Heating copper(II) sulfate pentahydrate will evaporate the water from the crystals, leaving anhydrous copper sulfate, a white powder. Anhydrous means “without water”. What mass of anhydrous copper(II) sulfate would be produced by heating 500.0 g of CuSO 4 · 5H 2 O?
The answer is
% CuSO 4 = 63.92%
% H 2 O = 36.08%
Mass of anhydrous copper(ii) sulfate = 319.6g
Im not sure how to solve this though
500 g CuSO4.5H2O x 0.6392 g CuSO4/1 g CuSO4.5H2O) = ? g CuSO4
CuSO4 = 63.5 + 32 + 4*16 = 159.5 grams/mol
5 H2O = 5 ( 2+16) = 90 grams / mol
so CuSO4 5H2O = 249.5 g/mol
so
CuSO4/total = 0.639 or 63.9 %
H2O/total = 0.361 or 36.1 %
To solve this problem, you need to calculate the amount of water present in the hydrated copper sulfate and then calculate the mass of anhydrous copper sulfate remaining after the water is evaporated.
Let's break down the steps:
1. Calculate the molar mass of CuSO4·5H2O:
Molar mass of CuSO4 = 63.55 g/mol (atomic mass of Cu) + 32.07 g/mol (atomic mass of S) + (4 * 16.00 g/mol) (4 oxygens) = 159.61 g/mol
Molar mass of H2O = 2 * 1.01 g/mol (atomic mass of H) + 16.00 g/mol (atomic mass of O) = 18.02 g/mol
Molar mass of CuSO4·5H2O = (159.61 g/mol) + (5 * 18.02 g/mol) = 249.69 g/mol
2. Calculate the number of moles of CuSO4·5H2O in 500.0 g:
Number of moles = Mass / Molar mass
Moles of CuSO4·5H2O = 500.0 g / 249.69 g/mol = 2.002 mol
3. Calculate the number of moles of H2O in CuSO4·5H2O:
Moles of H2O = Moles of CuSO4·5H2O * (5 * 18.02 g/mol / 249.69 g/mol) = 2.002 mol * 0.3613 = 0.723 mol
4. Calculate the mass of H2O:
Mass of H2O = Moles of H2O * Molar mass of H2O = 0.723 mol * 18.02 g/mol = 13.04 g
5. Calculate the mass of anhydrous CuSO4:
Mass of anhydrous CuSO4 = Mass of CuSO4·5H2O - Mass of H2O = 500.0 g - 13.04 g = 486.96 g
Therefore, the mass of anhydrous copper(II) sulfate produced by heating 500.0 g of CuSO4·5H2O is 486.96 g.
To solve this problem, you need to understand the concept of percent composition and use it to calculate the masses of copper(II) sulfate and water in the given compound.
Here's how you can approach this problem step-by-step:
Step 1: Determine the molar mass of the compound.
The molar mass of CuSO4·5H2O can be obtained by adding the atomic masses of all its constituent elements:
Cu (Copper) = 63.55 g/mol
S (Sulfur) = 32.07 g/mol
O (Oxygen) = 16.00 g/mol
H (Hydrogen) = 1.01 g/mol
O (Oxygen) = 16.00 g/mol
Since there are 4 oxygen atoms in the sulfate group and 10 hydrogen atoms in the water molecules, the molar mass of CuSO4·5H2O is calculated as follows:
Molar mass = (63.55 g/mol) + (32.07 g/mol) + (4 × 16.00 g/mol) + (5 × (2 × 1.01 g/mol) + 5 × 16.00 g/mol)
= 159.61 g/mol + 64.08 g/mol + 20.20 g/mol + 80.00 g/mol
= 323.89 g/mol
Step 2: Calculate the percent composition of copper(II) sulfate.
The percent composition is the mass percentage of each element in the compound. You can calculate it using the formula:
% composition = (mass of element / molar mass of compound) × 100
For copper(II) sulfate:
% CuSO4 = (mass of CuSO4 / molar mass of CuSO4·5H2O) × 100
Substituting the given values:
% CuSO4 = [(63.55 g/mol) / (323.89 g/mol)] × 100
= 19.63% (approximately)
Step 3: Calculate the percent composition of water.
Using the same formula as above, for water:
% H2O = (mass of H2O / molar mass of CuSO4·5H2O) × 100
Substituting the given values:
% H2O = [(5 × (2 × 1.01 g/mol) + 5 × 16.00 g/mol) / (323.89 g/mol)] × 100
= 80.37% (approximately)
Step 4: Calculate the mass of anhydrous copper(II) sulfate.
Since heating the compound removes the water, the percentage of water (36.08%) can be subtracted from 100% to find the percentage of anhydrous copper(II) sulfate:
% anhydrous CuSO4 = 100% - % H2O
= 100% - 36.08%
= 63.92% (approximately)
To calculate the mass of anhydrous copper(II) sulfate, multiply the total mass of the compound (500.0 g) by the percentage of anhydrous CuSO4:
Mass of anhydrous copper(II) sulfate = (63.92/100) × 500.0 g
= 319.6 g
Therefore, by heating 500.0 g of CuSO4·5H2O, you would obtain 319.6 g of anhydrous copper sulfate.