The compound adrenaline contains 56.79% C, 6.56% H, 28.37% O, and 8.28% N by mass. What is the empirical formula for adrenaline?
i worked the problem out and got C8O3N
but its not right..what did i do wrong?
(56.79/12.01) x 100
(6.56/1.01) x 100
(28.37/16) x 100
after, divide divide it by the smallest number you come up with.
thanks i forgot the H :)
Hehe. Your welcome.
Or you can repost your work and let us look for the error.
To determine the empirical formula for adrenaline, you need to find the simplest whole number ratio of the elements in the compound. Let's calculate the empirical formula step by step.
1. Start with the given percentages of each element in the compound:
C = 56.79%
H = 6.56%
O = 28.37%
N = 8.28%
2. Convert the percentages to grams:
Assume you have a 100g sample of the compound, so you'll have:
C = 56.79g
H = 6.56g
O = 28.37g
N = 8.28g
3. Convert the masses to moles using the molar masses:
C: molar mass of carbon (C) = 12.01 g/mol
H: molar mass of hydrogen (H) = 1.008 g/mol
O: molar mass of oxygen (O) = 16.00 g/mol
N: molar mass of nitrogen (N) = 14.01 g/mol
C moles = 56.79g / 12.01 g/mol = 4.73 mol
H moles = 6.56g / 1.008 g/mol = 6.51 mol
O moles = 28.37g / 16.00 g/mol = 1.77 mol
N moles = 8.28g / 14.01 g/mol = 0.59 mol
4. Divide each number of moles by the smallest number of moles:
To simplify the ratio, divide all the moles by 0.59 (the smallest number of moles):
C moles / 0.59 = 4.73 mol / 0.59 = 8.02
H moles / 0.59 = 6.51 mol / 0.59 = 11.03
O moles / 0.59 = 1.77 mol / 0.59 = 3.00
N moles / 0.59 = 0.59 mol / 0.59 = 1.00
The simplified ratio is approximately:
C: 8
H: 11
O: 3
N: 1
Therefore, the empirical formula for adrenaline is C8H11NO3, which is different than the formula C8O3N you calculated.