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Molarity
Preparing a solution
Dilution
Solubility rules
Complete & Net Ionic Equations
Colligative properties
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Heat Flow
Energy diagrams
Thermochemical equations
Heating/ Cooling curves
Specific Heat Capacity
Calorimetry
Hess's Law
Enthalpies of formation
Bond enthalpies
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Collision Theory
Rate Comparisons
Integrated Rate Law
Differential Rate Law
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Equilibrium
Equilibrium Expression
ICE Tables
Calculating K
K vs Q
Le Chatelier's Principle
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Definitions
Conjugate Acids & Base Pairs
Autoionization of water
pH Scale
Strong Acids/ Bases
Ka and Kb
Buffer
Titrations
Indicators
pH salts
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Entropy
Gibb's Free Energy
G and Temperature
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Oxidation numbers
Half Reactions
Balancing Redox reactions
Voltaic cells
Cell potential (standard conditions)
Cell potential (non-standard)
Electrolysis
Quantitative Electrochemistry
Empirical Formula
Related Examples and Practice Problems
Additional Worked Out Examples/ Practice
Identifying classification types: Differentiation between elements, compounds or mixtures and homogeneous and heterogenous mixtures
Separation techniques: Selected and explaining limitation of appropriate separation
Relating Properties to Composition: Predicting classification based on descriptive properties
and more …
Topic Summary & Highlights
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Core Concept
The empirical formula of a compound represents the simplest whole-number ratio of the elements within it. Unlike the molecular formula, which shows the exact number of each atom in a molecule, the empirical formula shows only the smallest ratio of elements.
Steps to Determine an Empirical Formula
Obtain the Masses of Each Element (from experimental data or percent composition):
If given percentages, assume you have a 100 g sample (so percentages convert directly to grams).
Convert Masses to Moles:
Use the atomic mass of each element to convert grams to moles.
Moles of element=Mass of element (g)Atomic mass of element (g/mol)\text{Moles of element} = \frac{\text{Mass of element (g)}}{\text{Atomic mass of element (g/mol)}}Moles of element=Atomic mass of element (g/mol)Mass of element (g)
Determine the Simplest Whole-Number Ratio:
Divide each element’s mole value by the smallest number of moles calculated for any element.
If the result is close to a whole number, round to the nearest whole number.
Adjust Ratios if Necessary:
If any ratios are not whole numbers (e.g., 1.5, 2.5), multiply all ratios by a common factor to obtain whole numbers.
Write the Empirical Formula:
Use the whole-number ratios as subscripts in the formula.
Step | Example Application (40.0% C, 6.7% H, 53.3% O) |
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1. Obtain the Masses of Each Element |
Assume a 100 g sample. Convert percentages to grams: - Carbon: 40.0 g - Hydrogen: 6.7 g - Oxygen: 53.3 g |
2. Convert Masses to Moles |
Use atomic masses to convert grams to moles: - Carbon: \( \frac{40.0 \, \text{g}}{12.01 \, \text{g/mol}} = 3.33 \, \text{mol} \) - Hydrogen: \( \frac{6.7 \, \text{g}}{1.01 \, \text{g/mol}} = 6.63 \, \text{mol} \) - Oxygen: \( \frac{53.3 \, \text{g}}{16.00 \, \text{g/mol}} = 3.33 \, \text{mol} \) |
3. Determine the Simplest Whole-Number Ratio |
Divide each mole value by the smallest number of moles: - Carbon: \( 3.33 / 3.33 = 1 \) - Hydrogen: \( 6.63 / 3.33 \approx 2 \) - Oxygen: \( 3.33 / 3.33 = 1 \) |
4. Adjust Ratios if Necessary | Ratios are whole numbers (1:2:1), so no adjustment needed. |
5. Write the Empirical Formula | The empirical formula is CH₂O. |