For each of the following scenarios, explain what effect the error will have on
ID: 526916 • Letter: F
Question
For each of the following scenarios, explain what effect the error will have on the percentage of water in the hydrate. Will the calculated value be greater than, less than, or the same as the true percentage? Explain your reasoning. You initially weigh (before heating) the crucible with the cover and the hydrate. After heating the sample, your lab partner weighs the content of the crucible without the cover. You weigh (before heating) the crucible with the cover and the hydrate. You begin to gently heat the crucible with the hydrate. Your lab partner is too anxious to leave the lab early. They begin to vigorously heat the sample and some of the hydrate splatters out of the crucible. You lower the heat and continue the experiment with this sample. A 35.64 g sample of a hydrate of barium bromide (BaBr_2 * XH_2O) is obtained. The sample is heated to drive off the water. The dry sample has a mass of 30.42 g of barium bromide. What is the formula for the hydrate?Explanation / Answer
Q1.
a
If the cover is not included, then the mass he will weight will be much lower
therefore, the difference in mass will be LARGER than the real one, therefore hydrate mass stoichiometric ratio will be higher
the ratio is higher (falsely)
b
If there is splatter of the cruibile, then assume some mass of the hydrate is also lost, therefore the final mass will be lower, meaning tha thte % will be incorrectly lower
Q2.
m = 35,64 g of sample
dry m,ass = 30.42
dm = 35.64-30.42 = 5.22 g
then
mol of BaBr2 = mass/MW = 5.22/297.14 = 0.0175674
mol of water = (35.64-30.42)/18 = 0.29 mol
ratio = 0.29/0.0175674 = 16
x = 16 H2O
BaBr2 * 16H2O
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.