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average atomic mass problems - Contenido educativo
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In this video, I'm going to explain how to solve problems with average atomic mass.
00:00:01
In the first one, we know the abundance of the natural isotopes of one element and their atomic masses.
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So we have this element. We don't know the element. That doesn't matter.
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And we know that it has three isotopes.
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And these numbers here are the abundance of each one of these isotopes.
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and then we also need to know the atomic mass of each one of these isotopes that
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are these numbers okay so to calculate the average atomic mass we will need to
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multiply the abundance of each isotope by its atomic mass so this is the first
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abundance in the first atomic mass, then the second isotope's abundance times its
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atomic mass, and we still have another stable isotope and its mass.
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Okay, so check. We have multiplied each abundance by its atomic mass, and we divide
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everything by 100. So right now we need a calculator, we do these numbers in the
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calculator, it's easy, and the average atomic mass of this element, we don't
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know which element is, it's not important, is 20.17 atomic mass units. In these
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problems we can use also this formula to calculate the abundance of an element,
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the isotopes of an element, if we know the average atomic mass, like in this case.
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Lithium has an elemental atomic mass of 6.941.
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This is the average atomic mass of the element.
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So it's this information here. So in this case we know this information.
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And it has two natural occurring isotopes.
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we are given or we know the masses of each one of these isotopes but we don't
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know the abundance of each one. So how do we do this? We are going to
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use the same formula only that in this case we know that the average atomic
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mass is 6.941 equals and then the abundance of each isotope we don't know
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so I'm going to say that the abundance of the isotope 6.0151 is
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for example X percent okay so I'm going to call this X because I don't know how
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much is that and then for the second isotope the mass is this number it will
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be like Y percent but we know that X and Y as there are only two isotopes the
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100 percent so Y is 100 minus X so I'm going to use this in this part so six
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7 times 100 minus X is the abundance of the second isotope so now I need to
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work out this equation to solve this equation for X so I'm the
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100 cos multiplying here I'm going to multiply this to remove the brackets
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this number multiplied by 100 is this number this number I'm going to move it
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here so i isolate the x the thing that we don't know i do this subtraction it's easy with my
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calculator there is a mistake here sorry this goes here so i have in this minus 7.5 equals
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now i do this and the number is minus 1.001 x okay so x is minus 7.5 divided by minus 1.001
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the number is 7.49 remember that it is percent that's the abundance of this isotope okay
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the isotope that has an average atomic mass of more or less six and then the abundance of the
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other isotope is 100 minus x and we can see that this isotope is much more abundant than
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the isotope with the lithium-7 than lithium-6 and you can notice that the average atomic mass
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is more like as a number that is closer to seven that it is the six so it's it's normal that
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lithium seven is much more abundant than lithium six i hope this helps see you in class
00:06:55
- Idioma/s:
- Autor/es:
- Juan Carlos Fajardo
- Subido por:
- Juan C. F.
- Licencia:
- Reconocimiento - Compartir igual
- Visualizaciones:
- 120
- Fecha:
- 14 de octubre de 2020 - 21:01
- Visibilidad:
- Público
- Centro:
- IES CLARA CAMPOAMOR
- Duración:
- 07′ 04″
- Relación de aspecto:
- 17:9 Es más ancho pero igual de alto que 16:9 (1.77:1). Se utiliza en algunas resoluciones, como por ejemplo: 2K, 4K y 8K.
- Resolución:
- 1920x1008 píxeles
- Tamaño:
- 43.74 MBytes
