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Chapter 2: Problem 95

The element chromium (Cr) consists of four naturally occurring isotopes with atomic masses 49.9460,51.9405,52.9407 , and \(53.9389 \mathrm{u}\). The relative abundances of these four isotopes are \(4.3,83.8,9.5,\) and \(2.4 \%\), respectively. From these data, calculate the atomic weight of chromium.

Short Answer

Expert verified
The atomic weight of chromium is approximately \(51.999 \mathrm{u}\).

Step by step solution

01

Convert relative abundances to decimals

To use the relative abundances in the atomic weight calculation, we need to convert them from percentages to decimals. We do this by dividing each percentage by 100: - 4.3% = 0.043 - 83.8% = 0.838 - 9.5% = 0.095 - 2.4% = 0.024
02

Multiply each isotope's atomic mass by its relative abundance

Now, we can multiply the atomic mass of each isotope by its relative abundance (in decimal form) to find the weighted contribution of each isotope to the atomic weight of chromium: - 49.9460 u * 0.043 = 2.147978 u - 51.9405 u * 0.838 = 43.526990 u - 52.9407 u * 0.095 = 5.0292665 u - 53.9389 u * 0.024 = 1.2943336 u
03

Add the weighted contributions together

Finally, we add the weighted contributions of each isotope to find the atomic weight of chromium: Atomic weight = 2.147978 u + 43.526990 u + 5.0292665 u + 1.2943336 u = 51.9985681 u
04

Round the result

Although the atomic weight we calculated is very precise, it is not necessary to report it with that many decimal places. Let's round it to 3 decimal places: Atomic weight = 51.999 u The atomic weight of chromium is approximately \(51.999 \mathrm{u}\).

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Most popular questions from this chapter

The natural abundance of \({ }^{3}\) He is \(0.000137 \% .\) (a) How many protons, neutrons, and electrons are in an atom of \({ }^{3} \mathrm{He}\) ? (b) Based on the sum of the masses of their subatomic particles, which is expected to be more massive, an atom of \({ }^{3}\) He or an atom of \({ }^{3} \mathrm{H}\) (which is also called tritium) \(?(\mathbf{c})\) Based on your answer to part (b), what would need to be the precision of a mass spectrometer that is able to differentiate between peaks that are due to \({ }^{3} \mathrm{He}^{+}\) and \({ }^{3} \mathrm{H}^{+}\) ?

Mass spectrometry is more often applied to molecules than to atoms. We will see in Chapter 3 that the molecular weight of a molecule is the sum of the atomic weights of the atoms in the molecule. The mass spectrum of \(\mathrm{H}_{2}\) is taken under conditions that prevent decomposition into \(\mathrm{H}\) atoms. The two naturally occurring isotopes of hydrogen are ${ }^{1} \mathrm{H}\( (atomic mass \)=1.00783 \mathrm{u}\(; abundance \)\left.99.9885 \%\right)\( and \){ }^{2} \mathrm{H}\( (atomic mass \)=2.01410 \mathrm{u}$; abundance \(\left.0.0115 \%\right)\). (a) How many peaks will the mass spectrum have? (b) Give the relative atomic masses of each of these peaks. (c) Which peak will be the largest, and which the smallest?

(a) What isotope is used as the standard in establishing the atomic mass scale? (b) The atomic weight of boron is reported as \(10.81,\) yet no atom of boron has the mass of \(10.81 \mathrm{u}\). Explain.

In a series of experiments, a chemist prepared three different compounds that contain only iodine and fluorine and determined the mass of each element in each compound: $$ \begin{array}{lcc} \hline \text { Compound } & \text { Mass of Iodine (g) } & \text { Mass of Fluorine (g) } \\ \hline 1 & 4.75 & 3.56 \\ 2 & 7.64 & 3.43 \\ 3 & 9.41 & 9.86 \\ \hline \end{array} $$ (a) Calculate the mass of fluorine per gram of iodine in each compound. (b) How do the numbers in part (a) support the atomic theory?

How many protons, neutrons, and electrons are in the following atoms? (a) ${ }^{84} \mathrm{Kr}\(, (b) \){ }^{200} \mathrm{Hg}\(, (c) \)^{59} \mathrm{Co}$, (d) \({ }^{55} \mathrm{Mn}\), (e) $^{239} \mathrm{U},(\mathbf{f})^{181} \mathrm{Ta}$.
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