Fifteen Examples | Problems #1 - 10 | Problems #21 - 35 | Return to Mole Table of Contents |
Determine empirical formula when given mass data
Determine empirical formula when given percent composition data
Determine identity of an element from a binary formula and a percent composition
Determine identity of an element from a binary formula and mass data
Determine molecular formula using the Ideal Gas Law
The example and problem questions only, no solutions
Problem #11: 6.9832 g of FeSO4 ⋅ xH2O is dissolved in water acidified with sulfuric acid. The solution is made up to 250. cm3. 25.00 cm3 of this solution required 25.01 cm3 of 0.0200 M KMnO4 to titrate completely. Calculate x.
Solution:
1) Fe2+ gets oxidized by the MnO4-. The products are Fe3+ and Mn2+
Fe2+ ---> Fe3+ + e-
5e- + 8H+ + MnO4- ---> Mn2+ + 4H2Owhich leads to:
8H+ + 5Fe2+ + MnO4- ---> 5Fe3+ + Mn2+ + 4H2O
2) The key is the 5 to 1 molar ratio between ferrous ion and permanganate. What I want to get is the number of moles of ferrous ion in the 25.00 cm3. That will get me to grams of FeSO4 in the solution. A subtraction will give me the water in the hydrate.
moles KMnO4 ---> (0.0200 mol/L) (0.02501 L) = 0.000500 molfor every one mole of permanganate, five moles of ferrous are oxidized.
0.000500 mol of MnO4¯ oxidizes 0.0025 of ferrous ion
3) That's the moles in 25.00 cm3. We originally had 250. cm3, so 0.0250 mol total of dissolved FeSO4 ⋅ xH2O
0.0250 moles of anhydrous FeSO4 weighs ---> 0.025 mol times 151.906 g/mol = 3.79765 gwater in the hydrate ---> 6.9832 g − 3.79765 g = 3.18555 g
moles of water ---> 3.18555 g / 18.015 g/mol = 0.17683 mol
4) I want this molar ratio:
0.0250 to 0.17683in smallest whole number terms where the FeSO4 part is 1:
1 to 7.0732
5) Close enough for this:
FeSO4 ⋅ 7H2O
Problem #12: A 81.4 gram sample of BaI2 ⋅ 2H2O was heated thoroughly in a porcelain crucible, until its weight remained constant. After heating, how many grams of the anhydrous compound remained?
Solution:
1) Dehydration of the hydrated barium iodide salt is shown by this reaction:
BaI2 ⋅ 2H2O(s) ---> BaI2(s) + 2H2O(g)
2) The following "dimensional analysis" set-up uses the concepts of mass-mass stoichiometry:
1 mol BaI2 ⋅ 2H2O 1 mol BaI2 391.1 g BaI2 81.4 g BaI2 ⋅ 2H2O x ––––––––––––––––– x ––––––––––––––– x –––––––––– = 74.5 g BaI2 427.1 g BaI2 ⋅ 2H2O 1 mol BaI2 ⋅ 2H2O 1 mol BaI2
3) A brief explanation of the steps:
(a) 81.4 g BaI2 ⋅ 2H2O ---> the starting mass(b) (1 mol BaI2 ⋅ 2H2O / 427.1 g BaI2 ⋅ 2H2O) ---> divide by the molar mass of BaI2 ⋅ 2H2O
(c) (1 mol BaI2 / 1 mol BaI2 ⋅ 2H2O) ---> the 1:1 molar ratio
(d) (391.1 g BaI2 / 1 mol BaI2) ---> multiply by the molar mass of BaI2
Problem #13: If the hydrated compound UO2(NO3)2 ⋅ 9H2O is heated gently, the water of hydration is lost. If you heat 4.05 g of the hydrated compound to dryness, what mass of UO2(NO3)2 will remain?
Solution:
Calculate how many moles of UO2(NO3)2 ⋅ 9H2O you have in 4.05 gWhen you heat it, you have only UO2(NO3)2 remaining. 1 mole of UO2(NO3)2 ⋅ 9H2O leaves 1 mole of UO2(NO3)2.
Multiply moles of UO2(NO3)2 by the molar mass of UO2(NO3)2 to get the mass.
Problem #14:
Problem #15: How many grams of water and anhydrous salt would you get when heating 9.42 g of Fe(NO3)3 ⋅ 9H2O?
Solution:
1) Determine mass percentages of Fe(NO3)3 and H2O:
mass of one mole of Fe(NO3)3 ⋅ 9H2O ---> 403.9902 gdecimal percent by mass of Fe(NO3)3 ---> 241.857 g / 403.9902 g = 0.59867
decimal percent by mass of water ---> 1 − 0.59867 = 0.40133
2) Determine masses of water and anhydrous salt:
anhydrous salt ---> (9.42 g) (0.59867) = 5.64 g
water ---> 9.42 g − 5.64 g = 3.78 g
Problem #16: Heating 0.695 g CuSO4 ⋅ nH2O gives a residue of 0.445 g. Determine the value of n.
Solution:
I copied this problem from an "answers" website because the person gave a correct solution in the question. That is, a correct solution right up until the end of the solution.
0.695 − 0.445 = 0.25 g of H2O0.25 g H2O / 18 g/mol = 0.014 mol
0.445 g CuSO4 / 159.5 g/mol = 0.0028 mol
0.0028 mol / 0.014 mol = 0.2
Here's the error:
So since there's no such thing has 0.2 of a molecule I round it to a whole number which in this case is 0 to get:0CuSO4 ⋅ 1H2O
The correct technique is to multiply by 5 so as to get a 1 in front of the CuSO4:
0.2CuSO4 ⋅ 1H2O times 5 equals this:CuSO4 ⋅ 5H2O
Comment: the student who made this mistake failed to see the coefficients of 0.2 and 1 as representing moles, focusing only on the ratio as molecules. It is perfectly fine to have a molar ratio of 0.2 to 1, which then becomes 1 to 5, which is the lowest whole-number ratio of moles.
Problem #17: Epsom salt is MgSO4 ⋅ nH2O. The hydrate was found to contain 71.4% oxygen. Calculate the number of water molecules associated with each formula unit of magnesium sulfate hydrate.
Solution:
The molar mass of MgSO4 is 120.366 g/molThere are 4 moles of O in MgSO4
The grams of O in one mole of MgSO4 is 63.9976 g
Let x = grams of oxygen from H2O
therefore, grams of H2O is this:
(x) (18.015 grams of H2O/15.9994 grams O)
which is 1.126x
grams of oxygen divided by total weight of compound equals 0.714 (from information given in the problem)
(63.9976 + x) / (120.366 + 1.126x) = 0.714
63.9976 + x = 0.714[120.366 + 1.126x]
63.9976 + x = 85.94 + 0.804x
0.196x = 21.9424
x = 111.951 grams (of oxygen in entire MgSO4 ⋅ nH2O)
(111.951 g) (1.000 mole H2O/15.9994 g O) = 6.997 moles H2O
MgSO4 ⋅ 7H2O
Problem #18: A sample of hydrated magnesium sulphate, MgSO4 · nH2O, is found to contain 51.1% water. What is the value of n?
Solution #1 (the traditional way):
1) Assume 100 g of the compound is present. This allows the percentages to be easily converted to masses:
MgSO4 ---> 48.9 g
H2O ---> 51.1 g
2) Convert the masses to moles:
MgSO4 ---> 48.9 g / 120.366 g/mol = 0.406261 mol
H2O ---> 51.1 g / 18.015 g/mol = 2.836525 mol
3) Divide through by smallest:
MgSO4 ---> 0.406261 mol / 0.406261 mol = 1
2.836525 mol / 0.406261 mol = 6.98
4) Empirical formula:
MgSO4 · 7H2O
Solution #2:
51.1% is water so 48.9% must be MgSO4Assume one mole of MgSO4 is present. This represents 48.9% of the hydrate.
120.366 is to 48.9 as x is to 51.1
x = 125.78 g <--- mass of water in the hydrate
125.78 g / 18.015 g/mol = 6.98
MgSO4 · 7H2O
Problem #19: If a 9.15 g sample of a hydrated salt produced 6.50 g of anhydrous salt (309.650 g/mol) and 2.65 g of water (18.015 g/mol), what is the molecular mass of the hydrated salt?
Solution:
6.50 g / 309.650 g/mol = 0.021 mol
2.65 g / 18.015 g/mol = 0.147 molWe want to know how many moles of water are present when one mole of the anhydrous salt is present.
0.147 is to 0.021 as x is to 1
x = 7
molecular mass = 309.650 g/mol + [(7) (18.015 g/mol)] = 435.755 g/mol
Problem #20: A hydrated compound has the formula MCl2 · 2H2O. In an experiment, the following data were determined:
mass of the hydrate: 1.000 g
mass of H2O: 0.185 g
Determine the identity of element M from these results.
Solution:
1) The mass of the anhydrate is:
1.000 − 0.185 = 0.815 g
2) There are 2 mol H2O in 1 mol MCl2 · 2H2O.
Therefore 0.185 g is to 36.0 g/mol as 0.815 is to xx = 158.6 g/mol <--- this is the formula weight of MCl2
3) Determine the atomic weight of M and identify it:
Two Cl weigh 71.0 gTherefore M is 87.6 g/mol (based on 158.6 − 71.0)
M is strontium
Bonus Problem: What is the correct chemical formula for the hydrated crystal Mn(NO3)4 · nH2O if analysis shows 511 g of Mn(NO3)4 and 334 g of H2O?
Solution:
1) Determine moles of each:
511 g / 302.954 g/mol = 1.6867 mol
334 g / 18.015 g/mol = 18.54 mol
2) Seek smallest whole-number ratio:
1.6867 mol / 1.6867 mol = 1
18.54 mol1.6867 mol = 11Mn(NO3)4 · 11H2O
Fifteen Examples | Problems #1 - 10 | Problems #21 - 35 | Return to Mole Table of Contents |
Determine empirical formula when given mass data
Determine empirical formula when given percent composition data
Determine identity of an element from a binary formula and a percent composition
Determine identity of an element from a binary formula and mass data