Suppose the CO gas evolved by a certain chemical reaction taking place at 30.0 °C is collected over water, and the final volume of gas in the collection tube is measured to be 128 mL.
Calculate the mass of CO that is in the collection tube.
To solve the problem we have to find the number of moles of CO that are in the sample. But, since the sample is collected over water, the CO is not alone. The sample also contains water vapor.
To solve problems like this one we usually use Dalton's Law to find the partial pressure of the gas of interest. And then the Ideal Gas Law to find the number of moles of CO.
According to Dalton's Law the total pressure exerted is equal to the sum of the partial pressures of the individual gases that form the mixture. So:
Ptotal = Pco + Pwater
We are not given the total pressure so we will assume that it is 1 atm (atmoshperical pressure). And if we look for the Vapor Pressure of water at 30 °C we will find that it is 31.82 mmHg (we have to convert it to atm also). So:
Ptotal = 1 atm
Pwater = 31.82 mmHg * 1 atm / 760 mmHg
Pwater = 0.0419 atm
Now we can find the partial pressure of CO:
Ptotal = Pco + Pwater
Pco = Ptotal - Pwater
Pco = 1 atm - 0.0419 atm
Pco = 0.9581 atm
So we found that the partial pressure of CO in our sample is 0.9581 atm. What does that mean? That if the molecules of CO were alone in the sample (occupying the 128 ml) they would have a pressure of 0.9581 atm.
Now we can use the Ideal gas Law to find the number of moles of CO:
Pco * V = n * R * T
Pco = 0.9581 atm
V = 128 mL = 128 mL * 1 L /(1000 mL) = 0.128 L
n= number of moles of CO
R = ideal gas constant = 0.082 atm*L/(mol*K)
T = 30.0 °C = (273.15 + 30.0) K = 303.15 K
Pco * V = n * R * T
n = Pco * V / (R * T)
n= 0.9581 atm * 0.128 L / (0.082 atm*L/(mol*K) *303.15 K)
n = 0.00493 moles of CO
So there are 0.00493 moles of CO in the sample. To finisht the problem we have to find the mass of CO using its molar mass.
molar mass of CO = 12 + 16 = 28 g/mol
mass of CO = number of moles of CO * molar mass of CO
mass of CO = 0.00493 moles of CO * 28 g/mol
mass of CO = 0.138 g = 0.14 g
So the answer to our problem using 2 SF is 0.14 g of CO.