Carboxylic Acids
6 marks · NormalA student wants to prepare a sample of sodium butanoate by reacting sodium metal with butanoic acid.
Draw the displayed formula for butanoic acid.
Write a balanced symbol equation (using molecular formulae) for the reaction between sodium metal and butanoic acid.
Calculate the mass of sodium metal required to react completely with 0.10 mol of butanoic acid. (Relative atomic mass: Na = 23)
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a) — · must show CH3CH2CH2COOH explicitly shown with all bonds
(b) 2Na + 2CH3CH2CH2COOH → 2CH3CH2CH2COONa + H2
(c) 1.15 g
Must show: All bonds shown for –COOH group
Common mistakes: Omitting bonds in COOH for drawing; Unbalanced equation; Using 0.10 mol Na instead of 0.05 mol
Show mark scheme and worked solution
Marking points
B1 (a) · 2 marks · Displayed formula: All atoms and bonds, carboxyl group shown with double and single bonds; ‘COOH’ group fully expanded
M1 (b) · 1 mark · Correct unbalanced reactants and products with formulae (Na, CH3CH2CH2COOH)
A1 (b) · 1 mark · Balanced equation (2:2:2:1) with correct products (CH3CH2CH2COONa and H2)
M2 (c) · 1 mark · Chooses correct stoichiometry from equation (2 mol butanoic acid: 1 mol Na)
A2 (c) · 1 mark · Correct final mass of sodium (1.15 g, accept 1.1–1.18 g) with units
Worked solution
a) Displayed formula for butanoic acid:
b) 2Na + 2CH3CH2CH2COOH → 2CH3CH2CH2COONa + H2
c) From the equation, 2 mol butanoic acid reacts with 2 mol Na (1:1). 0.10 mol acid needs 0.05 mol Na: mol Na. Mass = moles × Ar = 0.05 × 23 = 1.15 g.