Answer:
C₅H₁₀O₄
Explanation:
The empirical formula is the simplest whole-number ratio of atoms in a compound.
The ratio of atoms is the same as the ratio of moles.
So, our job is to calculate the molar ratio of C:H:O.
Assume 100 g of deoxyribose.
1. Calculate the mass of each element.
Then we have 44.8 g C, 7.5 g H, and 47.7 g O.
2. Calculate the moles of each element
[tex]\text{Moles of C} = \text{44.8 g C} \times \dfrac{\text{1 mol C}}{\text{12.01 g C}} = \text{3.730 mol C}\\\\\text{Moles of H} = \text{7.5 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H }} = \text{7.44 mol H}\\\\\text{Moles of O} = \text{47.7 g O} \times \dfrac{\text{1 mol O}}{\text{16.00 g O }} = \text{2.981 mol O}[/tex]
3. Calculate the molar ratio of the elements
Divide each number by the smallest number of moles
C:H:O = 3.730:7.44:2.981 = 1.251:2.50:1 = 5.005:9.98:4 ≈ 5:10:4
4. Write the empirical formula
EF = C₅H₁₀O₄
