Download formula math & the mole pwpt

Formula Math & The
Mole
I. Percent Composition
–Gives the percent, by mass,
of the elements in a
compound
–Grams of element x 100
grams of compound
Two ways to calculate:
1) Chemical formula is given
2) Masses, but no formula, are given
Calculating percent composition
when the formula is given:
1.
2.
3.
4.
5.
Write the chemical formula.
Count the number of atoms of each element
in the compound.
Multiply the # of atoms by the mass number
from the periodic table (rounded to the
nearest tenth) for each element.
Add answers together and round to the
nearest tenth to determine grams of
compound.
Use formula above to calculate %
composition for each element. Round final
answer to nearest hundredth.
Example # 1
What is the percent composition of water?
One mole of water is 18.0152 grams.
In that compound, there are two moles of H atoms and 2
x 1.008 = 2.016 grams. That's how many grams of
hydrogen are present in one mole of water.
There is also one mole of oxygen atoms weighing 16.00
grams in the mole of water.
To get the percentage of hydrogen, divide the 2.016 by
18.015 and multiply by 100, giving 11.19%.
For oxygen it is 16.00 ÷ 18.015 = 88.81%.
Example # 2
Step 1: Determine the total mass of each element in the
molar mass
1 mol Ca x (40.08gCa/1 mol Ca) = 40.08 g Ca
2 mol N x (14.01gN/1 mol N) = 28.02 g N
6 mol O x (16.00gO/1 mol O) = 96.00 g O
1 mol of Ca(NO3)2 = 164.1 g (molar mass)
Step 2: Calculate percent by multiplying the mass ratio by
100%
%Ca = (40.08gCa/164.1 Ca(NO3)2) x 100 = 24.42%
%N = (28.02gN/164.1 Ca(NO3)2) x 100 = 17.07%
%O = (96.00gO/164.1 Ca(NO3)2) x 100 = 58.50%
Example # 3
CO2 = 12 + (16 X 2) = 44
molecular weight (mass).
C = 12/44 = 27.3%
O = 32/44 = 72.7%
Calculating percent
composition when masses,
but no formula are given:
1) Find the sum of the masses of the
elements that make up the compound.
2) Divide the mass of each element by
the total mass.
3) Multiply by 100 to determine %
composition of each element ( rounded
to the nearest hundredth).
Example #1
what is the percent composition of a compound
formed when 27.07 grams of calcium reacts
with 47.93 grams of chlorine?
27.07 + 47. 93 = 75 g
27.07/75 = .36 x 100 = 36% calcium
47.93/75 = .64 x 100 = 64% chlorine
Example #2
What is the percent composition of a
compound when 12.5 grams of carbon
reacts with 16.3 grams of oxygen?
12.5 + 16.3 = 28.8 grams
12.5/28.8 = .43 x 100 = 43% carbon
16.3/28.8 = .57 x 100 = 57% oxygen
Example #3
Seventy-five grams of nitrogen gas react with oxygen
gas to form one hundred fifty grams of a compound.
What is the percent composition of each element?
150 – 75 = 75 grams of oxygen
75/150 = .5 x 100 = 50% oxygen
75/150 = .5 x 100 = 50% nitrogen
In Chemistry, we commonly
measure:
–Mass (grams)
–Volume (L or mL)
–Particles (counted)
We can relate each of
these measurements
to a single quantity
called the “Mole”.
The Mole is the SI
unit for the
“amount of
something”
Why use it?
To estimate the number
of particles that are too
small or too numerous to
actually count.
Volume
1.0 mole = 22.4 Liters of a gas, at
standard temperature and pressure
(STP)
Std. temp. = 0º C or 273K
Std. pressure = 1 atmosphere (atm)
or 760 mm Hg (Torr)
Particles
1 mole = 6.022 x 1023 particles of
a pure substance.
* Element = atom
* Molecular = molecule (mlc)
*Ionic = formula unit (fmu)
Mass
1.0 mole = ____ grams
of a pure substance.
(“Molar Mass” of that
substance)
Calculating molar mass:
1. Write the formula.
2. Count the number of atoms of each
element in the formula.
3. For each element, multiply the # of
atoms by its mass number (rounded to the
nearest tenth) from the periodic table.
4. Add all products and round to the
nearest tenth.
What is the Molar Mass of:
Carbon tetrachloride
1. First write the correct molecular formula for carbon tetrachloride.
CCl4
2. Add the molar masses of all the atoms in the molecule.
Look at the periodic table and get these numbers.
molar mass of C = 12.01 g/mole
molar mass of Cl = 35.45 g/mole
CCl4 has 1 mole of C and 4 moles of Cl.
So the molar mass of CCl4 = 1(12.01 g/mole) + 4(35.45 g/mole) =
153.81 g/mole
Oxygen
Oxygen is a diatomic element: O2
15.9994 x 2 = 31.9988
since you are looking for molar mass, units will
be in grams per mole
Aluminum
carbonate
Aluminum Carbonate is represented by
the formula Al2(CO3)3.
Two Al (2x 27g/mol)+three C (3 x
12g/mol)+ nine O (9 x 16) =
234g/mol!
III. Unit Conversions
Remember: The mole is related
to volume, particles, and mass.
This allows you to convert from
one unit to another.
To convert between units of the
same substance:
1. Read the entire problem.
2. Identify what is given (what you have)
and what is unknown (what you want).
3. Convert what is given into moles.
4. Convert from moles to the unknown.
5. Solve. Round to the correct number of
significant figures and include proper
units.
Remember….
1 mole =
23
6.022 x 10 particles =
22.4 liters of gas, at STP
______ grams (Molar
Mass)
Example
On interwrite
EXAMPLE 2
How many moles of Magnesium is 3.01x1022
atoms of magnesium?
3.01x1022
atoms Mg
1 mol Mg
X
6.02x1023 atoms of Mg
= 5.00x10-2 mol Mg
Example:
How many molecules are in
38.69 grams of chlorine gas?
Example:
What is the volume, in L, at
STP, occupied by 0.600
moles of sulfur dioxide gas?
Empirical Formula:
•Gives the lowest whole number ratio of
elements in a compound
•From a molar perspective, the empirical
formula represents the lowest whole number
ratio of moles of each element in a compound
•Subscripts cannot be reduced
Molecular Formula:
Is a whole number multiple of the
empirical formula
Identifying Empirical vs. Molecular:
Example # 1 : Ba3N2
Example # 2 : Ca3(PO4)2
Example # 3: C6H12O6
Example # 4 : H2O2
Calculating Empirical Formula:
1.
Change % sign to grams (assume a 100 gram
sample). If grams are already given, go to step 2.
2.
Convert from grams to moles.
3.
Create a “mole ratio” by dividing all answers by the
smallest answer.
1.
If all answers are whole numbers, you are
finished. Write formula for compound.
2.
If all answers ARE NOT whole numbers,
see step 4.
4.
Multiply all answers from step 3 by “a number” that
will give the lowest whole number ratio of elements.
Write formula for compound.
Example # 1:
What is the E.F. of a compound composed of 32.00%
Carbon, 42.66% Oxygen, 18.67% Nitrogen, and
6.67% Hydrogen?
Grams C = 100 * 0.320 = 32.0 g; moles C = 32.0 g/(12.0 g/mole) = 2.67
moles
Grams O = 100 * 0.4266 = 42.66 g; moles O = 42.66 g/(16.0 g/mole) =
2.67 moles
Grams N = 100 * 0.1867 = 18.67 g; moles N = 18.67 g/(14.0 g/mole) =
1.33 moles
Grams H = 100 * 0.0667 = 6.67 g; moles H = 6.67 g/(1.0 g/mole) = 6.67
moles
Divide by the lowest number of moles (1.33):
C: 2.67/1.33 = 2.0
O: 2.67/1.33 = 2.0
N: 1.33/1.33 = 1.0
H: 6.67/1.33 = 5.0
Empirical formula = C2H5NO2
Example # 2:
What is the E.F. of a compound composed of 25.9%
Nitrogen and 74.1% Oxygen?
If 25.9% of the compound is nitrogen and 74.1% of the compound is
oxygen, then a compound with a mass of 100 g has 25.9 g of nitrogen
and 74.1 g of oxygen.
mol N 
mol O 
25.9 g N
1 mol N

 1.85 mol N
1
14.0067 g N
74.1 g O
1 mol O

 4.63 mol O
1
15.9994 g O
This would mean that the ratio of nitrogen to oxygen
is N1.85O4.63.
 We can divide each number in the ratio of N O by
1.85 4.63
1.85 to get N1O2.50.
Since we cannot have 2.50 atoms of
oxygen, we must multiply through each number by 2 to
even it out, getting N2O5 as our empirical formula.
Example # 3:
What is the E.F. of a compound composed of 36.6
grams Carbon and 9.2 grams Hydrogen?
• To determine the molecular
formula, you need to know the
empirical formula mass (you
calculate) and the molecular
formula mass (always given)
M.F. mass = whole # multiple
E.F. mass
Calculating molecular formulas:
1. Calculate E.F. mass
2. Divide M.F. mass by E.F. mass
3. Multiply E.F. by whole number
answer from step 2.
4. Write M.F.
Example # 1:
What is the M.F. of a compound with a mass of 92
grams and an E.F. of NO2?
Step 1: Determine the empirical formula molar mass
N has 1 atom
1 X 14.0067= 14.0067g/mol
O has 2 atoms 2X 16.00=
32.00g/mol
46.0067 g/mol
Step 2: Divide the experimentally determined molar mass of
succinic acid by the mass of the empirical formula to determine n
n=
Experimental Molar mass of NO2
Molar mass of NO2
n= 92g/mol
46.0067 g/mol
n= 2.000
Step 2: Multiple subscripts by 2 (n)
N 2O4
Example # 2:
What is the M.F. of a compound with the mass
of 150 grams and an E.F. of CH2O?
CH2O = (1 x 12.01 g/mol) + (2 x 1.008 g/mol) + (1 x 16.00 g/mol)
150 grams/30.026 = 5
CH2O x 5 = C5H10O5
= 30.026
Example # 3
What is the M.F. of a compound
with a mass of 132 grams if that
compound is composed of 54.6%
Carbon, 13.6% Hydrogen, and
31.8% Nitrogen?
STOP
STOP