Atomic Mass: Videos & Practice Problems

Atomic mass is the weighted average mass of an element's isotopes, reflecting the masses of protons, neutrons, and electrons. It's typically found on the periodic table below the element's symbol, like hydrogen's 1.008. These values aren't whole numbers due to averaging different isotopic masses. Atomic mass can be expressed in grams per mole or atomic mass units (amu), with 1 amu equaling:

$1.66\times {10}^{-27}\mathrm{kg}$

To calculate atomic mass without a periodic table, use the atomic mass formula, which multiplies each isotope's mass by its fractional abundance (the percent abundance divided by 100). When only two isotopes are involved and no fractional abundances are given, their sum equals one since they represent 100% of the element's natural abundance. Understanding atomic mass is crucial for grasping the composition and characteristics of elements.

Atomic Mass

Atomic mass is a crucial concept in chemistry, representing the mass of an element that encompasses the masses of its subatomic particles: protons, neutrons, and electrons. This value can be easily found on the periodic table, where each element is listed with its corresponding atomic mass. For instance, hydrogen, denoted by the symbol H, has an atomic mass of approximately 1.008.

It is important to note that atomic masses are often not whole numbers. This is because the atomic mass of an element is calculated as an average of the masses of its isotopes, which are variants of the element that have the same number of protons but different numbers of neutrons. Consequently, the atomic mass reflects the weighted average of these isotopes based on their natural abundance.

Atomic mass can be expressed in various units, with grams per mole being the most common. Additionally, it can be represented in atomic mass units (amu) or Daltons. To provide context, one atomic mass unit is defined as approximately \(1.66 \times 10^{-27}\) kilograms.

When examining the periodic table, each element is identified by its symbol, with the atomic mass displayed below it. The number above the symbol represents the atomic number, denoted by the variable \(Z\), which indicates the number of protons in the nucleus of an atom. Understanding these components is essential for interpreting the periodic table and grasping the fundamental properties of elements.

Greatest Atomic Mass Example

In this example, we are tasked with identifying an element from the first column of the periodic table that has the greatest atomic mass. The first column, known as the alkali metals, includes elements such as lithium (Li), sodium (Na), potassium (K), rubidium (Rb), cesium (Cs), and francium (Fr). Each element has an associated atomic mass, typically represented as a non-whole number, which can be measured in grams per mole, atomic mass units, or daltons.

Upon examining the options, we find that barium (Ba) is located in the second column, thus it is not a valid choice. Aluminum (Al), found in the thirteenth column, is also excluded. Next, we consider cesium (Cs), which is indeed in the first column and has an atomic mass of approximately 132.91. This value is significant as it indicates a relatively high atomic mass compared to other elements in the same column.

Francium (Fr), located at the bottom of the first column, is another candidate with a higher atomic mass than cesium. However, francium is highly unstable and typically represented with a whole number due to its limited isotopes. Therefore, while cesium appears to be the highest among the options provided, francium would ultimately have the greatest atomic mass if included in the comparison.

In summary, when evaluating elements based on their atomic masses, it is crucial to reference their positions in the periodic table, understand the significance of atomic mass values, and recognize the implications of stability and isotopes on these measurements. Thus, cesium (Cs) is the best choice from the given options, representing the element with the greatest atomic mass from the first column.

Which of the following choices has the greatest atomic mass?

• A. Element A (0.283 kg)
• B. Element B (3.20 x 10²⁴ amu)
• C. Element C (0.350 kg)
• D. Element D (4.14 x 10²⁶ Da)

Answer: (D)

Calculating Atomic Mass

To determine the atomic mass of an element when a periodic table is unavailable, one can calculate it using the isotopic masses and their corresponding percent abundances. The isotopic masses refer to the masses of all isotopes of the element, while the percent abundances indicate how much of each isotope is naturally occurring. These percent abundances can also be referred to as natural abundances.

To facilitate calculations, percent abundances can be converted into fractional abundances by dividing the percentage by 100. For example, a percent abundance of 25% would convert to a fractional abundance of 0.25.

The atomic mass can be calculated using the following formula:

\[ \text{Atomic Mass} = (\text{Isotope Mass}_1 \times \text{Fractional Abundance}_1) + (\text{Isotope Mass}_2 \times \text{Fractional Abundance}_2) + \ldots \]

In this formula, you multiply the mass of each isotope by its fractional abundance and sum the results. If an element has more than two isotopes, simply continue adding terms for each isotope. This method allows for the calculation of atomic mass even in the absence of a periodic table, providing a practical approach to understanding elemental properties.

Atomic Mass Comparison Example

To calculate the atomic mass of lithium, we consider its two naturally occurring isotopes: lithium-6 and lithium-7. Each isotope has a specific isotopic mass and a natural abundance expressed as a percentage. The first step in this calculation is to convert the percent abundances into decimal form by dividing by 100. For lithium-6, with a natural abundance of 7.59%, this results in a fractional abundance of 0.0759. For lithium-7, with a natural abundance of 92.41%, the fractional abundance becomes 0.9241.

Next, we apply the atomic mass formula, which is the weighted average of the isotopic masses based on their abundances. The formula can be expressed as:

Atomic Mass = (Isotopic Mass of Li-6 × Fractional Abundance of Li-6) + (Isotopic Mass of Li-7 × Fractional Abundance of Li-7)

Substituting the values, we have:

Atomic Mass = (6.015122 AMU × 0.0759) + (7.016004 AMU × 0.9241)

Calculating this gives an approximate atomic mass of 6.940037056 AMU. When rounding, this value is typically reported as 6.941 AMU, which aligns with the value found on the periodic table for lithium. This atomic mass represents the average of the isotopic masses of lithium, reflecting the relative abundances of its isotopes.

Calculating Atomic Mass

To calculate the atomic mass of an element when the fractional abundances of its isotopes are not provided, it is essential to understand that the sum of the fractional abundances of all isotopes equals 1. This means that if you have two isotopes, their fractional abundances can be represented as x and 1 - x, where x is the fractional abundance of the first isotope, and 1 - x is the fractional abundance of the second isotope.

When dealing with atomic mass calculations involving two isotopes, you can set up an equation based on the known atomic masses of the isotopes. The formula for calculating the atomic mass M of the element can be expressed as:

M = (x \cdot M_1) + ((1 - x) \cdot M_2)

In this equation, M_1 and M_2 are the atomic masses of the two isotopes. By substituting the known values into the equation, you can solve for x, which represents the fractional abundance of the first isotope. The fractional abundance of the second isotope can then be easily determined as 1 - x.

This method is particularly useful in scenarios where only two isotopes are involved, as it simplifies the calculations and allows for a clear understanding of how the isotopes contribute to the overall atomic mass of the element.

Calculating Abundance Example

Iron has two naturally occurring isotopes: Iron-54 and Iron-56, with isotopic masses of 53.939615 atomic mass units (AMU) and 55.934942 AMU, respectively. To determine the percent abundance of Iron-56, we start by defining the fractional abundance of Iron-54 as x. Since the total abundance must equal 100%, the fractional abundance of Iron-56 can be expressed as 1 - x.

The average atomic mass of iron, as listed on the periodic table, is 55.845 AMU. We can set up the equation based on the average atomic mass formula:

\[55.845 = (53.939615 \cdot x) + (55.934942 \cdot (1 - x))\]

Expanding this equation gives:

\[55.845 = 53.939615x + 55.934942 - 55.934942x\]

Combining like terms results in:

\[55.845 = 55.934942 - 1.995327x + 53.939615x\]

Rearranging the equation to isolate x leads to:

\[-0.089942 = -1.995327x\]

Dividing both sides by -1.995327 yields:

\[x \approx 0.04508\]

This value represents the fractional abundance of Iron-54. To find the fractional abundance of Iron-56, we calculate:

\[1 - x = 1 - 0.04508 \approx 0.95492\]

To convert this fractional abundance into a percentage, we multiply by 100:

\[0.95492 \times 100 \approx 95.492\%\]

Rounding this value gives us a percent abundance of approximately 95.5% for Iron-56. This calculation illustrates how isotopic masses and their abundances can be used to derive the natural distribution of isotopes in elements.

Neon possesses three naturally occurring isotopes. ²⁰Ne has a percent abundance of 90.48% and an isotopic mass of 19.99244 amu, ²¹Ne has a percent abundance of 0.27% and an isotopic mass of 20.99384 amu, and ²²Ne has a percent abundance of 9.25%. What is the isotopic mass of the ²²Ne isotope?

• A. 20.9182 amu
• B. 15.119 amu
• C. 23.001 amu
• D. 21.98768 amu

Answer: (D)

Three isotopic forms of potassium exist: ³⁹K, ⁴⁰K and ⁴¹K. Potassium has an atomic mass of 39.0983 amu. Potassium-40 has an isotopic mass of 39.9640 amu and natural abundance of 0.0117%. Potassium-41 has an isotopic mass of 40.9618 amu and natural abundance of 6.7302%. What is the isotopic mass of Potassium-39?

• A. 106.905 amu
• B. 38.9637 amu
• C. 106.903 amu

Answer: (B)

Here’s what students ask on this topic:

What is atomic mass?

Atomic mass, also known as atomic weight, is the average mass of atoms of an element, calculated using the relative abundance of isotopes in a naturally-occurring element. It is expressed in atomic mass units (amu), where one atomic mass unit is defined as one-twelfth of the mass of a carbon-12 atom, which has six protons and six neutrons.

Since an element can have several isotopes—atoms with the same number of protons but different numbers of neutrons—the atomic mass is not a whole number. It takes into account the masses of each isotope and their relative abundance. For example, chlorine has two common isotopes, chlorine-35 and chlorine-37. The atomic mass of chlorine is a weighted average of these isotopes, reflecting the fact that chlorine-35 is more abundant than chlorine-37. This concept is crucial for understanding the composition of elements and their isotopes, as well as for calculations in chemistry involving the amount of substance, like stoichiometry.

How do you calculate atomic mass?

Atomic mass, often referred to as atomic weight, is calculated by taking into account the masses of each isotope of an element and their relative abundance. Each isotope has a certain natural percent abundance and a specific mass number. To calculate the atomic mass of an element, you multiply the mass of each isotope by its relative abundance (expressed as a decimal), then sum these values.

Here's a step-by-step guide:

1. Convert the percentage of each isotope into decimal form by dividing by 100.
2. Multiply the decimal abundance of each isotope by its isotope mass (the number of protons and neutrons in the nucleus).
3. Add the results from step 2 for all isotopes of the element.

The final sum will give you the atomic mass of the element. This value is usually close to the mass number of the most abundant isotope but takes into account the presence of other isotopes and their respective masses. Atomic mass is typically reported in atomic mass units (amu), where 1 amu is defined as 1/12 the mass of a carbon-12 atom.

How do you calculate average atomic mass?

To calculate the average atomic mass of an element, you'll need to consider all the naturally occurring isotopes of that element and their respective abundances. Here's how you do it step by step:

1. List Isotopes: Start by listing each isotope of the element.
2. Find Mass of Each Isotope: Note the atomic mass of each isotope, usually given in atomic mass units (amu).
3. Determine Natural Abundance: Find the natural abundance of each isotope, often expressed as a percentage.
4. Convert Percentages to Decimals: Convert the percentage of each isotope's natural abundance into a decimal by dividing by 100.
5. Multiply Mass and Abundance: For each isotope, multiply the atomic mass by the decimal abundance.
6. Sum the Products: Add up the products from the previous step for all isotopes.

The result will be the average atomic mass of the element. This value is the weighted average based on the abundance of each isotope, which reflects the mass that would typically be found for an atom of that element on Earth.

How do you calculate the atomic mass of isotopes?

To calculate the atomic mass of isotopes, you need to consider the mass of each isotope and its relative abundance. Here's a step-by-step process:

1. Identify the Isotopes: Determine which isotopes of the element you're dealing with. For example, chlorine has two stable isotopes: chlorine-35 and chlorine-37.
2. Find the Mass of Each Isotope: Look up the atomic mass of each isotope, usually given in atomic mass units (amu). For chlorine-35, it's approximately 34.969 amu, and for chlorine-37, it's about 36.966 amu.
3. Determine the Relative Abundance: This is the percentage of each isotope found in nature. For chlorine, about 75.77% is chlorine-35 and 24.23% is chlorine-37.
4. Calculate the Weighted Average: Multiply the atomic mass of each isotope by its relative abundance (expressed as a decimal), then add these values together.

For chlorine:

$(34.969\text{amu}\times 0.7577)+(36.966\text{amu}\times 0.2423)=26.4957\text{amu}+8.9571\text{amu}=35.4528\text{amu}$

So, the atomic mass of chlorine is approximately 35.4528 amu.

What is the relative atomic mass?

Relative atomic mass, often abbreviated as A_r, is a dimensionless quantity (meaning it has no unit) that represents the average mass of atoms of an element, measured in atomic mass units (amu), compared to one-twelfth of the mass of an atom of carbon-12. This average takes into account the different masses of the isotopes of the element and their relative abundance in nature.

To calculate the relative atomic mass of an element, you multiply the mass of each isotope by its relative natural abundance (expressed as a decimal), and then sum these values. For example, chlorine has two main isotopes, ³⁵Cl and ³⁷Cl, with relative abundances of about 75% and 25% respectively. The relative atomic mass of chlorine would be calculated as follows:

$(35\text{amu}\bullet 0.75)+(37\text{amu}\bullet 0.25)=35.5\text{amu}$

Therefore, the relative atomic mass of chlorine is approximately 35.5. This value is often found on the periodic table and is used in chemical calculations involving moles and molar mass.