When elemental sulfur, S8, is heated with AgF, a gas forms that contains only sulfur and fluorine. The following tests were done to characterize this gaseous product.

Empirical formula…..

AgF(s) + S8(s) –> SxFy(g) + Ag2S(s)

PV = nRT

PV = mRT / M

PM = (m/V)RT

PM = DRT

M = DRT / P

M = 0.807 g/L x 62.36 Ltorr/molK x 308K / 152 torr

M = 102.0 g/mol ………… molar mass of SxFy gas

SxFy(g) + H2O(l) —> HF(aq) + ?

0.480L…. 80.0mL…… 0.080M

PV = nRT

n = PV / (RT)

n = 125 torr x 0.480L / 62.36 Ltorr/molK / 301K

n = 0.00320 mol SxFy

0.080L x (0.080 mol HF / 1L) x (1 mol F / 1 mol HF) = 0.00640 mol F

Conclusion:

0.00320 mol SxFy

0.00640 mol F

2 mol F for each mole of SxFy

partial formula: SxF2

32.1x + 2(19.0) = 102.0 ………. molar mass calculation, solve for x

x = 2

Molecular formula = S2F2

Empirical formula = SF

PV = nRT

n = g/MW

152 mm Hg = 0.200 atm

0.200 atm x 1.00 L = (0.807 g/MW) 0.082057 x (35 + 273)

0.200 = 20.3957/MW

MW = 101.98

SxFz

PV=nRT

P = 0.1645 atm (125 mm Hg/760)

V = 0.480 L

T = 301 K

n = 0.1645 x 0.480 / (0.082057 x 301) = 0.00320 mole

0.080 L x 0.080 m/L = 0.0064 mole of HF

0.0032 mole of gas –> 0.0064 mole of HF

SxF2

MW = 101.98

Sx = 101.98 – 37.996 = 63.98

S2F2 = the unknown gas

The empirical formula is SF

The molecular formula is S2F2

S=32.065

F=18.998

n / V = P / RT = (152 mmHg) / ((62.36367 L mmHg/K mol) x (35 + 273) K) = 0.00791337 mol/L

(0.807 g/L) / (0.00791337 mol/L) = 101.979 g/mol

n = PV / RT = (125 mmHg) x (0.480 L) / ((62.36367 L mmHg/K mol) x (28 + 273) K) = 0.003196 mol

(0.003196 mol) x (101.979 g/mol) = 0.32592 g total

(0.080 mol/L HF) x (0.0800 L) x (1 mol F / 1 mol HF) x (18.9984032 g F/mol) = 0.12159 g F

(0.32592 g total) – (0.12159 g F) = 0.20433 g S

(0.12159 g F) / (0.32592 g total) x (101.979 g/mol) / (18.9984032 g F/mol) = 2.00

(0.20433 g S) / (0.32592 g total) x (101.979 g/mol) / (32.0655 g S/mol) = 1.99

Round to the nearest whole numbers to find the molecular formula:

S2F2

So the empirical formula is:

SF

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