1. What is the empirical formula of a compound that contains 52.7% K and 47.3% Cl? 2. What is the empirical formula of a compound that contains 13.0% Mg and 87.0% Br? 3. What is th... 1. What is the empirical formula of a compound that contains 52.7% K and 47.3% Cl? 2. What is the empirical formula of a compound that contains 13.0% Mg and 87.0% Br? 3. What is the molar mass of a compound whose empirical formula is CO₂? 4. Given the structural formula of butane (C4H10), what is the empirical formula of the compound?

Understand the Problem
The image presents a series of chemistry questions related to empirical formulas and molar mass. Let's break down each question:
- Empirical Formula Calculation: Determine the empirical formula of a compound given the percentages of Potassium (K) and Chlorine (Cl). This involves converting percentages to moles and finding the simplest whole number ratio.
- Empirical Formula Calculation: Determine the empirical formula of a compound given the percentages of Magnesium (Mg) and Bromine (Br). This involves converting percentages to moles and finding the simplest whole number ratio.
- Molar Mass Calculation: Calculate the molar mass of a compound given its empirical formula (CO₂). This requires using the periodic table to find the atomic masses of Carbon and Oxygen.
- Empirical Formula from Structural Formula: Determine/confirm the empirical formula of Butane given its structural formula.
Answer
2. KCl, 3. MgBr₂, 4. 44.01 g/mol, 5. C₂H₅
The empirical formula for the compounds are: 2. KCl 3. MgBr₂ 4. 44.01 g/mol 5. C₂H₅
Answer for screen readers
The empirical formula for the compounds are: 2. KCl 3. MgBr₂ 4. 44.01 g/mol 5. C₂H₅
More Information
The empirical formula represents the simplest whole number ratio of elements in a compound. The molar mass is the mass of one mole of a substance.
Tips
A common mistake is not dividing by the smallest mole value to get the simplest ratio.
Sources
- Question #f5bf1 - Socratic - socratic.org
- Steven analyzed a compound and found it to be 52.7% Potassium ... - homework.study.com
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