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(solution) Help please by tomorrow!! I only need help with question number 3


Help please by tomorrow!! I only need help with question number 3


Isotopes and Weighted Averages

 

My Goals for this Lesson: Identify the structural differences between the isotopes of elements.

 

Determine the weighted average mass of an element when given the appropriate

 

data. I?m preparing to identify differences between isotopes and calculate weighted average masses.

 

Fill in the blanks using the lesson page.

 

You already know that the number of protons in an atom?s nucleus

 

an atom as a

 

specific element. For example, all oxygen atoms? nuclei contain eight

 

. However, like all

 

other naturally occurring elements, oxygen nuclei can contain different numbers of

 

.

 

Atoms of the

 

element that have

 

masses with varying numbers of neutrons

 

are called

 

.

 

There are two main ways to represent a specific isotope in your notes, or other writing.

 

Method One:

 

Add the mass number of the isotope, with a hyphen, to the end of the element's name.

 

Using Method 1, write the isotope for a carbon with 6 protons and 6 neutrons.

 

How many protons and neutrons does Uranium-235 have? Method Two:

 

The mass number is given as a superscript to the upper left side of the element's symbol, with

 

the atomic number as a subscript to the lower left side of the symbol.

 

Using Method 2, write the isotope for a carbon with 6 protons and 8 neutrons.

 

Using Method 2, write the isotope for hydrogen with a mass of 2.

 

Let's Practice

 

Use what you have learned about atomic number, mass number, and isotopes to answer the

 

following questions. Refer to the periodic table for each element?s atomic number.

 

Fill in the chart using what you have learned in the lesson. Check your answers in the

 

lesson Interactive.

 

Isotope

 

# Protons

 

# Electrons

 

# Neutrons

 

Chlorine-37

 

Br with a mass

 

number of 80

 

Beryllium-9

 

Ar with a mass

 

number of 38

 

Average Atomic Mass Fill in the blanks using the lesson page.

 

You may have noticed that the atomic mass values given on the periodic table

 

have

 

values even though the mass numbers for individual atoms

 

are

 

numbers. This is a result of averaging the mass values for all the

 

element?s

 

.

 

Most elements found in nature occur as a mixture of two or more

 

. The percentages of each

 

isotope in the mixture are the same in every sample of a given element, no matter where on

 

Earth the samples are collected. Because pure elements are made up of mixtures of

 

, it is

 

important to take into account the percentage and mass of each

 

, especially when some

 

make up a greater percentage of the mixture than other

 

. To do this, the

 

average atomic mass of each element is calculated using a weighted average.

 

What is a weighted average?

 

A weighted average accounts for the percent abundance and mass of each isotope in an

 

element.

 

Imagine that a class of 100 students took a test worth 100 points. If all of the students scored 80

 

out of 100 on the test, the average score for the class would be 80. If just one of the students

 

scored 90 out of 100 on the test, what would happen to the class average? The class average

 

would be slightly higher than 80 because of that one higher score. Similarly, most carbon atoms

 

are the carbon-12 isotope. One out of every 100 carbon atoms is the slightly heavier carbon-13

 

isotope. This small amount of carbon-13 raises the average mass of carbon from exactly 12.000

 

u to the slightly higher mass of 12.010 u. This is the mass that you see on the periodic table.

 

The average atomic mass of an element can be calculated by multiplying the mass of each

 

isotope by its relative abundance (the percentage represented in decimal form) and adding

 

the results.

 

Isotope A [percent abundance × mass] + Isotope B [percent abundance × mass] + continue for

 

each isotope = average atomic mass

 

Example

 

Isotope

 

Oxygen-16

 

Oxygen-17

 

Oxygen-18

 

This information will be Mass

 

Percent Abundance

 

16.0 u

 

99.762%

 

17.0 u

 

0.038%

 

18.0 u

 

0.200%

 

given to you in the problem. Percent Value changed to decimal.

 

0.99762

 

0.00038

 

0.00200 [0.99762 × 16.0 u] + [0.00038 × 17.0 u] + [0.00200 × 18.0 u] = 16.0 u

 

This is the work you will need to show for credit on these types of problems.

 

(Note that the percentages do not count toward significant figures because they are counted, not

 

measured.)

 

Let's Practice

 

Question One: Copper has two naturally occurring isotopes. Copper-63, with an atomic mass of

 

62.94 u, makes up 69.17 percent of the sample, and Copper-65, with an atomic mass of 64.93 u,

 

makes up the other 30.83 percent. Calculate the weighted average atomic mass of copper based

 

on the given information. Please give your answer in a total of four significant figures.

 

Isotope

 

Mass

 

Percent Abundance

 

Percent Value changed to decimal. Copper-63

 

Copper-65

 

This information is given to you in the problem.

 

[Abundance (decimal form) × atomic mass (u)] + [Abundance (decimal form) × atomic mass (u)]

 

= weighted average atomic mass (u)

 

[

 

x

 

u] + [

 

x

 

u] =

 

u

 

Question Two: Three isotopes of argon (Ar) occur in nature. Argon-36 (35.968 u) makes up

 

0.337 percent of the sample, Argon-38 (37.963 u) makes up 0.063 percent of the sample, and

 

Argon-40 (39.962 u) makes up the other 99.6 percent of the sample. Calculate the weighted

 

average atomic mass of argon, giving your answer in a total of five significant figures.

 

Isotope

 

Mass

 

Percent Abundance

 

Percent Value changed to decimal. This information is given to you in the problem.

 

[Abundance (decimal form) × atomic mass (u)] + [Abundance (decimal form) × atomic mass (u)]

 

+ [Abundance (decimal form) × atomic mass (u)] = weighted average atomic mass (u)

 

[

 

x

 

u] + [

 

x

 

u] + [

 

x

 

u] =

 

u Question Three: Imagine that scientists discovered a new element, given the symbol X. If 93.7

 

percent of the sample is an isotope with a mass of 272.98 u and the rest of the sample is an

 

isotope with a mass of 275.96 u, what is the average atomic mass of the sample? Please give

 

your answer in a total of five significant figures.

 

Isotope

 

Mass

 

Percent Abundance

 

Percent Value changed to decimal.

 

100 ? 93.7 = _______

 

This information is given to you in the problem.

 

[Abundance (decimal form) × atomic mass (u)] + [Abundance (decimal form) × atomic mass (u)]

 

= weighted average atomic mass (u)

 

[

 

x

 

u] + [

 

x

 

u] =

 

u

 

The average atomic masses that are given on the periodic table are important to chemists

 

because they represent the masses of the naturally occurring mixture of isotopes for each

 

element. This means that when we use the average atomic mass from the periodic table in a

 

calculation, we are taking into account that a sample of a given element is made up of a specific

 

mixture of its isotopes. Because these mixtures of isotopes occur naturally, we will use the

 

average atomic mass values from the periodic table in most of the calculations in this course.

 


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