Examkin

Soru

Zorluk: KolayPercentage Composition and Percentage Purity Calculations

What is the percentage by mass of water of crystallization in copper(II) tetraoxosulfate(VI) pentahydrate (CuSO4⋅5H2O\text{CuSO}_4 \cdot 5\text{H}_2\text{O})? [Cu=64,S=32,O=16,H=1][\text{Cu} = 64, \text{S} = 32, \text{O} = 16, \text{H} = 1]

  1. A
    7.2%7.2\%
  2. 36.0%36.0\%Cevap
  3. C
    56.3%56.3\%
  4. D
    64.0%64.0\%

Cevap

The percentage by mass of water of crystallization in CuSO4⋅5H2O\text{CuSO}_4 \cdot 5\text{H}_2\text{O} is 36.0%36.0\%.
The total molar mass of CuSO4⋅5H2O\text{CuSO}_4 \cdot 5\text{H}_2\text{O} is 250 g/mol250\text{ g/mol} and the mass contributed by the five water molecules is 90 g/mol90\text{ g/mol}. Dividing 9090 by 250250 and multiplying by 100%100\% yields 36.0%36.0\%.

Adım Adım Çözüm

1
Calculate the molar mass of water (H2O\text{H}_2\text{O}) and the total mass of five moles of water.
Molar mass of H2O=(2×1)+16=18 g/mol\text{H}_2\text{O} = (2 \times 1) + 16 = 18\text{ g/mol}. For 5H2O5\text{H}_2\text{O}, mass =5×18=90 g/mol= 5 \times 18 = 90\text{ g/mol}.
Water of crystallization in the formula consists of five water molecules per formula unit.
2
Calculate the total molar mass of hydrated copper(II) tetraoxosulfate(VI) (CuSO4⋅5H2O\text{CuSO}_4 \cdot 5\text{H}_2\text{O}).
Molar mass =64+32+(4×16)+90=64+32+64+90=250 g/mol= 64 + 32 + (4 \times 16) + 90 = 64 + 32 + 64 + 90 = 250\text{ g/mol}.
The percentage composition must be based on the complete formula weight of the hydrated compound.
3
Calculate the percentage by mass of water of crystallization.
Percentage of H2O=(90250)×100%=36.0%\text{Percentage of } \text{H}_2\text{O} = \left(\frac{90}{250}\right) \times 100\% = 36.0\%.
Percentage composition by mass is the mass of the component divided by the total molar mass of the compound multiplied by 100.

Anahtar Kavram

Percentage Water of Crystallization in Hydrated Salts
Bu soruyu puanla

Konular

Percentage Composition and Percentage Purity Calculations

Bu konudan daha fazla soru