This includes numerical problems such as conversions (e.g. between mass, moles, volumes, pH, etc), Avogadro’s number, making and diluting solutions of different concentrations, balancing equations, stoichiometry problems, limiting reagent problems, problems related to equilibria, and thermodynamics.
These are skills you should have learned in Chem. Prin. lecture and labs, Organic labs, Instrumental lecture and labs, and Physical chemistry.
Here are 4 exams (with answers) that have been used in the past, which indicate the expected level of ability.
They are similar to the questions (with answers) listed on the following web page from the American Chemical Society (ACS).
http://www.acs.org/content/acs/en/education/students/highschool/olympiad/pastexams.html
You should focus only on the numerical and calculation type of the ACS questions above.
For example…
Local Exam 2011 Questions # 7, 8, 9, 12, 14, 18, 20, 21, 36, 52.
Local Exam 2001 Questions # 10, 11, 12, 13, 17, 24, 27, 28, 29, 34
The link currently works, but if it changes the info can be found from the ACS page using the following progression:
American Chemical Society > Education > Students > High School > Chemistry Olympiad > Past Exams
Here are four more illustrative examples (solved answers are at the end).
(EASY) How many grams of KCl do we need to make 1.0 L of a 1.0M KCl solution?
(MEDIUM) If 10 mL of a 1.0 M KCl solution is added to 1.0 L of water, then what is the concentration of KCl?
(HARDER) A 1.50 mL sample of sulfuric acid is titrated with 1.47M sodium hydroxide. The endpoint is reached with 23.70 mL. What is the molarity of the sulphuric acid?
(HARDEST) A mineral containing iron(II) sulfide but no other sulfides is treated with excess hydrochloric acid to produce hydrogen sulfide. If a 3.15 g sample of the mineral yields 448mL of hydrogen sulfide gas (measured at 0.0oC and 760 mmHg), what is the mass percentage of iron(II) sulfide in the sample?
You will require a calculator. Other electronic devices, like cell phones or iPads, are not acceptable.
ANSWERS
(EASY) How many grams of KCl do we need to make 1.0 L of a 1.0 M KCl solution?
Moles of KCl we need is obtained from knowing that 1M = 1 mole per liter, so 1L of 1M is a total of 1 mole.
How many grams of KCl contain 1 mole ? The molecular weight of a compound tells you how many grams are required for 1 mole of that compound. The MW for KCl is 74.6 g/mol. You can get this from the periodic table: K = 39.1 and Cl = 35.5).
Therefore 74.6 g of KCl contains 1 mole of KCl.
So 74.6g of KCl dissolved in 1.0 L of solvent will generate 1.0L of a 1.0M solution.
75g is the correct answer to 2 sig.figs.
(MEDIUM) If 10 mL of a 1.0 M KCl solution is added to 1.0 L of water, then what is the concentration of KCl?
Total number of moles of KCl is 10/1000 x 1 = 0.010 moles of KCl
This number of moles is in 1.01 L of water, so the concentration (in moles per liter) is:
0.01 / 1.01 = 0.0099 M (or 9.9mM) correctly to 2 sig. figs.
(HARDER) A 1.50 mL sample of sulfuric acid is titrated with 1.47M sodium hydroxide. The endpoint is reached with 23.70 mL. What is the molarity of the sulphuric acid?
You need to know the formulae for sulphuric acid and sodium hydroxide. (H2SO4 and NaOH):
And write their reaction equation H2SO4 + NaOH => Na2SO4 + H2O
And make sure it is balanced. H2SO4 + 2NaOH => Na2SO4 + 2H2O
The number of moles of NaOH is 23.7/1000 x 1.47 = 0.034839 moles.
From the balanced equation we note that there is 1 mole of H2SO4 for every 2 moles of NaOH ,
So number of moles of H2SO4 is 0.034839 / 2 = 0.0174195 moles.
If there is 0.0174195 moles in 1.5mL of H2SO4, then the Molarity is 0.0174195 x 1000/1.5 = 11.6 M correctly to 3 sig. figs.
(HARDEST )A mineral containing iron(II) sulfide but no other sulfides is treated with excess hydrochloric acid to produce hydrogen sulfide. If a 3.15 g sample of the mineral yields 448mL of hydrogen sulfide gas (measured at 0.0oC and 760 mmHg), what is the mass percentage of iron(II) sulfide in the sample?
FeS => H2S
PV = nRT, rearrange so n = PV/RT
(R is in units of L.atm.mol-1.K-1 so remember to convert pressure to atm, Temp to K, and Volume to L)
P= 760/760, V = 448/1000, R = 0.082, T = 0+273.17
Therefore n = 0.02 moles of H2S, which also means 0.02 moles of FeS.
The MW of FeS is (55.85 + 32.06) = 87.91 g.mol-1
So the mass of FeS is 0.020 x 87.91 = 1.76g
The Mass Percentage of Fes is 100 x 1.76/3.15 = 56% to 2 sig. figs.