Percentage Purity of magnesium oxide

Question

1.10 g of impure magnesium oxide (MgO) was dissolved in 25.0 mL of 3.20 mol L–1 hydrochloric acid. The resulting solution was diluted to 250 mL in a volumetric flask. 20.0 mL of this diluted solution required 24.9 mL of 0.102 mol L–1 sodium hydroxide to reach the end-point. Find the percentage purity of the magnesium oxide sample.

The equations for the reactions are:

MgO(s) + 2H+(aq) → Mg2+(aq) + H2O(l)

H+(aq) + OH–(aq) → H2O(l)

Answer 1

In Chemistry, this is known as a back-titration problem. What this means is that the HCl that reacts with the MgO is present in excess, and then whatever HCl is left over AFTER reacting with MgO is titrated with NaOH to determine amount in excess. Subtracting this amount from original amount HCl used gives us the amount HCl that reacted with the MgO.

mols HCl originally present = 25.0 ml x 1 L / 1000 ml x 3.20 mol / L = 0.080 mols HCl

Back titration equation: NaOH + HCl ==> NaCl + H2O

mols NaOH used = 24.9 ml x 1 L / 1000 ml x 0.102 mol / L = 0.00254 mols

mols HCl in excess = 0.00254 mols NaOH x 1 mol HCl / mol NaOH x 12.5 = 0.03175 mols HCl in excess

Why did we multiply by 12.5? Because diluted the sample to 250 mls and then only assayed 20 mls. 250/20 = 12.5 so there will be 12.5 x as many moles in the 250 mls as in the 20 mls.

mols HCl used to react with MgO = 0.080 - 0.03175 mols =0.04825 mols HCl used to react with MgO

Since the original reaction between MgO and HCl is

MgO + 2HCl ==> MgCl₂ + O₂, the ratio of HCl to MgO is 2:1

moles MgO originally present = 0.04825 / 2 = 0.0241 mols MgO

To find % purity (%MgO in original sample), we need the grams of MgO present:

molar mass MgO = 40.3 g / mol

grams MgO present = 0.0241 mols MgO x 40.3 g / mol = 0.971 g MgO

Percent purity = 0.971 g MgO / 1.10 g (x100%) = 88.3% pure

(please be sure to check all of the math)

Answer 2

Hi Tayla, I can provide some assistance with this task.

First, let's determine the number of moles of HCl that reacted with the impure MgO:

Moles of HCl used = (Concentration of HCl) * (Volume of HCl used) = (3.20 mol/L) * (25.0 mL) = 0.08 mol

To ascertain the moles of MgO, let's refer to the balanced chemical equation for the MgO and HCl reaction. According to this equation, 1 mole of MgO reacts with 2 moles of HCl.

Therefore, the moles of MgO in the impure sample can be calculated as:

Moles of MgO = 0.08 mol / 2 = 0.04 mol

Next, let's determine the moles of MgO present in the 20 mL of the diluted solution that reacted with NaOH to reach the endpoint. We'll use the equation for the reaction between H+ and OH-:

H+ (aq) + OH- (aq) --> H2O (l)

This balanced equation indicates that one mole of H+ reacts with one mole of OH-. Since 20 mL of the diluted solution required 24.9 mL of 0.102 mol/L NaOH to reach the endpoint, we can calculate the moles of H+ in the solution:

Moles of H+ in the diluted solution = (0.102 mol/L) * 0.0249 L = 0.00254 mol

Because each mole of H+ corresponds to one mole of MgO, the moles of MgO in the 20.0 mL of diluted solution are also 0.00254 mol. Now, let's find the percentage purity of the magnesium oxide sample:

Percentage Purity = (Moles of pure MgO / Moles of impure MgO) × 100 = (0.00254 mol / 0.040 mol) = 6.35%

Thus, the percentage purity of the magnesium oxide sample is approximately 6.35%. This indicates that the impure magnesium oxide sample contains roughly 6.35% pure magnesium oxide.

I hope that was helpful!