Answer to Question #292289 in Chemistry for Master56

Question

On analysis of ammonium salt of alkanoic acid give 60.5% of carbon, 6.5% of hydrogen. If 0.309g of the salt yielded 0.0313g of Nitrogen. Determine the empirical formula of the salt.

Answer 1

Solution:

ammonium salt of alkanoic acid = RCOONH₄ = C_xH_yN_zO_n

Thus ammonium salt of alkanoic acid consists of carbon, hydrogen, nitrogen and oxygen.

Сalculate the mass of each element given that the mass of the salt is 0.309 g:

Mass of C = w(C) × Mass of salt = 0.605 × 0.3090 g = 0.1869 g

Mass of H = w(H) × Mass of salt = 0.065 × 0.3090 g = 0.0201 g

Mass of N = 0.0313 g

Mass of O = Mass of salt - Mass of C - Mass of H - Mass of N

Mass of O = 0.3090 g - 0.1869 g - 0.0201 g - 0.0313 g = 0.0707 g

The molar mass of carbon (C) is 12.0107 g mol^-1

The molar mass of hydrogen (H) is 1.00784 g mol^-1

The molar mass of nitrogen (N) is 14.0067 g mol^-1

The molar mass of oxygen (O) is 15.999 g mol^-1

Convert to moles:

Moles of C = (0.1869 g C) × (1 mol C / 12.0107 g C) = 0.01556 mol C

Moles of H = (0.0201 g H) × (1 mol H / 1.00784 g H) = 0.01994 mol H

Moles of N = (0.0313 g N) × (1 mol O / 14.0067 g N) = 0.00223 mol N

Moles of O = (0.0707 g O) × (1 mol O / 15.999 g O) = 0.00442 mol O

Divide all moles by the smallest of the results:

C: 0.01556 / 0.00223 = 6.978 = 7

H: 0.01994 / 0.00223 = 8.942 = 9

N: 0.00223 / 0.00223 = 1.000 = 1

O: 0.00442 / 0.00223 = 1.982 = 2

The empirical formula of the ammonium salt of alkanoic acid is C₇H₉NO₂

Possible compound formula: C₆H₅COONH₄ - ammonium salt of benzoic acid.

Answer: The empirical formula of the salt is C₇H₉NO₂