What mass of calcium citrate, Ca3(C6 H5 O7)2, would a tablet manufacturer need to use to provide the same amount of Ca that you found in your tablet? (show calculations)
The mass of Calcium for the experiment was .7374g
We assumed that any solid that formed when Na2CO3 was added was CaCO3 . This is not necessarily true. There may have been other cation in the tablet that could also have reacted with Na2CO3 to form a precipitate. Give any two other cations that might have have been present in the tablet that would have formed a precipitate when you added Na2CO3 solution. There are many correct answers to this question.
Are you sure that was 0.7374 g Ca? or was it CaCO3. The answer below is for 0.7374 g Ca.
0.7374 x (molar mass Ca citrate/atomic mass Ca) = ?
For other metals look for other carbonates that are insoluble; ;i.e., Mg might be one.
To calculate the mass of calcium citrate needed to provide the same amount of calcium found in the tablet, we need to determine the molar mass of calcium citrate and use it to convert the mass of calcium in the tablet.
The molar mass of calcium citrate, Ca3(C6H5O7)2, can be calculated by adding up the molar masses of each element in the formula:
Molar mass of Ca = 40.08 g/mol
Molar mass of C = 12.01 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of O = 16.00 g/mol
Now, let's calculate the molar mass of calcium citrate:
(3 x Molar mass of Ca) + (2 x Molar mass of C) + (10 x Molar mass of H) + (14 x Molar mass of O)
= (3 x 40.08 g/mol) + (2 x 12.01 g/mol) + (10 x 1.01 g/mol) + (14 x 16.00 g/mol)
= 120.24 g/mol + 24.02 g/mol + 10.10 g/mol + 224.00 g/mol
= 378.36 g/mol
Now, since the molar mass of calcium citrate is 378.36 g/mol, we can set up a proportion to find the mass of calcium citrate needed:
Mass of calcium citrate / Molar mass of calcium citrate = Mass of calcium / Molar mass of calcium
Mass of calcium citrate = (Mass of calcium / Molar mass of calcium) x Molar mass of calcium citrate
Substituting the given values:
Mass of calcium citrate = (0.7374 g / 40.08 g/mol) x 378.36 g/mol
Mass of calcium citrate = 13.607 g
So, a tablet manufacturer would need to use approximately 13.607 grams of calcium citrate to provide the same amount of calcium found in your tablet.
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Regarding the second question, here are two examples of other cations that might have been present in the tablet and would have formed a precipitate when Na2CO3 solution was added:
1. Iron (III) cation - Fe3+
2. Copper (II) cation - Cu2+
These cations could potentially react with Na2CO3 to form precipitates such as Fe2(CO3)3 and CuCO3. However, it is important to note that there could be various other cations present in the tablet, and these examples are just a couple of possibilities.