Practice term test #1

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Term test #1

Time for writing: 90 minutes

Total Marks: 48

Name: __________________
ID No.___________________

Read each question carefully. Show all work. No partial credit can be given if work is not completely shown.

1. [6] Supply the missing name or formula:

tin(IV) sulfate

Sn(SO4)2

calcium hypochlorite

Ca(OCl)2

iron (II) sulfide

FeS

aluminum oxide or aluminum(III) oxide Al2O3

potassium permanaganate  KMnO4

ammonium hydrogen sulfate or ammonium bisulfate NH4HSO4

2. [9] Write the balanced reaction for each of the following reactions. For those reactions that occur in solution, write down a net ionic reaction, if one exists. For any redox reaction, also include the oxidation and reduction reactions. If no reaction occurs write NO REACTION. Classify each reaction as REDOX, METATHESIS or DECOMPOSITION WITHOUT REDOX. Show the phases for all of the reactants and products.

i) A dilute solution of aqueous ammonia was added to an aqueous solution of copper (II) sulfate.

Cu2+ (aq) + 2OH- (aq) --> Cu(OH)2 (s)

(metathesis)

ii) Hydrogen chloride gas is bubbled through an aqueous solution of sodium carbonate.

CO32- (aq) + 2H+ (aq) --> CO2(g) + H2O (l)
(decomposition without redox)

iii) Solid gallium (III) sulfite is strongly heated in a vacuum.

2Ga(SO3)3 (s) --> Ga2O3 (s) + 3SO2 (g)
(decomposition without redox)

3. [4] Give the formal oxidation state for the underlined atom in each of the following chemical species:

i) I2O4

+4

ii) Na3VO4

+5

iii) KO2

-1/2

iv) CoF63-

+3

4. [4] Calculate the number of electrons in 1.00 mL of water at 298 K.

1.00 mL of water has a mass of 1.00 g at 298K (data sheet)

n(water) = 1.00 g/18.02 g/mol = 0.0555 mol

N(water molecules) = 0.0555 mol X (6.022 X 1023 mol-1) = 3.34 X 1022

each water molecule has 8 + 2(1) = 10 electrons

hence, total in 1.00 g is 3.34 X 1023

5. [5] Lead carbonate, 24.3 g, was combined with 0.350 mol of HCl in sufficient water to bring the total volume to 0.750 L.

a) Give the balanced net ionic equation for the reaction that occurs.

PbCO3 (s) + 2H+ (aq) + 2Cl- (aq) --> CO2 (g) + H2O(l) + PbCl2 (s)

 

b) Find the concentrations of the ions at the completion of the reaction.

n PbCO3 (s) = 24.3 g / 267.2 g/mol = 0.0909 mol

This consumes 2 X 0.0909 mol of HCl = 0.182 mol of HCl

The remaining HCl is: 0.350 mol - 0.182 mol = 0.168 mol

Similarily, 2 moles of Cl- are consumed for each mole of PbCl2 formed.

[H+] = [Cl-] = 0.168 mol/0.750 L = 0.224 mol/L

 

6. [6] Traumatic acid is composed of carbon, hydrogen and oxygen, and has a molecular weight of 198.26 g/mol. The combustion of 3.080 mg of traumatic acid produced 7.519 mg of carbon dioxide and 2.520 mg of water. Determine the molecular formula for traumatic acid.

First find the masses of C, H and O in the sample:

mass of carbon = (mass fraction of C in CO2)(mass of CO2)
= ( (12.011 g/mol)/(44.02 g/mol) ) X (7.519 mg) = 2.0519 mg

mass of hydrogen = ( (2 X 1.008 g/mol) / (18.02 g/mol) ) X 2.520 mg = 0.2824 mg

mass of oxygen = total sample mass - mass of C - mass of H = 3.080 mg - 2.0519 mg - 0.2824 mg = 0.7457 mg

Divide these masses by the atomic weights of C, H and O respectively to give the moles of each, and then divide each of these by the lowest number of moles, to give the empirical formula:

n (carbon) = 1.71 X 10-4 mol
n (hydrogen) = 2.80 X 10-4 mol
n (oxygen) = 4.68 X 10-5 mol

Dividing each by 4.68 X 10-5 gives:

C3.66H6O1

Multiply by three to get whole numbers

Thus, the empirical formula is C11H18O3

This has a molar mass of ~198 g/mol which is the same as the molecular mass, hence there is only one empirical formula in the molecular formula.

Thus, the molecular formula is C11H18O3

 

7. [8] Fill in each blank with the appropriate answer.

i) nitric acid is a/an OXIDIZING (oxidizing/reducing) acid.

ii) the highest oxidation state that phosphorus can achieve is +5

iii) The number of moles of oxygen atoms in 0.0083 mol of uranyl nitrate hydrate UO2(NO3)2. 6H2O 0.116 mol

iv) Most main group metals lose electrons in order to achieve a noble gas configuration, have a full valence shell etc.

v) Describe the following ionic compounds as water soluble or water insoluble:

a) scandium chloride SOLUBLE

b) cesium carbonate SOLUBLE

c) Pb(OH)2 INSOLUBLE

d) NH4ClO4 SOLUBLE

8. [6] A 0.336 g sample of impure NaClO2 was treated with 25.00 mL of 0.113 M Ti4+ solution. The excess Ti4+ was back-titrated with 16.87 mL of 0.0183 M Mn2+ solution.

The relevant unbalanced reactions are:

ClO2- + Ti4+ + H2O --> Ti3+ + ClO3- + H+

Mn2+ + Ti4+ + H2O --> MnO2 + Ti3+ + H+

 

a) Balance the equations for the two reactions.

ClO2- + 2Ti4+ + H2O --> 2Ti3+ + ClO3- + 2H+

Mn2+ + 2Ti4+ + 2H2O --> MnO2 + 2Ti3+ + 4H+

b) Find the number of moles of NaClO2.

moles of Ti4+ = 0.025 L X 0.113 mol/L = 0.00283 mol
moles of Mn2+ used in back titration = (0.01687 L)(0.0183 mol/L) = 0.000308 mol

this reacts with 2(0.000308 mol) of Ti4+ = 0.000617 mol (this was the excess Ti4+)

thus, the moles of NaClO2 = 1/2(0.00283 mol - 0.000617 mol) = 0.00111 mol

c) Calculate the mass percent of NaClO2 ,in the original impure sample.

mass percent is the mass of 0.00111 mol of NaClO2 divided by the total sample mass, i.e.:

mass% = ( (0.00111 mol X 90.4 g/mol) / (0.336 g) ) X 100 % = 29.8 %

Bonus question: [2] For use in semiconductor devices, silicon crystals are prepared in which only one atom out of every billion is an impurity. If such a crystal weighs one mg, how many foreign atoms does it contain?

n (Si) = (1 X 10-3 g) /( 28.086 g/mol) = 3.56 X 10-5 mol

The number of impurity atoms is one billionth of the number of atoms in this many moles:

N (impurity) = ( (3.56 X 10-5 mol)(6.022 X 1023 /mol) ) / (1 X 109)

= 2.14 X 1010