Respuesta :
Answer:
C3H6O
Explanation:
Data obtained from the question. This includes:
Carbon (C) = 0.62069 g
Hydrogen (H) = 0.103448 g
Oxygen (O) = 0.275862 g
The empirical formula can be obtained as follow:
Step 1:
Divide by their molar mass.
C = 0.62069 / 12 = 0.0517
H = 0.103448 / 1 = 0.103448
O = 0.275862 / 16 = 0.0172
Step 2:
Divide by the smallest number.
C = 0.0517 / 0.0172 = 3
H = 0.103448 / 0.0172 = 6
O = 0.0172 / 0.0172 = 1
Step 3:
Writing the empirical formula.
The empirical formula is C3H6O
Answer:
The empirical formula is C3H6O
Explanation:
Step 1: Data given
A sample of ethyl butyrate contains:
0.62069 g of carbon
0.103448 g of hydrogen
0.275862 g of oxygen
Atomic mass of carbon = 12.01 g/mol
Atomic mass of hydrogen = 1.01 g/mol
Atomic mass of oxygen = 16.0 g/mol
Step 2: Calculate moles
Moles = mass / molar mass
Moles carbon = 0.62069 grams / 12.01 g/mol
Moles carbon = 0.0517 moles
Moles hydrogen = 0.103448 grams / 1.01 g/mol
Moles hydrogen = 0.1024 moles
Moles oxygen = 0.275862 grams / 16.0 g/mol
Moles oxygen = 0.0172 moles
Step 3: Calculate the mol ratio
We divide by the smalllest amount of moles
C: 0.0517 moles / 0.0172 moles = 3
H: 0.1024 moles / 0.0172 moles = 6
O: 0.0172 moles / 0.0172 moles = 1
The empirical formula is C3H6O