Empirical/Molecular Formula Practice Worksheet Directions: Find the empirical AND molecular formulas given the percent compositions or masses. SHOW YOUR WORK to receive full credit 1) 26,4% Carbon 3.3 % Hydrogen 70.3 % Oxygen Molar Mass: 91.0 g/mol Empirical Formula: Molecular Formula: 2) 81.8 grams Carbon 18.2 grams Hydrogen Molar Mass: 132,0 g/mol Empirical Formula: Molecular Formula: 3) 20.2 % Sodium 37.6 % Sulfur 42.2% Oxygen Molar Mass: 682.8 g/mol Empirical Formula: Molecular Formula: Molar Mass: 1362.5 g/mol 4) 8.81 g Carbon 91.2 g Chlorine Empirical Formula: Molecular Formula:
The Correct Answer and Explanation is :
It seems there was an error because I used an incorrect length for the atomic masses in relation to the number of elements in the compounds. Let me correct that and provide you with the answers and a more detailed explanation.
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Here are the answers to the problems with explanations for each:
1) 26.4% Carbon, 3.3% Hydrogen, 70.3% Oxygen
- Empirical Formula: ( \text{CH}_2\text{O} )
- Molecular Formula: ( \text{C}_2\text{H}_4\text{O}_2 ) Explanation: The empirical formula is derived by finding the simplest ratio of atoms. The percent composition is first converted to moles, and the smallest mole ratio is identified to give the empirical formula. The molecular formula is then found by comparing the molar mass to the empirical formula’s mass.
2) 81.8 grams Carbon, 18.2 grams Hydrogen
- Empirical Formula: ( \text{CH}_3 )
- Molecular Formula: ( \text{C}9\text{H}{27} ) Explanation: After converting the given masses of Carbon and Hydrogen into moles, the empirical formula is obtained by simplifying the mole ratio. The molecular formula is calculated by comparing the molar mass provided to the empirical formula mass.
3) 20.2% Sodium, 37.6% Sulfur, 42.2% Oxygen
- Empirical Formula: ( \text{Na}_2\text{SO}_4 )
- Molecular Formula: ( \text{Na}{10}\text{S}{22}\text{O}_{20} ) Explanation: The empirical formula for this compound comes from determining the ratio of sodium, sulfur, and oxygen in the simplest whole numbers. The molecular formula is found by comparing the molar mass provided to the empirical formula mass.
4) 8.81 g Carbon, 91.2 g Chlorine
- Empirical Formula: ( \text{CCl}_{12} ) Explanation: This case only provided masses of Carbon and Chlorine. The empirical formula is derived by converting the masses into moles and simplifying the ratio.
These calculations involve converting percent compositions or masses into moles, determining the simplest mole ratios (for the empirical formula), and then using the given molar mass to find the molecular formula.