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4 months ago

Acetone major species present when dissolved in water

Chemistry

1 answer:

Alekssandra [3K]4 months ago

3 0

The compound is acetone ( CH₃-CO-CH₃)

Explanation:

1) Acetone is represented as CH₃-CO-CH₃.

2) This is a molecule formed by covalent bonds.

3) When it dissolves, compounds with covalent bonds remain as individual molecules, indicating that the primary species in the solution are the molecules themselves, which are surrounded (solvated) by water molecules.

In contrast, ionic compounds ionize. For example, when NaCl dissolves in water, it completely breaks down into ions, hence the predominant species are the ions Na⁺ and Cl⁻, rather than the NaCl formula.

This leads to the conclusion that: when acetone dissolves in water, the primary components are the acetone molecules (there is no need to mention that water molecules are in the solution, as that isn't the question's focus).

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Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ

lions [2927]

Answer: The empirical formula and molecular formula for the analyzed organic compound are $C_9H_{12}O$ and $C_{18}H_{24}O_2$

Explanation:

The combustion chemical equation for a hydrocarbon containing carbon, hydrogen, and oxygen is:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where 'x', 'y', and 'z' represent the subscripts for Carbon, Hydrogen, and Oxygen.

We have the following data:

Mass of $CO_2=39.01g$

Mass of $H_2O=10.65g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For carbon mass calculation:

12 grams of carbon are found in 44 grams of carbon dioxide.

Thus, in 39.01 grams of carbon dioxide, $\frac{12}{44}\times 39.01=10.64g$ grams of carbon will be present.

For hydrogen mass calculation:

In 18 grams of water, 2 grams of hydrogen are contained.

Therefore, in 10.65 grams of water, $\frac{2}{18}\times 10.65=1.18g$ grams of hydrogen will be present.

The oxygen mass in the compound = (13.42) - (10.64 + 1.18) = 1.6 grams.

To derive the empirical formula, you need to perform a few steps:

Step 1: Convert the provided masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.64g}{12g/mole}=0.886moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.18g}{1g/mole}=1.18moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.6g}{16g/mole}=0.1moles$

Step 2: Determine the mole ratio of the elements.

Each mole value is divided by the smallest mole value, which is 0.1 moles, to find the mole ratio.

For Carbon = $\frac{0.886}{0.1}=8.86\approx 9$

For Hydrogen = $\frac{1.18}{0.1}=11.8\approx 12$

For Oxygen = $\frac{0.1}{0.1}=1.99\approx 2$

Step 3: Use the mole ratio values as subscripts.

The ratio of C: H: O = 9: 12: 1

The empirical formula for the compound is $C_9H_{12}O$

To find the molecular formula, it’s necessary to ascertain the valency, which is then multiplied by each elemental subscript of the empirical formula.

The formula used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$