[huskywolf]

Hi,

If I have a tablet with 830 mg MgCaCO₃ + 40mg MgCO₃ ,how many moles of CO₃ is in the tablet?

Is it 0.83g/60g mol^-1 =0.0138 mol
+           0.04g/60g mol^-1 = 0.0007  ,
so the answer is 0.0138 + 0.0007 =0.0145 Please?

[sjb]

Hi,

If I have a tablet with 830 mg MgCaCO₃ + 40mg MgCO₃ ,how many moles of CO₃ is in the tablet?

Is it 0.83g/60g mol^-1 =0.0138 mol
+           0.04g/60g mol^-1 = 0.0007  ,
so the answer is 0.0138 + 0.0007 =0.0145 Please?

Not quite. How many moles of MgCO₃ (assuming a typo in the first instance) are there in 830 mg? How many moles of CO₃^2- ?

[huskywolf]

Sorry a mistake in the first part,

 830 mg CaCO3 + 40mg MgCO3 ,how many moles of CO3 is in the tablet?

n(CaCO3)=0.83g/100g mol^-1 = 0.0083 mol

and

0.04g/84gmol^-1 = 0.000476 mol

so final answer is  0.0088 mol ? please

[DrCMS]

Hi,

If I have a tablet with 830 mg MgCaCO₃ + 40mg MgCO₃ ,how many moles of CO₃ is in the tablet?

Is it 0.83g/60g mol^-1 =0.0138 mol
+           0.04g/60g mol^-1 = 0.0007  ,
so the answer is 0.0138 + 0.0007 =0.0145 Please?

No, you can not just divide the weights of calcium carbonate and magnesium carbonate by the weight of the carbonate ion to get the moles of carbonate, what happened to the weight of calcium and magnesium?

You posted  while I was writing.

Sorry a mistake in the first part,

 830 mg CaCO3 + 40mg MgCO3 ,how many moles of CO3 is in the tablet?

n(CaCO3)=0.83g/100g mol^-1 = 0.0083 mol

and

0.04g/84gmol^-1 = 0.000476 mol

so final answer is  0.0088 mol ? please

Yes that is the correct answer and the correct way to work it out.

[huskywolf]

Great , thanks for the help