# The compound adrenaline contains 56.79% carbon, 6.56% hydrogen, 28.3% oxygen, and 8.28% nitrogen by mass, what is its empirical formula?

Oct 24, 2015

Moles = mass/molar mass

#### Explanation:

First, as each percentage is over 100, multiply the values of carbon, hydrogen, oxygen and nitrogen by 100. You will obtain:

56.79 grams of carbon
6.56 grams of hydrogen
28.3 grams of oxygen
8.28 grams of nitrogen

Moles of carbon: 56.79/12.0107 = 4.728 moles
Moles of hydrogen: 6.56/1.00794 = 6.51 moles
Moles of oxygen: 28.3/15.9994 = 1.77 moles
Moles of nitrogen: 8.28/14.0067 = 0.59 moles

Divide all moles by the smallest value. In this case, it's 0.59 moles.

4.728/0.59 = 8
6.51/0.59 = 11
1.77/0.59 = 3

Thus, the emperical formula is C8H11NO3

Oct 24, 2015

${C}_{8} {H}_{11} {O}_{3} N$

#### Explanation:

With this kind of problem, it is always safe to assume that you have 100g sample. It makes the mathematical equation easier.

Assume 100g sample. Therefore,
wt of C = 56.79 g
wt of H = 6.56 g
wt of O = 28.30 g
wt of N = 8.28 g

Since molecular and empirical formulas deal with number of moles rather than weight, you need to convert these number of grams into number of moles. How? By multiplying the weight with the corresponding atomic weight.

Hence,

56.79 cancel ("grams C")* "1 mole"/(12.0107 cancel ("grams C")) = "4.728 mol C"

6.56 cancel ("grams H")* "1 mole"/(1.00794 cancel ("grams H")) = "6.508 mol H"

28.30 cancel ("grams O")* "1 mole"/(15.9994 cancel ("grams O")) = "1.769 mol O"

8.28 cancel ("grams N")* "1 mole"/(14.0067cancel ("grams N")) = "0.591 mol N"

Get the element with the smallest number of moles, (in this case, N) and use it as your denominator in determining the number of atoms for the other elements.

Thus,

$N = \text{0.591 mol" / "0.591 mol} = 1$

$C = \text{4.728 mol" / "0.591 mol} = 8$

$H = \text{6.508 mol" / "0.591 mol} = 11.01 \cong 11$

$O = \text{1.769 mol" / "0.591 mol} = 2.99 \cong 3$

Therefore, the empirical formula for adrenaline is ${C}_{8} {H}_{11} {O}_{3} N$. Please take note, empirical formulas are estimations only and should not be confused with the molecular formula.