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Blababa [14]
2 years ago
5

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?
Chemistry
1 answer:
Lera25 [3.4K]2 years ago
3 0

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

\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}

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

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The molar concentration is 1.11M.

<h3>What is molar concentration?</h3>

The phrase "molar concentration" (also known as "molarity," "amount concentration," or "substance concentration") refers to the amount of a substance per unit volume of solution and is used to describe the concentration of a chemical species, specifically a solute, in a solution. The most frequent measure of molarity in chemistry is the number of moles per liter, denoted by the unit symbol mol/L or mol/dm3 in SI units. A solution with a concentration of 1 mol/L is referred to as 1 molar, or 1 M.

<h3>Given : </h3>

Volume of the solution = 2L

Mass of glucose given = 200g

Concentration of glucose= ?

<h3>Formula use: </h3>

Molarity = no. of moles of solute / volume of the solution (L)

Moles of solute = given mass of solute / molar mass of the solute

<h3>Solution: </h3>

No. of moles of solute( glucose ) = 200 / 180 = 1.11 moles'

Molarity = 1.11 / 2 = 0.5555 mol L ^(-1)

Therefore, the molar concentration of glucose in the solution = 0.555 mol L ^(-1)

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