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The question in the image asks to determine the empirical formulas for compounds based on their percentage composition. Let's solve each part step by step.

### Steps to determine empirical formula:
1. Assume you have 100 grams of the compound. This converts the percentage directly into grams.
2. Convert the mass of each element to moles by dividing by the atomic mass (molar mass).
3. Find the ratio of the moles of each element by dividing each by the smallest number of moles.
4. Round these ratios to the nearest whole number to get the subscripts in the empirical formula.

Let's begin:

### (a) 43% C and 57% O
- Mass of C = 43 g, Mass of O = 57 g
- Moles of C = \( \frac{43}{12.01} \approx 3.58 \) mol (atomic mass of C is 12.01 g/mol)
- Moles of O = \( \frac{57}{16.00} \approx 3.56 \) mol (atomic mass of O is 16.00 g/mol)
- Ratio: C : O = 3.58 : 3.56 ≈ 1:1
- Empirical formula: **CO**

### (b) 26.7% Cr, 40.9% K, and 32.7% O
- Mass of Cr = 26.7 g, Mass of K = 40.9 g, Mass of O = 32.7 g
- Moles of Cr = \( \frac{26.7}{52.00} \approx 0.514 \) mol (atomic mass of Cr is 52.00 g/mol)
- Moles of K = \( \frac{40.9}{39.10} \approx 1.046 \) mol (atomic mass of K is 39.10 g/mol)
- Moles of O = \( \frac{32.7}{16.00} \approx 2.044 \) mol
- Divide by the smallest number of moles (0.514):
  - Cr : K : O = \( \frac{0.514}{0.514} \) : \( \frac{1.046}{0.514} \) : \( \frac{2.044}{0.514} \)
  - Cr : K : O ≈ 1 : 2 : 4
- Empirical formula: **K₂CrO₄**

### (c) 32.8% C, 14.7% N, 42.0% O, and the remainder H
- Remainder of H: 100% - (32.8% + 14.7% + 42.0%) = 10.5%
- Masses: C = 32.8 g, N = 14.7 g, O = 42.0 g, H = 10.5 g
- Moles of C = \( \frac{32.8}{12.01} \approx 2.73 \) mol
- Moles of N = \( \frac{14.7}{14.01} \approx 1.05 \) mol
- Moles of O = \( \frac{42.0}{16.00} \approx 2.63 \) mol
- Moles of H = \( \frac{10.5}{1.008} \approx 10.42 \) mol
- Divide by the smallest number of moles (1.05):
  - C : N : O : H = \( \frac{2.73}{1.05} \) : \( \frac{1.05}{1.05} \) : \( \frac{2.63}{1.05} \) : \( \frac{10.42}{1.05} \)
  - C : N : O : H ≈ 2.6 : 1 : 2.5 : 9.9
  - Approximate ratio: C₂N₁O₂H₁₀ (rounded to nearest whole numbers)
- Empirical formula: **C₂H₁₀NO₂**

### (d) 31.9% K, 28.9% Cl, and the remainder O
- Remainder of O: 100% - (31.9% + 28.9%) = 39.2%
- Masses: K = 31.9 g, Cl = 28.9 g, O = 39.2 g
- Moles of K = \( \frac{31.9}{39.10} \approx 0.816 \) mol
- Moles of Cl = \( \frac{28.9}{35.45} \approx 0.815 \) mol
- Moles of O = \( \frac{39.2}{16.00} \approx 2.45 \) mol
- Divide by the smallest number of moles (0.815):
  - K : Cl : O = \( \frac{0.816}{0.815} \) : \( \frac{0.815}{0.815} \) : \( \frac{2.45}{0.815} \)
  - K : Cl : O ≈ 1 : 1 : 3
- Empirical formula: **KClO₃**

Let me know if you'd like further details or have any questions!

### Related questions:
1. What is the difference between empirical and molecular formulas?
2. How do you determine molecular formulas if the molar mass is provided?
3. Why do empirical formulas only represent the simplest ratio of atoms in a compound?
4. How can you convert an empirical formula to a molecular formula?
5. What are some common compounds with the same empirical and molecular formula?

**Tip:** Always verify your results by checking the sum of percentages. If it's not 100%, there may be a remainder for one of the elements.

Determine the empirical formula for each compound whose percentage composition is shown below:

a. 43% C and 57% O
b. 26.7% Cr, 40.9% K, and 32.7% O
c. 32.8% C, 14.7% N, 42.0% O, and the remainder H
d. 31.9% K, 28.9% Cl, and the remainder O

Learn how to calculate the empirical formula for compounds based on percentage composition with step-by-step instructions. This example involves mole calculations, finding elemental ratios, and applying the Law of Definite Proportions. Suitable for high school chemistry students (Grades 10-12).

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