Question

An organic compound has % composition of c=48% h=8% n=28% calculate the emperical formula of the compound

Answer 1

Answer:

firstly divide the percentage composition of element from it's atomic weight . for carbon =48/12 (4). for hydrogen 8/1 (8) ,,. for nitrogen 28/14(2). and then after dividing the smallest value in all of three is 2 then all value divided by 2 ,. for carbon 4/2..... value is 2. for hydrogen 8/2 ..... value is 4, for nitrogen 2/2 ..... value is 1 so the emprical formula is C2H4N

Answer 2

Answer : The empirical formula of a compound is,

C2H14N1

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 = $\frac{given mass of C}{moler mass of C} =\frac{48g}{12g/mole} = 4moles$

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

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

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 = C2H14N1

Therefore, the empirical formula of a compound is, C2H14N1

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