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Chapter 2: Problem 49

Ammonia is a gas with a characteristic pungent odor. It is sold as a water solution for use in household cleaning. The gas is a compound of nitrogen and hydrogen in the atomic ratio 1: 3 . A sample of ammonia contains \(7.933 \mathrm{~g}\) \(\mathrm{N}\) and \(1.712 \mathrm{~g} \mathrm{H}\). What is the atomic weight of \(\mathrm{N}\) relative to \(\mathrm{H}\) ?

Short Answer

Expert verified
The atomic weight of nitrogen is 14 times that of hydrogen.

Step by step solution

01

Write down the given values

We have a sample of ammonia with nitrogen (N) mass of \(7.933\, \text{g}\) and hydrogen (H) mass of \(1.712\, \text{g}\). The atomic ratio of \( \text{N:H} \) is \( 1:3 \).
02

Assume atomic masses for calculation

Assume the atomic weight of hydrogen (H) as \( x \) and the atomic weight of nitrogen (N) as \( y \). The ratio \( y:x \) needs to be determined.
03

Set up the ratio equation

Given \( \text{N:H} = 1:3 \), this translates to \( y:x = 1:3 \). Thus, \( y = 3x \).
04

Calculate relative atomic weights using sample masses

The given mass of nitrogen in the sample is \(7.933 \text{g}\) and hydrogen is \(1.712 \text{g}\). Since \( y = 3x \), the relative atomic weight can be calculated by dividing the masses in similar proportions as per the ratio. So, \( \frac{7.933}{1.712} = \frac{y}{3x} \).
05

Solve the equation

Solving \( \frac{7.933}{1.712} = \frac{y}{3x} \) gives \( y = 3 \times \frac{7.933}{1.712} \times x \). After finding the value of \( y/x \), simplify the expression to get \( y:x = 14:3 \).
06

Conclusion

Therefore, the atomic weight of nitrogen is approximately 14 times that of hydrogen, relative to hydrogen's atomic weight.

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Key Concepts

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

Atomic Mass
Atomic mass is the mass of a single atom of a chemical element, fundamentally defined by the protons and neutrons present in the atom's nucleus. It is typically expressed in atomic mass units (amu), where the atomic mass of hydrogen (H) is approximately 1 amu as it contains only one proton. Nitrogen (N) has an atomic mass of around 14 amu, reflecting its greater number of nucleons.
If we consider ammonia (NH extsubscript{3}), which consists of nitrogen and hydrogen atoms, understanding the atomic masses is crucial for many calculations, as it directly impacts how we determine the weight ratio of elements in compounds.
  • Atomic mass is critical for calculating molecular weights and determining stoichiometric ratios in chemical reactions.
  • The known atomic mass of hydrogen is a foundational reference for comparing other elements.
  • Through calculations, we identified nitrogen's atomic mass as roughly 14 times that of hydrogen, which aligns with periodic table data.
Recognizing atomic mass aids in predicting the behavior, interaction, and weight ratios of elements in chemical reactions.
Chemical Composition
The chemical composition of a substance refers to the identity and proportion of elements that constitute that substance. Ammonia, chemically represented as NH extsubscript{3}, is composed of nitrogen (N) and hydrogen (H) in a fixed ratio. Understanding the chemical composition is crucial to grasp how the molecular structure influences the properties of a substance.

In ammonia, the ratio of nitrogen to hydrogen is crucial for understanding its chemical makeup and properties. As it consists of one nitrogen atom bonded to three hydrogen atoms, this specific composition is responsible for its ability to act effectively as a cleaning agent or fertilizer.
  • Each ammonia molecule contains one nitrogen atom and three hydrogen atoms.
  • This 1:3 atomic ratio is key to ammonia鈥檚 structural integrity and physical properties like solubility and volatility.
  • The chemical composition dictates how ammonia interacts with other compounds and substances.
Grasping the concept of chemical composition enables us to predict and manipulate chemical dynamics and reactions, essential in both academic and industrial chemistry.
Nitrogen-Hydrogen Ratio
The nitrogen-hydrogen ratio is a fundamental concept for understanding the stoichiometry of ammonia. In ammonia, the nitrogen to hydrogen ratio is always 1:3. This stoichiometric ratio implies that every one unit of nitrogen atom requires three units of hydrogen atoms for the compound to form. This ratio determines the mole-based relationships seen in chemical equations and reactions.

In terms of atomic ratio, the nitrogen-hydrogen ratio reflects their proportional contribution to the compound's overall mass and relevant calculations in stoichiometry.
  • This ratio is a constant defining feature of ammonia.
  • Knowing the fixed ratio, one can calculate the relative masses or moles needed in reactions involving ammonia.
  • The ratio is intrinsically linked to the properties of ammonia, affecting how it combines and functions chemically.
By comprehending the nitrogen-hydrogen ratio, we gain insights into ammonia鈥檚 formation and its predictable behavior in various chemical processes.

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Most popular questions from this chapter

The mass-to-charge ratio for the positive ion \(\mathrm{F}^{+}\) is \(1.97 \times 10^{-7} \mathrm{~kg} / \mathrm{C}\). Using the value of \(1.602 \times 10^{-19} \mathrm{C}\) for the charge on the ion, calculate the mass of the fluorine atom. (The mass of the electron is negligible compared with that of the ion, so the ion mass is essentially the atomic weight.)

A power plant is driven by the combustion of a complex fossil fuel having the formula \(\mathrm{C}_{11} \mathrm{H}_{7} \mathrm{~S}\). Assume the air supply has a \(\mathrm{N}_{2} / \mathrm{O}_{2}\) molecular ratio of 3.76: 1.00 and the \(\mathrm{N}_{2}\) remains unreacted. In addition to the water produced, assume the fuel's \(\mathrm{C}\) is completely converted to \(\mathrm{CO}_{2}\) and its sulfur content is converted to \(\mathrm{SO}_{2}\) a. Including \(\mathrm{N}_{2}\) supplied in the air, write a balanced combustion equation for the complex fuel, assuming \(100 \%\) stoichiometric combustion (i.e., when there is no excess oxygen in the products). Except in the case of \(\mathrm{N}_{2}\), use only integer coefficients. b. Including \(\mathrm{N}_{2}\) supplied in the air, write a balanced combustion equation for the complex fuel, assuming \(133 \%\) stoichiometric combustion (i.e., when there is excess oxygen present in the products). Except in the case of \(\mathrm{N}_{2}\), use only integer coefficients.

Balance the following equations. a. \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}+\mathrm{H}_{3} \mathrm{PO}_{4} \longrightarrow \mathrm{Ca}\left(\mathrm{H}_{2} \mathrm{PO}_{4}\right)_{2}\) b. \(\mathrm{MnO}_{2}+\mathrm{HCl} \longrightarrow \mathrm{MnCl}_{2}+\mathrm{Cl}_{2}+\mathrm{H}_{2} \mathrm{O}\) c. \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}+\mathrm{I}_{2} \longrightarrow \mathrm{NaI}+\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}\) d. \(\mathrm{Al}_{4} \mathrm{C}_{3}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{Al}(\mathrm{OH})_{3}+\mathrm{CH}_{4}\) e. \(\mathrm{NO}_{2}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{HNO}_{3}+\mathrm{NO}\)

Natural carbon, which has an atomic weight of 12.011 amu, consists of carbon- 12 and carbon- 13 isotopes. Given that the mass of carbon-13 is 13.00335 amu, what would be the average atomic weight (in amu) of a carbon sample prepared by mixing equal numbers of carbon atoms from a sample of natural carbon and a sample of pure carbon-13?

There are \(1.699 \times 10^{22}\) atoms in \(1.000 \mathrm{~g}\) of chlorine. Assume that chlorine atoms are spheres of radius \(0.99 \AA\) and that they are lined up side by side in a 0.5 -g sample. How many miles in length is the line of chlorine atoms in the sample?
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