Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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How was the empirical formula mass of 141.94g/mol found?
What are some easy steps I can use so I can remember for next time I calculate?
![### Determining Empirical and Molecular Formulas II
A white powder is analyzed and found to contain 43.64% phosphorus and 56.36% oxygen by mass. The compound has a molar mass of 283.88 g/mol. What are the compound’s empirical and molecular formulas?
#### Solution
**Where are we going?**
To find the empirical and molecular formulas for the given compound.
**What do we know?**
- Percent of each element
- Molar mass of the compound is 283.88 g/mol
**What information do we need to find the empirical formula?**
- Mass of each element in 100.00 g of compound
- Moles of each element
**How do we get there?**
**What is the mass of each element in 100.00 g of compound?**
- Phosphorus (P): 43.64 g
- Oxygen (O): 56.36 g
**What are the moles of each element in 100.00 g of compound?**
\[ \text{Moles of P} = 43.64 \, \text{g} \times \frac{1 \, \text{mol P}}{30.97 \, \text{g P}} = 1.409 \, \text{mol P} \]
\[ \text{Moles of O} = 56.36 \, \text{g} \times \frac{1 \, \text{mol O}}{16.00 \, \text{g O}} = 3.523 \, \text{mol O} \]
**What is the empirical formula for the compound?**
Dividing each mole value by the smaller one gives:
\[ \frac{1.409}{1.409} = 1 \text{ P} \quad \text{and} \quad \frac{3.523}{1.409} = 2.5 \text{ O} \]
This yields the formula \( \text{PO}_{2.5} \). Since compounds must contain whole numbers of atoms, the empirical formula should contain only whole numbers. To obtain the simplest set of whole numbers, we multiply both numbers by 2 to give the empirical formula \( \text{P}_2\text{O}_5 \).
**What is the molecular formula for the compound?**
Compare the empirical formula mass to the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F40e25f29-5021-499e-9c49-7da7bd92383c%2F1aad2569-65f8-4870-8fb1-8041cf0a1f33%2Fnawqf4j.png&w=3840&q=75)
Transcribed Image Text:### Determining Empirical and Molecular Formulas II
A white powder is analyzed and found to contain 43.64% phosphorus and 56.36% oxygen by mass. The compound has a molar mass of 283.88 g/mol. What are the compound’s empirical and molecular formulas?
#### Solution
**Where are we going?**
To find the empirical and molecular formulas for the given compound.
**What do we know?**
- Percent of each element
- Molar mass of the compound is 283.88 g/mol
**What information do we need to find the empirical formula?**
- Mass of each element in 100.00 g of compound
- Moles of each element
**How do we get there?**
**What is the mass of each element in 100.00 g of compound?**
- Phosphorus (P): 43.64 g
- Oxygen (O): 56.36 g
**What are the moles of each element in 100.00 g of compound?**
\[ \text{Moles of P} = 43.64 \, \text{g} \times \frac{1 \, \text{mol P}}{30.97 \, \text{g P}} = 1.409 \, \text{mol P} \]
\[ \text{Moles of O} = 56.36 \, \text{g} \times \frac{1 \, \text{mol O}}{16.00 \, \text{g O}} = 3.523 \, \text{mol O} \]
**What is the empirical formula for the compound?**
Dividing each mole value by the smaller one gives:
\[ \frac{1.409}{1.409} = 1 \text{ P} \quad \text{and} \quad \frac{3.523}{1.409} = 2.5 \text{ O} \]
This yields the formula \( \text{PO}_{2.5} \). Since compounds must contain whole numbers of atoms, the empirical formula should contain only whole numbers. To obtain the simplest set of whole numbers, we multiply both numbers by 2 to give the empirical formula \( \text{P}_2\text{O}_5 \).
**What is the molecular formula for the compound?**
Compare the empirical formula mass to the
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