Magnesium phosphate is used to minimize the effects of vitamin E deficiency. A 3.80L solution has a magnesium ion concentration of 0.798mol/L. Calculate the mass of solute needed to prepare this solution.
1 month ago
Start with a dissociation equation for magnesium phosphate. Dissociation occurs when an ionic compound breaks up into its ions.
Mg3(PO4)2 (s) --> 3Mg2+(aq) + 2PO43-(aq)
The next step is to list all the information you know from the question:
Volume = 3.80 L
Concentration = 0.798 mol/ L Mg2+
m = ? g Mg3(PO4)2
Molar mass = 262.87 g/mol Mg3(PO4)2
There are two ways to solve this question.
The first way is to use unit analysis (arrange the question so that all the units will 'cancel' out):
3.80 L x (0.798 mol Mg2+/L) x (1 mol Mg3(PO4)2 / 3 mol Mg2+) x (262.87 g/mol Mg3(PO4)2) = 265.708...g Mg3(PO4)2
The second way is to use formulas:
1.) concentration = moles/ volume
Manipulate this formula to find moles (moles = concentration x volume)
moles of Mg2+ = 3.80 L x 0.798 mol Mg2+/L = 3.0324 mol Mg2+
2.) mole ratio (from the balanced chemical formula)
3.0324 mol Mg2+ x (1 mol Mg3(PO4)2 / 3 mol Mg2+) = 1.0108 mol Mg3(PO4)2
3.) number of moles = (mass of substance/ molar mass)
Manipulate the formula to find mass of substance (mass of substance = number of moles x molar mass)
mass of Mg3(PO4)2 = 1.0108 mol Mg3(PO4)2 x 262.87 g/mol Mg3(PO4)2 =265.708...g Mg3(PO4)2
The final answer is the mass of magnesium phosphate required is 266 g Mg3(PO4)2 (this is considering the fact there is three significant digits from the question so the question is rounded to three digits).