Question

Epinephrine (adrenaline) is 59.0% carbon, 7.1% hydrogen, 26.2% oxygen, and 7.7% nitrogen by mass. What is the empirical formula of epinephrine if its molar mass is 180 g/mol?

Answer 1

empirical formula is the simplest ratio of whole numbers of components making up a compound

the percentage composition of each element has been given so we are going to find the mass for 100 g of the compound

mass C - 59.0 g

H - 7.1 g

O - 26.2 g

N - 7.7g

number of moles

C - 59.0 g/ 12 g/mol

H - 7.1 g/ 1 g/mol

O - 26.2 g/ 16 g/mol

N - 7.7 g / 14 g/mol

C = 4.9 mol H= 7.1 mol O= 1.6 mol N =0.55 mol

divide by the least number of moles

C - 4.9 / 0.55 = 8.9

H - 7.1 / 0.55 = 12.9

O -1.6 / 0.55 = 2.9

N - 0.55/ 0.55 = 1.0

the number of atoms of each element rounded off

C - 9

H - 13

O - 29

N - 1

the empirical formula is - C₉H₁₃O₃N

Answer 2

Answer : The empirical formula of a compound is, $C_{9}H_{13}O_3N$

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 = 59 g

Mass of H = 7.1 g

Mass of O = 26.2 g

Mass of N = 7.7 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of O = 16 g/mole

Molar mass of N = 14 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{59g}{12g/mole}=4.9moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{7.1g}{1g/mole}=7.1moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{26.2g}{16g/mole}=1.6moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{7.7g}{14g/mole}=0.55moles$

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

For C = $\frac{4.9}{0.55}=8.9\approx 9$

For H = $\frac{7.1}{0.5}=12.9\approx 13$

For O = $\frac{1.6}{0.5}=2.9\approx 3$

For N = $\frac{0.55}{0.55}=1$

The ratio of C : H : O : N = 9 : 13 : 3 : 1

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

The Empirical formula = $C_{9}H_{13}O_3N_1$  = $C_{9}H_{13}O_3N$

Therefore, the empirical formula of a compound is, $C_{9}H_{13}O_3N$