1. Section 12.1: The arithmetic of equations
Stochiometry I
Section 12.1: The arithmetic of equations

2. Objectives
Upon completing this presentation, the student will be able to:
Explain how balanced equations apply to both chemistry and everyday situations.
Interpret balanced chemical equations in terms of moles, representative particles, mass, and gas volume at STP.
Identify the quantities that are always conserved in chemical reactions.

3. Introduction
When we bake cookies, we need to know the amounts of materials we need.
How much flour?
How much sugar?
How many eggs?
How much milk?
How much butter?
How much salt?
How much baking powder?
How many chocolate chips?

4. CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
Introduction
Likewise, a chemist needs to know the number of reactant molecules are needed to produce a given number of product molecules.
For example:
We use one molecule of methane gas, CH4(g), with two molecules of oxygen gas, O2(g) to produce one molecule of carbon dioxide gas, CO2(g), and two molecules of liquid water, H2O(l).
This is written as:
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
This is a balanced equation.

5. Using Balanced Chemical Equations
Chemists use balanced chemical equations as a basis to calculate how much reactant is needed or product is formed in a reaction.
The calculation of quantities in chemical reactions is a subject of chemistry called stoichiometry.
It is a form of chemical bookkeeping.
The reactants are the assets and the products are the debits.
Everything must balance out at the end.

6. Interpreting Chemical Equations
A balanced chemical equation can be interpreted in terms of many different quantities.
numbers of atoms
numbers of molecules
moles
mass
volume
We will learn to interpret chemical equations in each of these terms.

7. Interpreting Chemical Equations
Interpreting in terms of number of atoms
At the atomic level, the balanced equation represents the number and type of each atom in a reaction.
The number of each kind of atom is the same on each side of the balanced equation.
For example:
N2(g) + 3 H2(g) → 2 NH3(g)
On the reactant side, there are two atoms of N combined with the six atoms of H.
On the product side, there are two atoms of N combined with the six atoms of H.

8. Interpreting Chemical Equations
Interpreting in terms of number of molecules
At the molecular level, the balanced equation represents the number and type of each molecule in a reaction.
We can determine the ratios of the molecules needed to produce the desired product by looking at the coefficients.
For example:
N2(g) + 3 H2(g) → 2 NH3(g)
With one nitrogen molecule,N2, and three hydrogen molecules, H2, we get two ammonia molecules, NH3.
The ratios of molecules for this reaction will always be 1:3:2 for N2:H2:NH3.

9. Interpreting Chemical Equations
Interpreting in terms of number of moles
We can expand the molecular interpretation to include moles.
We can determine the ratios of the moles needed to produce the desired product by looking at the coefficients.
For example:
N2(g) + 3 H2(g) → 2 NH3(g)
With one mole of N2 and three moles of H2 give us two moles of NH3.
The ratios of moles for this reaction will always be 1:3:2 for N2:H2:NH3.

10. Interpreting Chemical Equations
Interpreting in terms of mass
A balanced chemical equation obeys the law of the conservation of mass.
Mass is neither created nor destroyed in a chemical reaction.
The mass of the products must equal the mass of the products.
For example:
N2(g) + 3 H2(g) → 2 NH3(g)
The mass of ammonia produced must equal the mass of hydrogen and nitrogen consumed.
1 mol of N2 (28 g) + 3 mol of H2 (6 g) is equal to 2 mol of NH3 (34 g).

11. Interpreting Chemical Equations
Interpreting in terms of volume
At STP (0°C and 1 atmosphere), 1 mol of gas occupies 22.4 L.
Therefore, the volume of the reactants and products are related to the number of mols of each.
For example:
N2(g) + 3 H2(g) → 2 NH3(g)
1 mol of N2 (22.4 L) and 3 mol of H2 (67.2 L) are consumed and produce 2 mol of NH3 (44.8 L).
Notice that the volume is not conserved like the mass.

12. Mass Conservation in Chemical Reactions
Mass and atoms are conserved in every chemical reaction.
Molecules, formula units, moles, and volumes are not necessarily conserved.
N2(g) + 3 H2(g) → 2 NH3(g) N2 +
3 H2 →
2 NH3
Conserved?
2 N atoms
6 H atoms
2 N atoms + 6 H atoms
Yes
28 g N2
6 g H2
34 g NH3
1 mol N2
3 mol H2
2 mol NH3 No
22.4 L N2
67.2 L H2
44.8 L NH3

13. 2 H2S(g) + 3 O2(g) → 2 SO2(g) + 2 H2O(g)
Conceptual Problem 12.1
Hydrogen sulfide, which smells like rotten eggs, is found in volcanic gases. The balanced equation for the burning of hydrogen sulfide is:
2 H2S(g) + 3 O2(g) → 2 SO2(g) + 2 H2O(g)
Interpret this equation in terms of
numbers of representative particles and moles.
masses of reactants and products.
The coefficients tell us the relative numbers of representative particles and moles. Therefore, the ratio of H2S:O2:SO2:H2O = 2:3:2:2

14. 2 H2S(g) + 3 O2(g) → 2 SO2(g) + 2 H2O(g)
Conceptual Problem 12.1
Hydrogen sulfide, which smells like rotten eggs, is found in volcanic gases. The balanced equation for the burning of hydrogen sulfide is:
2 H2S(g) + 3 O2(g) → 2 SO2(g) + 2 H2O(g)
Interpret this equation in terms of
numbers of representative particles and moles. H2S:O2:SO2:H2O = 2:3:2:2
masses of reactants and products.
The coefficients tell us the relative numbers of representative particles and moles. Therefore, the ratio of H2S:O2:SO2:H2O = 2:3:2:2

15. 2 H2S(g) + 3 O2(g) → 2 SO2(g) + 2 H2O(g)
Conceptual Problem 12.1
Hydrogen sulfide, which smells like rotten eggs, is found in volcanic gases. The balanced equation for the burning of hydrogen sulfide is:
2 H2S(g) + 3 O2(g) → 2 SO2(g) + 2 H2O(g)
Interpret this equation in terms of
numbers of representative particles and moles. H2S:O2:SO2:H2O = 2:3:2:2
masses of reactants and products.
The balanced equation obeys the law of the conservation of mass. Therefore, the mass of reactants is equal to the mass of products.
mass = (number of mols)(molar mass)
⇒ mH2S = (2)(34) g = 68 g
⇒ mO2 = (3)(32) g = 96 g
⇒ mreactants = mH2S + mO2 = 72 g + 96 g = 164 g

16. 2 H2S(g) + 3 O2(g) → 2 SO2(g) + 2 H2O(g)
Conceptual Problem 12.1
Hydrogen sulfide, which smells like rotten eggs, is found in volcanic gases. The balanced equation for the burning of hydrogen sulfide is:
2 H2S(g) + 3 O2(g) → 2 SO2(g) + 2 H2O(g)
Interpret this equation in terms of
numbers of representative particles and moles. H2S:O2:SO2:H2O = 2:3:2:2
masses of reactants (164 g) and products.
The balanced equation obeys the law of the conservation of mass. Therefore, the mass of reactants is equal to the mass of products.
mass = (number of mols)(molar mass)
⇒ mH2S = (2)(34) g = 68 g
⇒ mO2 = (3)(32) g = 96 g
⇒ mreactants = mH2S + mO2 = 72 g + 96 g = 164 g

17. 2 H2S(g) + 3 O2(g) → 2 SO2(g) + 2 H2O(g)
Conceptual Problem 12.1
Hydrogen sulfide, which smells like rotten eggs, is found in volcanic gases. The balanced equation for the burning of hydrogen sulfide is:
2 H2S(g) + 3 O2(g) → 2 SO2(g) + 2 H2O(g)
Interpret this equation in terms of
numbers of representative particles and moles. H2S:O2:SO2:H2O = 2:3:2:2
masses of reactants (164 g) and products.
The balanced equation obeys the law of the conservation of mass. Therefore, the mass of reactants is equal to the mass of products.
mass = (number of mols)(molar mass)
⇒ mSO2 = (2)(64) g = 128 g
⇒ mH2O = (2)(18) g = 36 g
⇒ mproducts = mSO2 + mH2O = 128 g + 36 g = 164 g

18. 2 H2S(g) + 3 O2(g) → 2 SO2(g) + 2 H2O(g)
Conceptual Problem 12.1
Hydrogen sulfide, which smells like rotten eggs, is found in volcanic gases. The balanced equation for the burning of hydrogen sulfide is:
2 H2S(g) + 3 O2(g) → 2 SO2(g) + 2 H2O(g)
Interpret this equation in terms of
numbers of representative particles and moles. H2S:O2:SO2:H2O = 2:3:2:2
masses of reactants (164 g) and products (164 g).
The balanced equation obeys the law of the conservation of mass. Therefore, the mass of reactants is equal to the mass of products.
mass = (number of mols)(molar mass)
⇒ mSO2 = (2)(64) g = 128 g
⇒ mH2O = (2)(18) g = 36 g
⇒ mproducts = mSO2 + mH2O = 128 g + 36 g = 164 g

19. Summary
A balanced chemical equation can be interpreted in terms of many different quantities.
numbers of atoms
numbers of molecules
moles
mass
volume
The quantities number of atoms and mass are conserved.
The quantities number of molecules, moles, and volume may not be conserved.