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DEPARTMENT OF CHEMISTRY AND CHEMICAL TECHNOLOGY
GENERAL CHEMISTRY
202-NYA-05 21, 22
TEST 1
24-SEPT-2012
INSTRUCTOR: I. DIONNE
Print your name: _____________________________________________
Answers
INSTRUCTIONS:
Answer all questions in the space provided.
1.
2.
3.
4.
5.
6.
7.
Duration of this test is 75 minutes.
No books or extra paper are permitted.
Answer the questions in ink in order to preserve the right to grieve.
In order to obtain full credit for your answers, you must clearly show your work.
Calculators may not be shared. Programmable calculators are not permitted.
Your attention is drawn to the College policy on cheating. This policy will be enforced.
A Periodic Table with constants is provided.
Problem 1:
/3
Problem 6:
/2
Problem 2:
/3
Problem 7:
/2
Problem 3:
/8
Problem 8:
/2
Problem 4:
/2
Problem 9:
/2
Problem 5:
/6
Problem 10:
/6
Total:
/36
(3 marks)
PROBLEM 1
A monatomic ion has a charge of +4. The nucleus of the ion has a mass number of 34. The
number of neutrons in the nucleus is equal to the number of electrons in Ca 2+ ion. Write the
A
X symbol for the ion and indicate the number of protons, neutrons, and electrons.
Z
34
16
+4
# protons =
16
______
S
symbol
18
# neutrons = ______
12
# electrons = ______
Mass number = # protons + # neutrons
The # neutrons = # electrons in Ca2+ ion = 18
The # protons = mass number − # neutrons = 34 − 18 = 16
Charge is +4, it therefore contains 12 electrons
PROBLEM 2
(3 marks)
Fill in the blanks:
Ra
(a) The symbol of the heaviest alkaline earth metal is ______.
14
(b) The atomic number of the lightest metalloid in Group 4A is ______.
(c) Group 1B consists of the coinage metals. The name of the coinage metal whose atoms
copper
have the fewest electrons is ________________.
79.90 amu
(d) The atomic mass of the halogen in Period 4 is _____________.
(e) The name of the monatomic ion formed when hydrogen gains an electron is the
hydride
________________
ion.
(f) The members of Group 8A are known as the ________________________.
noble gases
Page |2
(8 marks)
PROBLEM 3
Name or write the formula for each of the following compounds.
(a) FeSO4
iron(II) sulfate
(b) lithium nitride
Li3N
(c) sodium carbonate
Na2CO3
(d) CoO
cobalt(II) oxide
(e) HIO4 (aq)
periodic acid
(f) hydrobromic acid
HBr (aq)
(g) disulfur decafluoride
S2F10
(h) PbCl2∙3H2O
lead(II) chloride trihydrate
PROBLEM 4
(2 marks)
Chlorine has two naturally occurring isotopes, 35Cl and 37Cl. The 37Cl isotope has a 24.23 %
abundance and an isotopic mass of 36.9659 amu. If chlorine has an atomic mass of 35.4527
amu, what is the isotopic mass of the other isotope?
The
35
Cl isotope has an abundance of: 100.00 − 24.23 = 75.77%
35.4527 amu = (0.7577)(isotopic mass of 35Cl) + (0.2423)(36.9659 amu)
35.4527 amu = (0.7577)(isotopic mass of 35Cl) + 8.9568376 amu
35.4527 amu − 8.9568376 amu = (0.7577)(isotopic mass of
35
Cl)
26.495862 amu = (0.7577)(isotopic mass of 35Cl)
isotopic mass of
35
Cl = 26.495862 / 0.7577 = 34.968804 amu
34.9688
The 35Cl isotope has an isotopic mass of _________________
amu.
Page |3
PROBLEM 5
(6 marks)
A compound was found to contain only three elements, C, H, and Cl. When a 1.50 g sample of
the compound was completely combusted in air, 3.52 g of CO2 was formed. In a separate
experiment the chlorine in a 1.00 g sample of the compound was converted to 1.27 g of AgCl.
Determine the empirical formula of the compound.
3.52 g CO2 / 44.01 g/mol = 0.07999818 mol CO2
Since all C is converted to CO2, the 1.50 g sample contains 0.07999818 mol C
0.07999818 mol C x 12.01 g/mol = 0.9606 g C
(0.9606 g C / 1.50 g sample) x 100 = 64.04 % C
1.27 g AgCl / 143.35 g/mol = 0.0088594 mol AgCl
Since all Cl is converted to AgCl, the 1.00 g sample contains 0.0088594 mol Cl
0.0088594 mol Cl x 35.45 g/mol = 0.31407 g Cl
(0.31407 g Cl / 1.00 g sample) x 100 = 31.41 % Cl
% H = 100.00 − (64.04 + 31.41) = 4.55 % H
Assume 100.0 g
64.04 g C / 12.01 g/mol = 5.3322 mol C
31.41 g Cl / 35.45 g/mol = 0.8860 mol Cl
4.55 g H / 1.008 g/mol = 4.5139 mol H
smallest ratio
C: 5.3322 / 0.8860 = 6.01
Cl: 0.8860 / 0.8860 = 1.00
H: 4.5139 / 0.8860 = 5.09
The empirical formula is C6H5Cl
C6H5Cl
The empirical formula is _________________
.
Page |4
(2 marks)
PROBLEM 6
One molecule of the antibiotic known as penicillin G has a mass of 5.342 x 10 −21 g. What is the
molar mass of penicillin G?
Molar mass = g/mol
one molecule has a mass of 5.342 x 10−21 g
One mole contains 6.022 x 1023 molecules
molar mass is: 5.342 x 10−21 g /molecules x 6.022 x 1023 molecules/mol = 3.217 x 103 g/mol
The molar mass of penicillin G is
3217 g/mol
_________________
.
PROBLEM 7
(2 marks)
Narceine is a narcotic in opium. It crystallizes from water solution as a hydrate that contains
10.8% water by mass. If the molar mass of narceine hydrate is 499.52 g/mol, determine x in
narceine∙xH2O.
10.8% = [(x)(mass of H2O) / (mass of compound) x 100
0.108 = (x) (18.016 g/mol) / 499.52 g/mol
x = 0.108 x 499.52 g/mol / 18.016 g/mol = 2.994
x is ___
3 .
Page |5
(2 marks)
PROBLEM 8
(a) What is the empirical formula for each of the following compounds?
i.
CH2O
C 3 H6 O 3
ii. Ga2(SO4)3
Ga2(SO4)3
(b) What is the molecular formula of a compound which has a molar mass of 32.05 g/mol and
the empirical formula NH2?
The mass of NH2 is 16.026 g/mol
This unit fits: 32.05 / 16.026 = 1.999 times = 2
2 x NH2 = N2H4
N 2H 4
The molecular formula is _________________
.
PROBLEM 9
(2 marks)
Write balanced equations for each of the following by inserting the correct coefficients in the
blanks:
(a)
16
__Cu(s)
8
+ __S8(s) →__Cu
2S(s)
(b)
__CH
2 2(g)
9 2(g) →__CO
10 2O(g) + __N
4
4
3NH2(g) + __O
2(g) + __H
Page |6
(6 marks)
PROBLEM 10
Phosphorus pentachloride reacts with water to give phosphoric acid and hydrochloric acid.
PCl5 (s) + H2O (l) → H3PO4 (aq) + HCl (aq)
If 191.56 g of PCl5 is mixed with 51.89 g of water.
(a) Calculate the theoretical yield of HCl (in grams).
(b) How many grams of the excess reactant are left over after the reaction is complete?
(c) What is the percent yield if 125.51 g of HCl are produced?
PCl5 (s) + 4H2O (l) → H3PO4 (aq) + 5HCl (aq)
191.56 g PCl5 /208.22 g/mol = 0.91999 mol PCl5
51.89 g H2O / 18.016 g/mol = 2.88022 mol H2O
If PCl5 is limiting: 5 x 0.91999 = 4.59995 mol HCl will form.
If H2O is limiting: 5/4 x 2.88022 = 3.600275 mol HCl will form.
Smallest amount of product formed is due to H2O being the limiting reagent.
Therefore 3.600275 mol x 36.458 g/mol = 131.3 g HCl will form
Excess reagent = amount of PCl5 present at start − amount of PCl5 used during the reaction
At start: 0.91999 mol
Amount used: 2.88022 mol H2O used up x 1/4 = 0.720055 mol PCl5 used up
Excess: 0.91999 mol − 0.720055 mol = 0.199935 mol PCl5
0.199935 mol x 208.22 g/mol = 41.64 g
Percent yield: (actual yield / theoretical yield) x 100
(125.51 g / 131.3 g) x 100 = 95.59 %
Theoretical yield of HCl
131.3 g
____________
g
41.64 g
Excess reactant left over ____________
g
% yield of HCl
95.59 %
____________
%
Page |7