A Sample Of A Compound Contains 4.86 G Of Magnesium, 12.85 G Of Sulfur And 9.70 G Of Oxygen What Is The Empirical Formula? Data: M.A. [Mg = 24, S = 32]
Introduction
In chemistry, determining the empirical formula of a compound is a crucial step in understanding its composition and properties. The empirical formula represents the simplest whole-number ratio of atoms of each element present in the compound. In this article, we will guide you through the process of determining the empirical formula of a compound using the given data.
Given Data
A sample of a compound contains 4.86 g of magnesium (Mg), 12.85 g of sulfur (S), and 9.70 g of oxygen (O). The atomic masses of Mg and S are given as 24 g/mol and 32 g/mol, respectively.
Step 1: Calculate the Number of Moles of Each Element
To determine the empirical formula, we need to calculate the number of moles of each element present in the sample. We can use the formula:
moles = mass / atomic mass
For Mg:
moles = 4.86 g / 24 g/mol = 0.2025 mol
For S:
moles = 12.85 g / 32 g/mol = 0.4016 mol
For O:
moles = 9.70 g / 16 g/mol (since the atomic mass of O is 16 g/mol) = 0.6113 mol
Step 2: Determine the Simplest Whole-Number Ratio
To determine the empirical formula, we need to find the simplest whole-number ratio of atoms of each element present in the compound. We can do this by dividing each number of moles by the smallest number of moles.
The smallest number of moles is 0.2025 mol (Mg). We can divide each number of moles by this value:
For Mg:
0.2025 mol / 0.2025 mol = 1
For S:
0.4016 mol / 0.2025 mol = 1.98 ≈ 2
For O:
0.6113 mol / 0.2025 mol = 3.02 ≈ 3
Step 3: Write the Empirical Formula
Now that we have the simplest whole-number ratio of atoms of each element, we can write the empirical formula of the compound.
The empirical formula is the simplest whole-number ratio of atoms of each element present in the compound. In this case, the empirical formula is:
MgS2O3
Conclusion
In this article, we have guided you through the process of determining the empirical formula of a compound using the given data. We have calculated the number of moles of each element, determined the simplest whole-number ratio, and written the empirical formula of the compound. The empirical formula represents the simplest whole-number ratio of atoms of each element present in the compound, and it is a crucial step in understanding the composition and properties of a compound.
Frequently Asked Questions
- What is the empirical formula of a compound? The empirical formula represents the simplest whole-number ratio of atoms of each element present in the compound.
- How do I determine the empirical formula of a compound? To determine the empirical formula, you need to calculate the number of moles of each element, determine the simplest whole-number ratio, and write the empirical formula.
- What is the difference between the empirical formula and the molecular formula? The empirical formula represents the simplest whole-number ratio of atoms of each element present in the compound, while the molecular formula represents the actual number of atoms of each element present in the compound.
References
- Chemistry: An Atoms First Approach, by Steven S. Zumdahl
- General Chemistry: Principles and Modern Applications, by Linus Pauling
Further Reading
- Empirical Formula: A Guide to Determining the Empirical Formula of a Compound
- Molecular Formula: A Guide to Determining the Molecular Formula of a Compound
- Stoichiometry: A Guide to Stoichiometry and Chemical Reactions
Introduction
In our previous article, we guided you through the process of determining the empirical formula of a compound using the given data. In this article, we will answer some frequently asked questions related to determining the empirical formula of a compound.
Q&A
Q: What is the empirical formula of a compound?
A: The empirical formula represents the simplest whole-number ratio of atoms of each element present in the compound.
Q: How do I determine the empirical formula of a compound?
A: To determine the empirical formula, you need to calculate the number of moles of each element, determine the simplest whole-number ratio, and write the empirical formula.
Q: What is the difference between the empirical formula and the molecular formula?
A: The empirical formula represents the simplest whole-number ratio of atoms of each element present in the compound, while the molecular formula represents the actual number of atoms of each element present in the compound.
Q: What is the significance of the empirical formula?
A: The empirical formula is significant because it represents the simplest whole-number ratio of atoms of each element present in the compound, which is essential for understanding the composition and properties of a compound.
Q: How do I calculate the number of moles of each element?
A: To calculate the number of moles of each element, you need to divide the mass of each element by its atomic mass.
Q: What is the atomic mass of an element?
A: The atomic mass of an element is the mass of one atom of that element, usually expressed in grams per mole (g/mol).
Q: How do I determine the simplest whole-number ratio of atoms of each element?
A: To determine the simplest whole-number ratio, you need to divide each number of moles by the smallest number of moles.
Q: What is the empirical formula of a compound with the following data: 2.5 g of carbon, 3.2 g of hydrogen, and 1.8 g of oxygen?
A: To determine the empirical formula, you need to calculate the number of moles of each element, determine the simplest whole-number ratio, and write the empirical formula.
First, calculate the number of moles of each element:
moles of C = 2.5 g / 12 g/mol = 0.2083 mol moles of H = 3.2 g / 1 g/mol = 3.2 mol moles of O = 1.8 g / 16 g/mol = 0.1125 mol
Next, determine the simplest whole-number ratio:
The smallest number of moles is 0.1125 mol (O). Divide each number of moles by this value:
For C:
0.2083 mol / 0.1125 mol = 1.85 ≈ 2
For H:
3.2 mol / 0.1125 mol = 28.44 ≈ 28
For O:
0.1125 mol / 0.1125 mol = 1
The empirical formula is C2H28O.
Q: What is the empirical formula of a compound with the following data: 5.6 g of nitrogen, 2.8 g of oxygen, and 1.2 g of fluorine?
A: To determine the empirical formula, you need to calculate the number of moles of each element, determine the simplest whole-number ratio, and write the empirical formula.
First, calculate the number of moles of each element:
moles of N = 5.6 g / 14 g/mol = 0.4 mol moles of O = 2.8 g / 16 g/mol = 0.175 mol moles of F = 1.2 g / 19 g/mol = 0.0632 mol
Next, determine the simplest whole-number ratio:
The smallest number of moles is 0.0632 mol (F). Divide each number of moles by this value:
For N:
0.4 mol / 0.0632 mol = 6.32 ≈ 6
For O:
0.175 mol / 0.0632 mol = 2.77 ≈ 3
For F:
0.0632 mol / 0.0632 mol = 1
The empirical formula is N6O3F.
Conclusion
In this article, we have answered some frequently asked questions related to determining the empirical formula of a compound. We have provided examples of how to calculate the number of moles of each element, determine the simplest whole-number ratio, and write the empirical formula. The empirical formula represents the simplest whole-number ratio of atoms of each element present in the compound, which is essential for understanding the composition and properties of a compound.
Frequently Asked Questions
- What is the empirical formula of a compound?
- How do I determine the empirical formula of a compound?
- What is the difference between the empirical formula and the molecular formula?
- What is the significance of the empirical formula?
- How do I calculate the number of moles of each element?
- What is the atomic mass of an element?
- How do I determine the simplest whole-number ratio of atoms of each element?
References
- Chemistry: An Atoms First Approach, by Steven S. Zumdahl
- General Chemistry: Principles and Modern Applications, by Linus Pauling
Further Reading
- Empirical Formula: A Guide to Determining the Empirical Formula of a Compound
- Molecular Formula: A Guide to Determining the Molecular Formula of a Compound
- Stoichiometry: A Guide to Stoichiometry and Chemical Reactions