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CHEM 1000 Lecture Notes - Magnesium Oxide, Antoine Lavoisier, Joseph Proust

Course Code
CHEM 1000
Hernan Humana

of 5
Chemistry Chapter Two:
Law of Conservation of Mass
In 1774, Antoine Lavoisier (1774-1794) performed an experiment in which he
heated a sealed glass vessel containing a sample of tin and some air. He found that
the glass before heating and the glass after heating were the same. Through further
investigation, he proved that the product of the reaction, tin calx, consisted of the
original tin together with a portion of the air. This proved that oxygen from air is
essential to combustion and it led to the formulation of the Law of Conservation of
Mass: the total mass of substance present after a chemical reaction is the same as
the total mass of substances before the reaction.
Problem: A 0.455g sample of Mg is allowed to burn in 2.315 g of O2 gas. The sole
produce is MgO. After the reaction, no magnesium remains and the mass of
unreacted oxygen is 2.015g. What mass of magnesium oxide is produced?
Mass before reaction: 0.455g magnesium + 2.315g oxygen = 2.770 g
2.770g mass after reaction =? g magnesium oxide after reaction + 2.015g oxygen
after reaction
? g magnesium oxide after reaction = 2.770 g mass after reaction – 2.015 g oxygen
after reaction = 0.755 g magnesium oxide after reaction.
Law of Constant Composition
In 1799, Joseph Proust: Law of constant composition or law of definite
proportions: All samples of a compound have the same composition – the same
proportions by mass of the constituent elements.
Consider H2O at 10.000g: 1.119 g H = 11.19 % Consider H2O at 27.000g: 3.021
g H = 11.19% 8.881 g O = 88.81%
23.979 g O= 88.81%
Problem: 0.455g of Mg reacted with an excess of oxygen to produce 0.755g of
magnesium oxide. According to the law of constant composition, the mast ratio:
0.455 g magnesium/0.755 g magnesium oxide should exist in all samples of
magnesium oxide. Thus, in 0.500 g of magnesium oxide, the mass of magnesium is:
? g magnesium = 0.500 g magnesium oxide x (0.455 g magnesium/0.755 g
magnesium oxide) = 0.301 g magnesium
Daltons Atomic Theory
1. Each chemical element is composed of minute, indivisible particles
called atoms. Atoms can be neither created nor destroyed during
chemical change.
2. All atoms of an element are alike in mass (weight) and other properties,
but the atoms of one element are different from those of all other
3. In each of their compounds, different elements combine in a simple
numerical ratio: for example, one atom of A to one of B (AB), or one
atom of A to two of B (AB2).
Daltons 1st assumption explained the law of conservation of mass and his 2nd and 3rd
assumption explained the law of constant composition. This theory led to the
prediction: the law of multiple proportions. If two elements form more than a
single compound, the masses of one element combined with the fixed ratio of the
second are in the ration of small whole numbers (e.g. CO2, N2O, H2O, NO3-)
Chemical Elements
A= mass number and Z= atomic number ***Z is often not specified when writing***
Example: An element with two stable isotopes has 35 protons. The separate isotopes
contain 44 and 46 neutrons. What is the element and write the symbol for the two
Solution: 35 protons = 35 electrons. Number 35 on the periodic table is Bromine (Br).
Isotopic Masses
Problem: Chlorine has two naturally occurring isotopes: Cl-35 with a natural
abundance of 75.77% and an isotopic mass of 34.969 amu and Cl-37 with a natural
abundance of 24.23% and an isotopic mass of 36.966 amu. What is the atomic mass
of chlorine?
# of
# of
protons +
# of
# of
protons – #
of electrons
Solution: [(75.77 / 100) x 34.969u]+ [(24.23 / 100) x 36.966u] = 35.45u
Problem: Bromine has two naturally occurring isotopes. One of them, bromine-79, has
a mass of 78.9183 u and an abundance of 50.69%. What must be the mass and
percent natural abundance of the other, bromine-81?
Solution: Percent abundance of 81Br is 100%-50.69% = 49.31% 81Br.
Atomic mass
= (fraction of atoms that are 79Br x mass of 79Br) + (fraction of
atoms that are 81Br x mass of 81Br) 79.904u= (0.5069 x 78.9183 u) + (0.4931
x mass o f 81Br) = 40.00 u +
(0.4931 x mass of 81Br) mass of 81Br
= [(79.904 u – 40.00 u) /0.4931]
= 80.92 u
Periodic Table
Group I: Alkali Metals; Group 2: Alkaline Earth Metals; Group VII: Halogens; Group
VIII: Noble Gases
Main group elements are Groups I, II, XIII-XVIII
Centre groups are transition metals, very bottom group are Lanthanides and
Group near thick black border are metalloids, elements with both metal and non-
metal properties.
Concept of the Mole and the Avogadro Constant
Mole: is the SI quantity that describes an amount of substance by relating it to a
number of particles of that substance. It is the amount of a substance that contains
the same number of elementary entities as there are atoms in exactly 12 g of pure
carbon-12. This is represented by the Avogadro constant, NA (6.02 x 10²³ mol-1).
Molar Mass
The molar mass, M, is the mass of one mole of a substance.
M(g/mol 12C)=A(g/atom 12C) x NA (atoms
12C/mol 12C) For any element atomic mass (amu) = molar mass