2.1 Atoms, Ions, Compounds, Inorganic Nomenclature, and Stoichiometry

2.1 Atoms, Ions, Compounds, Inorganic Nomenclature, and Stoichiometry

Quick Answer: The Inorganic Chemistry TOS opens with two short areas: atoms, ions, and molecules with their formulas (6 items), and mass relationships in chemical reactions (6 items). Expect isotope and atomic-mass calculations, naming of ionic and molecular compounds, the mole and molar mass, percent composition, empirical and molecular formulas, and limiting-reagent and percent-yield problems. These skills also feed every quantitative subject later in the exam.

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Atoms, Isotopes, and Atomic Mass

An atom of element X is described by its atomic number $Z$ (protons) and mass number $A$ (protons plus neutrons), written $^{A}_{Z}X$. Isotopes have the same $Z$ but different numbers of neutrons. The atomic mass on the periodic table is the abundance-weighted average of the isotopic masses:

$\bar{M} = \sum_i f_i\,m_i$

Worked example: Chlorine is 75.78% $^{35}Cl$ (34.969 u) and 24.22% $^{37}Cl$ (36.966 u):

$\bar{M} = (0.7578)(34.969) + (0.2422)(36.966) = 26.50 + 8.953 = 35.45\text{ u}$

Ions form when atoms gain or lose electrons: cations are positive ($Na^+$, $Fe^{3+}$) and anions negative ($Cl^-$, $O^{2-}$). Molecules are discrete groups of covalently bonded atoms ($H_2O$, $P_4$). Ionic solids such as $NaCl$ have no discrete molecules, so their formula gives the simplest ion ratio (a formula unit).

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Writing and Naming Inorganic Compounds

Ionic Compounds

Write the cation first, then balance charges with subscripts. Metals that form more than one cation take a Roman numeral (Stock system); older names use -ous (lower charge) and -ic (higher charge).

Common Polyatomic Ions

Oxyanion endings follow oxygen count: -ate has more oxygen than -ite; per-…-ate has the most and hypo-…-ite the fewest.

Binary Molecular Compounds, Acids, and Hydrates

• Binary molecular compounds use Greek prefixes: $N_2O_5$ is dinitrogen pentoxide, $SF_6$ sulfur hexafluoride, $P_4O_{10}$ tetraphosphorus decoxide.
• Binary acids in water take hydro-…-ic acid: $HCl(aq)$ is hydrochloric acid, $H_2S(aq)$ hydrosulfuric acid.
• Oxoacids follow the anion: -ate becomes -ic acid ($H_2SO_4$, sulfuric acid; $HNO_3$, nitric acid) and -ite becomes -ous acid ($H_2SO_3$, sulfurous acid; $HClO$, hypochlorous acid).
• Hydrates state the water count: $CuSO_4 \cdot 5H_2O$ is copper(II) sulfate pentahydrate.

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The Mole, Molar Mass, and Percent Composition

The mole contains exactly $N_A = 6.02214076 \times 10^{23}$ entities; this value has been exact since the 2019 SI redefinition. Molar mass (g/mol) numerically equals the formula mass in u.

$n = \frac{m}{M}, \qquad N = n\,N_A, \qquad \%\,X = \frac{(\text{atoms of X})\,M_X}{M_{\text{compound}}} \times 100$

Example: In $Fe_2O_3$ ($M = 159.69$ g/mol), %Fe $= \frac{2(55.845)}{159.69} \times 100 = 69.94\%$.

Empirical and Molecular Formulas

1. Convert mass percentages (per 100 g) to moles of each element.
2. Divide by the smallest number of moles; round to whole-number ratios, multiplying through if needed (ratios such as 1.5 or 1.33 signal a multiplier of 2 or 3).
3. Divide the molar mass by the empirical-formula mass to find the multiplier.

Worked example: A compound is 40.00% C, 6.71% H, and 53.29% O, with $M = 180.16$ g/mol.

The empirical formula is $CH_2O$ (30.03 g/mol). Since $180.16 / 30.03 = 6.00$, the molecular formula is $C_6H_{12}O_6$.

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Balanced Equations, Limiting Reagents, and Yield

A balanced equation conserves atoms and charge, and its coefficients give mole ratios. To find the limiting reagent, convert each reactant to moles and work out how much product each could make; the reactant that gives the least product limits the reaction.

Worked example: $N_2 + 3H_2 \rightarrow 2NH_3$ with 50.0 g $N_2$ and 10.0 g $H_2$:

• $n(N_2) = 50.0 / 28.014 = 1.785$ mol, which would need $3 \times 1.785 = 5.355$ mol $H_2$
• $n(H_2) = 10.0 / 2.016 = 4.960$ mol, which is less than 5.355 mol, so $H_2$ limits
• Theoretical $NH_3 = 4.960 \times \frac{2}{3} = 3.307$ mol $= 3.307 \times 17.031 = 56.3$ g
• If 45.0 g $NH_3$ is actually isolated, the percent yield is $\frac{45.0}{56.3} \times 100 = 79.9\%$

Common traps: using mass ratios instead of mole ratios; forgetting diatomic elements ($O_2$, $N_2$, $Cl_2$); and reporting more significant figures than the data support.