Question

The sugar deoxyribose is an important component of DNA. Deoxyribose is 44.8% 7.5% H, and 47.7% O by mass. What is the empirical formula for deoxyribose?

Answer 1

C₅H₁₀O₄

Step-by-step 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

`\text{Moles of C} = \text{44.8 g C} * \frac{\text{1 mol C}}{\text{12.01 g C}} = \text{3.730 mol C}\\\\\text{Moles of H} = \text{7.5 g H} * \frac{\text{1 mol H}}{\text{1.008 g H }} = \text{7.44 mol H}\\\\\text{Moles of O} = \text{47.7 g O} * \frac{\text{1 mol O}}{\text{16.00 g O }} = \text{2.981 mol O}`

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