A solution of copper(II) sulfate that contains 15.00% CuSO4 by mass has a density of 1.169 g/mL. A 2 — Stoichiometry / Titration Chemistry Question
Problem Context
A solution of copper(II) sulfate that contains 15.00% CuSO4 by mass has a density of 1.169 g/mL. A 25.0 mL portion of this solution was reacted with excess concentrated ammonia to form a dark blue solution. When cooled, filtered and dried, 6.127 g of a dark blue solid were obtained. A 0.195 g sample of the solid was analyzed for ammonia by titrating with 0.1036 M hydrochloric acid solution, requiring 30.63 mL to reach the equivalence point. A 0.150 g sample was analyzed for copper (II) by titrating with 0.0250 M EDTA, (which reacts with Cu2+ in a 1:1 ratio). The endpoint was reached after 24.43 mL of the EDTA were added. A 0.200 g sample was heated at 110 ˚C to drive off water, producing 0.185 g of the anhydrous material.
Determine the molarity of Cu2+ ions in the original solution.
Model Answer
1.169 g / mL × 0.1500 g CuSO4 / 1.000 g solution × 1 mol / 159.62 g = 1.099 M
Find the number of moles of Cu2+ in the 25.0 mL portion.
Model Answer
1.099 mol / L × 0.0250 L = 0.0275 mol CuSO4
Calculate the percentages by mass in the prepared compound of;
i. NH3
Model Answer
0.1036 mol / L × 0.03063 L × 1 mol NH3 / 1 mol HCl = 0.003173 mol NH3
0.003173 mol NH3 × 17.034 g / mol = 0.05405 g NH3
% NH3 = (0.05405 g NH3 / 0.195 g sample) × 100 = 27.7%
ii. Cu2+
Model Answer
0.02443 L EDTA × 0.0250 mol / L × 1 mol Cu2+ / 1 mol EDTA = 6.11 × 10^-4 mol Cu2+
6.11 × 10^-4 mol Cu2+ × 63.55 g / mol = 0.03881 g Cu
% Cu = (0.03881 g Cu / 0.150 g sample) × 100 = 25.9%
iii. H2O
Model Answer
0.200 g compound - 0.185 g anhydrous compound = 0.015 g H2O
% H2O = (0.0150 g H2O / 0.200 g sample) × 100 = 7.5%
iv. SO4 2-
Model Answer
% SO4 = 100 - (27.7 + 25.88 + 7.5) = 38.92 %
Use the results in part c. to determine the formula of the compound.
Model Answer
Assume 100 g; calculate moles using molar mass; divide all results by the smallest number
27.7 g NH3 × (1 mol / 17.03 g) = 1.627 / 0.405 = 4.02
25.88 g Cu × (1 mol / 63.55 g) = 0.407 / 0.405 = 1.00
7.5 g H2O × (1 mol / 18.02 g) = 0.416 / 0.405 = 1.03
38.9 g SO4 × (1 mol / 96.1 g) = 0.405 / 0.405 = 1
Based on these results, we can see the molecular formula is Cu(NH3)4SO4·H2O whose molar mass is 245.28 g/mol
Assuming that Cu2+ is the limiting reactant in the synthesis, determine the percent yield.
Model Answer
0.0275 mol × 245.28 g / mol is the theoretical yield = 6.76 g
% yield = (6.127 g / 6.76 g) × 100 = 90.7%