Question

What is the empirical formula for a compound that contains 1.67 mole Carbon and 5.01 Moles of Hydrogen *​

Answer 1

Answer:

• The empirical formula is C₃H₃O.
• The empirical formula is C₃H₃O.Assume we have 100 g of the compound.
• The empirical formula is C₃H₃O.Assume we have 100 g of the compound.Then we have 65.5 g of C, 5.5 g of H, and 29.0 g of O.
• The empirical formula is C₃H₃O.Assume we have 100 g of the compound.Then we have 65.5 g of C, 5.5 g of H, and 29.0 g of O.Moles of C = 65.5 g C ×
• The empirical formula is C₃H₃O.Assume we have 100 g of the compound.Then we have 65.5 g of C, 5.5 g of H, and 29.0 g of O.Moles of C = 65.5 g C × 1
• The empirical formula is C₃H₃O.Assume we have 100 g of the compound.Then we have 65.5 g of C, 5.5 g of H, and 29.0 g of O.Moles of C = 65.5 g C × 1 mol C
• The empirical formula is C₃H₃O.Assume we have 100 g of the compound.Then we have 65.5 g of C, 5.5 g of H, and 29.0 g of O.Moles of C = 65.5 g C × 1 mol C12.01
• The empirical formula is C₃H₃O.Assume we have 100 g of the compound.Then we have 65.5 g of C, 5.5 g of H, and 29.0 g of O.Moles of C = 65.5 g C × 1 mol C12.01g C
• The empirical formula is C₃H₃O.Assume we have 100 g of the compound.Then we have 65.5 g of C, 5.5 g of H, and 29.0 g of O.Moles of C = 65.5 g C × 1 mol C12.01g C = 5.45 mol C
• The empirical formula is C₃H₃O.Assume we have 100 g of the compound.Then we have 65.5 g of C, 5.5 g of H, and 29.0 g of O.Moles of C = 65.5 g C × 1 mol C12.01g C = 5.45 mol CMoles of H = 5.5 g H ×
• The empirical formula is C₃H₃O.Assume we have 100 g of the compound.Then we have 65.5 g of C, 5.5 g of H, and 29.0 g of O.Moles of C = 65.5 g C × 1 mol C12.01g C = 5.45 mol CMoles of H = 5.5 g H × 1
• The empirical formula is C₃H₃O.Assume we have 100 g of the compound.Then we have 65.5 g of C, 5.5 g of H, and 29.0 g of O.Moles of C = 65.5 g C × 1 mol C12.01g C = 5.45 mol CMoles of H = 5.5 g H × 1 mol H
• The empirical formula is C₃H₃O.Assume we have 100 g of the compound.Then we have 65.5 g of C, 5.5 g of H, and 29.0 g of O.Moles of C = 65.5 g C × 1 mol C12.01g C = 5.45 mol CMoles of H = 5.5 g H × 1 mol H1.008
• The empirical formula is C₃H₃O.Assume we have 100 g of the compound.Then we have 65.5 g of C, 5.5 g of H, and 29.0 g of O.Moles of C = 65.5 g C × 1 mol C12.01g C = 5.45 mol CMoles of H = 5.5 g H × 1 mol H1.008g H
• The empirical formula is C₃H₃O.Assume we have 100 g of the compound.Then we have 65.5 g of C, 5.5 g of H, and 29.0 g of O.Moles of C = 65.5 g C × 1 mol C12.01g C = 5.45 mol CMoles of H = 5.5 g H × 1 mol H1.008g H = 5.5 mol H
• The empirical formula is C₃H₃O.Assume we have 100 g of the compound.Then we have 65.5 g of C, 5.5 g of H, and 29.0 g of O.Moles of C = 65.5 g C × 1 mol C12.01g C = 5.45 mol CMoles of H = 5.5 g H × 1 mol H1.008g H = 5.5 mol HMoles of O = 29.0 g O ×
• The empirical formula is C₃H₃O.Assume we have 100 g of the compound.Then we have 65.5 g of C, 5.5 g of H, and 29.0 g of O.Moles of C = 65.5 g C × 1 mol C12.01g C = 5.45 mol CMoles of H = 5.5 g H × 1 mol H1.008g H = 5.5 mol HMoles of O = 29.0 g O × 1
• The empirical formula is C₃H₃O.Assume we have 100 g of the compound.Then we have 65.5 g of C, 5.5 g of H, and 29.0 g of O.Moles of C = 65.5 g C × 1 mol C12.01g C = 5.45 mol CMoles of H = 5.5 g H × 1 mol H1.008g H = 5.5 mol HMoles of O = 29.0 g O × 1 mol O
• The empirical formula is C₃H₃O.Assume we have 100 g of the compound.Then we have 65.5 g of C, 5.5 g of H, and 29.0 g of O.Moles of C = 65.5 g C × 1 mol C12.01g C = 5.45 mol CMoles of H = 5.5 g H × 1 mol H1.008g H = 5.5 mol HMoles of O = 29.0 g O × 1 mol O16.00
• The empirical formula is C₃H₃O.Assume we have 100 g of the compound.Then we have 65.5 g of C, 5.5 g of H, and 29.0 g of O.Moles of C = 65.5 g C × 1 mol C12.01g C = 5.45 mol CMoles of H = 5.5 g H × 1 mol H1.008g H = 5.5 mol HMoles of O = 29.0 g O × 1 mol O16.00g O
• The empirical formula is C₃H₃O.Assume we have 100 g of the compound.Then we have 65.5 g of C, 5.5 g of H, and 29.0 g of O.Moles of C = 65.5 g C × 1 mol C12.01g C = 5.45 mol CMoles of H = 5.5 g H × 1 mol H1.008g H = 5.5 mol HMoles of O = 29.0 g O × 1 mol O16.00g O = 1.81 mol H
• The empirical formula is C₃H₃O.Assume we have 100 g of the compound.Then we have 65.5 g of C, 5.5 g of H, and 29.0 g of O.Moles of C = 65.5 g C × 1 mol C12.01g C = 5.45 mol CMoles of H = 5.5 g H × 1 mol H1.008g H = 5.5 mol HMoles of O = 29.0 g O × 1 mol O16.00g O = 1.81 mol HMoles of C:Moles of H:Moles of O = 5.45:5.5:1.81 = 3.01:3.0:1 ≈ 3:3:1
• The empirical formula is C₃H₃O.Assume we have 100 g of the compound.Then we have 65.5 g of C, 5.5 g of H, and 29.0 g of O.Moles of C = 65.5 g C × 1 mol C12.01g C = 5.45 mol CMoles of H = 5.5 g H × 1 mol H1.008g H = 5.5 mol HMoles of O = 29.0 g O × 1 mol O16.00g O = 1.81 mol HMoles of C:Moles of H:Moles of O = 5.45:5.5:1.81 = 3.01:3.0:1 ≈ 3:3:1The empirical formula is C₃H₃O.

Answer 2

Answer:

C1H3 IN QUESTION WE HAVE MOLE RATIO WE NEED TO FIND LEAST MOLE RATIO (EMPIRICAL FORMULA) SO DIVIDE BOTH VALUE WITH LEAST NO WHICH IS 1.67 SO WE GET

C 1 H 3

(EMPIRICAL FORMULA)

Explanation: