Question

A sample of an unknown compound is vaporized at 200°C. The gas produced has a volume of 1920 mL at a pressure of 1.00 atm, and it weighs 2.50 g. Assuming the gas behaves as an ideal gas, calculate the molar mass of the compound.

Answer 1

50.6 grams/mole

Step-by-step explanation:

Lets use the ideal gas law PV = nRT,

where n is moles, V is volume, R is the gas constant, T is the temperature (in kelvin only) and P is the pressure.

Since we know V, P, and T, we can find n, the moles of the gas produced. Rearrange for n:

PV = nRT

n = PV/RT

We'll use a value for R, the gas constant, of 0.0821 L*atm/(K*mol)

Enter the data:

Since the units must match those of the gas constant we chose, we need to covert ml to liters: 1920ml = 1.920L

We also need to convert 200C to 473.15K

n = (1.00atm)*(1.920L)/(0.0821 L*atm/(K*mol))*(473.15K)

Pressure, temperature, and volume units all cancel, leaving only moles:

n = (1.00)*(1.920)/(0.0821/(mol))*(473.15)

n = 0.0495 moles

We have 0.0495 moles of a gas that has a mass of 2.50g. Divide to find grams/mole, the molar mass:

(2.50g)/(0.0495 moles) = 50.6 grams/mole

Answer 2

To calculate the molar mass of the unknown compound, use the ideal gas law equation (MM = mRT/PV). Plug in the given values and calculate the molar mass to be approximately 39.9 g/mol.

Step-by-step explanation:

To calculate the molar mass of the unknown compound, we can use the ideal gas law equation:

MM = (mRT)/(PV)

Where MM is the molar mass, m is the mass of the gas, R is the ideal gas constant (0.0821 L·atm/mol·K), T is the temperature in Kelvin, P is the pressure in atm, and V is the volume in liters.

Using the given values, the equation becomes:

MM = (2.50g * 0.0821 L·atm/mol·K * (200+273)K) / (1.00atm * 1.920L)

Simplifying the equation gives a molar mass of approximately 39.9 g/mol.