What is the empirical formula of a compound containing 40% carbon, 6.7% hydrogen, and 53.3% oxygen?
Practice Questions
1 question
Q1
What is the empirical formula of a compound containing 40% carbon, 6.7% hydrogen, and 53.3% oxygen?
CH2O
C2H4O2
C3H6O3
C4H8O4
Converting percentages to moles gives C: 40/12 = 3.33, H: 6.7/1 = 6.7, O: 53.3/16 = 3.33. The simplest ratio is 1:2:1, giving the empirical formula CH2O.
Questions & Step-by-step Solutions
1 item
Q
Q: What is the empirical formula of a compound containing 40% carbon, 6.7% hydrogen, and 53.3% oxygen?
Solution: Converting percentages to moles gives C: 40/12 = 3.33, H: 6.7/1 = 6.7, O: 53.3/16 = 3.33. The simplest ratio is 1:2:1, giving the empirical formula CH2O.
Steps: 7
Step 1: Start with the percentages of each element in the compound: 40% Carbon (C), 6.7% Hydrogen (H), and 53.3% Oxygen (O).
Step 2: Convert the percentages to grams. Assume you have 100 grams of the compound. This means you have 40 grams of C, 6.7 grams of H, and 53.3 grams of O.
Step 3: Convert grams to moles for each element using their atomic masses: Carbon (C) has an atomic mass of about 12 g/mol, Hydrogen (H) is about 1 g/mol, and Oxygen (O) is about 16 g/mol.
Step 4: Calculate the moles of each element: For Carbon, 40 grams / 12 g/mol = 3.33 moles; for Hydrogen, 6.7 grams / 1 g/mol = 6.7 moles; for Oxygen, 53.3 grams / 16 g/mol = 3.33 moles.
Step 5: Find the simplest ratio of the moles by dividing each mole value by the smallest number of moles calculated. The smallest number is 3.33.
Step 6: Divide each mole value by 3.33: C: 3.33/3.33 = 1, H: 6.7/3.33 ≈ 2, O: 3.33/3.33 = 1.
Step 7: Write the empirical formula using the simplest ratio: C1H2O1, which is commonly written as CH2O.
