# Stoichiometry of Chemical Equations and Formulas - PowerPoint by morgossi7a3

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```									Mole Calculations
The Mole

• Mole – measurement of the amount of
a substance.
– We know the amount of different
substances in one mole of that substance.
Atomic Mass Unit

• Mass of 1 mole of compound
• Found by adding the atomic weights
of each atom of each element that
makes up a compound.
• H2 O
• There are 2 atoms of hydrogen and 1
atom of oxygen.
AMU

• H = 2 x 1.01 = 2.02
• O = 1 x 16 = 16
_____
18.02 = amu

1 mole of any substance = the amu of
that substance
AMU

• Molar mass = the sum of the molar
masses of atoms of the elements in the
formula
• Calculated the same way
The Mole

• One mole of any substance has
• 6.022 x 1023 atoms, molecules, ions
The Mole

• For gases only – 1 mole of a gas
occupies 22.4 L
Converting Between Units

amu  1 mole  6.022 x 1023 atoms, mlc, ions
(expressed
in grams)
Conversions

• Change 5.0 grams of sodium chloride
to moles of sodium chloride.
• Change 0.45 moles of barium chlorate
to grams of barium chlorate.
• Determine the number of atoms in 15
grams of water.
Empirical Formulas
Empirical Formulas

• The simplest whole number ratio of
moles of each element in the
compound.

• H2 O
• NaCl
Empirical Formulas

• Usually given in percentages of each
element in the compound.
• Based on 100% of the compound.
• Can be compared to a 100 gram
sample of the substance.

• 11.2% Hydrogen
• 88.8% Oxygen
• 1. Change percents to grams

11.2% H = 11.2 g H
88.8% O = 88.8 g O
• 2. Convert grams to moles. (We know
an empirical formula is the lowest whole
number ratio of moles)

11.2 g H | 1 mol = 11.089 mol
| 1.01 g

88.8g O | 1 mol = 5.550 mol
| 16 g
Cannot have decimals, the numbers must be
whole numbers. Divide by the lowest.

11.2 g H | 1 mol = 11.089 mol / 5.550 = 2
| 1.01 g

88.8g O | 1 mol = 5.550 mol / 5.550 = 1
| 16 g
• H2O = empirical formula
• 36.84% N
• 63.16% O
• 35.98% Al
• 64.02% S
Molecular Formula
• Specifies the actual number of atoms
of each element in one molecule or
formula unit of the substance.
• The molar mass of acetylene is 26.04
g/mol and the mass of the empirical
formula, CH, is 13.02 g/mol
• 1. The problem will give you a molar
mass of the compound.
• 2. Calculate the empirical formula as
usual.
• 40.68% carbon
• 5.08% hydrogen
• 54.24% oxygen

• Molar mass = 118.1 g/mol
• Change % to grams

• 40.68 g C | 1 mol      = 3.387 mol
| 12.01 g

• 5.08 g H | 1 mol       = 5.030 mol
| 1.01 g

• 54.24 g O | 1 mol      = 3.390 mol
| 16 g
• 3.387 mol / 3.387 mol = 1 x2     =2
• 5.030 mol / 3.387 mol = 1.5 x2   =3
• 3.390 mol / 3.387 mol = 1 x2     =2

• Empirical Formula = C2H3O2
• To find the molecular formula:

• (EFamu)x = MM (molar mass)
• C2H3O2 = 59.05 = amu

• (59.05)x = 118.1
•X=2

• (EF)x2 = (C2H3O2)x2 = C4H6O4
• 65.45% C
• 5.45% H
• 29.09% O

• MM = 110.0 g/mol
• 49.98 g C
• 10.47 g H

• MM = 58.12 g/mol
• 46.68% N
• 53.32% O

• MM = 60.01 g/mol

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