Answer to Question #230148 in Chemistry for Jessica

Question

A compound of P and F was analyzed as follows: Heating 0.2324 g of the compound in a 378-cm^3 container turned all of it to gas, which had a pressure of 97.3 mmHg at 77 degrees C. Then the gas was mixed with calcium chloride solution, which turned all of the F to 0.2631 g of CaF2. Determine the molecular formula of the compound

Answer 1

Solution:

The molar mass of CaF₂ is 78.07 g/mol

(0.2631 g CaF₂) × (1 mol CaF₂ / 78.07 g CaF₂) = 0.00337 mol CaF₂

2F^- + Ca²⁺ → CaF₂

(0.00337 mol CaF₂) × (2 mol F / 1 mol CaF₂) = 0.00674 mol F

The molar mass of F is 18.9984 g/mol

(0.00674 mol F) × (18.9984 g F / 1 mol F) = 0.12805 g F

1 mmHg = 0.00131578947 atm

(97.3 mmHg) × (0.00131578947 atm / 1 atm) = 0.128 atm

1 L = 1000 mL = 1000 cm³

(378 cm³) × (1L / 1000 cm³) = 0.378 L

T = t + 273

T = 77 + 273 = 350 K

PV = nRT

n(P_xF_y) = PV / RT

n(P_xF_y) = (0.128 atm × 0.378 L) / (0.0821 L atm mol^-1 K^-1 × 350 K) = 0.001684 mol

n(P_xF_y) = 0.001684 mol

n(P_xF_y) = m(P_xF_y) / M(P_xF_y)

M(P_xF_y) = m(P_xF_y) / n(P_xF_y) = (0.2324 g) / (0.001684 mol) = 138 g/mol

m(P_xF_y) = m(P) + m(F)

0.2324 g = m(P) + 0.12805 g

m(P) = 0.2324 - 0.12805 = 0.10435

m(P) = 0.10435 g

The molar mass of P is 30.9737 g/mol

(0.10435 g P) × (1 mol P / 30.9737 g P) = 0.003369 mol P

n(P) : n(F) = 0.003369 mol : 0.00674 mol = 1 : 2

Therefore, the empirical formula of P_xF_y is PF₂

The molar mass of PF₂ is 68.9705 g/mol

The molecular formula of P_xF_y is (PF₂)_n

n = M(P_xF_y) / M(PF₂) = (138) / (68.9705) = 2

Therefore, the molecular formula of P_xF_y is (PF₂)₂ or P₂F₄

Answer: The molecular formula of the compound is P₂F₄