Chemistry 110 Assignment: Molecular Formulas and Reactions Analysis

This document presents solutions to a chemistry assignment, addressing a variety of concepts including balancing chemical equations, identifying reaction types (acid-base, single displacement), and determining ionic and covalent bond characteristics. It provides examples of Lewis structures, and naming and formula writing for various compounds like aluminium sulfate, calcium phosphate, dinitrogen pentoxide, and magnesium fluoride. The assignment also involves molar mass calculations, stoichiometry problems to determine the mass of reactants and products, and calculating the percent yield of a reaction. The solution covers a wide range of topics, including atomic weight, molecular formula determination, and the geometry around central atoms in molecules like ammonia. The document offers a comprehensive overview of fundamental chemistry principles and their application in solving problems related to chemical reactions and compound properties.

CHEMISTRY 110
Assignment
[DATE]

Question 1
Balanced chemical equation
C5 H12 +8 O2 → 5C O2+6 H2 O
Sum of coefficients = 1 + 8 + 5 + 6 = 20
Question 2
Here, an acid is reacting with the base and thus, this reaction is an acid-base reaction.
Question 3
In order to form a stable ion, magnesium atom losses two electrons and forms Mg+2.
Question 4
The molecule of HC O3
−¿¿ has two equivalent Lewis structure or resonance structure as shown
below.
Question 5
Name of compound = Aluminium sulfate
Formula of compound ¿ A l2 ¿
Here,
1

Molar mass of Al = 27
Molar mass of S = 32.1
Molar mass of O = 16
Hence,
Molar mass of Aluminium sulfate = 2 (27) + 3 {32.1 + (16*4)} = 342.3 gram per mole
Question 6
3.75 moles of Cl represent = n grams of Cl
Mass of Cl = 35.5 gram per mole
1 mole has 35.5 gram of Cl
And thus, 3.75 moles would have n = 35.5*3.75 = 133.125 gram of Cl
Question 7
% composition of oxygen∈N H 4 N O3=?
Mass of O = 16
Mass of N = 14
Mass of H = 1
Total mass of 1 mole of N H3 N O3= ( 2∗14 ) + ( 4∗1 ) + ( 16∗3 ) =80 grams
Mass of O in compound = 16*3 = 48
% composition of oxyge n= 48
80∗100 %=60 %
Question 8
6.023∗1023atoms of Ca have 40 grams of Ca
2

1 atom of Ca would have = ( 40
6.023∗1023
)grams of Ca
18.066∗1023 atoms of Cawould have
Gram of calcium= 18.066∗1023
6.023∗1023 ∗40=120 gram of Ca
Question 9
Reaction
Here, Zn replace the hydrogen from chlorine to form ZnCl2. Hence, it would be a single
displacement reaction.
Question 10
Answer: 24.03050 which is atomic weight of Mg.
Question 11
Name of compound = Calcium phosphate
Formula of compound ¿ C a2 ¿
Question 12
Formula of compound¿ N2 O5
Name of compound = Dinitrogen pentoxide
Question 13
Bond angle around a central in ammonia = 107o
3

Question 14
A chemical equation is said to be balanced when the number of atoms presents in the
reactants (left side of equation) is exactly the same as the number of atoms of produced (right
side of equation).
Example
Unbalanced equation
CaCO3 + HCl →Ca C l2+ H 2 O+C2 O
Atoms of reactant side = 2
Atoms of product side = 3
Balanced equation
CaCO3 +2 HCl→ CaC l2 +H2 O+ C2 O
Atoms of reactant side = 3
Atoms of product side = 3
Question 15
Ionic bond – This type of bond forms when one atom donates an electron to another electron
to make it more stable, then the bond is termed as ionic bond. The atom which discharges
electron becomes positively charged ion. and the atom which receives the electron becomes
negatively charged ion.
4

Sodium atom donates one electron to chlorine atom which means they have formed an ionic
bond an as a result of these two ions are formed i.e. Na+ and Cl-.
Question 16
Electron dot-structure of H2 P O4
−¿¿
Question 17
Type of bond in HBr = Covalent
Question 18
Geometry around a central atom in CB r4=Tetraheral
Question 19
Name of compound = Magnesium Fluoride
Formula of compound ¿ Mg F2
Question 20
Carbon % in unknown compound = 40%
Hydrogen % in unknown compound = 6.67%
5

Oxygen % in unknown compound = 53.33%
Molar mass of compound = 150 gram per mole
Molecular formula =?
In 100 gram of the compound there would be
¿ ( 40
12.011 moles of C , 6.67
1.00794 moles of H , 53.33
16 moles of O )
Now ,
C : H : 0= ( 3.33: 6.617:3.33 )
Empirical formula = C H2 O
Further,
( Empirical formula )n =Molecular formula
150.00 g per mole=n ( 12.011+ ( 2∗1.00794 ) +16.00 ) g per mole
150=30.03n
n 5
Now,
Molecular formula=5∗( C H2 O ) =C5 H10 O5
Molecular formula =C5 H10 O5
Question 21
Yield of water =?
Water in product = 5 gram
Oxygen amount = 150 gram
Molar mass of oxygen = 32 gram per mol
6

Mol of oxygen = Mass/Molar mass = 150/32 = 4.688 moles
Balanced equation
H2 +2O2 → 2 H2 O
Mole of water formed through reaction = (2/1) *4.688 = 9.375 moles
Molar mass of water = (2*1.008) + (1*16) = 18.016 g/mol
Mass of water = Number of mols of water * Molar mass = 9.375 * 18.02 = 1.689*10^2 gram
% yield = (Actual mass/ Theoretical mass) *100 = (5/1.689*10^2) *100 = 2.96%
Yield of water would be 2.96%.
Question 22
Formula of compound ¿ FeC l3
Name of compound = Iron trichloride OR Iron (III) chloride
Question 23
(D) Barium sulfate
Question 24
Name of compound = Chromium (III) bromide
Formula of compound ¿ CrB r3
Question 25
Formula of compound ¿ NaN O3
Name of compound = Sodium Nitrate
7