# Understanding the Law of Conservation of Mass Power Point (1)

```Understanding the Law of
Conservation of Mass
Who Discovered this Law?
•
•
•
•
•
•
1789, France
Antoine Lavoisier
Nobleman
Statesman
Scientist
Used one of the first analytical mass balances to
prove this law.
• Executed on the guillotine during the French
Revolution.
• He is known as the “Father of Chemistry”
because he made it a quantitative science.
What does the Law of
Conservation of Mass State?
• During any chemical reaction, matter is
neither created nor destroyed. Mass is
conserved from reactants to products.
• Therefore,
MASS REACTANTS = MASS PRODUCTS
What does the law really mean?
CH4 (g)
+
methane gas
2O2 (g)
oxygen gas

CO2 (g)
+
carbon dioxide gas
2H2O (l)
water
Reactants: methane gas (CH4) and oxygen gas (O2)
Products: carbon dioxide gas (CO2) and water (H2O)
“” means yields.
Little numbers (subscripts) – tell how many of a particular type of
atom are inside of a molecule.
(ex. 4 hydrogen atoms per methane molecule)
Big numbers (coefficients) – tell how many of each particle is
involved in the reaction.
(ex. 2 molecules of oxygen react with 1 molecule of methane)
How would you draw this reaction as
particles and show conservation of mass?
CH4 (g)
+
methane gas
2O2 (g)
oxygen gas

CO2 (g)
+
carbon dioxide gas
2H2O (l)
water
How does this picture show that particles and therefore mass are conserved
from reactant’s side to product’s side?
What is all that really happens to the particles in a chemical reaction?
Can atoms of one type be changed into (transformed) atoms of another type
during a chemical reaction?
particle diagrams
in
If you have the reaction:
A2 + B2 A3B
You would show conservation by drawing
+

3A2 + 1B2  2A3B
Do not simply add stray “atoms” to molecules. It changes
them to a different substance.
Solving simple math problems involving the
Law of Conservation of Mass
– copy both solution and answer!
MASS REACTANTS = MASS PRODUCTS
•
If 16 grams of CH4 reacts completely with
64 grams of O2, what mass of products
should form?
CH4 (g)
+
methane gas

2O2 (g)
oxygen gas
16g + 64g
80 g
CO2 (g)
+
carbon dioxide gas
=
=
Xg
X
2H2O (l)
water
Another Problem…
•
If 32 grams of CH4 reacts completely with
128 g of O2, and 88 g of CO2 forms, how
many grams of H2O form?
CH4 (g)
+
methane gas
2O2 (g)
oxygen gas

CO2 (g)
+
carbon dioxide gas
32 g + 128 g = 88 g
160 g = 88 g
72 g = X
+
+
2H2O (l)
water
X
X
Another Problem…
•
If 8 grams of CH4 react completely with
oxygen, and 22 g of CO2 and 9 g of H2O
form, how much oxygen (O2) was
consumed?
CH4 (g)
+
methane gas
8g
8g

2O2 (g)
+
+
oxygen gas
X
X
X
=
=
=
CO2 (g)
+
carbon dioxide gas
22 g
23 g
2H2O (l)
water
+ 9g
31 g
Last Problem
•
4 grams of CH4 reacts with 20 g of O2. The
CH4 is used up completely but there is some
O2 left over. Given that 20 grams of product
was formed, how much oxygen was used up?
CH4 (g)
+
methane gas
2O2 (g)
oxygen gas

CO2 (g)
+
carbon dioxide gas
2H2O (l)
water
4g + X =
20 g
X = 16 grams oxygen used up
and 20 g oxygen available – 16 grams used =
4 g oxygen left over.
```