Download Terms to Know

Terms to Know
Percent composition – relative
amounts of each element in a
compound
Empirical formula – lowest wholenumber ratio of the atoms of an
element in a compound
An 8.20 g piece of magnesium
combines completely with 5.40 g of
oxygen to form a compound. What
is the percent composition of this
compound?
1. Calculate the total mass
2. Divide each given by the total mass
and then multiply by 100%
3. Check your answer: The
percentages should total 100%
Answer
The total mass is 8.20 g + 5.40 g =
13.60 g
Divide 8.2 g by 13.6 g and then
multiply by 100% = 60.29412 = 60.3%
Divide 5.4 g by 13.6 g and then
multiply by 100% = 39.70588 = 39.7%
Check your answer: 60.3% + 39.7% =
100%
Calculate the percent composition
of propane (C3H8)
1. List the elements
2. Count the atoms
3. Multiply the number of atoms of
the element by the atomic mass of the
element (atomic mass is on the
periodic table)
4. Express each element as a
percentage of the total molar mass
5. Check your answer
Answer
Total molar mass = 44.0 g/mol
36.0 g C = 81.8%
8.0 g H = 18.2%
Calculate the mass of carbon in
52.0 g of propane (C3H8)
1. Calculate the percent composition
using the formula (See previous
problem)
2. Determine 81.8% of 82.0 g
Move decimal two places to the
left (.818 x 82 g)
3. Answer = 67.1 g
Calculating Empirical Formulas
Microscopic – atoms
Macroscopic – moles of atoms
Lowest whole-number ratio may not
be the same as the compound
formula
Example: The empirical formula of
hydrogen peroxide (H2O2) is HO
Empirical Formulas
The first step is to find the mole-to-mole
ratio of the elements in the compound
If the numbers are both whole numbers,
these will be the subscripts of the elements
in the formula
If the whole numbers are identical,
substitute the number 1
Example: C2H2 and C8H8 have an empirical
formula of CH
If either or both numbers are not whole
numbers, numbers in the ratio must be
multiplied by the same number to yield
whole number subscripts
What is the empirical formula of a
compound that is 25.9% nitrogen
and 74.1% oxygen?
1. Assume 100 g of the compound, so that
there are 25.9 g N and 74.1 g O
2. Convert to mole-to-mole ratio:
Divide each by mass of one mole
25.9 g divided by 14.0 g = 1.85 mol N
74.1 g divided by 16.0 g = 4.63 mol O
3. Divide both molar quantities by the
smaller number of moles
4. 1.85/1.85 = 1 mol N
4.63/1.85 = 2.5 mol O
5. Multiply by a number that converts
each to a whole number (In this case,
the number is 2 because 2 x 2.5 = 5,
which is the smallest whole number )
2 x 1 mol N = 2
2 x 2.5 mol O = 5
Answer: The empirical formula is
N 2O 5
Determine the Empirical Formulas
1. H2O2
2. CO2
3. N2H4
4. C6H12O6
5. What is the empirical formula of a
compound that is 3.7% H, 44.4% C,
and 51.9% N?
Answers
Compound
1. H2O2
2. CO2
3. N2H4
4. C6H12O6
5. HCN
Empirical Formula
HO
CO2
NH2
CH2O
Calculating Molecular Formulas
The molar mass of a compound is a
simple whole-number multiple of the
molar mass of the empirical formula
The molecular formula may or may
not be the same as the empirical
formula
Calculate the molecular formula of
the compound whose molar mass
is 60.0 g and empirical formula is
CH4N.
1. Using the empirical formula, calculate
the empirical formula mass (efm)
(Use the same procedure used to calculate
molar mass.)
2. Divide the known molar mass by the
efm
3. Multiply the formula subscripts by this
value to get the molecular formula
Answer
Molar mass (efm) is 30.0 g
60.0 g divided by 30.0 g = 2
Answer: C2H8N2
Practice Problems
1) What is the empirical formula of a compounds
that is 25.9% nitrogen and 74.1% oxygen?
2) Calculate the empirical formula of a compound
that is 32.00% C, 42.66% O, 18.67% N, and 6.67%
H.
3) Calculate the empirical formula of a compound
that is 42.9% C and 57.1% O.
Practice Problems
4) What is the molecular formula for each compound:
a) CH2O: 90 g
b) HgCl: 472.2 g
c) C3H5O2: 146 g