Chemistry, asked by sneha354, 1 year ago

an organic compound has the following percentage composition: C= 48%, H= 28%, N= 28%. calculate the empirical formula of the compound.

Answers

Answered by gaurv8511
15
C2H14N..... EMPIRICAL FORMULA
Answered by BarrettArcher
40

Answer : The empirical formula of a compound is, C_2H_{14}N_1

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 = 48 g

Mass of H = 28 g

Mass of N = 28 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Step 1 : convert given masses into moles.

Moles of C = \frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{48g}{12g/mole}=4moles

Moles of H = \frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{28g}{1g/mole}=28moles

Moles of N = \frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{28g}{14g/mole}=2moles

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

For C = \frac{4}{2}=2

For H = \frac{28}{2}=14

For N = \frac{2}{2}=1

The ratio of C : H : N = 2 : 14 : 1

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

The Empirical formula = C_2H_{14}N_1

Therefore, the empirical formula of a compound is, C_2H_{14}N_1

Similar questions