Copper (II) sulfate pentahydrate, CuSO4 ·5 H2O, (molar mass 250 g/mol) can be dehydrated by repeated heating in a crucible. Which value is closest to the percentage mass of water lost from the total mass of salt in the crucible when the crucible undergoes repetitive heatings until a constant mass is reached?36%26%13%25%

Question

Chemistry

Orr

24 April, 06:15

Copper (II) sulfate pentahydrate, CuSO4 ·5 H2O, (molar mass 250 g/mol) can be dehydrated by repeated heating in a crucible. Which value is closest to the percentage mass of water lost from the total mass of salt in the crucible when the crucible undergoes repetitive heatings until a constant mass is reached?36%26%13%25%

+5

Answers (1)

Know the Answer?

Nathalia Rowe · Answer 1

The water lost is 36% of the total mass of the hydrate

Explanation:

Step 1: Data given

Molar mass of CuSO4*5H2O = 250 g/mol

Molar mass of CuSO4 = 160 g/mol

Step 2: Calculate mass of water lost

Mass of water lost = 250 - 160 = 90 grams

Step 3: Calculate % water

% water = (mass water / total mass of hydrate) * 100 %

% water = (90 grams / 250 grams) * 100% = 36 %

We can control this by the following equation

The hydrate has 5 moles of H2O

5*18. = 90 grams

(90/250) * 100% = 36%

(160/250) * 100% = 64 %

The water lost is 36% of the total mass of the hydrate