• 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.

• During any chemical reaction, matter is neither created nor destroyed. Mass is conserved from reactants to products.
• Therefore,
MASS
REACTANTS
= MASS
PRODUCTS

CH
4
(g) + 2 O
2
(g)  CO
2
(g) + 2 H
2
O (l) methane gas oxygen gas carbon dioxide gas water
Reactants : methane gas (CH
4
) and oxygen gas (O
2
)
Products : carbon dioxide gas (CO
2
) and water (H
2
O)
“  ” 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?
CH
4
(g) + 2O
2
(g)  CO
2
(g) + 2H
2
O (l) methane gas oxygen gas carbon dioxide gas 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?

Note about showing “conservation
particle diagrams
If you have the reaction:
A
2
+ B
2
 A
3
B
You would show conservation by drawing
+ 
3A
2
+ 1B
2
 2A
3
B
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 CH
4
64 grams of O should form?
2 reacts completely with
, what mass of products
CH
4
(g) + 2 O
2
(g)  CO
2
(g) + 2 H
2
O (l) methane gas oxygen gas carbon dioxide gas water
16g + 64g = X g
80 g = X

• If 32 grams of CH
4 reacts completely with
128 g of O
2
, and 88 g of CO
2 forms, how many grams of H
2
O form?
CH
4
(g) + 2 O
2
(g)  CO
2
(g) + 2 H
2
O (l) methane gas oxygen gas carbon dioxide gas water
32 g + 128 g = 88 g + X
160 g = 88 g + X
72 g = X

• If 8 grams of CH
4 react completely with oxygen, and 22 g of CO
2 and 9 g of H
2 form, how much oxygen (O
2 consumed?
) was
O
CH
4
(g) + 2 O
2
(g)  CO
2
(g) + 2 H
2
O (l) methane gas oxygen gas carbon dioxide gas water
8 g + X = 22 g + 9 g
8 g + X = 31 g
X = 23 g

• 4 grams of CH
CH
4
O
2
4 reacts with 20 g of O
2
. The is used up completely but there is some left over. Given that 20 grams of product was formed, how much oxygen was used up?
CH
4
(g) + 2 O
2
(g)  CO
2
(g) + 2 H
2
O (l) methane gas oxygen gas carbon dioxide gas water
4g + X = 20 g
X = 16 grams oxygen used up and 20 g oxygen available – 16 grams used =
4 g oxygen left over.