Molar Mass
Do Now
• You have a bag of blue and red marbles.
You’re trying to figure out the mass of the bag,
but you don’t have a balance/ scale.
• What might you need to know to solve this
problem? How might you be able to find the
mass?
Do Now- con’t.
• Your bag contains 10 blue marbles and 15 red
marbles.
• What else may we need to know?
Do Now- con’t.
• Your bag contains 10 blue marbles and 15 red
marbles.
• Each blue marble weighs 2 grams, and each
red marbles weighs 3 grams.
• What is the mass of the bag?
Mass on the Molecular Level
• Just like we weren’t able to directly weigh the bag of
marbles, it’s impossible to “weigh” one molecule.
– But, we can find the mass of a molecule by adding up
the mass of its individual atoms
• Molar Mass: the mass (in grams) of one mole
of molecules
– This is based on the atomic mass in the periodic
table
Determining Molar Mass
1. From the formula, write down each of the different
elements that are present.
MgCl2:
Mg
Cl
2. Identify how many atoms of each element are present
using the subscripts (no subscript = just one atom)
Mg: 1
Cl: 2
3. Multiply the number of atoms of each element by
that element’s atomic mass, then find the total mass
by adding all the elements’ masses together
Mg: 1 x 24.3050 g = 24.3050g
Cl: 2 x 35.4527 g = + 70.9054g
95.2104g
Practice!
• Determine the molar mass of H2SO4 (sulfuric
acid).
Practice!
• Determine the molar mass of H2SO4 (sulfuric
acid).
H2SO4
H: 2 x 1.0079g = 2.0158g
S: 1 x 32.066g = 32.066g
O: 4 x 15.999 = 63.996g
98.0778g
Practice!
• What is the molar mass of (H3O)3PO4?
Practice!
• What is the molar mass of (H3O)3PO4?
• Note: when you have parentheses in the
formula, the subscript outside them is
distributed (multiplied) by all of the subscripts
inside!
Practice!
• What is the molar mass of (H3O)3PO4?
H: (3 x 3) = 9
O: (1 x 3) + 4 = 7
P: 1
H: 9 x 1.0079 = 9.0711
O: 7 x 15.999 = 111.993
P: 1 x 30.974 = 30.974
152.038g
What exactly is a mole?
• Avogadro’s Number: 6.022 x 1023
• The number of carbon atoms in exactly 12g of pure 12C
• 1 mole of ANYTHING = 6.022 x 1023 units of that
substance
• How many teachers are in 1 mole of teachers?
• How many pencils are in 1 mole of pencils?
• How many candy bars are in 1 mole of candy bars?
What exactly is a mole?
• Avogadro’s Number: 6.022 x 1023
• The number of carbon atoms in exactly 12g of pure 12C
• 1 mole of ANYTHING = 6.022 x 1023 units of that
substance
• How many teachers are in 1 mole of teachers?
• How many pencils are in 1 mole of pencils?
• How many candy bars are in 1 mole of candy bars?
6.022 x 1023!!!
If we know moles…
• The molar mass is the grams of that substance that it
takes to make up 1 mole
• So, if we know how many moles we started with, we
can determine the mass of our substance
• OR,
• We also know the number of particles in one mole
(Avogadro’s number)
• So, we can determine the number of molecules of our
substance
Our map…
Moles
Grams
Molar Mass
# of Particles
Avogadro’s
Number
Going between Grams and Moles
We have 24g of carbon. We know that the molar
mass is 12.011g/mol (there are 12.011 grams
of carbon in one mole)
• SET UP THE CONVERSION! Start with what you
know. The molar mass is our conversion factor.
Going between Grams and Moles
We have 24g of carbon. We know that the molar
mass is 12.011g/mol (there are 12.011 grams
of carbon in one mole)
• SET UP THE CONVERSION! Start with what you
know. The molar mass is our conversion factor.
24g C x 1 mole = 24g C mol = 1.998 mol C
1 12.011g C 12.011g C
Going between Grams and Moles
• We have 3 moles of sulfur. How many grams
do we have?
Going between Grams and Moles
• We have 3 moles of sulfur. How many grams
do we have?
3 mol S x 32.066g S = 96.198 g S
1
1 mol S
Higher Order Questions
• What is another real-life application of
stoichiometry?
• What is happening on a molecular level when
two substances react to form new products?
THINK-WRITE-PAIR-SHARE!
Pick one topic to contemplate with your partner.
CO2(g) + 2LiOH(s) -> Li2CO3(s) + H2O(l)
• In a spacecraft, the carbon dioxide exhaled by
astronauts can be removed by its reaction
with lithium hydroxide, LiOH, according to the
above chemical equation.
How many grams of carbon dioxide are
produced if the average person breathes out
20 mol of CO2 per day?
You Practice!
Convert from grams to moles:
• 125g phosphorus (P)
• 75g NaCl
Convert from moles to grams:
• 5 moles of carbon (C)
• 7 moles of HBr
You Practice!
Convert from grams to moles:
• 125g phosphorus (P) = 4.04 mol P
• 75g NaCl = 1.28 mol NaCl
Convert from moles to grams:
• 5 moles of carbon (C) = 60.055g C
• 7 moles of HBr = 566.4g HBr
Going Between Moles and Particles
• We have 15 x 1023 carbon atoms. How many
moles is this?
• SET UP THE CONVERSION! Start with what you
know. Avogadro’s number is our conversion
factor.
Going Between Moles and Particles
• We have 15 x 1023 carbon atoms. How many
moles is this?
• SET UP THE CONVERSION! Start with what you
know. Avogadro’s number is our conversion
factor.
15 x 1023 C atoms x
= 2.49 mol C
1 mol
6.022 x 1023 atoms
Going Between Moles and Particles
• We have 5.5 mol of NaCl. How many
molecules is this?
Going Between Moles and Particles
• We have 5.5 mol of NaCl. How many
molecules is this?
5.5 mol NaCl x 6.022 x 1023 molecules
1 mol
= 33.121 x 1023 molecules NaCl
= 3.3121 x 1024 molecules NaCl (correct
scientific notation)
Going Between Grams and Particles
• This is just another conversion! We don’t know
the relationship between grams and particles,
so what could we go to first?
Going Between Grams and Particles
• This is just another conversion! We don’t know
the relationship between grams and particles,
so what could we go to first?
• MOLES!!! Moles are like our base in metric
conversions- when in doubt, convert to moles!
Going Between Grams and Particles
• We have 600 x 1023 carbon atoms. How many
grams is this?
Going Between Grams and Particles
• We have 600 x 1023 carbon atoms. How many
grams is this?
600 x 1023 atoms C x
1 mol C
x
6.022 x 1023 atoms C
12.011g C= 119.7g C
1 mol C
Going Between Grams and Particles
• We have 30g of pure copper. How many atoms
of copper is this?
Going Between Grams and Particles
• We have 30g of pure copper. How many atoms
of copper is this?
30g Cu x 1 mol Cu x 6.022 x 1023 atoms Cu =
63.546g Cu 1 mol Cu
= 2.8 x 1023 atoms Cu
Density
• Density: a physical property of a substance,
representing the mass per unit volume
d = m/V
m = mass (in grams)
V = volume (in mL)
Note: 1mL = 1cm3
We can use the density to determine the mass if it
isn’t given!
Using Density to Determine Mass
• Determine the mass of mercury (Hg) in 30mL
if the density of mercury is 13.55g/mL.
Set up the equation! Fill in what you know, then
solve.
Using Density to Determine Mass
• Determine the mass of mercury (Hg) in 30mL
if the density of mercury is 13.55g/mL.
d = m/V
13.55g = m
mL
30mL
13.55g x 30 = m
406.5g = m
Using Density to Determine Mass
• Turpentine has a density of 0.85g/mL. How
many grams are in 75mL?
Using Density to Determine Mass
• Turpentine has a density of 0.85g/mL. How
many grams are in 75mL?
d = m/V
0.85g = m
mL
75mL
0.85g x 75 = m
63.75g = m
Exit Ticket #1
Exit Ticket #1
Solve the following problems, showing all work for
full credit.
1. What is the molar mass of acetic acid
(HC2H3O2)? (1 point)
2. How many pennies are in one mole of pennies?
(1 point)
3. How many moles are in 50g of magnesium?
Show your work! (2 points)
4. How many grams are in 3.5mol of CaO? Show
your work! (3 points)
5. How many molecules are in 15g of HF? Show
your work! (3 points)
Percent Composition
• You always bring a lunch to school. You
wanted to know how heavy your lunch was, so
you decided to weigh each of the foods you
packed. Your sandwhich was 550g. Your apple
was 200g. Your cookie was 50g. Your yogurt
was 200g. What percent of the mass of your
lunch was from the sandwich?
Percent Composition
• Percent Composition: a given element’s portion
of the molecule’s mass
= molar mass of the element
x 100%
molar mass of the whole molecule
1. Find the mass of that element in the molecule
2. Find the total mass of the molecule
3. Enter values into the above equation
**If you add up the percentages for all elements in the
compound, you should get 100%
Percent Composition: Example
• What percent of the molecule’s mass is
hydrogen in H2SO4?
Percent Composition: Example
• What percent of the molecule’s mass is hydrogen
in H2SO4?
H: 2 x 1.0079g = 2.0158g
S: 1 x 32.066g = 32.066g
O: 4 x 15.999g = 63.996g
= 98. 078g
g H x 100% = 2.0158g x 100% = 0.0206 x 100%
g H2SO4
98.078g
= 2.06%
Percent Composition: Example 2
• Still using H2SO4, determine the percentage of
the mass for sulfur and oxygen.
Percent Composition: Example 2
• Still using H2SO4, determine the percentage of
the mass for sulfur and oxygen.
• S: 32.066g/ 98. 078g x 100% = 32.69% S
• O: 63.996g/ 98. 078g x 100% = 65.25% O
Empirical Formulas
• Empirical Formula: the simplest whole-number ratio of
atoms in a compound
1. Assume that you have 100g of substance
2. Determine the number of grams of each element
from the percent composition (ex. 50% carbon = 50g
carbon)
3. Convert each element from grams to moles
4. Divide the moles of each element by the smallest
value to get the simplest whole-number ratio
•
•
•
•
If you have a decimal of 0.5, multiply EVERYTHING by 2
If you have a decimal of 0.333, multiply EVERYTHING by 3
If you have a decimal of 0.25, multiply EVERYTHING by 4
If the number is VERY close to a whole number (ex.
0.999), it is acceptable to round
Empirical Formula Example
• Determine the empirical formula for a
compound consisting of the following:
71.65% Cl
24.27% C
4.07% H
Empirical Formula Example
• Determine the empirical formula for a
compound consisting of the following:
71.65% Cl
24.27% C
4.07% H
Cl: 71.65g x 1mol/35.45g = 2.021mol/ 2.021 = 1
C: 24.27g x 1mol/ 12.01g = 2.021mol/ 2.021 = 1
H: 4.07g x 1mol/1.008g = 4.04mol/ 2.021 = 2
ClCH2
Molecular Formula
• Molecular Formula: the exact formula of a
molecule
– May or may not be the same as the empirical formula
– You need to know the molar mass of the compound!
1. Determine the empirical formula
2. Find the molar mass of the empirical formula
3. Divide the molar mass of the compound by the
molar mass from the empirical formula
4. Multiply the subscripts in the empirical formula
by the above ratio
Molecular Formula Example
• Earlier, we determined the empirical formula
was ClCH2. Given that the molar mass of the
compound is 98.96g/mol, determine the
molecular formula.
Molecular Formula Example
• Earlier, we determined the empirical formula
was ClCH2. Given that the molar mass of the
compound is 98.96g/mol, determine the
molecular formula.
• ClCH2: 35.45g + 12.01g + 2.02g = 49.48g/mol
• 98.96g/mol = 2
49.48g/mol
• (ClCH2)2 = Cl2C2H4
Practice
• Determine the empirical and molecular
formulas for a compound containing 40.0%
carbon, 6.7% hydrogen, and 53.5% oxygen.
The compound has a molar mass of
180.156g/mol.
Practice
Determine the empirical and molecular formulas for a
compound containing 40.0% carbon, 6.7% hydrogen, and
53.5% oxygen. The compound has a molar mass of
180.156g/mol.
C: 40.0g x 1mol/12.0107g = 3.33mol/ 3.33 = 1
H: 6.7g x 1mol/ 1.00794g = 6.65mol/ 3.33 = 2
O: 53.3g x 1mol/15.9994g = 3.33mol/ 3.33 = 1
Empirical: CH2O = 30.026g/mol
180.156 = 6 : CH2O  Molecular: C6H12O6
30.026
Exit Ticket #2
Exit Ticket #2- Show ALL Work!
1. What is the percent composition of oxygen in H2O? (2
points)
2. Which of the following is an empirical formula? (1
point)
a. C6H12O6 b. C2H4NO c. C2H6 d. All of the above.
3. A compound’s empirical formula is C3H7 and its
molecular weight is 86 g/mole. Find its molecular
formula. (2 points)
4. A compound consists of 72.2% magnesium and 27.8%
nitrogen by mass. What is its empirical formula? (2
points)
5. A compound is 58.8 % C, 9.9 % H, and 31.3 % O. If its
molecular mass is 306 g/mole, what is the molecular
formula? (3 points)
Exit Ticket #3
Exit Ticket #3
• Balance the following reactions. Write a
coefficient 1 wherever needed so I can tell you
didn’t just leave it blank. (2 points each)
1. S8 + NO2  SO2 + N2
2. C3H8 + O2  CO2 + H2O
3. C6H6 + HNO3  C6H5NO2 + H2O
4. C5H10 + O2  CH2O
5. N2 + C2H6  N2H4 + C2H2
Stoichiometry
• Stoichiometry: the area of chemistry that
considers the quantities of materials
consumed and produced in chemical reactions
• The coefficients in a balanced reaction tell you
the RATIO of MOLES for each compound that
are needed for the reaction
Reading the Reaction Equation
• Cl2 + 2 KBr  2 KCl + Br2
– How many moles of Cl2 will react with 2 moles of
KBr?
• P4 + 5 O2 + 6 H2O 4 H3PO4
– How many moles of O2 are needed to produce 4
moles of H3PO4?
• 2 Fe + 3 Cl2  2 FeCl3
– How many moles of Fe will produce 2 moles of
FeCl3?
Reading the Reaction Equation
• Cl2 + 2 KBr  2 KCl + Br2
– How many moles of Cl2 will react with 2 moles of
KBr? 1 mole
• P4 + 5 O2 + 6 H2O 4 H3PO4
– How many moles of O2 are needed to produce 4
moles of H3PO4? 5 moles
• 2 Fe + 3 Cl2  2 FeCl3
– How many moles of Fe will produce 2 moles of
FeCl3? 2 moles
Stoichiometry
P4 + 5 O2 + 6 H2O 4 H3PO4
What is the mole ratio between O2 and H3PO4?
(How many moles of O2 per how many moles
of H3PO4?)
Stoichiometry
P4 + 5 O2 + 6 H2O  4 H3PO4
What is the mole ratio between O2 and H3PO4?
(How many moles of O2 per how many moles
of H3PO4?)
There are 5 mol of O2 per 4 mol of H3PO4
5 mol O2
or 4 mol H3PO4
4 mol H3PO4
5 mol O2
Stoichiometry
P4 + 5 O2 + 6 H2O  4 H3PO4
What is the mole ratio between P4 and O2?
What is the mole ratio between P4 and H2O?
What is the mole ratio between P4 and H3PO4?
Stoichiometry
P4 + 5 O2 + 6 H2O  4 H3PO4
What is the mole ratio between P4 and O2?
1mol P4/ 5 mol O2
What is the mole ratio between P4 and H2O?
1mol P4/ 6mol H2O
What is the mole ratio between P4 and H3PO4?
1mol P4/4mol H3PO4
Stoichiometry
P4 + 5 O2 + 6 H2O  4 H3PO4
What is the mole ratio between O2 and H2O?
What is the mole ratio between H2O and H3PO4?
Stoichiometry
P4 + 5 O2 + 6 H2O  4 H3PO4
What is the mole ratio between O2 and H2O?
5mol O2/ 6mol H2O
What is the mole ratio between H2O and H3PO4?
6mol H2O/ 4mol H3PO4
Moles to Moles
• We use the mole ratio (from the balanced
reaction) as a conversion factor
P4 + 5 O2 + 6 H2O 4 H3PO4
• If we started with 6mol of O2, how many
moles of H3PO4 would we produce? (Set up
the conversion!)
Moles to Moles
P4 + 5 O2 + 6 H2O 4 H3PO4
• If we started with 6mol of O2, how many
moles of H3PO4 would we produce? (Set up
the conversion!)
6mol O2 x 4mol H3PO4 = 24mol H3PO4 = 4.8mol
1
5mol O2
5mol
 4.8 mol of H3PO4
Moles to Moles to Grams
P4 + 5 O2 + 6 H2O 4 H3PO4
• We determined that if we start with 6mol O2,
we produce 4.8mol H3PO4. How many grams is
this? (Continue with the same conversion setup! Round molar mass to the nearest whole
number.)
Moles to Moles to Grams
P4 + 5 O2 + 6 H2O 4 H3PO4
• We determined that if we start with 6mol O2,
we produce 4.8mol H3PO4. How many grams is
this?
6mol O2 x 4mol H3PO4 x 98g H3PO4 = 470.4g
1
5mol O2
1mol H3PO4
Grams to Moles to Moles to Grams
P4 + 5 O2 + 6 H2O 4 H3PO4
• If we start with 25g of O2, how many grams of
H3PO4 will we produce? (Round molar masses
to the nearest whole number)
Grams to Moles to Moles to Grams
P4 + 5 O2 + 6 H2O 4 H3PO4
• If we start with 25g of O2, how many grams of
H3PO4 will we produce?
25g O2 x 1mol O2 x 4mol H3PO4 x 98g H3PO4 =
1
32g
5mol O2 1mol H3PO4
=61.25g H3PO4
Exit Ticket #4
Limiting Reactant
• Limiting Reactant: the reactant that limits the
amount of product that can be formed in a
reaction
• Hint: when the word “excess” is used in a
problem, it means there is MORE than enough
of that reactant, so that reactant will not be
limiting
Finding the Limiting Reactant
• For each of your reactants, multiply the
number of MOLES you start with by the
coefficient in front of that compound in the
balanced reaction
• Whichever compound has the smaller number
is the limiting reactant
Finding the Limiting Reactant
1 C3H8 + 5 O2  3 CO2 + 4 H2O
You have 3mol C3H8 of and 1.8mol of O2. What is
the limiting reactant?
Finding the Limiting Reactant
1 C3H8 + 5 O2  3 CO2 + 4 H2O
You have 3mol C3H8 of and 1.8mol of O2. What is
the limiting reactant?
C3H8: 3mol x 1 = 3mol  Limiting
O2: 1.8mol x 5 = 9mol
Finding the Limiting Reactant
1 C3H8 + 5 O2  3 CO2 + 4 H2O
You have 7.6mol C3H8 of and 2.3mol of O2. What
is the limiting reactant?
Finding the Limiting Reactant
1 C3H8 + 5 O2  3 CO2 + 4 H2O
You have 7.6mol C3H8 of and 2.3mol of O2. What
is the limiting reactant?
C3H8: 7.6mol x 1 = 7.6mol  Limiting
O2: 2.3mol x 5 = 11.5mol
Finding the Limiting Reactant
1 C3H8 + 5 O2  3 CO2 + 4 H2O
You have 15.9mol C3H8 of and 3.3mol of O2.
What is the limiting reactant?
C3H8: 15.9mol x 1 = 15.9mol
O2: 3.1mol x 5 = 15.5mol  Limiting
Finding the Limiting Reactant
1 C3H8 + 5 O2  3 CO2 + 4 H2O
You have 7.5mol C3H8 of and 1.5mol of O2. What
is the limiting reactant?
C3H8: 7.5mol x 1 = 7.5mol
O2: 1.5mol x 5 = 7.5mol  No limiting reactant
Finding the Limiting Reactant
1 C3H8 + 5 O2  3 CO2 + 4 H2O
You have 45g C3H8 of and 40g of O2. What is the
limiting reactant?
Finding the Limiting Reactant
1 C3H8 + 5 O2  3 CO2 + 4 H2O
You have 45g C3H8 of and 40g of O2. What is the
limiting reactant?
*Since we’re given grams, we must convert to moles!
Finding the Limiting Reactant
1 C3H8 + 5 O2  3 CO2 + 4 H2O
You have 45g C3H8 of and 40g of O2. What is the
limiting reactant?
C3H8: 45g x 1mol = 1.02mol x 1 = 1.02mol (limiting)
44.097g
O2: 40g x 1mol = 1.25mol x 5 = 6.25mol
31.998g
Determining Yield
• Theoretical Yield: the amount of product that a
reaction produces if it reaches completion under
perfect conditions
– Once you know the limiting reactant, you can
calculate how much of your product the reaction
should produce
– To set up the conversion, use whatever amount of the
limiting reactant was given to you, then set up the
conversion from there
Determining Yield
1 C3H8 + 5 O2  3 CO2 + 4 H2O
You have 3mol C3H8 of and 1.8mol of O2. How
many grams of H2O will be produced?
C3H8: 3mol x 1 = 3mol  Limiting O2: 1.8mol x 5 = 9mol
Determining Yield
1 C3H8 + 5 O2  3 CO2 + 4 H2O
You have 3mol C3H8 of and 1.8mol of O2. How
many grams of H2O will be produced?
C3H8: 3mol x 1 = 3mol  Limiting
O2: 1.8mol x 5 = 9mol
3mol C3H8 x 4mol H2O x 18g H2O = 216g H2O
1
1mol C3H8 1mol H2O
Determining Yield
1 C3H8 + 5 O2  3 CO2 + 4 H2O
You have 7.6mol C3H8 of and 2.3mol of O2. How
many grams of CO2 are produced?
C3H8: 7.6mol x 1 = 7.6mol  Limiting
=11.5mol
O2: 2.3mol x 5
Determining Yield
1 C3H8 + 5 O2  3 CO2 + 4 H2O
You have 7.6mol C3H8 of and 2.3mol of O2. How
many grams of CO2 are produced?
C3H8: 7.6mol x 1 = 7.6mol  Limiting
=11.5mol
O2: 2.3mol x 5
7.6mol C3H8 x 3mol CO2 x 44g CO2 = 1,003g H2O
1
1mol C3H8 1mol CO2
Determining Yield
1 C3H8 + 5 O2  3 CO2 + 4 H2O
You have 200g C3H8 of and 200g of O2. How many
grams of H2O are produced?
Determining Yield
1 C3H8 + 5 O2  3 CO2 + 4 H2O
You have 200g C3H8 of and 200g of O2. How many
grams of H2O are produced?
C3H8: 200g x 1mol = 4.55mol x 1 = 4.55mol: Limiting
44g
O2: 200g x 1mol = 6.25mol x 5 = 31.25mol
32g
200g C3H8 x 1mol C3H8 x 4mol H2O x 18g H2O = 327.3g
1
44g C3H8 1mol C3H8 1mol H2O
Percent Yield (Recovery)
• Percent Yield (Percent Recovery): the percent
of the theoretical yield that was obtained
during an experiment
– Will always be less than 100%
Percent Yield = Actual Yield x 100%
Theoretical Yield
Percent Yield (Recovery)
Previously, we calculated that 216g H2O would
be produced if we started with 3mol C3H8 and
1.8mol of O2. If we recovered 200g, what was
our percent yield (percent recovery)?
Percent Yield (Recovery)
Previously, we calculated that 216g H2O would
be produced if we started with 3mol C3H8 and
1.8mol of O2. If we recovered 200g, what was
our percent yield (percent recovery)?
% Yield = 200g x 100 = 92.6%
216g
Percent Yield (Recovery)
Previously, we calculated that 1,003g H2O would
be produced if we started with 7.6mol C3H8
and 2.3mol of O2. If we recovered 578g, what
was our percent yield (percent recovery)?
% Yield = 578g x 100 = 57.6%
1,003g
Summary of Stoichiometry
1. Balance the reaction.
2. Find the limiting reactant by multiplying
starting moles by the coefficients in the
reaction.
3. Determine the theoretical yield
(conversions).
4. If given actual yield, determine percent yield.
Making Solutions: Molarity
• Molarity: the concentration of a solution,
measured in moles of solute per liter of
solvent
M = mol solute / L solvent
Solute: what you’re dissolving
Solvent: what does the dissolving (usually a
liquid)
Making Solutions
• If you have 10mol of HCl dissolved in 0.75L of
water, what is the molarity of the solution?
Making Solutions
• If you have 10mol of HCl dissolved in 0.75L of
water, what is the molarity of the solution?
M = mol/ L
M = 10mol/ 0.75L = 13.3M
Making Solutions
• If you have a 6M solution of HCl that is 300mL,
how many moles of HCl are dissolved?
Making Solutions
• If you have a 6M solution of HCl that is 300mL,
how many moles of HCl are dissolved?
M = mol/ L
300mL x 1L/ 1,000mL = 0.3L
6M = mol/ 0.3L
*0.3L = *0.3L
 1.8mol
Making Solutions
• You’re trying to make a 1.25M solution of HCl.
If you have 500g of HCl, what volume must
the solution be in L? In mL?
Making Solutions
• You’re trying to make a 1.25M solution of HCl.
If you have 500g of HCl, what volume must
the solution be in L? In mL?
M = mol/L
500g x 1mol/ 36g = 13.9mol
1.25M = 13.9mol/ L
11.1L
11.1L x 1,000mL/ 1L = 11, 100L
Dilutions
• Dilution: adding solvent to a solution to
decrease the molarity (concentration)
• The product of the initial volume and molarity
= the product of the final volume and molarity
M1V1 = M2V2
M = concentration (molarity)
V = volume
Dilutions Example
• You have a 250mL stock solution of 5M HCl.
You need to make a dilution to 1.5M HCl.
What must the final volume be? How much
water will you have to add?
Dilutions Example
• You have a 250mL stock solution of 5M HCl. You
need to make a dilution to 1.5M HCl. What must
the final volume be? How much water will you
have to add?
M 1 V1 = M 2 V2
5M*250mL = 1.5MV2
1,250M*mL = 1.5MV2
1.5M
1.5M
833.3mL = V2
Molarity of Sugar in Sodas Activity
• Your task is to determine the molarity
(concentration) of sugar in various beverages. To
do this, you must determine:
– The amount of sugar in one serving
– The volume of one serving
• We will make the following assumptions:
– All sugar in the beverage is fructose: C6H12O6
– There are 30mL per 1 fl. oz.
• When you finish, read the article on sugar and
tooth decay, and relate the molarity of sugar that
you found to the article.
Exit Ticket #6
Exit Ticket #6
Show all work, including the original equation. The molar mass of HCl is
36.5g/mol. Round your answers to the nearest tenth.
1. How many mol of HCl are in a 2L solution with a
concentration of 6.5M? (1 point)
2. How many grams of HCl are required to make a 2M,
350mL solution? (2 points)
3. What volume of HCl is needed to create a 3.5M solution
from a starting mass of 150g? (3 points)
4. You dilute a 5.5M solution of HCl from 100mL to 250mL.
What is the final concentration? (1 point)
5. You leave a 200mL solution of 3M HCl out overnight,
and some of the water evaporates. If the final volume is
125mL, what is the final concentration? (2 points)
6. You have a 3M solution of HCl. You add water to the
solution until you reach a final concentration of 1.2M. If
you started with 175mL, how much water did you add?
(3 points)