Problem 50
Question
Apply How many moles of \(\mathrm{K}, \mathrm{C},\) and 0 atoms are there in 1 \(\mathrm{mol}\) of \(\mathrm{K}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) ?
Step-by-Step Solution
Verified Answer
2 moles of K, 2 moles of C, and 4 moles of O.
1Step 1: Understand the Compound Structure
The compound \( \text{K}_2\text{C}_2\text{O}_4 \) represents the chemical potassium oxalate. The subscript numbers indicate the number of atoms of each type in one formula unit of the compound. Specifically, there are 2 potassium (\(\text{K}\)) atoms, 2 carbon (\(\text{C}\)) atoms, and 4 oxygen (\(\text{O}\)) atoms in each \( \text{K}_2\text{C}_2\text{O}_4 \) unit.
2Step 2: Determine Moles of Atoms in 1 Mole of Compound
In 1 mole of \( \text{K}_2\text{C}_2\text{O}_4 \), there are: - 2 moles of \( \text{K} \) atoms (since each formula unit contains 2 \( \text{K} \) atoms),- 2 moles of \( \text{C} \) atoms (since each formula unit contains 2 \( \text{C} \) atoms),- 4 moles of \( \text{O} \) atoms (since each formula unit contains 4 \( \text{O} \) atoms).
Key Concepts
Understanding Chemical FormulasThe Art of Mole CalculationThe Process of Atom Counting
Understanding Chemical Formulas
Chemical formulas are like the recipes for molecules. They provide essential information about the number and type of each atom present in a molecule. For example, in the compound potassium oxalate, which is represented as \( \text{K}_2\text{C}_2\text{O}_4 \), the subscript numbers indicate how many atoms of each element are included in the formula unit.
These subscripts are crucial because they help you determine the composition of the molecule:
These subscripts are crucial because they help you determine the composition of the molecule:
- The subscript after potassium (\( \text{K} \)) indicates there are 2 potassium atoms.
- After carbon (\( \text{C} \)), the subscript shows there are 2 carbon atoms.
- For oxygen (\( \text{O} \)), the subscript means there are 4 oxygen atoms.
The Art of Mole Calculation
The mole is a fundamental unit in chemistry, serving as a bridge between the atomic and macroscopic scales. One mole contains Avogadro's number of entities, equivalent to \( 6.022 \times 10^{23} \). When it comes to chemical compounds like potassium oxalate \( (\text{K}_2\text{C}_2\text{O}_4) \), knowing the molecular formula allows you to calculate how many moles of each atom are present per mole of compound.
For instance:
For instance:
- Since there are 2 \( \text{K} \) atoms per molecule, 1 mole of \( \text{K}_2\text{C}_2\text{O}_4 \) contains 2 moles of potassium atoms.
- Similarly, with 2 \( \text{C} \) atoms in each molecule, it equals 2 moles of carbon atoms per mole of compound.
- Finally, the 4 \( \text{O} \) atoms per unit means 4 moles of oxygen atoms in 1 mole of compound.
The Process of Atom Counting
Atom counting involves determining the actual number of atoms in a sample, given the mole values. Using the mole concept, you can convert moles into atoms, facilitating a deeper understanding of a compound's structure.
Here's how to break it down:
Here's how to break it down:
- To find out how many potassium atoms are in 1 mole of \( \text{K}_2\text{C}_2\text{O}_4 \), multiply Avogadro's number \( (6.022 \times 10^{23}) \) by the 2 moles of \( \text{K} \) atoms present per mole of compound.
- Repeat the same for carbon: 2 moles of \( \text{C} \) times \( 6.022 \times 10^{23} \). This gives the exact number of carbon atoms.
- For oxygen, multiply 4 moles by Avogadro's number to determine the total number of oxygen atoms.
Other exercises in this chapter
Problem 47
Describe how to determine the molar mass of a compound.
View solution Problem 48
Identify the conversion factors needed to convert between the number of moles and the mass of a compound.
View solution Problem 52
Calculate Calcium carbonate is the calcium source for many vitamin tablets. The recommended daily allowance of calcium is 1000 \(\mathrm{mg}\) of \(\mathrm{Ca}^
View solution Problem 53
Design a bar graph that will show the number of moles of each element present in 500 g of a particular form of dioxin \(\left(\mathrm{C}_{12} \mathrm{H}_{4} \ma
View solution