Question

Kevlar is the high-strength polymer material. It has the following composition: 70.58% carbon, 11.76% nitrogen, 4.23% hydrogen, and 13.43% oxygen. Calculate the empirical formula for Kevlar.(Enter the elements in the order: C, H, N, O.)

Answer 1

Answer : The empirical formula for kevlar is,
`C_7H_5NO`

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 = 70.58 g

Mass of H = 4.23 g

Mass of N = 11.76 g

Mass of O = 13.43 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}}= (70.58g)/(12g/mole)=5.88moles`

Moles of H =
`\frac{\text{ given mass of H}}{\text{ molar mass of H}}= (4.23g)/(1g/mole)=4.23moles`

Moles of N =
`\frac{\text{ given mass of N}}{\text{ molar mass of N}}= (11.76g)/(14g/mole)=0.84moles`

Moles of O =
`\frac{\text{ given mass of O}}{\text{ molar mass of O}}= (13.43g)/(16g/mole)=0.84moles`

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

For C =
`(5.88)/(0.84)=7`

For H =
`(4.23)/(0.84)=5.03\approx 5`

For N =
`(0.84)/(0.84)=1`

For O =
`(0.84)/(0.84)=1`

The ratio of C : H : N : O = 7 : 5 : 1 : 1

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

The Empirical formula =
`C_7H_5N_1O_1` =
`C_7H_5NO`

Therefore, the empirical formula for kevlar is,
`C_7H_5NO`