Question

A student, Ken, is given a mixture containing two carbonate compounds . The mixture includes MgCO3 and (NH4)2CO3 . The mixture is 64.16% CO3 is by mass. What is the mass percent of MgCO3 in the mixture?

Answer 1

The mass percent of MgCO3 in the mixture is 18.33%.

In the mixture we have:

MgCO3 + (NH4)2CO3 but the percentage by mass of carbonate CO3 is 64.16%

Hence,

Total mass = 100 grams

Let the mass of MgCO3 = x grams

Let the mass of (NH4)2CO3 = y grams

So we have,       x + y = 100 ------------(1)

If the mass of CO3 = 64.16% = 64.16 g

But the molar mass of CO3 = 60 g/mol

Then,

Number of moles of CO3 = 64.16 g/60 g.mol-1 = 1.069 moles

Since we obtained the number of moles of CO3 from CO3 in MgCO3 and (NH4)2CO3.

Therefore,

moles MgCO3 + moles (NH4)2CO3 = 1.069

mass of MgCO3 present/molar mass MgCO3 + mass of (NH4)2 CO3 present/molar mass = 1.069

So,

x/84 + y/96 = 1.069---------(2)

From equation 1;

 y =100 - x  

x/84 + (100-x)/96 = 1.069

x = 18.33 g

Mass of MgCO3 = 18.33 g

Mass percent of MgCO3 = 18.33%

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