Table of contents

1. Intro to General Chemistry3h 48m
2. Atoms & Elements4h 16m
3. Chemical Reactions4h 10m
4. BONUS: Lab Techniques and Procedures1h 38m
5. BONUS: Mathematical Operations and Functions48m
6. Chemical Quantities & Aqueous Reactions3h 53m
7. Gases3h 49m
8. Thermochemistry2h 37m
9. Quantum Mechanics2h 58m
10. Periodic Properties of the Elements3h 9m
11. Bonding & Molecular Structure3h 29m
12. Molecular Shapes & Valence Bond Theory1h 57m
13. Liquids, Solids & Intermolecular Forces2h 23m
14. Solutions3h 1m
15. Chemical Kinetics2h 53m
16. Chemical Equilibrium2h 29m
17. Acid and Base Equilibrium5h 1m
18. Aqueous Equilibrium4h 47m
19. Chemical Thermodynamics1h 50m
20. Electrochemistry2h 42m
21. Nuclear Chemistry2h 36m
22. Organic Chemistry5h 7m
23. Chemistry of the Nonmetals2h 39m
24. Transition Metals and Coordination Compounds3h 16m

3. Chemical Reactions

Empirical Formula

General Chemistry

3. Chemical Reactions

Empirical Formula

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Multiple Choice

What is the empirical formula for a compound that contains 39.5% carbon, 2.21% hydrogen, and 58.3% chlorine?

CHCl

C2H2Cl2

CH2Cl

C2HCl3

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Verified step by step guidance

Convert the percentage of each element to grams, assuming you have 100 grams of the compound. This means you have 39.5 grams of carbon, 2.21 grams of hydrogen, and 58.3 grams of chlorine.

Convert the mass of each element to moles by dividing by their respective molar masses: Carbon (C) has a molar mass of approximately 12.01 g/mol, Hydrogen (H) is about 1.01 g/mol, and Chlorine (Cl) is about 35.45 g/mol.

Calculate the moles of each element: \( \text{moles of C} = \frac{39.5}{12.01} \), \( \text{moles of H} = \frac{2.21}{1.01} \), \( \text{moles of Cl} = \frac{58.3}{35.45} \).

Determine the simplest whole number ratio of moles of each element by dividing each by the smallest number of moles calculated in the previous step.

Use the whole number ratios to write the empirical formula. If necessary, multiply the ratios by a common factor to get whole numbers for each element in the formula.