Question

an organic compound containing C,H, N gave the following analysis C=40%,H=13.33%,N=46.67%.its empirical formula would be

Answer 1

here is ur answer...

Answer 2

Answer: The empirical formula for the organic compound will be $CH_4N$

Explanation:

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

• Step 1: Converting percentages to masses

If percentage are given then we are taking total mass is 100 grams.  Thus, the mass of each element is equal to the percentage given.

Mass of C = 40 g

Mass of H = 13.33 g

Mass of N = 46.67 g

• Step 2: Converting given masses into moles.

$\text{Moles of C}=\frac{\text{Given mass of C}}{\text{Molar mass of C}}= \frac{40g}{12g/mole}=3.33moles$

$\text{Moles of H}=\frac{\text{Given mass of H}}{\text{Molar mass of H}}= \frac{13.33g}{1g/mole}=13.33moles$

$\text{Moles of N}=\frac{\text{Given mass of N}}{\text{Molar mass of N}}= \frac{46.67g}{14g/mole}=3.33mole$

• Step 3: Converting the moles into mole ratios

For calculating the mole ratio, we divide each value of moles by the smallest number of moles calculated which is 3.33 moles.

For C = $\frac{3.33}{3.33}=1$

For H = $\frac{13.33}{3.33}=4$

For N= $\frac{3.33}{3.33}=1$

• Step 4: Now writing the mole fraction as the subscripts of the elements.

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

Hence the empirical formula is $CH_{4}N$