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Chapter 2: Problem 101

Correct each of the following formulas: (a) Tetraphosphorus decoxide is \(\mathrm{P}_{4} \mathrm{O}_{6}\). (b) Diboron trioxide is \(\mathrm{B}_{3} \mathrm{O}_{2}\). (c) Phosphorus trifluoride is \(\mathrm{F}_{3} \mathrm{P}\).

Short Answer

Expert verified
(a) \( \mathrm{P}_{4} \mathrm{O}_{10} \), (b) \( \mathrm{B}_{2} \mathrm{O}_{3} \), (c) \( \mathrm{PF}_{3} \).

Step by step solution

01

Understand each compound name

First, understand the compound names and their corresponding prefixes and elements. For example, 'Tetraphosphorus decoxide' indicates 4 phosphorus (P) atoms and 10 oxygen (O) atoms.
02

Correct the formula for Tetraphosphorus decoxide (a)

Tetraphosphorus (tetra- indicates 4) and decoxide (deca- indicates 10) is written as \( \mathrm{P}_{4}O_{10} \). The correct formula should be \( \mathrm{P}_{4} \mathrm{O}_{10} \).
03

Correct the formula for Diboron trioxide (b)

Diboron (di- indicates 2) and trioxide (tri- indicates 3) is written as \( \mathrm{B}_{2}O_{3} \). The correct formula should be \( \mathrm{B}_{2} \mathrm{O}_{3} \).
04

Correct the formula for Phosphorus trifluoride (c)

Phosphorus trifluoride (tri- indicates 3 fluorine atoms) should be written as P for phosphorus and 3 F atoms, hence \( PF_{3} \). The correct formula should be \( \mathrm{PF}_{3} \).

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Key Concepts

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

Molecular Compounds
Molecular compounds are formed by atoms of two or more different elements bonded together. They are also known as covalent compounds, due to the type of bonds they form. These compounds typically consist of nonmetal elements. In molecular compounds, atoms share electrons to achieve a stable balance of attraction and repulsion.
Examples of molecular compounds include water (H鈧侽), carbon dioxide (CO鈧�), and methane (CH鈧�).
To understand and write the formulas of these compounds correctly, it's important to know how to read their names and understand the prefixes that indicate the number of atoms involved.
Chemical Nomenclature
Chemical nomenclature is the system of naming chemical compounds. For molecular compounds, the names provide information about the different elements and the number of atoms of each element in a molecule.
Prefixes are used to show the number of atoms:
  • Mono- (1)
  • Di- (2)
  • Tri- (3)
  • Tetra- (4)
  • Penta- (5)
  • Hexa- (6)
  • Sept- (7)
  • Octa- (8)
  • Nona- (9)
  • Deca- (10)

For example, 'Tetraphosphorus decoxide' means the compound has four phosphorus (P) atoms and ten oxygen (O) atoms.
By understanding these prefixes, you can correctly interpret and write the chemical formulas of molecular compounds.
Chemical Formulas
Chemical formulas show the elements in a compound and the ratio of these elements. They are derived from the compound's name using the principles of chemical nomenclature.
Let's correct the formulas given in the exercise:
(a) Tetraphosphorus decoxide should be written as \[\text{P}_4\text{O}_{10}\] because 'tetra-' means 4 and 'deca-' means 10.
(b) Diboron trioxide is \[\text{B}_2\text{O}_3\] because 'di-' means 2 and 'tri-' means 3.
(c) Phosphorus trifluoride is \[\text{PF}_3\] since 'tri-' indicates 3 fluorine atoms.
Knowing how to decode these names and write the correct formulas is essential in chemistry to avoid mistakes.

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Most popular questions from this chapter

Many chemical names are similar at first glance. Give the formulas of the species in each set: (a) Ammonium ion and ammonia (b) Magnesium sulfide, magnesium sulfite, and magnesium sulfate (c) Hydrochloric acid, chloric acid, and chlorous acid (d) Cuprous bromide and cupric bromide

You are working in the laboratory, preparing sodium chloride. Consider the following results for three preparations of the compound: Case \(1: 39.34 \mathrm{~g} \mathrm{Na}+60.66 \mathrm{~g} \mathrm{Cl}_{2} \longrightarrow 100.00 \mathrm{~g} \mathrm{NaCl}\) Case \(2: 39.34 \mathrm{~g} \mathrm{Na}+70.00 \mathrm{~g} \mathrm{Cl}_{2} \longrightarrow\) $$ 100.00 \mathrm{~g} \mathrm{NaCl}+9.34 \mathrm{~g} \mathrm{Cl}_{2} $$ Case \(3: 50.00 \mathrm{~g} \mathrm{Na}+50.00 \mathrm{~g} \mathrm{Cl}_{2} \longrightarrow\) $$ 82.43 \mathrm{~g} \mathrm{NaCl}+17.57 \mathrm{~g} \mathrm{Na} $$ Explain these results in terms of the laws of conservation of mass and definite composition.

State the mass law(s) demonstrated by the following experimental results, and explain your reasoning: Experiment 1: A student heats \(1.00 \mathrm{~g}\) of a blue compound and obtains \(0.64 \mathrm{~g}\) of a white compound and \(0.36 \mathrm{~g}\) of a colorless gas. Experiment 2: A second student heats \(3.25 \mathrm{~g}\) of the same blue compound and obtains \(2.08 \mathrm{~g}\) of a white compound and \(1.17 \mathrm{~g}\) of a colorless gas.

Give the name and formula of the compound formed from each pair of elements: (a) cesium and bromine; (b) sulfur and barium; (c) calcium and fluorine.

Show, with calculations, how the following data illustrate the law of multiple proportions: Compound 1: 77.6 mass \(\%\) xenon and 22.4 mass \(\%\) fluorine Compound 2: 63.3 mass \(\%\) xenon and 36.7 mass \(\%\) fluorine
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