Question

Your lab partner accidentally mixed some sodium chloride with your sample of Epsom salts (MgSO₄･7H₂O). You want to make a standard solution of magnesium ion, and this is the only sample of a magnesium salt you have. To determine the amount of magnesium salt in the mixture, you heat 100.00 g to drive off the water of hydration and find that the anhydrous mixture has a mass of 56.75 g. How many grams of the original salt mixture must you add to 1.000 L of water to make a 0.100 M solution of Mg²⁺ ion? g

Answer 1

Answer:

$\large \boxed{\text{16.68 g}}$

Explanation:

Let x = the mass of MgSO₄·7H₂O

and y = the mass of NaCl

and z = the mass of MgSO₄

Then, you have two conditions:

$\begin{array}{lrcl}(1) & x + y & = &100.00 & \\(2) & z + y &= & 56.75 & \\\end{array}$

That's two conditions but three unknowns. You need a third condition.

Fortunately, you are both an excellent chemist and mathematician. You know that

Mᵣ:          247.47             120.37

          MgSO₄·7H₂O ⟶ MgSO₄ + 7H₂O

m/g:             x                        z

$\dfrac{x}{z} = \dfrac{247.47}{120.37}\\\\x = z \times \dfrac{247.47}{120.37} = 2.056z$

This is your third condition.

1. Calculate the mass of MgSO₄

$\begin{array}{lrcll}(1) & x + y & =& 100.00 & \\(2) & z + y & = &56.75& \\(3) & x& = & 2.056z & \\(4) & 2.056z + y &= & 100.00 & \text{Substituted (3) into (1)}\\ & 1.056z & = & 43.25 & \text{Subtracted (2) from (4)}\\ & z & = & 40.96 &\text{Divided each side by 1.056} \\\end{array}\\\text{The mixture contains 40.96 g of MgSO _{4} }.$

2. Calculate the moles of MgSO₄ needed for the solution

$c = \dfrac{n}{V}\\n & = &Vc & = & \text{1.000 L}\times \dfrac{\text{0.1000 mol}}{\text{1 L}} = \text{0.1000 mol}\\\\\text{You need 0.1000 mol of MgSO}_{4}$

3. Calculate the mass of MgSO₄

$m = \text{0.1000 mol} \times \dfrac{\text{120.37 g}}{\text{1 mol}} = \text{12.04 g}\\\\\text{You need 12.04 g of MgSO}_{4}$

4.Calculate the mass of the mixture required.

$\text{Mass of mixture} = \text{12.04 g MgSO}_{4} \times \dfrac{\text{56.75 g mixture}}{\text{40.96 g MgSO}_{4}} = \textbf{16.68 g mixture}\\\text{You need \large \boxed{\textbf{16.68 g}} of the mixture.}$