1. Understanding the Law of Conservation of Mass

2. 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.

3. 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

4. What does the law really mean?
CH4 (g) O2 (g)  CO2 (g) H2O (l)
methane gas oxygen gas carbon dioxide gas 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)

5. How would you draw this reaction as particles and show conservation of mass?
CH4 (g) O2 (g)  CO2 (g) H2O (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?

6. Note about showing “conservation” in particle diagrams
If you have the reaction:
A2 + B2 A3B
You would show conservation by drawing 
3A B2  2A3B
Do not simply add stray “atoms” to molecules. It changes them to a different substance.

7. MASS REACTANTS = MASS PRODUCTS
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) O2 (g)  CO2 (g) H2O (l)
methane gas oxygen gas carbon dioxide gas water
16g g = X g
80 g = X

8. 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) O2 (g)  CO2 (g) H2O (l)
methane gas oxygen gas carbon dioxide gas water
32 g g = g X
160 g = g X
72 g = X

9. 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) O2 (g)  CO2 (g) H2O (l)
methane gas oxygen gas carbon dioxide gas water
8 g X = g g
8 g X = g
X = g

10. 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) O2 (g)  CO2 (g) H2O (l)
methane gas oxygen gas carbon dioxide gas water
4g X = g
X = 16 grams oxygen used up
and 20 g oxygen available – 16 grams used =
4 g oxygen left over.