City College, Chemistry Department
Chemistry 10301, sections E*, Prof. T. Lazaridis
First Midterm exam, Oct 3, 2011
Last Name: _____________________________________________
First Name: _____________________________________________
SSN last 4: ______________________________________________
Note: There are 6 questions in this exam. Fill in your answer in the blank space provided
immediately following each question. Half a point will be subtracted every time you report
a numerical result with an incorrect number of significant figures. A copy of the periodic
table is attached. Good luck!
1. (30)
a. (4) Give the atomic symbols for the following elements:
gold :
Au
sodium:
Na
copper:
Cu
nitrogen:
N
b. (4) What is the chemical formula of potassium nitrate?
KNO3
c. (4) What is the name of the compound HClO4 ?
perchloric acid
d. (4) What is the molar mass of the above compound ?
1.008 + 35.45 + 4X16.00= 100.46 g/mol
e. (4) Give the symbol of an isotope with 22 protons and 26 neutrons.
48
22
Ti
€
or
48
Ti
f. (4) What is the name of the compound CH3CH2COOH ?
propanoic acid
g. (6) Express the density of 0.80 g/mL in lb/in3
[1 in = 2.54 cm, 1 lb = 453.6 g]
3
0.80 g/mL X (1mL/cm3) X (1 lb/453.6g) X (2.54 cm/in) = 0.029 lb/in3
2. (20) Balance the following chemical equations:
a. (5)
N2O5 +
b. (5)
CH4 +
c. (5)
d. (5)
H2O
-----> 2 HNO3
4 Br2 --->
CBr4
+
4 HBr
2 C5H10O2 + 13 O2 ---> 10 CO2 +
NCl3 + 3 H2O ----->
10 H2O
NH3 + 3 HOCl
3. (10) Determine the percentage composition of the lithium chloride
monohydrate.
LiCl*H2O
Molar mass= 6.941+35.45+2X1.008+16.00= 60.41 g/mol
Li: 6.941/60.41 X 100% = 11.49%
Cl: 35.45/60.41 X 100% = 58.68%
H: 2X1.008/60.41 X 100% = 3.337%
O: 16.00/60.41 X 100% = 26.49%
4. (10) Complete combustion of octane proceeds as follows:
2 C8H18 (l) + 25 O2 (g) ----->
16CO2 (g) + 18 H2O (l)
a) (4) How many moles of O2 are needed to burn 1.50 mol of octane?
1.50 mol octane X 25/2 = 18.8 mol O2
b) (6) How many grams of O2 are needed to burn 1.50 g of octane?
Molar mass: C8H18: 8X12.01+18X1.008= 114.22 g/mol O2: 32.00 g/mol
1.50 g octane / (114.22 g/mol) X 25/2 X 32.00 g O2/mol = 5.25 g O2
5. (15) Combustion analysis of 2.78 mg of a compound containing C,H, and
O produces 6.32 mg of CO2 and 2.58 mg of H2O.
a. (10) What is the empirical formula of the compound?
6.32 mg CO2 X 12.01/44.01 = 1.72 mg C
2.58 mg H2O X 2X1.008/18.02= 0.289 mg H
2.78-1.72-0.289 = 0.77 mg O
Convert to moles and then divide by smallest:
C: 1.72 X 10-3 g/12.01 g/mol = 0.143 mol
/0.048 = 2.98 ≈ 3
H: 0.289 X 10-3 g/1.008 g/mol = 0.287 mol
/0.048 = 5.98 ≈ 6
O: 0.77 X 10-3 g/16.00 g/mol = 0.048 mol
/0.048 = 1
So, empirical formula is C3H6O
b. (5) If the molar mass of the compound is 116.6 g/mol, what is the
molecular formula?
formula mass= 3X12.01+6X1.008+16.00= 58.08 ,
This is half of the molar mass.
So, molecular formula: C6H12O2
6. (15) Consider the production of bromobenzene from benzene and
bromine:
C6H6 + Br2  C6H5Br + HBr
The reacting mixture contains 30.0 g of benzene and 65.0 g of bromine.
a. (5) Which compound is the limiting reactant?
Molar Masses: C6H6: 78.11 , Br2: 2X79.90= 159.9 , C6H5Br: 157.00 g/mol
Convert to moles:
C6H6 : 30.0/78.11 = 0.384 mol
Br2 : 65.0/159.8 = 0.407 mol
Since stoichiometry is 1:1, C6H6 is limiting
b. (5) What is the theoretical yield for bromobenzene?
0.384 mol C6H6  0.384 mol C6H5Br X 157.00 g/mol = 60.3 g
c. (5) You conduct the experiment and you obtain 56.7 g of
bromobenzene. What was the percent yield?
56.7/60.3 X 100% = 94%