Question

Determine the empirical formula for a compound that contains C, H and O. It contains 52.14% C and 34.73% O by mass. what is the molecular formula if the molar mass is 92 g/mol?

Answer 1

Answer:

C4O2H12

Explanation:

The carbon molecular weight is 12 g/mol, the oxygen molecular weight is 16 g/mol, and he hydrogen molecular weight is 1 g/mol. So 0.5214*92 g/mol = 48g/mol, then the number of atoms of carbon must be 48/12 = 4; and 0.3473*92 g/mol = 32, so 32/16=2 = the number of atoms of oxygen. The 12 restant atoms must be from oxygen (12).

Answer 2

Answer: The molecular formula for the given organic compound is $C_4H_{12}O_2$

Explanation:

We are given:

Percentage of C = 52.14 %

Percentage of O = 34.73 %

Percentage of H = (100 - 52.14 - 34.73) = 13.13 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 52.14 g

Mass of O = 34.73 g

Mass of H = 13.13 g

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

• Step 1: Converting the given masses into moles.

Moles of Carbon =$\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{52.14g}{12g/mole}=4.345moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{34.73g}{16g/mole}=2.171moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{13.13g}{1g/mole}=13.13moles$

• 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 2.171 moles.

For Carbon = $\frac{4.345}{2.171}=2$

For Oxygen = $\frac{2.171}{2.171}=1$

For Hydrogen = $\frac{13.13}{2.171}=6.05\approx 6$

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

The ratio of C : O : H = 2 : 1 : 6

The empirical formula for the given compound is $C_2H_6O$

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 = 92 g/mol

Mass of empirical formula = 46 g/mol

Putting values in above equation, we get:

$n=\frac{92g/mol}{46g/mol}=2$

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

$C_{(2\times 2)}H_{(6\times 2)}O_{(1\times 2)}=C_4H_{12}O_2$

Hence, the molecular formula for the given organic compound is $C_4H_{12}O_2$