How many grams of magnesium cyanide are required to make this solution?

Question

How many grams of magnesium cyanide are required to make 275 mL of 0.075 molarity solution?

Answer 1

I like to always start these problems with this formula and plug in the information:

M = mol / L (M = molarity, mol = number of moles, and L = liters)

0.075 = mol / 0.275 L

(0.275) 0.075 = mol / 0.275 L (0.275)

0.02 = mol

Then convert the number or moles (mol) to the mass (g)

0.2 mol x (76.34 g / 1 mol) = 1.6 g Mg (CN)₂

Answer 2

Magnesium cyanide is an ionic composed of...magnesium and cyanide! Magnesium is Mg in the second column the periodic table, meaning it loses two electrons to form Mg²⁺. Cyanide is a polyatomic ion composed of one carbon and one nitrogen with a negative one charge: CN^1-.

Is any of this relevant? Yes! These come together in a ratio of 1:2 to form a neutral compound: Mg(CN)₂. The molar mass would be the sum of the molar masses of the elements: 24.31 + 2(12.01 + 14.01) g/mol = 76.35 g/mol. If we didn't get the ratio of ions, we'd get a totally different molar mass, which would change our final answer significantly.

To form 0.075 molarity, you'd need 0.075 mol of solute in 1 L of solution.

Here, we only need 275 mL (0.275 L) of solution, so we'd need less than 0.075 mol of solute:

x mol / 0.275 L = 0.075 M

x mol = (0.075 M)(0.275 L) = 0.020625 mol Mg(CN)₂.

Finally, we're looking for grams of Mg(CN)₂, so we convert using molar mass:

0.020625 mol Mg(CN)₂ * 76.35 g/mol = 1.6 g Mg(CN)₂ (2 significant figures)