Download Formula Mass

Chapter 3
Molecules, Compounds, and
Chemical Equations
3.7 Formula Mass versus Molar
mass

Formula mass
◦ The average mass of a
molecule or formula unit in
amu
◦ also known as molecular
mass or molecular weight
(MW)
◦ whole = sum of the parts!

Molar mass
◦ Total mass of a
compound in gram per 1
mol of its molecules or
formula unit
Molar Mass of Compounds

the relative masses of molecules can be calculated from atomic
masses
Formula Mass = 1 molecule of H2O
= 2(1.01 amu H) + 16.00 amu O = 18.02 amu
Why multiplying 2?

since 1 mole of H2O contains 2 moles of H and 1 mole of O
Molar Mass = 1 mole H2O
= 2(1.01 g H) + 16.00 g O = 18.02 g
so the Molar Mass of H2O is 18.02 g/mole
3
Molar Mass of Na2SO4
Calculate the molar mass of Na2SO4.
Element Number
of Moles
Atomic Mass
Total Mass
in each
element
Na
S
O
Total mass in 1 mol Na2SO4
4
Example – Using Molar mass

Consider 10.8 g of dry ice, CO2
 What is the molar mass of CO2?
 Write the two possible conversion of this molar mass
 Calculate the number of CO2 molecules
3.8 Mass Percent Composition


1.
2.

Percentage of each element in a compound
◦
By mass
Can be determined from
the formula of the compound
the experimental mass analysis of the compound
The percentages may not always total to 100% due to rounding
part
Percentage 
 100%
whole
E.g Calculate the mass percent composition of calcium chloride
6
Example - Mass Percent as a
Conversion Factor

Find the mass of table salt containing 2.4 g of Na. Sodium chloride
is 39.0% by mass.

Benzaldehyde is 79.2% carbon. What mass of benzaldehyde
contains 19.8 g of C?
Chemical Formulas and Elemental
Composition
chemical formulas have inherent in them relationships between
numbers of atoms and molecules
◦ or moles of atoms and molecules
 these relationships can be used to convert between amounts of
constituent elements and molecules
◦ like percent composition
Vol A  Grams A  Moles A  Moles B  Grams B

Grams A  Moles A  Moles B  Grams B
8
Example
Butane (C4H10) is the liquid fuel in lighters.
 a.
Determine the number of atoms ratio between carbon and 1
molecule of C4H10

b. Determine the number moles ratio between C and 1 mol of
C4H10
c.
How many grams of carbon are present within a lighter
containing 7.5 mL of butane? The density of quid butane is
0.601 g/mL
Empirical Formula
simplest, whole-number ratio of the atoms of elements in a
compound
 can be determined from elemental analysis
◦ masses of elements formed when decompose or react compound
 combustion analysis
◦ percent composition

10
Steps in determine the Empirical
formula





Step 1: Obtain the mass of each element (in grams)
E.G 100% = 100g therefore mass percent is the same numerical
value in grams
Step 2: Determine the numbers moles of each atom present
◦ Use molar mass of each element
Step 3: Divide the smallest moles by numbers moles of each atom
to obtain the closet integer as possible.
◦ if result is within 0.1 of whole number, round to whole number
Step 4: If the result ended with 0.5, 0.33, 1.125, 1.50 etc… then
multiply with a factor to get the nearest integer as possible.
E.g 1.5 x 2 = 3.0 atoms
1.33 x 3 = 3. 99 = 4 atoms
Step 5: Write the result (number atoms) from step 4 as a subscript
for the appropriate element.
Example

Determine the empirical formula of stannous fluoride, which
contains 75.7% Sn (118.70g/mol) and the rest fluorine (19.00
g/mol)

An unknown sample gives the following mass percent:
17.5% Na, 39.7% Cr and 42.8% O. What is the empirical
formula?
Molecular Formulas


The molecular formula is a multiple of the empirical formula
To determine the molecular formula you need to know the
empirical formula and the molar mass of the compound
Multiple (n) =
molecular mass
empirical formula mass
Molecular formula = empirical formula
x n
where n = 1, 2, 3, 4
13
Example

Benzopyrene has a molar mass of 252 g/mol and an empirical
formula of C5H3. What is its molecular formula? (C = 12.01
g/mol, H=1.01g/mol)
Example

Laboratory analysis of aspirin determined the following mass
percent composition. Find the empirical formula.
C = 60.00%
H = 4.48%
O = 35.53%
15
Determining Empirical Formulas:
Elemental Analysis
Combustion Analysis: A compound of unknown composition
(containing a combination of carbon, hydrogen, and possibly
oxygen) is burned with oxygen to produce the volatile combustion
products CO2 and H2O, which are separated and weighed by an
automated instrument called a gas chromatograph.
hydrocarbon + O2(g)
carbon
hydrogen
xCO2(g) + yH2O(g)
Combustion Analysis





Unknown formula: CxHyOx (Oxygen can be replaced with other
nonmetal)
gCO2  moles CO2  moles C  gC
gH2O  moles H2O  moles H  gH
g O = g sample – (g H + g C)
◦ gO  moles O
Follow steps in determine the empirical formula and molecular
formula
Example

Combustion of a 0.8233 g sample of a compound containing only
carbon, hydrogen, and oxygen produced the following:
CO2 = 2.445 g
H2O = 0.6003 g
Determine the empirical formula of the compound
18
Example

Upon combustion, a compound containing only carbon and
hydrogen produced 1.60g CO2 and 0.819g H2O. Find the empirical
formula
19
Chemical Reactions
Reactions involve chemical changes in matter resulting in new
substances
 Reactions involve rearrangement and exchange of atoms to produce
new molecules
◦ Elements are not transmuted during a reaction

20
Chemical Equations
A chemical equation gives
• the formulas of the reactants on the left of the arrow.
• the formulas of the products on the right of the arrow.
Reactants
O2 (g)
Product
CO2 (g)
C(s)
21
Symbols Used in Equations
Symbols in chemical
equations show
TABLE
• the states of the
reactants.
• the states of the
products.
• the reaction conditions.
22
Chemical Equations are Balanced
In a balanced
chemical reaction
• no atoms are lost or
gained.
• the number of reacting
atoms is equal to the
number of product
atoms.
23
Balancing Chemical Equations
A balanced chemical equation shows that the law of conservation
of mass is adhered to.
In a balanced chemical equation, the numbers and kinds of atoms
on both sides of the reaction arrow are identical.
2Na(s) + Cl2(g)
2NaCl(s)
left side:
right side:
2 Na
2 Cl
2 Na
2 Cl
Balancing Chemical Equations
1. Write the unbalanced equation using the correct chemical
formula for each reactant and product.
H2(g) + O2(g)
H2O(l)
2. Find suitable coefficients—the numbers placed before
formulas to indicate how many formula units of each
substance are required to balance the equation.
2H2(g) + O2(g)
2H2O(l)
3. Reduce the coefficients to their smallest whole-number
values, if necessary, by dividing them all by a common
denominator.
2H2(g) + O2(g)
2H2O(l)
Balancing Chemical Equations
4. Check your answer by making sure that the numbers and
kinds of atoms are the same on both sides of the equation.
2H2(g) + O2(g)
2H2O(l)
left side:
right side:
4H
2O
4H
2O
Balancing Chemical Equations
Do not change subscripts when you balance a chemical equation.
You are only allowed to change the coefficients.
H2(g) + O2(g)
2H2(g) + O2(g)
2H2O(l)
Balanced properly
H2O(l)
unbalanced
H2(g) + O2(g)
H2O2(l)
Chemical equation changed!
Examples
Balance the coefficients from reactants to products.
A.
__N2(g) + __H2(g)
A.
B. __Co2O3(s) + __ C(s)
__ NH3(g)
__Co(s) + __CO2(g)
Write a balanced equation for the reaction between
a.
carbon dioxide and potassium hydroxide to form potassium
carbonate and water.
b.
The combustion of gaseous ethane (C2H6)
28