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Chapter 3: Problem 34

Hydrogen has two stable isotopes, \({ }_{1}^{1} \mathrm{H}\) and \({ }_{1}^{2} \mathrm{H},\) and sulfur has four stable isotopes, \({ }_{16}^{32} \mathrm{~S},{ }_{16}^{33} \mathrm{~S},{ }_{16}^{34} \mathrm{~S},\) and \({ }_{16}^{36} \mathrm{~S}\). How many peaks would you observe in the mass spectrum of the positive ion of hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S}^{+}\) ? Assume no decomposition of the ion into smaller fragments.

Short Answer

Expert verified
You would observe 6 peaks in the mass spectrum of the positive ion of hydrogen sulfur, \(\mathrm{H}_{2} \mathrm{~S}^{+}\).

Step by step solution

01

Identify Isotopes

First, identify the stable isotopes of both hydrogen (H) and sulfur (S). Hydrogen has two stable isotopes: \({ }_{1}^{1} \mathrm{H}\) and \({ }_{1}^{2}\mathrm{H}\). Sulfur has four stable isotopes: \({ }_{16}^{32} \mathrm{S}\), \({ }_{16}^{33} \mathrm{S}\), \({ }_{16}^{34} \mathrm{S}\), and \({ }_{16}^{36}\mathrm{S}\).
02

Calculate Combinations

Hydrogen sulfide has the formula \(\mathrm{H}_{2} \mathrm{~S}^{+}\). This means it consists of 2 hydrogen atoms and one sulfur atom. For each hydrogen atom we have 2 possible isotopes, and for the sulfur atom we have 4 different isotopes. The total number of combinations of these isotopes that can form hydrogen sulfide, then, is the product of the number of isotopes: \(2 \times 2 \times 4 = 16\). It means there are 16 different isotopic combinations that can form \(\mathrm{H}_{2} \mathrm{~S}^{+}\).
03

Determine Unique Masses

However, not all of these combinations will result in a unique mass. Therefore, count the number of unique masses. Here, considering the masses of isotopes: 1. Hydrogen sulfide can have the hydrogen atoms either of mass 1 or 2. This contributes to 2 mass varies: \(2 \times 1 = 2\) and \(2 \times 2 = 4\). 2. For sulfur atom, there are 4 isotopes of mass 32, 33, 34, 36 each. It means when counting unique masses, one might end up with hydrogen sulfide of masses \(32+2=34, 32+4=36, 33+2=35, 33+4=37, 34+2=36, 34+4=38, 36+2=38, 36+4=40\). After removing the duplicate mass, one gets: 34, 35, 36, 37, 38, and 40. This are 6 unique masses.

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Key Concepts

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

Understanding Isotopes
Isotopes are different forms of the same element. They have the same number of protons but different numbers of neutrons.
This results in variations in their atomic masses but not in their chemical properties. For hydrogen, we have two stable isotopes:
  • \( {}_{1}^{1} \mathrm{H} \) - with one proton and no neutrons.
  • \( {}_{1}^{2} \mathrm{H} \) (deuterium) - with one proton and one neutron.
Sulfur, on the other hand, has four common isotopes:
  • \( {}_{16}^{32} \mathrm{S} \)
  • \( {}_{16}^{33} \mathrm{S} \)
  • \( {}_{16}^{34} \mathrm{S} \)
  • \( {}_{16}^{36} \mathrm{S} \)
These isotopes alter the mass of compounds they form without changing their chemical behavior. The presence of isotopes is crucial in mass spectrometry as they lead to various peaks corresponding to different mass-to-charge ratios.
Hydrogen Sulfide and Its Formula
Hydrogen sulfide is a compound represented by the chemical formula \( \mathrm{H}_{2} \mathrm{S} \). It consists of two hydrogen atoms bonded to a sulfur atom.
This simple compound is known for its characteristic rotten egg smell and is often found in natural gas and crude petroleum. When studying hydrogen sulfide in the context of isotopes, we look at its possible combinations with hydrogen鈥檚 two isotopes and sulfur鈥檚 four isotopes.
These combinations inform us about the "molecular weight variations鈥� possible in hydrogen sulfide. It is also important for predicting the peaks in a mass spectrum, where each combination represents a potential unique mass.
Decoding Mass Spectrum Peaks
Mass spectrum peaks help us understand the composition of a sample. Each peak represents a unique mass-to-charge ratio, often denoted by the isotopic combinations in molecules.
In the mass spectrum of hydrogen sulfide \( \mathrm{H}_{2} \mathrm{S}^{+} \), we calculate how many unique peaks appear based on its isotopic combinations:
  • Each hydrogen atom can be \( {}_{1}^{1} \mathrm{H} \) or \( {}_{1}^{2} \mathrm{H} \). That gives us 2 choices per hydrogen atom.
  • Sulfur offers 4 isotopic choices.
Combining these options, we initially find 16 possible combinations \( (2 \times 2 \times 4) \).
However, several combinations can share the same mass. Calculating these gives 6 unique mass values in this example:
  • Masses of 34, 35, 36, 37, 38, and 40.
Each mass corresponds to a distinct peak in the mass spectrum, providing insight into the isotopic composition of the molecule.

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Most popular questions from this chapter

On what law is stoichiometry based? Why is it essential to use balanced equations in solving stoichiometric problems?

Potash is any potassium mineral that is used for its potassium content. Most of the potash produced in the United States goes into fertilizer. The major sources of potash are potassium chloride \((\mathrm{KCl})\) and potassium sulfate \(\left(\mathrm{K}_{2} \mathrm{SO}_{4}\right)\). Potash production is often reported as the potassium oxide \(\left(\mathrm{K}_{2} \mathrm{O}\right)\) equivalent or the amount of \(\mathrm{K}_{2} \mathrm{O}\) that could be made from a given mineral. (a) If \(\mathrm{KCl}\) costs \(\$ 0.055\) per \(\mathrm{kg}\), for what price (dollar per \(\mathrm{kg}\) ) must \(\mathrm{K}_{2} \mathrm{SO}_{4}\) be sold in order to supply the same amount of potassium on a per dollar basis? (b) What mass (in \(\mathrm{kg}\) ) of \(\mathrm{K}_{2} \mathrm{O}\) contains the same number of moles of \(\mathrm{K}\) atoms as \(1.00 \mathrm{~kg}\) of \(\mathrm{KCl} ?\)

When heated, lithium reacts with nitrogen to form lithium nitride: $$ 6 \mathrm{Li}(s)+\mathrm{N}_{2}(g) \longrightarrow 2 \mathrm{Li}_{3} \mathrm{~N}(s) $$ What is the theoretical yield of \(\mathrm{Li}_{3} \mathrm{~N}\) in grams when \(12.3 \mathrm{~g}\) of \(\mathrm{Li}\) are heated with \(33.6 \mathrm{~g}\) of \(\mathrm{N}_{2} ?\) If the actual yield of \(\mathrm{Li}_{3} \mathrm{~N}\) is \(5.89 \mathrm{~g}\), what is the percent yield of the reaction?

Industrially, nitric acid is produced by the Ostwald process represented by the following equations: \(\begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 2 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow & \mathrm{HNO}_{3}(a q)+\mathrm{HNO}_{2}(a q) \end{aligned}\) What mass of \(\mathrm{NH}_{3}\) (in g) must be used to produce 1.00 ton of \(\mathrm{HNO}_{3}\) by the above procedure, assuming an 80 percent yield in each step? \((1\) ton \(=2000 \mathrm{lb}\); \(1 \mathrm{lb}=453.6 \mathrm{~g} .)\)

Tin(II) fluoride \(\left(\mathrm{SnF}_{2}\right)\) is often added to toothpaste as an ingredient to prevent tooth decay. What is the mass of \(\mathrm{F}\) in grams in \(24.6 \mathrm{~g}\) of the compound?
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