Chemistry 101
Chapter 9
Chemical Quantities
Formula and molecular weight: formula weight (FW) of a compound is the sum of the
atomic weights in atomic mass units (amu) of all atoms in the compound’s formula (for both
ionic and covalent compounds). The molecular weight (MW) is the same as the formula
weight; however, it is only used for the covalent compounds.
MW of H2SO4:
2(1 amu) for H + 1(32 amu) for S + 4(16 amu) for O = 98 amu
MW or FW of AlCl3:
1(27 amu) for Al + 3(35.5 amu) for Cl = 133.5 amu
Mole (mol): is the amount of substance that contains as many atoms, molecules, or ions as
there are atoms in exactly 12g of carbon-12. Mole is the formula weight of a substance
expressed in grams.
FW of NaCl = 58.5 amu → 58.5g of NaCl = 1 mole of NaCl
MW or FW of AlCl3 = 133.5 amu → 133.5g of AlCl3 = 1 mole of AlCl3
Avogadro’s number (6.022×
×1023): number of formula units in a mole.
1 mole of hydrogen atoms = 6.022×1023 atoms of hydrogen
1 mole of water molecules = 6.022×1023 molecules of water
1 mole of Na+ ions = 6.022×1023 ions of Na+
Molar mass: is the mass of one mole of the substance expressed in grams. We can say that it
is the formula weight of a compound expressed in grams.
Formula weight of H2O = 18 amu → molar mass = 18 g (mass of 1 mole H2O)
Formula weight of NaCl = 58.5 amu → molar mass = 58.5 g (mass of 1 mole NaCl)
Stoichiometry: the study of mass relationships in chemical reactions. The coefficients in a
balanced equation refer to the relative numbers of moles, particles (atoms, molecules, ions),
and volume, not grams.
2H2O(g) → O2(g) + 2H2(g)
2
1
2
2 moles
1 mole
2moles
2 liters
1 liter
2 liters
2 molecules 1 molecule 2 molecules
2 grams
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1 gram
Chemistry 101
2 grams
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Note: We use the coefficients of a balanced chemical equation for the following conversions.
Therefore, there is only one step:
Mole A ↔ Mole B
Volume A ↔ Volume B
Number of particles A ↔ Nomber of particles B
Example:
2H2 + O2 → 2H2O
How many moles of H2O are produced from the reaction of 56 moles of hydrogen?
56 mole H 2 ×
Example:
2 moles H2O
= 56 moles H2O
2 moles H2
CH4 + 2O2 → 2H2O + CO2
How many molecules of O2 are needed to produce 2×1026 molecules H2O?
Note: For other conversions, we use the following diagram:
mass
volume
volume
A
mole
B
mole
Particle
Particle
(atom)
(molecule)
(atom)
(molecule)
mass
Note: for some conversions we need to use two steps:
Mole A ↔ Volume B
Mass A ↔ Mole B or Volume A
Number of particles A ↔ Mole B or Volume A
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Chemistry 101
CSUB
Example:
2H2 + O2 → 2H2O
We have a sample of 74 grams of oxygen. Find the volume (in liter) that this sample can
occupy.
Note: at STP condition (0ºC as the standard temperature and 1 atm as the standard pressure),
one mole of any gas occupies a volume of 22.4 L (or 22.4 dm3 or 22400 cc).
74g O2 ×
Example:
1 mole O2 22.4 L O2
×
= 52 L O2
32 g O2 1 mole O2
CH4 + 2O2 → 2H2O + CO2
A sample of 32 grams of CH4 is reacted woth oxygen. How many moles of CO2 are
produced?
Note: for some conversions we need to use three steps:
Mass A ↔ Mass B
Mass A ↔ Volume B or # of particles B
Number of particles A ↔ Volume B
Example:
How many molecules of H2 are needed to produce 51 g of H2O?
2H2 + O2 → 2H2O
51 grams H 2 O ×
Example:
1 mole H 2 O
2 moles H2
6.02 × 10 23 molecules H2
×
×
= 1.5 × 10 25 molecules H2
2 grams H 2 O 2 moles H 2 O
1 mole H2
CH4 + 2O2 → 2H2O + CO2
A sample of 46.0 grams of CH4 is reacted with oxygen. How many grams of H2O are
produced?
Limiting reagent: is the reactant that is used up first, leaving an excess of another reagent(s)
unreacted.
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Chemistry 101
CSUB
Note: The limiting reagent can control a reaction. Whenever, the limiting reagent is used up
the reaction will be stopped. Therefore, to determine how much product can be formed a
given mixture of reactants, we have to look for the reactant that is limiting.
How to find the limiting reagent: 1. Write and balance the equation for the reaction. 2.
Convert known masses of reactants to moles. 3. Using the numbers of moles of reactants to
determine which reactant is limiting.
Note: If the coefficients of reactants in a balanced reaction are not the same, we need to
divide the number of the moles of each reactant by its coefficient.
Note: Always use the amounts of a limiting reagent to find the amount of the product.
Example:
2H2 + O2 → 2H2O
If 39 grams oxygen reacts with 87 grams hydrogen, which reactant is limiting. How many
grams of H2O will be formed?
39g O2 ×
1 mole O2
= 1.2 moles O2
32 g O2
87g O2 ×
1 mole H2
= 44 moles H2
2 g H2
Because the coefficients of reactants (hydrogen and oxygen) are not the same, we have to
divide the number of moles of each reactant by its coefficient:
1.2 mole O2
= 1.2 moles O2
1
44 mole H2
= 22 moles H2
2
Therefore, oxygen is the limiting reagent. To find the amount of the product, we need to use
the amount of the limiting reagent (which is oxygen):
39 grams O 2 ×
1 mole O 2
2 moles H 2 O 18 grams H 2 O
×
×
= 44 grams H 2 O
32 grams O 2
1 moles O 2
1 mole H 2 O
Percent Yield:
precent yield =
actual yield
× 100
theoretical yield
Actual yield: the mass of product formed in a chemical reaction (experimental).
Theoretical yield: the mass of product that should form in a chemical reaction according to
the stoichiometry of the balance equation (it is always more than actual yield).
Dr. Behrang Madani
Chemistry 101
CSUB