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Chapter 2: Problem 68

William Prout (1815) proposed that all other atoms are built up of hydrogen atoms, suggesting that all elements should have integral atomic masses based on an atomic mass of one for hydrogen. This hypothesis appeared discredited by the discovery of atomic masses, such as 24.3 u for magnesium and 35.5 u for chlorine. In terms of modern knowledge, explain why Prout's hypothesis is actually quite reasonable.

Short Answer

Expert verified
Prout's hypothesis is quite reasonable when considering the isotopic composition of elements. Each isotope of an element has an integral atomic mass, which aligns with the concept that they're built up of hydrogen atoms. The non-integral atomic masses presented in the periodic table are due to these masses being a weighted average of the masses of the various isotopes of that element.

Step by step solution

01

Understand Isotopes

Firstly, it needs to be understood that most elements in nature exist as a mixture of isotopes. Isotopes are atoms of the same element that have the same number of protons but different numbers of neutrons, meaning they have different atomic masses.
02

Understand Calculation of Atomic Mass

The atomic mass of an element is not determined by the mass of one atom, but is an average value from all its isotopes taking in consideration their abundance in nature. Hence, the atomic mass is a weighted average, which may not be an integer.
03

Discuss Examples

Taking the examples: Chlorine exists primarily as two isotopes, one having atomic mass 35 and the other 37. Considering their natural abundance, the average atomic mass comes as 35.5 u. Similarly, Magnesium's most common isotopes have masses of 24, 25, and 26. The weighted average atomic mass of these isotopes is 24.3 u.
04

Final Explanation

Therefore, looking at isotopes and the calculation of atomic masses, Prout's hypothesis holds true. Every element can be concluded as being composed of multiple hydrogen atoms, as the atomic masses are integral for each individual isotope. The non-integral atomic masses from the periodic table are due to the weighted average of isotopic masses.

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Most popular questions from this chapter

Atoms are spherical and so when silver atoms pack together to form silver metal, they cannot fill all the available space. In a sample of silver metal, approximately \(26.0 \%\) of the sample is empty space. Given that the density of silver metal is \(10.5 \mathrm{g} / \mathrm{cm}^{3}\), what is the radius of a silver atom? Express your answer in picometers.

The two species that have the same number of electrons as \(^{3}\) th S are (a) \(^{32} \mathrm{Cl} ;\) (b) \(^{34} \mathrm{S}^{+} ;\) (c) \(^{33} \mathrm{P}^{+} ;\) (d) \(^{28} \mathrm{Si}^{2-}\) (e) \(^{35} S^{2-} ;(f)^{40} A r^{2+} ;(g)^{40} C a^{2+}\)

An isotope with mass number 44 has four more neutrons than protons. This is an isotope of what element?

Without doing detailed calculations, determine which of the following samples has the greatest number of atoms: (a) a cube of iron with a length of \(10.0 \mathrm{cm}\) \(\left(d=7.86 \mathrm{g} / \mathrm{cm}^{3}\right)\) (b) \(1.00 \mathrm{kg}\) of hydrogen contained in a \(10,000 \mathrm{L}\) balloon (c) a mound of sulfur weighing \(20.0 \mathrm{kg}\) (d) a 76 lb sample of liquid mercury \((d=13.5 \mathrm{g} / \mathrm{mL})\)

Refer to the periodic table inside the front cover and identify (a) the element that is in group 11 and the sixth period (b) an element with atomic number greater than 50 that has properties similar to the element with atomic number 18 (c) the group number of an element \(\mathrm{E}\) that forms an ion \(\mathrm{E}^{2-}\) (d) an element \(M\) that you would expect to form the ion \(\mathrm{M}^{3+}\)
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