Chemistry: 5. Azulene is a beautiful blue hydrocarbon. If 0.106 g of the compound is burned in oxygen, 0.364 g of CO2 and 0.0596 g of H2O are isolated. Determine its empirical and molecular formulas if the molar mass of azulene was found to be 128.2 g. Is it an alkane

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5. Azulene is a beautiful blue hydrocarbon. If 0.106 - id50105657026, naiomireyes74p2aybs

Ответ:

abigail2403

21.09.2020

Chemistry

The empirical and molecular formula for the given organic compound is CH and $C_{10}H_{10}$ and it is not an alkane.

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon and hydrogen follows:

$C_xH_y+O_2\rightarrow CO_2+H_2O$

where, 'x' and 'y' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=0.364g

Mass of H_2O=0.0596g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 0.364 g of carbon dioxide, $\frac{12}{44}\times 0.364=0.0993g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 0.0596 g of water, $\frac{2}{18}\times 0.0596=0.0067g$ of hydrogen will be contained.

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.0993g}{12g/mole}=0.00828moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.0067g}{1g/mole}=0.0067moles$ ]

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0067 moles.

For Carbon = $\frac{0.00828}{0.0067}=1.23\approx 1$

For Hydrogen = $\frac{0.0067}{0.0067}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 1

The empirical formula for the given compound is CH

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

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

We are given:

Mass of molecular formula = 128.2 g/mol

Mass of empirical formula = 13 g/mol

Putting values in above equation, we get:

$n=\frac{128.2g/mol}{13g/mol}=9.86\approx 10$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 10)}H_{(1\times 10)}=C_{10}H_{10}$

The general formula of an alkane is $C_nH_{(2n+2)}$ , where n = any natural number

Here, n = 10 and it does not satisfy being an alkane

Hence, the empirical and molecular formula for the given organic compound is CH and $C_{10}H_{10}$ and it is not an alkane.

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5. Azulene is a beautiful blue hydrocarbon. If 0.106 g of the compound is burned in oxygen, 0.364 g of CO2 and 0.0596 g of H2O are isolated. Determine its empirical and molecular formulas if the molar mass of azulene was found to be 128.2 g. Is it an alkane

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