2-13 12 ANALYSIS OF A DOUBLE SALT NAME 1315 Section_ Pre-Lab Assignment, Week 1

Question

2-13 12 ANALYSIS OF A DOUBLE SALT NAME 1315 Section_ Pre-Lab Assignment, Week 1 Note: The data used here are for ilustrative purposes. The data you obtain wil be dflerent A. Record all mass measurements made in the sulfate determination. Report the mass of double salt used and the mass of barium sulfate obtained. EXAMPLE: The beaker weighs 97.033 g: the beaker and double salt weigh 98 111 g: the mass of double salt is 1.078 g: the filter paper weighs 0.956 g: the fiter paper and barium sulate weigh 2.178 g; the mass of barium sulfato is 1.222 g PRACTICE: The beakor weighs 103.285 g, the beaker and double salt weigh 104.397g the fillor paper weighs 0972 g: the fiter paper and barium sulfate weigh 2.318 g Use the above results to determine the mass of sulfate in the barlum sulfate sample. EXAMPLE: 0.5030g PRACTICE Use the above results to determine the mass percent of sulfate in EXAMPLE: 46.66% PRACTICE: the double salt. B. Record all mass measurements made in the water determination. Report the mass of double salt used and the mass of water lost on heati ng. EXAMPLE The crucible weighs 11.034 g, the crucible and double salt weigh 12.245 g: the mass of double salt is 1.212 g: the crucible and anhydrous double salt (after heating) weigh 12.001 g: the mass of water lost on hoating is 0.245 g PRACTICE: The crucible weighs 9.691 g: the crucible and double salt weigh 10.816 g: the crucible and anhydrous double salt weigh 10.675 g. Use the above results to determine the mass percent of water in the double salt EXAMPLE: 20.2 % PRACTICE

Explanation / Answer

A) Mass of beaker = 103.285 g.

Mass of beaker plus double salt = 104.397 g.

Mass of filter paper = 0.972 g.

Mass of filter paper plus barium sulfate = 2.318 g.

Mass of double salt = (mass of beaker plus double salt) – (mass of beaker) = (104.397 g) – (103.285 g) = 1.112 g …..(1)

Mass of barium sulfate = (mass of filter paper plus barium sulfate) – (mass of filter paper) = (2.318 g) – (0.972 g) = 1.346 g ……(2)

Molar mass of barium sulfate = (1*137.327 + 1*32.065 + 4*15.999) g/mol = 233.388 g/mol.

Mole(s) of barium sulfate corresponding to 1.346 g barium sulfate = (1.346 g)/(233.388 g/mol) = 5.7672*10-3 mole.

We know that

1 mole barium sulfate = 1 mole sulfate ion.

Therefore, 5.7672*10-3 mole barium sulfate = 5.7672*10-3 mole sulfate ion.

Molar mass of sulfate ion, SO42- = (1*32.065 + 4*15.999) g/mol = 96.061 g/mol.

Mass of sulfate in 1.346 g barium sulfate = (5.7672*10-3 mole)*(96.061 g/mol) = 0.5540 g ? 0.554 g (ans).

Mass percent of sulfate in the double salt = (mass of sulfate)/(mass of double salt)*100 = (0.554 g)/(1.112 g)*100 = 49.820% ? 49.82% (ans).

B) Mass of crucible = 9.691 g.

Mass of crucible plus double salt = 10.816 g.

Mass of crucible plus anhydrous double salt = 10.675 g.

Mass of double salt = (mass of crucible plus double salt) – (mass of crucible) = (10.816 g) – (9.691 g) = 1.125 g …..(1)

Mass of anhydrous double salt = (mass of crucible plus anhydrous double salt) – (mass of crucible) = (10.675 g) – (9.691 g) = 0.984 g ……(2)

Mass of water in the double salt = (mass of double salt) – (mass of anhydrous double salt) = (1.125 g) – (0.984 g) = 0.141 g.

Mass percent water in the double salt = (mass of water)/(mass of double salt)*100 = (0.141 g)/(1.125 g)*100 = 12.533% ? 12.53% (ans).

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