Download Stoichiometry2

FORMULA MASS AND MOLES
When the formula mass of a molecule or compound is determined by the addition of its
component atomic masses and expressed in grams, it is called the gram- formula mass.
The gram-formula mass is called one mole of that material. The term gram- molecular
mass can also be used when it is known that the material is a molecular substance and not
an ionic lattice like NaCl or NaOH.
GAS VOLUMES AND MOLECULAR MASS
Because the volume of a gas may vary depending on the conditions of temperature and
pressure, a standard is set for comparing gases. The standard conditions of temperature
and pressure (abbreviated STP) are 0°C and 760 mm of mercury pressure. The molecular
mass of a gas expressed in grams and under standard conditions occupies 22.4 liters. This
is an important relationship to remember! The 22.4 liters is referred to as the molar
volume. Two men are associated with this relationship.
Gay-Lussac's Law states that, when only gases are involved in a reaction, the volumes of
the reacting gases and the volumes of the gaseous products are in ratio to each other as
small whole numbers. steam
Avogadro's Law, which explains Gay-Lussac's, states that equal volumes of gases under
the same conditions of temperature and pressure contain equal number of molecules. This
means that 1 mole of any gas at STP occupies 22.4 liters.
DENSITY AND MOLECULAR MASS
Since the density of a gas is usually given in grams/liter of a gas at STP, we can use the
molar volume to molecular mass relationship to determine the molecular mass of a gas
from the gases density.
The density of a gas can also be determined if the molecular mass of the gas is known.
Since the molecular mass of the gas at STP occupies 22.4 liters, dividing the molecular
mass by 22.4 liters will produce the mass per unit liter or density.
MASS-VOLUME PROBLEMS
A typical mass-volume problem is as follows: How many liters of oxygen gas at STP can
be prepared from the decomposition of 42.6 grams of sodium chlorate?
Step 1: Write the balanced equation for the reaction.
2NaClO3  2NaCl + 3O2
Step 2: Convert the given quantity to moles. (If quantity is in grams, divide by
the molar mass; if the quantity is in liters, divide by 22.4 liters/mole)
42.6 grams NaClO3 ÷ 106.5 grams/mole = 0.4 mole NaClO3
Step 3: Convert moles of the given substance to moles of the unknown substance
by creating an equality from the balanced equation.
0.4 mole NaClO3 x 3 moles O2 = 0.6 moles O2
2NaClO3
Step 4: Convert the moles of the unknown to either liters or grams depending on
the question.
(If the quantity is to be expressed in grams, then multiply by the molar
mass. If the quantity is to be expressed in liters, then multiply by 22.4
liters/mole.)
0.6 moles O2 x 22.4 liters/mole = 13.4 liters O2
MASS-MASS PROBLEMS
A typical problem concerning just mass relationships is as follows: What mass of oxygen
can be obtained from heating 100 grams of potassium chlorate?
Step 1: Write the balanced equation for the reaction.
2 KCIO3 → 2 KCl + 3 O2
Step 2: Convert the given quantity to moles. (If the quantity is in grams, divide by the
molar mass)
100 grams KCIO3 x 1 mole/122 grams KCIO3 = .815 mole KCIO3
Step 3: Convert moles of the given substance to moles of the unknown substance by
creating an equality from the balanced equation.
3 moles O2
.815 mole KCIO3 x ――――― = 1.22 moles O2
2 moles KCIO3
Step 4: Convert the moles of the unknown to grams.
1.22 moles O2 x 32 grams O2 /mole O2 = 39 grams O2
VOLUME-VOLUME PROBLEMS
This type of problem involves only volume units and therefore can make use of GayLussac's Law: "When gases combine, they combine in simple whole number ratios."
These simple numbers are the coefficients of the equation. A typical problem concerning
just volume relationships is as follows:
What volume of NH3 is produced when 22.4 L of nitrogen gas are made to combine with
a sufficient quantity of hydrogen gas under the appropriate conditions?
Step 1: Write the balanced equation for the reaction.
N2 + 3 H2 → 2 NH3
Step 2: Convert the given quantity to moles. (Divide the given quantity by 22.4 L per
mole)
1 mole N
2
22.4 L N2 x ―――――― = 1 mole N2
22.4 L N2
Step 3: Convert moles of the given substance to moles of the unknown substance by
creating an equality from the balanced equation. above them as denominators.
2 moles NH3
1 mole N2 x ―――――
1 moles N2
= 2 moles NH3
Step 4: Convert the moles of the unknown to liters.
2 moles NH3 x 22.4 L NH3 / 1 mole NH3 = 44.8 liters NH3