7. A student conducts an experiment on a different hydrate. The empty crucible is heated to a constant
mass of 24.330. A sample of the unidentified hydrate is placed in the crucible, and the total mass is
31.571 g. The crucible and hydrate are heated to constant mass, which is measured at 29.002 g. The
molar mass of the anhydrous salt is 159.609 g.
a. Calculate the formula of this hydrate. Note that in the formula of a hydrate, salt•xH2O, the
coefficient “x” must be a whole number.
b. Give a reason your calculated value of “x” is less than the whole number to which you
rounded.
8. A student dehydrates a sample of magnesium sulfate heptahydrate. The anhydrous salt to water
ratio is 1:6.913. The correct formula is salt•7H2O.
a. Calculate the student’s percent error.
b. What is the physical meaning of the negative sign in the answer?
moles of salt=(29.002-24.330)/159.609
Moles of hydrate:(31.571-29.002)/18
now find the ratio of hydratemoles to salt moles:
x=moleshydrate/molessalt
To solve these problems, we'll need to use the concept of hydrates and stoichiometry. Let's break down the questions step by step:
7. To calculate the formula of the hydrate, we need to determine the ratio of the moles of water to the moles of the anhydrous salt. Here's how you can do it:
1. Calculate the mass of the anhydrous salt by subtracting the mass of the empty crucible from the total mass of the crucible and hydrate:
Mass of anhydrous salt = Total mass - Mass of empty crucible
= 31.571 g - 24.330 g
= 7.241 g
2. Calculate the number of moles of anhydrous salt using its molar mass:
Moles of anhydrous salt = Mass of anhydrous salt / Molar mass of anhydrous salt
= 7.241 g / 159.609 g/mol
≈ 0.045 moles
3. Calculate the mass of water lost by subtracting the mass of the anhydrous salt from the constant mass measured after heating:
Mass of water lost = Constant mass after heating - Mass of anhydrous salt
= 29.002 g - 7.241 g
= 21.761 g
4. Calculate the number of moles of water lost by dividing the mass of water lost by the molar mass of water (18.015 g/mol):
Moles of water lost = Mass of water lost / Molar mass of water
= 21.761 g / 18.015 g/mol
≈ 1.207 moles
5. Finally, determine the mole ratio between the anhydrous salt and water by dividing the moles of water lost by the moles of anhydrous salt:
Mole ratio = Moles of water lost / Moles of anhydrous salt
= 1.207 moles / 0.045 moles
≈ 26.82
Since the coefficient "x" in the formula of a hydrate must be a whole number, we round 26.82 to the nearest whole number. In this case, the formula of the hydrate is salt•27H2O.
Now let's move on to question 8.
8. To calculate the student's percent error, we'll compare their calculated ratio of anhydrous salt to water with the correct ratio. Here's how you can do it:
The correct ratio given is 1:6.913, which means 1 part anhydrous salt to 6.913 parts water.
1. Calculate the student's ratio of anhydrous salt to water by taking the reciprocal of the given ratio:
Student's ratio = 1 / 6.913
≈ 0.1446
2. Calculate the percent error using the formula:
Percent error = | (Correct value - Experimental value) / Correct value | * 100%
Percent error = | (0.1446 - 1) / 1 | * 100%
= | -0.8554 / 1 | * 100%
≈ 85.54%
The negative sign in the answer indicates that the student's experimental value (0.1446) is lower than the correct value (1). It suggests that the student's calculated ratio is significantly smaller compared to the expected value, resulting in a large percent error.
I hope this explanation helps you understand the steps involved in solving these problems!