In this week\'s experiment you will determine the molar mass of carbon dioxide b

Question

In this week's experiment you will determine the molar mass of carbon dioxide by measuring the mass of an Erlenmeyer flask full of the gas. The following calculations are intended to familiarize you with the general procedure:

The mass of an empty Erlenmeyer flask and stopper was determined to be 51.64 grams. When filled with distilled water, the mass was 298.8 grams. The atmospheric pressure was measured to be 0.9226 atm, the room temperature was 25.00oC. At this temperature, the vapor pressure of water is 23.80 torr -- but assume a 50% relative humidity as outlined in the procedure for this experiment.

What is the number of moles of air in the flask?

Moles of air in the flask = mol

If the average molar mass of the gases present in air is 28.960 g mol-1, what is the mass of air in the flask?

Mass of air in the flask = grams

What would be the mass of carbon dioxide in the flask?

Mass of carbon dioxide = grams

What would be the mass of the flask (and stopper) when filled with carbon dioxide gas? In this case the CO2 will be saturated with water vapor so assume 100% humidity as outlined in the experiment.

Mass of flask and stopper filled with CO2 gas = grams

Explanation / Answer

mass = 308-51.64 = 256.36g

for water assume 1g=1ml

so 251.03ml to m^3 we divide by 10^6

so VOLUME= 251.03/10^6 = 2.5103 * 10^-4m^3

Pressure at25^oC = 23.8/ 760 =0.0313 atm

Partial pressure = 0.0313 * .50 = 0.01565 atm

Pressure due to air alone= 0.9226-0.01565 = 0.90695 atm

atm to Pa ( Nm^-2) we multiply by normal atm P ( 1.10 * 10^5)

So Pressure due to the air alone = 0.90695* (1.01*10^5) = 9.16 *10^4 Pa

PRESSURE = 9.16 *10^4 Pa

TEMPERATURE = 25+273.15 =298.15K

MOLAR GAS CONSTANT (R) = 8.314 J K^-1 mol^-1

u have P, V, R and T

so using PV=nRT

n= PV/RT


n = (9.16 *10^4) * (2.5103 * 10^-4) /(8.314* 298.15)

n= 0.00927 moles

Mass of air = molar mass * moles

mass of air = 28.96 * 0.00927 = 0.2686g

Molar mass for CO2 = 12 + ( 16*2) = 44 g/mol


Pressure due to CO2 alone = 0.9226 - ( 0.0313) = 0.8913 atm because humidity = 100%

Pressure due to CO2 alone in Pa (Nm^-2) = 0.8913 * (1.01*10^5) = 8.913 * 10 ^4 Pa

PRESSURE = 8.913 * 10 ^4 Pa

VOLUME is the same 2.5103 * 10^-4m^3

MOLAR GAS CONSTANT = 8.314 J K^-1 mol^-1

TEMPERATURE = 292.25K

n = PV/RT

(8.913 * 10 ^4) * (2.5103 * 10^-4m^3)
n= ----------------------------------------...
(8.314* 298.15)

n= 0.009

Mass of CO2 = molar mass of CO2 * moles

Mass of CO2 = 44 * 0.009 = 0.3971 g

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