Notes
Naming rules
| Compound type | How to name | Example |
|---|---|---|
| binary ionic | metal + nonmetal-ide | NaCl, sodium chloride |
| transition metal ionic | metal(Roman numeral) + anion | $\text{CuCl}_2$, copper(II) chloride |
| polyatomic ion | use the ion's name | $\text{Na}_2\text{SO}_4$, sodium sulfate |
| molecular | prefixes + -ide | $\text{N}_2\text{O}_4$, dinitrogen tetroxide |
| binary acid | hydro- + root + -ic acid | HCl, hydrochloric acid |
| oxyacid | -ate becomes -ic, -ite becomes -ous | $\text{HNO}_2$, nitrous acid |
- Ionic formulas: charges must total zero. Use parentheses for more than one polyatomic ion: $\text{Mg(OH)}_2$.
- Key ions: $\text{NH}_4^+$, $\text{OH}^-$, $\text{NO}_3^-$, $\text{SO}_4^{2-}$, $\text{CO}_3^{2-}$, $\text{PO}_4^{3-}$.
Percent composition
- $\displaystyle \% \text{ element} = \frac{\text{mass of element in formula}}{\text{molar mass}} \times 100\%$.
Empirical and molecular formulas
- Percent to grams (assume 100 g), grams to moles, divide by the smallest, make whole numbers.
- Molecular formula $= n \times$ empirical formula, where $\displaystyle n = \frac{\text{molar mass}}{\text{empirical mass}}$.
Topics in this unit
Each topic has its own short notes and practice.
Practice
Sulfate is 2-, so copper must be 2+.
$\text{Mg}^{2+}$ needs two $\text{OH}^-$ ions, and parentheses keep the hydroxide together.
Two nonmetals, so use prefixes: penta = 5. (P is phosphorus, not potassium.)
$\displaystyle \frac{16.00}{18.02} \times 100\% = 88.8\%$.
Both reduce to $\text{CH}_2$.
Mg: $\displaystyle \frac{2.43}{24.31} = 0.100$ mol. O: $\displaystyle \frac{1.60}{16.00} = 0.100$ mol. Ratio 1 : 1.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A lab analyzes a gas used as a fuel. It is 85.6% carbon and 14.4% hydrogen by mass, and its molar mass is 42.08 g/mol. (C = 12.01, H = 1.01)
(a) Assume a 100.0 g sample. Find the moles of carbon and of hydrogen.
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C: $\displaystyle \frac{85.6 \text{ g}}{12.01 \text{ g/mol}} = 7.13$ mol. H: $\displaystyle \frac{14.4 \text{ g}}{1.01 \text{ g/mol}} = 14.3$ mol.
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(b) Find the empirical formula.
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Divide by the smallest: C $\displaystyle \frac{7.13}{7.13} = 1$, H $\displaystyle \frac{14.3}{7.13} = 2.00$. Empirical formula $\text{CH}_2$.
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(c) Find the molecular formula.
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Empirical mass $= 12.01 + 2(1.01) = 14.03$ g/mol. $\displaystyle n = \frac{42.08}{14.03} = 3.00$. Molecular formula $\text{C}_3\text{H}_6$.
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