Solving Chemistry Problem II: 0.005 M Carbonate Solution

[higherme]

My question is:

Assuming the weight percent of CaCO3 is 50%, calculate the mass of unknown sample required to prepare a 15 mL solution that contains 0.005 M carbonate.

Well i know that in the sample of unknown, there half of them are CaCO3. I don't know what do to next...help please

Thank you

 

[physstudent1]

if you mean the mass percent of CaCO3 is 50% in an unknown sample then, try getting to how many moles of CaCO3 you would need to make a .005 M solution of Carbonate in 15mL

 

[higherme]

that means i would need:
0.015 L x 0.005 mol/L = 7.5 E -5 moles of carbonate which means i neeed 7.5 E-5 moles of CaCO3

then... ?

 

[eli64]

keep going, how do you get mass (grams) from moles? a great equation to know is MM (molar mass) = g/mol

then you have mass of CaCO3, if its 50% (or 1/2) of the mass (x) your unknown sample ...