Question

Atomic mass of magnesium

Would you expect the atomic mass to be higher lower the same as the expected value if the vapor pressure of water had been ignored? Explain in a way i can understand please

Answer 1

Ans. Balanced Reaction:       Mg(s) + 2 HCl(aq) -----------> MgCl₂(aq) + H₂(g)

For the determination of molar mass of Mg-

Step 1. The amount (volume) of H₂ gas liberated is measured. Let it be V liters.

Step 2. From the volume of H₂ gas liberated, and pressure & temperature of the lab, the number of moles of H₂ gas is calculated using ideal gas equation, PV = nRT.

Let the number of moles of H₂ calculated be M1 moles.

Step 3: In balanced equation above, 1 mol Mg produces 1 mol H₂ gas. So, moles of Mg taken during experiment = M1 mol, too.

Step 4: Molar mass of Mg = Mass of Mg sample taken / M1 moles         - equation 2

# ERROR when water vapor (vapor pressure) is NOT accounted:

The total volume of gas after reaction includes volume of H₂ gas present plus the volume of water vapor in the vessel.

That is,

            Actual volume of gas = Volume of H₂ gas liberated + Volume of water vapor

So, if you don’t account water vapor, your assumed the volume of H₂ gas equal to “Volume of H₂ gas liberated + Volume of water vapor”. That is, the volume of H₂ gas taken is greater than actual volume of H₂ gas liberated.

If you put a “higher than actual volume of H­₂” in ideal gas equation, you get higher number of moles of gas because V is proportional to number of moles.

Now, if you put “higher than actual number of moles” in equation 2, the calculated molar mass of Mg would be lesser than the actual because molar mass is inversely proportional to molar mass.

Therefore, the calculated molar mass would be lesser than the actual value.