the smallest quantity of energy that can be emitted (or absorbed) in the form of electromagnetic radiation. The energy (E) of a single quantum of energy is given by: E = hν …………………..(4) h is Planck's constant = 6.63 x10 -34 J.s ν is the frequency of radiation. v =

8 www.soran.edu.iq Example: The yellow light given off by a sodium vapor lamp has a wavelength of (589nm), what is the energy of a single photon of this radiation. Solution: Step (1): you are given wavelength in this problem. Equation (4) shows the relationship between energy and frequency. However, there is a relationship between frequency and wavelength (see equation 3). Substituting for the frequency in equation (4), we have: ………5 Step (2): since the speed of light is in units of (m/s), we must convert the wavelength from units of (nm) to (m).

9 www.soran.edu.iq x = 5.89 x 10 -7 m. Step (3): Calculate the value of (E) by substituting the known quantities into equation (5). E = = 3.38 x 10 -19 J.


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