Question

3.00 g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 136. g/mol, is burned completely in excess oxygen, and the mass of the products carefully measured product mass water .59 g Use this information to find the molecular formula of X.

Answer 1

The question is incomplete, here is the complete question:

3.00 g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 136. g/mol, is burned completely in excess oxygen, and the mass of the products carefully measured:

        Product                  Mass

Carbon dioxide               7.77 g

       Water                       1.59 g

Use this information to find the molecular formula of X.

Answer: The molecular formula for the given organic compound is $C_8H_8O_2$

Explanation:

We are given:

Mass of $CO_2=7.77g$

Mass of $H_2O=1.59g$

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 7.77 g of carbon dioxide, $\frac{12}{44}\times 7.77=2.12g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 1.59 g of water, $\frac{2}{18}\times 1.59=0.176g$ of hydrogen will be contained.

Mass of oxygen in the compound = (3.00) - (2.12 + 0.176) = 0.704 g

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

• Step 1: Converting the given masses into moles.

Moles of Carbon =$\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{2.12g}{12g/mole}=0.176moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.176g}{1g/mole}=0.176moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.704g}{16g/mole}=0.044moles$

• 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 0.044 moles.

For Carbon = $\frac{0.176}{0.044}=4$

For Hydrogen = $\frac{0.176}{0.044}=4$

For Oxygen = $\frac{0.044}{0.044}=1$

• Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 4 : 4 : 1

The empirical formula for the given compound is $C_4H_4O$

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 = 68 g/mol

Mass of empirical formula = 136 g/mol

Putting values in above equation, we get:

$n=\frac{136g/mol}{68g/mol}=2$

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

$C_{(4\times 2)}H_{(4\times 2)}O_{(1\times 2)}=C_8H_8O_2$

Hence, the molecular formula for the given organic compound is $C_8H_8O_2$