Question

The percntage proportion of carbon, hydrogen, nitrogen, oxygen in an organic compound are 62.07%, 10.34%, 14.0%, 13.59%, respectively . Find its empirical formula. If its molecular mass is 144 gram/mol, find its molecular formula

Answer 1

there will be 2 answers because the value of n can be 1.26 that is approximately equal to 2 or 1

Answer 2

Answer : The molecular of the compound is, $C_6H_{12}NO$

Solution : Given,

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 62.07 g

Mass of H = 10.34 g

Mass of N = 14.0 g

Mass of O = 13.59 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{62.07g}{12g/mole}=5.17moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{10.34g}{1g/mole}=10.34moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{14.0g}{14g/mole}=1moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{13.59g}{16g/mole}=0.85moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{5.17}{0.85}=6$

For H = $\frac{10.34}{0.85}=12$

For N = $\frac{1}{0.85}=1$

For O = $\frac{0.85}{0.85}=1$

The ratio of C : H : N : O = 6 : 12 : 1 : 1

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_6H_{12}N_1O_1$

The empirical formula weight = 6(12) + 12(1) + 1(14) + 1(16) = 114 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}=\frac{144g/mole}{114g/eq}=1.26=1$

Molecular formula = $(C_6H_{12}N_1O_1)_n=(C_6H_{12}N_1O_1)_1=C_6H_{12}N_1O_1=C_6H_{12}NO$

Therefore, the molecular of the compound is, $C_6H_{12}NO$