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Chapter 45: Problem 8

Potassium found in nature contains two isotopes. One isotope constitutes \(93.4\) percent of the whole and has an atomic mass of \(38.975 \mathrm{u}\); the other \(6.6\) percent has a mass of \(40.974 \mathrm{u}\). Compute the atomic mass of potassium as found in nature. The atomic mass of the material found in nature is obtained by combining the individual atomic masses in proportion to their abundances. The \(38.975\) u material is \(93.4 \%\), while the \(40.974 \mathrm{u}\) material is \(6.6 \%\). Hence, in combination: Atomic mass \(=(0.934)(38.975 \mathrm{u})+(0.066)(40.974 \mathrm{u})=39.1 \mathrm{u}\)

Short Answer

Expert verified
The atomic mass of potassium in nature is 39.1 u.

Step by step solution

01

Convert Percentage to Decimal

To calculate the weighted average, first convert the percentage of each isotope to a decimal by dividing by 100. Thus, convert 93.4% to 0.934 and 6.6% to 0.066.
02

Calculate Weighted Average

Multiply the atomic mass of each isotope by their respective abundance in decimal form: for the first isotope, it's \(0.934 \times 38.975\; \mathrm{u}\), and for the second isotope, it's \(0.066 \times 40.974\; \mathrm{u}\).
03

Perform Multiplications

Calculate each product: \(0.934 \times 38.975 = 36.41765\; \mathrm{u}\) and \(0.066 \times 40.974 = 2.704284\; \mathrm{u}\).
04

Sum the Weighted Products

Add the results of the weighted products from Step 3 to find the atomic mass of potassium: \(36.41765\; \mathrm{u} + 2.704284\; \mathrm{u} = 39.121934\; \mathrm{u}\).
05

Round the Result

Round the calculated atomic mass to the appropriate degree of precision, typically to one decimal place, giving 39.1 for the atomic mass of potassium in nature.

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Key Concepts

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

Isotopes
Isotopes are fascinating forms of the same element, sharing the same number of protons but differing in the number of neutrons. This difference in neutrons results in different mass numbers for each isotope.
Potassium, a common element found in nature, exists as two isotopes with distinct atomic masses and properties. Despite being the same element chemically, these isotopes of potassium contribute differently when calculating its average atomic mass.
Understanding isotopes is key to grasping the concept of average atomic mass, as this calculation depends on the properties and presence of each isotope in a natural sample.
Weighted Average
The idea of a weighted average is central to finding the atomic mass of elements with multiple isotopes. Unlike a simple average, a weighted average takes into account the relative abundances of each isotope.
* In this context, each isotope's contribution to the average is amplified by its proportion in the mixture. * This ensures that isotopes that are more abundant have a greater influence on the final atomic mass.
Weighted averages give a more realistic representation of the atomic mass as it appears in nature, rather than treating each isotope equally.
Percent Abundance
Percent abundance tells us how much of each isotope is found in nature compared to the total amount of that element. It is usually expressed as a percentage of the whole.
* For potassium, the isotopic composition specifies that 93.4% is one isotope and 6.6% is another. * These percentages are crucial because they serve as the weights in the weighted average calculation.
To use these correctly, we always need to convert the percentage to a decimal (by dividing by 100) before performing any calculations, ensuring accurate results.
Atomic Mass Calculation
Calculating the atomic mass means finding a value that accurately represents the combination of all isotopes in a naturally occurring sample. The formula for this calculation is:
\[ \text{Atomic Mass} = (\text{abundance isotope 1}) \times (\text{mass isotope 1}) + (\text{abundance isotope 2}) \times (\text{mass isotope 2}) + \ldots \]
In the case of potassium, we consider:
  • The first isotope: \(0.934 \times 38.975\; \mathrm{u}\)
  • The second isotope: \(0.066 \times 40.974\; \mathrm{u}\)
By summing these products, we arrive at the total weighted atomic mass, which is then typically rounded to one decimal for practical uses, here giving us 39.1 u as the atomic mass of potassium.

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Most popular questions from this chapter

Neon-23 beta decays in the following way: $$ { }_{10}^{23} \mathrm{Ne} \rightarrow{ }_{11}^{23} \mathrm{Na}+{ }_{-1}^{0} e+{ }_{0}^{0} \bar{v} $$ where \(\bar{v}\) is an antineutrino, a particle with no charge and almost no mass. Depending on circumstances, the energy carried away by the antineutrino can range from zero to the maximum energy available from the reaction. Find the minimum and maximum KE that the beta particle \(_{-1}^{0} e\) can have. Pertinent atomic masses are \(22.9945\) u for \({ }^{23} \mathrm{Ne}\), and \(22.9898\) u for \({ }^{23} \mathrm{Na}\). The mass of the beta particle is \(0.00055 \mathrm{u}\). Note that the given reaction is a nuclear reaction, while the masses provided are those of neutral atoms. To calculate the mass lost in the reaction, subtract the mass of the atomic electrons from the atomic masses given. We have the following nuclear masses: Which gives a mass loss of \(22.9945-22.9898=0.0047 \mathrm{u}\). Since \(1.00\) u corresponds to \(931 \mathrm{MeV}\), this mass loss corresponds to an energy of \(4.4 \mathrm{MeV}\). The beta particle and antineutrino share this energy. Hence, the energy of the beta particle can range from zero to \(4.4 \mathrm{MeV}\).

Technetium-99 (忙?r) has an excited state that decays by emission of a gamma ray. The half-life of the excited state is 360 min. What is the activity, in curies, of \(1.00 \mathrm{mg}\) of this excited isotope? Because we have the half-life \(\left(t_{1 / 2}\right)\) we can determine the decay constant since \(\lambda t_{1 / 2}=0.693 .\) The activity of a sample is \(\lambda N .\) In this case, $$ \lambda=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{21600 \mathrm{~s}}=3.21 \times 10^{-5} \mathrm{~s}^{-1} $$ We also know that \(99.0 \mathrm{~kg}\) of Tc contains \(6.02 \times 10^{26}\) atoms. A mass \(m\) will therefore contain \([m /(99.0 \mathrm{~kg})]\left(6.02 \times 10^{26}\right)\) atoms. In our case, \(m=1.00 \times 10^{-6} \mathrm{~kg}\), and so $$ \begin{aligned} \text { Activity } &=\lambda N=\left(3.21 \times 10^{-5} \mathrm{~s}^{-1}\right)\left(\frac{1.00 \times 10^{-6} \mathrm{~kg}}{99.0 \mathrm{~kg}}\right)\left(6.02 \times 10^{26}\right) \\ &=1.95 \times 10^{14} \mathrm{~s}^{-1}=1.95 \times 10^{14} \mathrm{~Bq} \end{aligned} $$

How many protons, neutrons, and electrons are there in \((a){ }^{3} \mathrm{He}\), (b) \(^{12} \mathrm{C}\), and \((c)^{206} \mathrm{~Pb}\) ? (a) The atomic number of He is 2; therefore, the nucleus must contain 2 protons. Since the mass number of this isotope is 3 , the sum of the protons and neutrons in the nucleus must equal 3; therefore, there is 1 neutron. The number of electrons in the atom is the same as the atomic number, 2 . (b) The atomic number of carbon is 6 ; hence, the nucleus must contain 6 protons. The number of neutrons in the nucleus is equal to \(12-6=6\). The number of electrons is the same as the atomic number, 6 . (c) The atomic number of lead is 82 ; hence, there are 82 protons in the nucleus and 82 electrons in the atom. The number of neutrons is \(206-82=124\).

Plutonium-239 decays by alpha emission. Write out the equation from the process.

Determine the parent nuclide (P) that decayed into thorium-234 as follows: $$ \mathrm{P} \rightarrow^{234} \mathrm{Th}+\alpha $$
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