Question

Calculate Copper has two isotopes: \(C u-63\) (abundance \(=69.2 \%\) mass \(=62.930\) amu) and \(C u-65\) (abundance \(=30.8 \%,\) mass \(=64.928\) amu). Calculate the atomic mass of copper.

Short answer

The atomic mass of copper can be calculated using the weighted average of its isotopes: Cu-63 with an abundance of 69.2% (0.692) and a mass of 62.930 amu, and Cu-65 with an abundance of 30.8% (0.308) and a mass of 64.928 amu. The atomic mass of copper is approximately \(63.548 \text{ amu}\), obtained by adding the weighted masses of both isotopes: \(43.547 \text{ amu}\) for Cu-63 and \(20.001 \text{ amu}\) for Cu-65.

Step-by-step solution

Convert the percentage abundance to decimal

We need to convert the abundance of each isotope from percentage to decimals. To do this, divide the percentage by 100: Abundance of Cu-63 = 69.2% = 0.692 Abundance of Cu-65 = 30.8% = 0.308

Calculate the weighted mass of each isotope

Next, multiply the mass of each isotope by their respective abundance, in decimals: Weighted mass of Cu-63 = mass of Cu-63 × abundance of Cu-63 = 62.930 amu × 0.692 ≈ 43.547 amu Weighted mass of Cu-65 = mass of Cu-65 × abundance of Cu-65 = 64.928 amu × 0.308 ≈ 20.001 amu

Calculate the atomic mass of copper

Add the weighted masses of both isotopes to find the atomic mass of copper: Atomic mass of copper = weighted mass of Cu-63 + weighted mass of Cu-65 ≈ 43.547 amu + 20.001 amu ≈ 63.548 amu The atomic mass of copper is approximately 63.548 amu.

Key concepts

Isotopic Abundance

When delving into the world of chemistry, the concept of Isotopic Abundance is fundamental. Atoms of the same element can have different numbers of neutrons, yielding isotopes with distinct masses.

Consider a jar filled with marbles, some red and some blue, where each color represents a different isotope of an element. Isotopic abundance is akin to understanding the proportion of red marbles to blue marbles in the jar. Similarly, isotopic abundance in chemistry refers to the proportion of each isotope present in a naturally occurring sample of an element.

The abundance is usually expressed as a percentage and can also be converted into a decimal for calculations. For instance, if the red marbles (isotope A) comprise 60% of the total, the isotopic abundance would be 0.60 when used in calculations.

Weighted Average Mass

Moving forward, the Weighted Average Mass of an element's isotopes plays an integral role in determining the element's atomic mass that we find on the periodic table. Think of it as if we are averaging the weight of our jar of red and blue marbles, taking into account not only how heavy each color of marble is, but how many of each color there are.

Mathematically, to find this average for an element, we multiply the mass of each isotope by its respective abundance (after converting the percentage to a decimal), and then sum these values together. This produces a 'weighted' average because it reflects the contribution of each isotope according to how prevalent it is, hence the term 'weighted'.

If we follow through with our example of copper isotopes, we calculate the average mass by taking the mass of each copper isotope and multiplying it by its abundance. The final atomic mass reflects the average mass of a copper atom as found in nature.

Chemistry Education

The importance of Chemistry Education comes into sharp focus when dealing with atomic mass calculations. Mastering such topics equips students with the understanding of matter at the most fundamental level.

A well-rounded chemistry education includes not only theoretical knowledge but also problem-solving skills, and the calculation of atomic mass is a quintessential exercise in combining both. Students learn to translate abstract concepts into practical applications.

Why Does It Matter?

For science students, grasping how to calculate weighted average mass is not just for solving textbook problems. It's essential for real-world scientific endeavors, such as developing new materials or pharmaceuticals, where knowing the exact composition of an element can influence the outcome of an experiment.