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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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Further Explanation:

The following reactions will undergo double displacement where the metal cations in each compound are exchanged and form new products.

For reaction 1, the compounds involved are nitrates and chlorides. To determine the states of the products, the solubility rules for nitrates and chlorides must be followed:

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Therefore, the products will have the following characteristics:

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For reaction 2, the compounds involved are phosphates and chlorides. The solubility rules for phosphates and chlorides are as follows:

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Hence, the products of the second reaction will have the following characteristics:

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Keywords: solubility rules, precipitation reaction

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Answer:

Explanation:

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kari74 [83]

Answer:

a

Explanation:

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at the second reaction when 1/2 H2(g) + 1/2 Br2 (g) ⇄ HBr
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we will make the first formula of Kc upside down:
1/7.04X10^-2 = [HBr]^2/[H2][Br2]
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