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Chapter 2: Problem 48

Gallium has two naturally occurring isotopes, \({ }^{69} \mathrm{Ga}\) (isotopic mass \(=68.9256\) amu, abundance \(=60.11 \%\) ) and \({ }^{71}\) Ga (isotopic mass \(=70.9247\) amu, abundance \(=39.89 \%\) ). Calculate the atomic mass of gallium.

Short Answer

Expert verified
The atomic mass of gallium is 69.72 amu.

Step by step solution

01

- Understand the given data

Identify the isotopic masses and their abundances. Isotopic mass of \({ }^{69} \mathrm{Ga}\) = 68.9256 amu Abundance of \({ }^{69} \mathrm{Ga}\) = 60.11% Isotopic mass of \({ }^{71} \mathrm{Ga}\) = 70.9247 amu Abundance of \({ }^{71} \mathrm{Ga}\) = 39.89%
02

- Convert percentages to decimals

Convert the percentage abundances to decimal form for calculations. Abundance of \({ }^{69} \mathrm{Ga}\) = 60.11% = 0.6011 Abundance of \({ }^{71} \mathrm{Ga}\) = 39.89% = 0.3989
03

- Multiply each isotopic mass by its fractional abundance

Calculate the product of each isotopic mass and its respective fractional abundance. For \({ }^{69} \mathrm{Ga}\): (68.9256 amu) \times 0.6011 = 41.4347 For \({ }^{71} \mathrm{Ga}\): (70.9247 amu) \times 0.3989 = 28.2841
04

- Add the calculated values

Add the results from Step 3 to find the weighted average atomic mass of gallium. 41.4347 + 28.2841 = 69.7188 amu

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

isotopic mass
Isotopic mass is the mass of a specific isotope of an element. Isotopes are different forms of the same element, having the same number of protons but different numbers of neutrons. This results in different masses for each isotope.

For example, Gallium has two isotopes:
  • 69Ga with an isotopic mass of 68.9256 amu
  • 71Ga with an isotopic mass of 70.9247 amu
abundance
Abundance refers to how common each isotope of an element is in nature. It is expressed as a percentage or a fraction of the total amount of the element. To calculate the atomic mass, you need to know the fractional abundance (convert percentage to decimal):

For example, the isotopic abundances for Gallium are:
  • 69Ga: 60.11% or 0.6011 (fractional abundance)
  • 71Ga: 39.89% or 0.3989 (fractional abundance)
weighted average
The atomic mass of an element is a weighted average of the masses of its isotopes. The term 'weighted average' means that each isotopic mass is multiplied by its fractional abundance before summing them to find the average. This process ensures that isotopes with higher abundances have a greater influence on the final atomic mass.

For Gallium, we calculated this as follows:
  • Multiply the isotopic mass of 69Ga (68.9256 amu) by its abundance (0.6011):\(68.9256 \times 0.6011 = 41.4347\)
  • Multiply the isotopic mass of 71Ga (70.9247 amu) by its abundance (0.3989):\(70.9247 \times 0.3989 = 28.2841\)


Finally, add these products to get the weighted average atomic mass:
\(41.4347 + 28.2841 = 69.7188\text{ }amu\)

This result represents the atomic mass of Gallium considering the natural abundances of its isotopes.

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Most popular questions from this chapter

Before the use of systematic names, many compounds had common names. Give the systematic name for each of the following: (a) Blue vitriol, \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) (b) Slaked lime, \(\mathrm{Ca}(\mathrm{OH})_{2}\) (c) Oil of vitriol, \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (d) Washing soda, \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) (e) Muriatic acid, HCl (f) Epsom salt, \(\mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\)

A compound of copper and sulfur contains \(88.39 \mathrm{~g}\) of metal and 44.61 g of nonmetal. How many grams of copper are in \(5264 \mathrm{~kg}\) of compound? How many grams of sulfur?

What monatomic ions would you expect potassium \((Z=19)\) and bromine \((Z=35)\) to form?

The elements in Groups \(1 \mathrm{~A}(1)\) and \(7 \mathrm{~A}(17)\) are all quite reactive. What is a major difference between them?

Describe Thomson's model of the atom. How might it account for the production of cathode rays?
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