Post a New Question

Chemistry. Percent mass...

posted by .

A 1.009 g mixture of the solid salts Na2SO4 (molar mass= 142.04) an Pb(No3)2 (molar mass= 331.20) form an aqueous solution with the precipitation of PbSO4 (molar mass= 303.26). The precipitate was filtered and dried and its was was found to be .471 g. The limiting reagent is Na2SO4.

How many moles and grams of Na2SO4 are in the reaction?

I found this answer to be .22 g or 1.55 x 10-3 mol.

How many moles and grams of Pb(NO3)2 reacted in the mixture?

I found this answer to be .514 g or 1.55 x 10-3 mol.

What is the percent by mass of each salt in the mixture?

I am confused on this part....


If you could check the answers that I gave, and help with the last part I would REALLY REALLY appreciate it! Thank you very much!

  • Chemistry. Percent mass... -

    Watch carefully because this can be confusing. Look carefully at what the question asks.


    How many moles and grams of Na2SO4 are in the reaction?
    1.55 x 10^-3 mol Na2SO4 is correct. I obtained 0.2206 g which I would round to 0.221 for grams Na2SO4. Note the questions asks for g Na2SO4 in the REACTION.

    How many moles and grams of Pb(NO3)2 reacted in the mixture?

    You are correct with 1.55 x 10^-3 mols Pb(NO3)2 and 0.514 g Pb(NO3)2 but note that the question asks for mols and grams that REACTED. It didn't ask for mols and grams Pb(NO3)2 that are present in the original 1.009 grams sample. Since Na2SO4 is the limiting reagent, that limits the amount of Pb(NO3)2 that reacts BUT some additional Pb(NO3)2 CAN be present.

    What is the percent by mass of each salt in the mixture?

    %Na2SO4 = (mass Na2SO4/mass sample)*100 = (0.2206/1.009)*100 = approximately 20%.

    %Pb(NO3)2 = (mass Pb(NO3)2/mass sample)*100 = [(1.009-0.221)/1.009] = about 80%

    Note that you could have determined % Pb(NO3)2 simply by 100 - %Na2SO4.
    But also note that you CANNOT determine %Pb(NO3)2 by (0.514/1.009)*100 BECAUSE the 0.514 g is the amount of Pb(NO3)2 that reacted and not the amount of Pb(NO3)2 in the sample at the beginning. If you do it this way you get about 51% and obviously the two percentages must add to 100%.

  • Chemistry. Percent mass... -

    •0.471g / 303.26g/mol = 1.553x10^-3 mol PbSO4
    • moles of Na2SO4 = 1.553x10^-3 —> 0.2206 g (total in the sample)
    • moles of Pb(NO3)2 reacting = 1.553x10^-3 —> 0.51435g
    • 1.009 g - 0.2206 g = 0.7884 g Pb(NO3)2 total in the sample
    • % Na2SO4 in the sample was: (100)(0.2206 g)/1.009g = ___?
    • % Pb(NO3)2 in the sample = [setup] = ___?

  • Chemistry. Percent mass... -

    That makes sense! I was confused because my answers werent adding up to 100% but you explained it. Thank you!!

  • Chemistry. Percent mass... -

    What is the molecular form of the equation for the reaction? and what is the net ionic equation for the reaction? Thanks!!

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question