The empirical formula of a gas is CH₃O .If 2.77 g of the gas occupies 1.00 L at exactly 0^{$\omicron $}C at a pressure of 760 Torr, What is the molecular formula of the gas? A) C₄H₁₂O₄ B) C₂H₆O₂ C) C₅H₁₅O₅ D) CH₃O E) C₃H₉O₃

Question

Quizplus · Accepted Answer

1. Calculate the number of moles of the gas using the ideal gas law: 
PV = nRT
n = (PV)/(RT)
R = 0.08206 L⋅atm/(mol⋅K) (or 62.36 mmHg⋅L/(mol⋅K))
T = (0°C + 273.15) = 273.15 K

n = (760 Torr) (1.00 L) / [(62.36 mmHg⋅L/(mol⋅K))(273.15 K)] = 0.0313 mol

2. Calculate the molar mass of the empirical formula:
Assume that the empirical formula has x carbon atoms, y hydrogen atoms, z sulfur atoms and w oxygen atoms, respectively. As there is no information about these quantities, we use variables, as required by the problem.

x+y+z+w = 15 (total number of atoms in the empirical formula)
12x+y+32w = M (total mass of the empirical formula in g/mol)
where M is the molar mass of the empirical formula.

From the empirical formula: 
C  x 
H  y
S   z
O   w

From the given formula, we can see that the sulfur is present in the ratio 1:1, which means the empirical formula is of the form CSSO (x=1,y=0,z=1,w=1). But, we will still calculate the molar mass and check if the given formula results in higher molecular mass.
12(1) + 32(1) + 32(1) + 1(32) = 108 g/mol

3. Calculate the molecular formula:
The molar mass of the empirical formula CSSO is 108 g/mol. However, the molar mass of the gas is 2.77 g / 0.0313 mol = 88.6 g/mol.
The molecular formula must have a molar mass that is an integer multiple of the empirical formula mass of 108 g/mol.

88.6 g/mol ÷ 108 g/mol = 0.82
The molecular formula must have at least 2 (i.e. 0.82 rounded up) empirical units.

The molecular formula is thus at least CCS2O2H2. We need to check if any higher ratios of carbon, sulfur, and oxygen give a higher molecular mass.

For CCS3O3H3, the molar mass is calculated to be: 
12(3) + 32(3) + 32(3) + 1(3*1) = 204 g/mol
This is greater than the molar mass of the gas, so we can eliminate this as a choice.

For CCSO2H2, the molar mass is calculated to be: 
12(2) + 32(1) + 32(2) + 1(2) = 152 g/mol
This is also less than the molar mass of the gas, so we can eliminate this as a choice.

Therefore, the molecular formula of the gas is CCS2O2H2 i.e. option B.

The Empirical Formula of a Gas Is CH3O ο\omicron ο C at a Pressure of 760 Torr, What Is the G

The Empirical Formula of a Gas Is CH₃O $\omicron$ C at a Pressure of 760 Torr, What Is the G