Mass Relationships in
Chemical Reactions
Atomic Mass
(0.9890) (12 amu) + (0.0110) (13.00335)
= 12.01 amu
Mole
Avogadro’s number: 6.022x1023
(atoms, molecules, particles)
Molar mass is numerically the same as
atomic mass of an element
Molar Mass
Molecular mass of a molecule is the sum of all atomic masses x how
many atoms
Ex) Find the molar mass of H2O
(1.01 x 2) + 16.00 = 18.02 amu
18.02 amu = 18.02g H2O
Formula mass of a compound is the sum of all atomic masses x how
many atoms
Ex) Find the molar mass of NaCl
22.99 + 35.45 = 58.44 amu
58.44 amu = 58.44g NaCl
Examples
1. Helium (He) is a valuable gas used in industry, low-temperature
research, deep-sea diving tanks and balloons. How many moles of
He atoms are in 6.46 g of He?
1. How many atoms are in 112g of Fe?
1. Methane (CH4) is the principal component of natural gas. How
many moles of CH4 are present in 6.07g of CH4?
4. How many hydrogen atoms are present in 25.6 g of urea
[(NH2)2CO], which is used as a fertilizer, in animal feed, and
in the manufacture of polymers?
Mass Spectrometry
Hydrate
Steps:
1. Find the mass of water or mass of anhydrate
2. Turn mass of water and mass of anhydrate into moles
(individually)
3. Find the mole ratio; mole of anhydrate
mole of water
:
mole of anhydrate
mole of anhydrate
4. Write the formula of the hydrate
Example
A calcium chlorite hydrate has a mass of 4.72 g. After heating for
several minutes the mass of the anhydrate is found to be 3.56 g.
Use this information to determine the formula for the hydrate.
Percent Composition
% comp. = n × molar mass of element
× 100
molar mass of compound
Ex) Find percent H and percent O in hydrogen peroxide (H2O2)
%H = 2 × 1.008g H × 100 = 5.926%
34.02g H2O2
%O = 2 × 16.00g O × 100 = 94.06%
34.02g H2O2
Empirical Formula
Steps:
1. Percent to mass
2. Mass to moles
3. Divide by small
4. Multiply ‘til whole
Example
1. Ascorbic acid (vitamin C) cures scurvy. It is composed of
40.92% C, 4.58% H, and 54.50% O by mass. Determine its
empirical formula.
Example
2. When ethanol is burned, carbon dioxide and water are given
off. Suppose that in one experiment the combustion of 11.5 g of
ethanol produced 22.0 g of CO2 and 13.5 g of H2O. Determine
the empirical formula for ethanol.
Molecular Formula
Steps:
1. Find empirical formula
2. Determine the molar mass of the empirical formula
3. Find multiplier (molar mass given/empirical molar mass)
4. Multiply empirical formula subscripts by the multiplier
Example
A sample of a compound contains 30.46% N and 69.54% O by mass, as
determined by a mass spectrometer. In a separate experiment, the
molar mass of the compound is found to be between 90 g and 95 g.
Determine the molecular formula and the accurate molar mass of the
compound.
Chemical Reactions and Equations
“Reactants
with”
“to
produce”
or “yield”
2 H2 (g) + O2 (g)
Reactants
2 H2O (l)
Product(s)
Balancing Chemical Reactions
Steps:
1. List elements present on
each side
2. Add coefficients to
balance (“met a non
hairy oxen”)
KClO3
O2+ KCl
Stoichiometry
• Mole method: coefficient in a reaction can be interpreted
as the number of moles
• Allows us to write conversion factors from a chemical
equation
Example
The food we eat is degraded, or broken down, in our bodies to provide
energy for growth and function. A general overall equation for this very
complex process represents the degradation of glucose (C6H12O6) to
carbon dioxide (CO2) and water (H2O): C6H12O6 + 6 O2
6 CO2 + 6 H2O
If 856 g of C6H12O6 is consumed by a person over a certain period, what
is the mass of CO2 produced?
Limiting Reagents
CO(g) + 2 H2(g)
CH3OH(g)
Example
Urea [(NH2)2CO] is prepared by reacting ammonia with carbon dioxide
in the following process: 2 NH3(g) + CO2(g)
(NH2)2CO(aq) + H2O(l)
In one process, 637.2g of NH3 are treated with 1142g CO2. Which of the
reactants is the limiting reagent? Calculate the mass of [(NH2)2CO]
formed. How much excess reagent (in grams) is left at the end of the
reaction?
Reaction Yield
Example
Titanium is a strong, lightweight, corrosion-resistant metal that is used
in rockets, aircraft, jet engines, and bicycle frames. It is prepared by the
reaction of titanium(IV) chloride with molten magnesium between
950°C and 1150°C: TICl4(g) + 2 Mg(l)
Ti(s) + 2 MgCl2(l)
In a certain industrial operation 3.54 x 107 g of TiCl4 are reacted with
1.13 x 107 g of Mg. Calculate the theoretical yield of Ti in grams.
Calculate the percent yield if 7.91 x 106 g of Ti are actually obtained.