1. First exam
Exercises

2. 1- Prefixes giga and deci represent, respectively:
First Exam/ Exercises
1- Prefixes giga and deci represent, respectively:
a) 10-9 and b) 106 and c) 103 and 10-3  d) 109 and 10-1.
2- If 586 g of bromine occupies 188 mL, what is the density of bromine in g/mL?
a) 3.12g/ml b) g/ml c) 3.63g/ml d) 3.08g/ml
d = m/V
m = 586 g, V = 188 mL
d = 586 / 188 = 3.12 g/ml

3. 1- Prefixes giga and deci represent, respectively:
First Exam/ Exercises
1- Prefixes giga and deci represent, respectively:
a) 10-9 and b) 106 and c) 103 and 10-3  d) 109 and 10-1.
2- If 586 g of bromine occupies 188 mL, what is the density of bromine in g/mL?
a) 3.12g/ml b) g/ml c) 3.63g/ml d) 3.08g/ml
nanometer contains how many micrometers?
a) 3.5x b) 3.5x c) 3.5x d) 3.5x 10-8
First convert from nm to m :
3.5 X 10-9m
Then from m to um
3.5 x 10-9 x 10+6 = 3.5 x 10-3

4. 1- Prefixes giga and deci represent, respectively:
First Exam/ Exercises
1- Prefixes giga and deci represent, respectively:
a) 10-9 and b) 106 and c) 103 and 10-3  d) 109 and 10-1.
2- If 586 g of bromine occupies 188 mL, what is the density of bromine in g/mL?
a) 3.12g/ml b) g/ml c) 3.63g/ml d) 3.08g/ml
nanometer contains how many micrometers?
a) 3.5x b) 3.5x c) 3.5x d) 3.5x 10-8
4- The element in group 2A and period 3 is:
a) Ga b) Be c) Al d) Mg

5. First Exam/ Exercises
5- Predict the formula of a compound formed from Sr and Cl?
a) Sr2Cl b) Sr2Cl c) SrCl d) SrCl
Sr+2
Cl-
SrCl2

6. 5- Predict the formula of a compound formed from Sr and Cl?
First Exam/ Exercises
5- Predict the formula of a compound formed from Sr and Cl?
a) Sr2Cl b) Sr2Cl c) SrCl d) SrCl
6- Which compound has the same empirical formula as C6H12O6?
a) C12H20O b) C2H8O c) C6H3O d) C12H24O12
C6H12O6 divided by 6 = CH2O
a- C12H20O4 divided by 4 = C3H5O
b- C2H8O4 divided by 2 = CH4O2
c- C6H3O6 divided by 3 = C2HO2
d- C12H24O12 divided by 12 = CH2O

7. 5- Predict the formula of a compound formed from Sr and Cl?
First Exam/ Exercises
5- Predict the formula of a compound formed from Sr and Cl?
a) Sr2Cl b) Sr2Cl c) SrCl d) SrCl
6- Which compound has the same empirical formula as C6H12O6?
a) C12H20O b) C2H8O c) C6H3O d) C12H24O12
7- An atom is
a) Smallest unit of matter that maintains its chemical identity.
b) The smallest unit of a compound.
c) Always made of carbon
d) Smaller than electron

8. a)Compounds are: N2, NH3 , H2O and CH4
First Exam/ Exercises
8-The right answer is:
a)Compounds are: N2, NH3 , H2O and CH4
b)Elements are: N2, H2O, H2 and F2
c) Molecules are: N2, H2, F2 and Cl2 and Fe2SO4
d) Compounds are: NH3, O2, H2O and CH4
9- Rubidium (Rb) and cesium (Cs) are members of which of the following categories?
a) Halogens b) Alkali metals c) Alkaline earth metals d) Noble gases
10- What is the number of protons, electrons and neutrons in the atom of
a) 32 protons, 34 electrons, 16 neutrons b) 16 protons, 16 electrons, 16 neutrons
c) 16 protons, 18 electrons, 16 neutrons d) 18 protons, 16 electrons, 32 neutrons

9. First Exam/ Exercises
10- What is the number of protons, electrons and neutrons in the atom of
a) 32 protons, 34 electrons, 16 neutrons b) 16 protons, 16 electrons, 16 neutrons
c) 16 protons, 18 electrons, 16 neutrons d) 18 protons, 16 electrons, 32 neutrons
P = 16, e = 16+2 = 18 , n = = 16

10. a)Compounds are: N2, NH3 , H2O and CH4
First Exam/ Exercises
8-The right answer is:
a)Compounds are: N2, NH3 , H2O and CH4
b)Elements are: N2, H2O, H2 and F2
c) Molecules are: N2, H2, F2 and Cl2 and Fe2SO4
d) Compounds are: NH3, O2, H2O and CH4
9- Rubidium (Rb) and cesium (Cs) are members of which of the following categories?
a) Halogens b) Alkali metals c) Alkaline earth metals d) Noble gases
10- What is the number of protons, electrons and neutrons in the atom of
a) 32 protons, 34 electrons, 16 neutrons b) 16 protons, 16 electrons, 16 neutrons
c) 16 protons, 18 electrons, 16 neutrons d) 18 protons, 16 electrons, 32 neutrons

11. 11- The correct systematic name for Cr2O3 is
First Exam/ Exercises
11- The correct systematic name for Cr2O3 is
a)Chromium (III) oxide b) Dichromium trioxide.
c) Chromium trioxide d) Chromium oxide.

12. First Exam/ Exercises Summery of naming compound Ionic Molecular
Cation: metal or NH4+
Anion: monotomic or polytomic
Nonmetal + nonmetal
Nonmetal + metalloid
Cation has only
one charge
Cation has more than
one charge
Pair Form one type
of compound
Pair Form more than
one type
of compound
Name first element
add ide to the name of second element
Alkali metal
Alkaline earth metal
Ag+, Al+3, Cd+2, Zn+2
Other metal cations
Name first element
add ide to the name of second element
Add the prefix (prefix mono usually omitted for the first element
Name metal first
Specify charge of metal cation with roman numeral (STOCK SYSTEM)
If monoatomic anion, add ide to the anion
If polyatomic anion use name of anion from previous table
Name metal first
If monoatomic anion, add ide to the anion
If polyatomic anion use name of anion from previous table

13. 11- The correct systematic name for Cr2O3 is
First Exam/ Exercises
11- The correct systematic name for Cr2O3 is
a)Chromium (III) oxide b) Dichromium trioxide.
c) Chromium trioxide d) Chromium oxide.
12- The correct formula for calcium sulfate is
a) CaHSO b) CaSO c) CaS d) Ca(HSO4)2
13- Gallium consists of % 69Ga with a mass of amu, and % 71Ga with a mass of amu, the average atomic mass of gallium is:
a) amu b) amu c) amu d) amu
Average atomic mass = (atomic mass x abundenace)Ga-69+ (atomic mass x abundenace)Ga-71
Average atomic mass = ( X ( /100))Ga-69 + ( x ( /100))Ga-71 =

14. 11- The correct systematic name for Cr2O3 is
First Exam/ Exercises
11- The correct systematic name for Cr2O3 is
a)Chromium (III) oxide b) Dichromium trioxide.
c) Chromium trioxide d) Chromium oxide.
12- The correct formula for calcium sulfate is
a) CaHSO b) CaSO c) CaS d) Ca(HSO4)2
13- Gallium consists of % 69Ga with a mass of amu, and % 71Ga with a mass of amu, the average atomic mass of gallium is:
a) amu b) amu c) amu d) amu
14- Which pair of Atoms would be most likely to form an ionic compound?
a) C and N b) K and Ca c) P and Ar d) Ba and S

15. 15- Two isotopes of the same element differ only in their
First Exam/ Exercises
15- Two isotopes of the same element differ only in their
a) Number of neutron b) Atomic number
c) Number of protons d) Number of electron
16- What is the mass in grams of one atom of iron (Fe)? 
a) 6.02  1023 g b) 1.66  g c) 9.3  g d) g
Mass of one atom = atomic mass/ avogadro number
= 56 / x 1023 = 9.3  g

16. 15- Two isotopes of the same element differ only in their
First Exam/ Exercises
15- Two isotopes of the same element differ only in their
a) Number of neutron b) Atomic number
c) Number of protons d) Number of electron
16- What is the mass in grams of one atom of iron (Fe)? 
a) 6.02  1023 g b) 1.66  g c) 9.3  g d) g
17- A g sample of MgCl2 is dissolved in enough water to give 750 mL of solution. What is the molarity of this solution?
a) 3.70  10-2 M b) 1.05  10-2 M c) 6.58  10-2 M d) 4.93  10-2 M
M = n/V
n = mass /molar mass
= / 95= mole
M = / (750 / 1000)= M = 6.58 X 10-2M

17. 15- Two isotopes of the same element differ only in their
First Exam/ Exercises
15- Two isotopes of the same element differ only in their
a) Number of neutron b) Atomic number
c) Number of protons d) Number of electron
16- What is the mass in grams of one atom of iron (Fe)? 
a) 6.02  1023 g b) 1.66  g c) 9.3  g d) g
17- A g sample of MgCl2 is dissolved in enough water to give 750 mL of solution. What is the molarity of this solution?
a) 3.70  10-2 M b) 1.05  10-2 M c) 6.58  10-2 M d) 4.93  10-2 M
18- Calculate the percent composition by mass of C in picric acid (C6H3N3O7)?
a) 1.3 % b) 18.3 % c) 31.4 % d) 48.9 %

18. Molar mass of C6H3N3O7 =229.114g/mol % C= n x molar mass of element
First Exam/ Exercises
18- Calculate the percent composition by mass of C in picric acid (C6H3N3O7)?
a) 1.3 % b) 18.3 % c) 31.4 % d) 48.9 %
Molar mass of C6H3N3O7 = g/mol
% C=
n x molar mass of element
molar mass of compound
x 100%
6x 12.01
229.11
x 100% = 31.4 %

19. 15- Two isotopes of the same element differ only in their
First Exam/ Exercises
15- Two isotopes of the same element differ only in their
a) Number of neutron b) Atomic number
c) Number of protons d) Number of electron
16- What is the mass in grams of one atom of iron (Fe)? 
a) 6.02  1023 g b) 1.66  g c) 9.3  g d) g
17- A g sample of MgCl2 is dissolved in enough water to give 750 mL of solution. What is the molarity of this solution?
a) 3.70  10-2 M b) 1.05  10-2 M c) 6.58  10-2 M d) 4.93  10-2 M
18- Calculate the percent composition by mass of C in picric acid (C6H3N3O7)?
a) 1.3 % b) 18.3 % c) 31.4 % d) 48.9 %

20. 19- How many molecules of ethane (C2H6) are present in 0.30 g of C2H6?
First Exam/ Exercises
19- How many molecules of ethane (C2H6) are present in 0.30 g of C2H6?
a) 1.20 x b) 1.00 x 1022 c) 8.03 x d) 6.02 x 1021
Molar mass of ethane = 2 x x 1 = 30 g/mole
First we should calculate the number of mole
Number of mole = mass / molar mass
Number of mole = 0.30 / 30 = 0.01 mole
We know that
Number of molecules = avogadro’s number x number of mole
= x 1023 x 0.01
= 6.02 x 1021 molecules.

21. 19- How many molecules of ethane (C2H6) are present in 0.30 g of C2H6?
First Exam/ Exercises
19- How many molecules of ethane (C2H6) are present in 0.30 g of C2H6?
a) 1.20 x b) 1.00 x 1022 c) 8.03 x d) 6.02 x 1021
20- The empirical formula of an organic compound with 85.7% C and 14.3% H is
a) CH b) CH2 c) C2H d) CH4
1- we change from % to g
85.7 g of C, 14.3 g of H
2- change from g to mole using
Divided by the smallest number
of mole which is 7.14
7.14
= 1
7.14
14.3
= 2 H: C:
85.7 12
= 7.14 mol of C
nc =
Thus the empirical formula is CH2
14.3 1
= 14.3 mol of H
nH =

22. 19- How many molecules of ethane (C2H6) are present in 0.30 g of C2H6?
First Exam/ Exercises
19- How many molecules of ethane (C2H6) are present in 0.30 g of C2H6?
a) 1.20 x b) 1.00 x 1022 c) 8.03 x d) 6.02 x 1021
20- The empirical formula of an organic compound with 85.7% C and 14.3% H is
CH b) CH2 c) C2H d) CH4
21- After balancing the following equation the coefficients are:
a C3H b O2 → c CO2 + d H2O
a) a=1, b=5, c=2, d= b) a=2, b=5, c=3, d=4
c) a =1, b=3, c=3, d= d) a=1, b=5, c=3, d=4

23. 21- After balancing the following equation the coefficients are:
First Exam/ Exercises
21- After balancing the following equation the coefficients are:
a C3H b O2 → c CO2 + d H2O
a) a=1, b=5, c=2, d= b) a=2, b=5, c=3, d=4
c) a =1, b=3, c=3, d= d) a=1, b=5, c=3, d=4
C3H O2 → CO2 + H2O
C C X3
C3H O2 → 3 CO2 + H2O
H H X4
C3H O2 → 3 CO H2O
O O = X 5
C3H O2 → 3 CO H2O

24. 19- How many molecules of ethane (C2H6) are present in 0.30 g of C2H6?
First Exam/ Exercises
19- How many molecules of ethane (C2H6) are present in 0.30 g of C2H6?
a) 1.20 x b) 1.00 x 1022 c) 8.03 x d) 6.02 x 1021
20- The empirical formula of an organic compound with 85.7% C and 14.3% H is
CH b) CH2 c) C2H d) CH4
21- After balancing the following equation the coefficients are:
a C3H b O2 → c CO2 + d H2O
a) a=1, b=5, c=2, d= b) a=2, b=5, c=3, d=4
c) a =1, b=3, c=3, d= d) a=1, b=5, c=3, d=4
22- Calculate the number of O atoms in 2.50 g of sucrose (C12H22O11)
a) 3.87 x b) 4.84 x c) 5.81 x d) 6.78 x 1022

25. 22- Calculate the number of O atoms in 2.50 g of sucrose (C12H22O11)
First Exam/ Exercises
22- Calculate the number of O atoms in 2.50 g of sucrose (C12H22O11)
a) 3.87 x b) 4.84 x c) 5.81 x d) 6.78 x 1022
First we calculate the number of mole
n = 2.5/ 342 = mole
Number of molecules = Avogadro's number x number of mole
= x 1023 x 0.007
= 4.4 x1022 molecules
From the chemical formula of sucrose C12H22O11
1 molecules of sucrose = 11 atom of O
4.4 x 1021 molecules = ? Atom of O
11 x 4.4 x 1021 = 4.84 x1021 atoms

26. 19- How many molecules of ethane (C2H6) are present in 0.30 g of C2H6?
First Exam/ Exercises
19- How many molecules of ethane (C2H6) are present in 0.30 g of C2H6?
a) 1.20 x b) 1.00 x 1022 c) 8.03 x d) 6.02 x 1021
20- The empirical formula of an organic compound with 85.7% C and 14.3% H is
CH b) CH2 c) C2H d) CH4
21- After balancing the following equation the coefficients are:
a C3H b O2 → c CO2 + d H2O
a) a=1, b=5, c=2, d= b) a=2, b=5, c=3, d=4
c) a =1, b=3, c=3, d= d) a=1, b=5, c=3, d=4
22- Calculate the number of O atoms in 2.50 g of sucrose (C12H22O11)
a) 3.87 x b) 4.84 x c) 5.81 x d) 6.78 x 1022

27. a) C6H8O4 b) C12H16O8 c) C9H12O6 d) C15H20O10
First Exam/ Exercises
23- An empirical formula of an organic compound is C3H4O2 , if the molecular weight of the compound is (360 g/mol), the molecular formula of the compound will be:
a) C6H8O b) C12H16O c) C9H12O d) C15H20O10
FIRST we calculate the molar mass of emperical formula
C3H4O2 = 72 g/mol
Ratio = 360 / 72 = 5
molecular formula = ratio x empirical formula
= 5 x C3H4O2 = C15H20O10

28. a) C6H8O4 b) C12H16O8 c) C9H12O6 d) C15H20O10
First Exam/ Exercises
23- An empirical formula of an organic compound is C3H4O2 , if the molecular weight of the compound is (360 g/mol), the molecular formula of the compound will be:
a) C6H8O b) C12H16O c) C9H12O d) C15H20O10
24- How many moles of FeSO4 are present in a 500 mg of this salt?
a) 1.64 x b) 2.30 x c) 3.29 x d) 4.93 x 10-3
Mass = 500mg = 0.5 g
n= mass / molar mass
= 0.5 / 152
= = 3.29 x 10-3 g

29. a) C6H8O4 b) C12H16O8 c) C9H12O6 d) C15H20O10
First Exam/ Exercises
23- An empirical formula of an organic compound is C3H4O2 , if the molecular weight of the compound is (360 g/mol), the molecular formula of the compound will be:
a) C6H8O b) C12H16O c) C9H12O d) C15H20O10
24- How many moles of FeSO4 are present in a 500 mg of this salt?
a) 1.64 x b) 2.30 x c) 3.29 x d) 4.93 x 10-3
25- How many milliliter of water must be added to 200 mL of 0.15M Na2CO3 to prepare 0.05 M Na2CO3?
a) 200 mL b) 300 mL c) 400 mL d) 450 mL
M1 V1 = M2 V2
0.15 X 200 = 0.05 X V2
V2 = 0.15 X 200 / 0.05 = 600 ml
Add water = = 400 ml

30. a) C6H8O4 b) C12H16O8 c) C9H12O6 d) C15H20O10
First Exam/ Exercises
23- An empirical formula of an organic compound is C3H4O2 , if the molecular weight of the compound is (360 g/mol), the molecular formula of the compound will be:
a) C6H8O b) C12H16O c) C9H12O d) C15H20O10
24- How many moles of FeSO4 are present in a 500 mg of this salt?
a) 1.64 x b) 2.30 x c) 3.29 x d) 4.93 x 10-3
25- How many milliliter of water must be added to 200 mL of 0.15M Na2CO3 to prepare 0.05 M Na2CO3?
200 mL b) 300 mL c) 400 mL d) 450 mL

31. First we have to determine the limiting reagent:
First Exam/ Exercises
26- What is the theoretical yield of chromium (Cr) that can be produced by the reaction of 40.0 g of Cr2O3 with 8.00 g of aluminum (Al) according to the chemical reaction?
2 Al + Cr2O3  Al2O Cr
a) 15.4 g b) g c) 19.4 g d) g
First we have to determine the limiting reagent:
First start with Al
1-Convert to mole :
n = 8 / 27 = mol
2- from equation
2mole Al ========= 2 mole Cr
0.296 mole Al =====? Mole Cr
2x = 2 x ?
Mole of Cr = mol
Mass = n x molar mass
= x 52
=15.4g
second start with Cr2O3
1-Convert to mole :
n = 40 / 152 = 0.263mol
2- from equation
1mole Cr2O3 ========= 2 mole Cr
0.263mole Cr2O3 =====? Mole Cr
2 x = 1 x ?
Mole of Cr = mol

32. First Exam/ Exercises
26- What is the theoretical yield of chromium (Cr) that can be produced by the reaction of 40.0 g of Cr2O3 with 8.00 g of aluminum (Al) according to the chemical reaction?
2 Al + Cr2O3  Al2O Cr
a) 15.4 g b) g c) 19.4 g d) g
27- If the actual yield for the experiment in question (26) produced 13.0g, what is the percentage yield?
a) 84.4% b) 75.0% c) 67% d) 96.4%

33. 2Na(s) + 2H2O(l) → H2(g) + 2NaOH(aq)
First Exam/ Exercises
26- What is the theoretical yield of chromium (Cr) that can be produced by the reaction of 40.0 g of Cr2O3 with 8.00 g of aluminum (Al) according to the chemical reaction?
2 Al + Cr2O3  Al2O Cr
a) 15.4 g b) g c) 19.4 g d) g
27- If the actual yield for the experiment in question (26) produced 13.0g, what is the percentage yield?
a) 84.4% b) 75.0% c) 67% d) 96.4%
28- The mass of Na that will react with excess of water to produce 16.0 g NaOH is
2Na(s) + 2H2O(l) → H2(g) + 2NaOH(aq)
a) 5.75 g b) 6.90 g c) 8.05 g d) 9.20 g

34. 2Na(s) + 2H2O(l) → H2(g) + 2NaOH(aq)
First Exam/ Exercises
28- The mass of Na that will react with excess of water to produce 16.0 g NaOH is
2Na(s) + 2H2O(l) → H2(g) + 2NaOH(aq)
a) 5.75 g b) 6.90 g c) 8.05 g d) 9.20 g
1-First make sure the equation is balanced
2- g to mole
n = mass / molar mass
= 16 / 40
= 0.4 mol
From equation
2 mole Na =======2mole NaOH
?mole Na ======== 0.4 Mole NaOH
2 X 0.4 = 2 X ?
Mole of Na = 0.4 mole
Mass = n x molar mass
= 0.4 x 23 = 9.2 g

35. 2Na(s) + 2H2O(l) → H2(g) + 2NaOH(aq)
First Exam/ Exercises
26- What is the theoretical yield of chromium (Cr) that can be produced by the reaction of 40.0 g of Cr2O3 with 8.00 g of aluminum (Al) according to the chemical reaction?
2 Al + Cr2O3  Al2O Cr
a) 15.4 g b) g c) 19.4 g d) g
27- If the actual yield for the experiment in question (26) produced 13.0g, what is the percentage yield?
a) 84.4% b) 75.0% c) 67% d) 96.4%
28- The mass of Na that will react with excess of water to produce 16.0 g NaOH is
2Na(s) + 2H2O(l) → H2(g) + 2NaOH(aq)
a) 5.75 g b) 6.90 g c) 8.05 g d) 9.20 g

36. First Exam/ Exercises
29- A mL of 0.25 M HCl is diluted with water to a total volume of mL. What is the final concentration in the resulting solution?
a) 0.125M b) M c) 0.05 M d) M
M1 V1 = M2 V2
0.25 X 100 = M2 X 200
M2= 0.25 X 100 / 200 = M

37. 30- What is the mass of carbon in 10 g carbon dioxide (CO2)?
First Exam/ Exercises
29- A mL of 0.25 M HCl is diluted with water to a total volume of mL. What is the final concentration in the resulting solution?
a) 0.125M b) M c) 0.05 M d) M
30- What is the mass of carbon in 10 g carbon dioxide (CO2)?
a) 4.1 g b) 2.73 g c) 6.82 g d) 5.45 g
Mole of CO2 = mass / molar mass
= 10 / 44 = mol
From the formula
1mole C ====== 1 mol CO2
? Mole C ====== mole CO2
Mole of C = mol
Mass of C = n x molar mass
= x 12 = 2.73 g.

38. 30- What is the mass of carbon in 10 g carbon dioxide (CO2)?
First Exam/ Exercises
29- A mL of 0.25 M HCl is diluted with water to a total volume of mL. What is the final concentration in the resulting solution?
a) 0.125M b) M c) 0.05 M d) M
30- What is the mass of carbon in 10 g carbon dioxide (CO2)?
a) 4.1 g b) 2.73 g c) 6.82 g d) 5.45 g

39. Thank you for listening