Question

A sample of raw mining ore contains a hydrated salt called copper sulfate tetrahydrate, CuSO4.4H2O, along with other impurities. If 10.854g of the ore loses 2.994g of water when heated strongly, what is the mass percentage of the salt CuSO4.4H2O in the raw ore sample?

Answer 1

Answer:

88.5 % of the raw ore is CuSO4*4H2O

Explanation:

Step 1: Data given

Mass of the ore = 10.854 grams

Mass of water = 2.994 grams

Step 2: Calculate moles H2O

Moles H2O = mass H2O / molar mass H2O

Moles H2O = 2.994 grams / 18.02 g/mol

Moles H2O = 0.166 moles

Step 3: Calculate moles CuSO4*4H2O

For 1 mol CuSO4*4H2O we have 4 moles H2O

For 0.166 moles H2O we have 0.166/4 = 0.0415 moles CuSO4*4H2O

Step 4: Calculate mass CuSO4*4H2O

Mass CuSO4*4H2O = moles * molar mass

Mass CusO4*4H2O = 0.0415 moles * 231.67 g/mol

Mass CuSO4*4H2O = 9.61 grams

Step 5: Calculate mass % of CuSO4*4H2O

Mass % = (9.61 grams / 10.854 grams )*100 %

Mass % = 88.5 %

88.5 % of the raw ore is CuSO4*4H2O