What mass of copper(II) sulfate would remain
after removing all the water from 78 g of
CuSO4 · 5H2O?
Answer in units of g.
by 
Anonymous
ANSWERS: 1
-
You should read your text book really, or ask your teacher! thats assuming your in school. Find the atomic mass of each element, Cu = 63.54 g/mol S = 32.06 g/mol O = 16 g/mol H = 1 g/mol Now find the molecular mass (g/mol) of the molecules CuSO4 and 5.H2O (5 water molecules) so CuSO4 = (63.54 + 32.06 + 4 x 16) = 159.6 g/mol 5 x (H2O) = 5 x (2 x 1 + 16) = 5 x (18) = 90 g/mol CuSO4.5H2O = 159.6 + 90 = 249.6 g/mol You want to know what percentage of the 72grams of CuSO4.5H2O is water. The water percentage of molecular weight of CuSO4.5H2O is: 90/249.6 = 0.36 or 36% so you will lose 36% of the weight 0.36 x 72 = 28.125g and be left with 72 - 28.125 = 43.875g CuSO4
RELATED QUESTIONS
Copyright 2023, Wired Ivy, LLC