Chapter 3: Problem 40
The element \(X\) forms the compound \(X O C l_{2}\) containing 59.6\% Cl. What is element X?
Short Answer
Expert verified
The element X in the compound \(X O C l_{2}\) is Aluminium (Al).
Step by step solution
01
Determine the percentage of each element in the compound
Start by breaking the compound \(X O C l_{2}\) into its components. Since the compound contains 59.6% chlorine (Cl), it must also contain 40.4% of element X and oxygen (O), because the percentages of all elements in a compound must add up to 100%.
02
Calculate the Percentage of Oxygen
Firstly, map out the atomic masses of each element present in the compound. The atomic mass of oxygen (O) is around 16g/mol and Chlorine (Cl) is 35.45g/mol. Since there are two Chlorines in the compound, the total mass contributed by Chlorine is \(2 \times 35.45g/mol = 70.9g/mol\). Substituting this value into the percentage formula \(\frac{70.9}{70.9+16+massX} \times 100% = 59.6%\), we can solve for \(massX\), which is the atomic mass of element X.
03
Solve for the Atomic Mass of Element X
Rearranging the formula in step 2 and substituting all known quantities, we get \(massX = (\frac{59.6}{100} \times (70.9 + 16)) - 70.9\). Calculating the result yields an approximate atomic mass of X of ~27g/mol.
04
Identify Element X
Consulting the periodic table, the element with atomic mass close to 27g/mol is Aluminium (Al). Itβs approximately 26.98g/mol which is close to the computed 27g/mol. Therefore, the element X in the compound \(X O C l_{2}\) is Aluminium (Al).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chemical Formula Interpretation
In chemistry, understanding a chemical formula allows us to understand the composition of a compound. Each formula is like a recipe, giving us the exact number and identity of different elements bonded together in a compound. For example, in the formula \( X O C l_{2} \), each letter represents an element from the periodic table. The subscript numbers indicate the number of each type of atom in a molecule.
Here's how it works:
In this specific case, we have a task of breaking down the compound \( X O C l_{2} \) to find out how much of each element is present in the molecular structure.
Here's how it works:
- "\( X \)" represents an unknown element, which is what we are trying to identify.
- "\( O \)" stands for Oxygen, a common non-metal.
- "\( Cl_{2} \)" means there are two Chlorine atoms included in the compound.
In this specific case, we have a task of breaking down the compound \( X O C l_{2} \) to find out how much of each element is present in the molecular structure.
Element Identification
Element identification involves pinpointing which element an unknown component in a compound might be. Through the use of atomic masses and the periodic table, we can achieve this.To identify element \( X \) in the compound \( X O C l_{2} \), we analyze its percentage composition. Since 59.6% of the compound is Chlorine, the remaining 40.4% is composed of element \( X \) and Oxygen. With this data, we can begin solving for the identity of \( X \).By understanding the formula setup and being aware of the typical atomic masses of elements, locating the unidentified element becomes manageable:
- Determine the percent composition rule that total element percentages must equal 100%.
- Use given percentages and atomic masses from the periodic table to solve for the unknown element's possible atomic mass.
- Cross-reference this atomic mass with known values on the periodic table to deduce the specific element.
Atomic Mass Calculation
Atomic mass calculation is a pivotal process in identifying unknown substances in chemistry. It's similar to a balancing act where the known and unknown masses must be compared to maintain equilibrium.The steps involve using a formula to combine known and unknown masses, which for the compound \( X O C l_{2} \) gives us:
- Calculate the total known mass of Chlorine: \( 2 \times 35.45 \text{g/mol} = 70.9 \text{g/mol} \).
- Apply the mass percent formula to find the mass of element \( X \):\[massX = \left( \frac{59.6}{100} \times (70.9 + 16) \right) - 70.9\]
- Perform algebraic rearrangements followed by calculations to isolate and solve for \( massX \).
- Utilize the periodic table to find the closest atomic weight to the calculated \( massX \).