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Lavoisier: Law of conservation of mass! Chemical Reaction Mass before = Mass after. Chapter 3: Stoichiometry. Scientific contributions cut short – Antoine Lavoisier was guillotined in 1794 at the age of 60. Today, considered to be father of modern chemistry. Chemical Equations. - PowerPoint PPT Presentation
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Lavoisier: Law of conservation of mass!
Chemical Reaction
Mass before = Mass after.
Chapter 3: Stoichiometry
Scientific contributions cut short – Antoine Lavoisier was guillotined in 1794at the age of 60. Today, considered to be father of modern chemistry.
1
Chemical equations are descriptions of chemical reactions.
There are 2 parts to an equation: reactants and products:
H2 + O2 H2O
To balance:
change only the stoichiometric coefficients
cannot change the chemical species
Stoichiometric coefficients: numbers in front of the chemical formulas; give ratio of
reactants and products.
Chemical Equations
2
Consider the unbalanced chemical equation:
H2 + O2 H2O
The equation H2 + O2 H2O2 is a balanced equation but for a different reaction.
DO NOT change subscripts!
Balancing options
H2 + 1/2 O2 H2O, or
2H2 + O2 2H2O, or
4H2 + 2O2 4H2O, etc.
Preferably, no fractions and smallest possible common coefficients used.
3
Example 1Write a balanced equation for the combustion of octane, C8H18.
We know that our reactants include C8H18 and O2(g), and the products will be CO2 and H2O.
But in what ratios will they react and produce? Let’s balance!
Let’s consider combustion reactions which involve the burning/oxidation of hydrocarbons to produce CO2 and
H2O always.
Hydrocarbons are organic compounds made up only of carbon and hydrogen.
For burning/oxidation to occur O2 has to be present! All combustion equations will include O2(g) as a reactant!
4
C8H18 + O2(g) CO2 + H2O
We need to increase C on the product side to 8, using a stoichiometric coefficient.
C8H18 + O2(g) 8CO2 + H2O
We need to increase H on the product side to 18, using a stoichiometric coefficient.
Element Reactants Products
C 8 1
H 18 2
O 2 3
Element Reactants Products
C 8 8
H 18 2
O 2 17
Start with C & H. Do O
last because it appears in 3 species.
5
C8H18 + O2(g) 8CO2 + 9H2O
We now need to increase O2 on the reactant side. It needs to change from 2 to 25. This can only be done using a fraction!
C8H18 + 12.5 O2(g) 8CO2 + 9H2O
Now to remove the fraction, and keep coefficients as small as possible we multiply ALL coefficients by 2!
2C8H18 + 25O2(g) 16CO2 + 18H2O
Element Reactants Products
C 8 8
H 18 9x2 = 18
O 2 16+9 =25
Element Reactants Products
C 8 8
H 18 18
O 12.5x2=25 25
===
6
C8H18 + O2 CO2 + H2OHINT: Balance O last (occurs in 3 species)
but C and H only appear in two!
Balance C
C8H18 + O2 8CO2 + H2O
Balance H
C8H18 + O2 8CO2 + 9H2O
Balance O
C8H18 + 25/2O2 8CO2 + 9H2O
2C8H18 + 25O2 16CO2 + 18H2O
We usually add the physical states of the reactants and products
2C8H18(l) + 25O2(g) 16CO2(g) + 18H2O(l)
Here is another set up of the same example:
7
In combustion reactions:
Rapid reactions that produce a flame.
Usually very “clean” and products are
predictable, but incomplete combustion
(leftover ash) can be very harmful.
8
Example 2
Write a balanced equation for the combustion of purine, C5H4N4.
C CO2 H H2O N N2
9
Patterns of chemical reactivity
Alkali metal + water
2Na + 2H2O 2NaOH + H2
2K + 2H2O 2KOH + H2
Elements in the same group of the periodic table react in a similar manner.
Lithium Sodium Potassium
10
Alkali earth metal + water
Mg + 2H2O Mg(OH)2 + H2
Ca + 2H2O Ca(OH)2 + H2
Note: These are the EXACT same products as for alkali metals, except for the stoichiometry!!!
Always look for patterns in chemistry!
11
2Mg(s) + O2(g) 2MgO(s)
Mg has combined with O2 to form MgO.Two reactants combine to form a single product.
They are characterized by having fewer products than reactants.
Metal + oxygen
Combination Reactions
12
C(s) + 2H2(g) CH4(g)
Non-metal + hydrogen
Ca(s) + Cl2(g) CaCl2(s)
Metal + halogen
2Li(s) + F2(g) 2LiF(s)
13
They are characterized by having fewer reactants than products.
2NaN3(s) 2Na(s) + 3N2(g)
The NaN3 is ignited and rapidly decomposes into Na and N2 gas
Metal azide
Metal carbonate
CaCO3(s) CaO(s) + CO2(g)
limestone,seashells
lime
Decomposition Reactions
14
15
Quantitative vs. Qualitative Chemical formulae and equations both have qualitative and
quantitative significance.
Chemical formula says what compounds you are working with and the equation gives a qualitative idea of how the reactants react, and what
products they produce.
N2 + 3H2 2NH3 (balanced)
Quality: We have nitrogen and hydrogen as reactants. The equation tells us that they undergo a combination reaction to form ammonia.
Subscripts in formulae and coefficients in balanced equations represent precise quantities.
Quantity: The product, NH3, is composed of 3H for every 1N atom.
The balanced equation shows that 2 molecules of NH3 are produced for every N2 and 3H2 that are used up. 16
Average Atomic MassREVISION FROM ISOTOPES!
Most elements occur as mixtures of isotopes.
To determine the average atomic mass of an element we must consider each isotopes mass as well as their respective
abundances.
Before we consider real isotopes, lets create an analogy using popcorn
I have a bowl of popcorn where 93% of the pieces weigh 1.14 mg, 2% weigh 1.09 mg and 5% weigh 1.03 mg.
Which popcorn piece mass is in the greatest and lowest abundance?
Which mass will dominate in calculating the average mass of the popcorn pieces?
So what do we expect the average mass to be? 17
If the bowl contains 1000 popcorn pieces in it. Then 930 of them would weigh 1.14 mg, 20 would weigh 1.09 mg and 50
would weigh 1.03 mg.
Let’s add those abundances together = 930 + 20 + 50
= 1000.
This shows that each isotope’s (popcorn-piece mass’) abundance must make up 100% , which is 1000 in this case.
In isotopes be sure to check that your abundances ALWAYS add up to 100% 18
......%100
%
%100
%2
2
1
1
mass
abundancemass
abundanceweightAverage
mgmgmgweightAverage 03.1
%100
%509.1
%100
%214.1
%100
%93
321
mgmgmgweightAverage 0515.00218.006.1
Now let us calculate the average weight of the popcorn pieces. (We are expecting it to
be just a little less than 1.14 mg.)
mgweightAverage 13.1
19
Example 3
Three isotopes of silicon occur in nature, calculate the average atomic weight of silicon.
28Si (92.21%) mass = 27.97693 amu
29Si (4.70%) mass = 28.97659 amu
30Si (3.09%) mass = 29.97376 amu
First make a logical prediction!
......%100
%
%100
%2
2
1
1
mass
abundancemass
abundanceweightAverage
20
Molecular and Formula Weights
Formula weight (Fr or FW) is the mass of a collection of atoms represented by a chemical formula.
So it is the sum of atomic masses (Ar) of the atoms in the chemical formula.
Fr (K2Cr2O7) = 2Ar(K) + 2Ar(Cr) + 7Ar(O)
= 2(39.10) + 2(52.00) + 7(16.00)
= 294.20 amu
For example, K2Cr2O7
Significant figures!!
21
If the chemical formula of a substance is the same as its molecular formula, then the formula weight is also called the
molecular weight.
Molecular weight (Mr or MW) is the mass of a collection of atoms represented by a chemical formula for a molecule.
MW(C6H12O6) =
What is the molecular formula for glucose?
Formula weight and molecular weight are virtually identical and are used interchangeably.
When dealing with an ionic compound (3D lattice) we do not use molecular formulas to name them, and so we could not calculate a molecular weight.
In this case we use the formula weight.22
Some compounds have water of crystallization. These are water molecules associated with the solid as it crystallizes
from solution.
CuSO4•5H2O5 water molecules per CuSO4 unit(Add them into molecular weight!)
Can often be driven off:
CuSO4•5H2O → CuSO4 + 5H2Oblue grey
anhydrous copper sulphate
Water of Crystallization
23
Mole: a convenient measure of enormous numbers.
NEW PUBLIC HOLIDAY!Mole Day: October 23 from 6:02am to 6:02pm, invented May 15, 1991.
6.022 x 1023 is known as Avogadro’s Number, NA
Defined as the amount of matter that contains as many objects (atoms/molecules etc.) as the number of atoms in 12g of 12C
12 g = 6.02214 x 1023 atoms
The Mole
24
1 mole of anything = 6.022 1023 of that thing.
1 mole 12C atoms = 6.022 1023 12C atoms. 1 mole H2O molecules = 6.022 1023 H2O molecules.
1 mole NO3- ions = 6.022 1023 NO3
- ions.
The Mole
Example 4
Calculate the number of carbon atoms in 0.350 mol of C6H12O6.
25
Example 4
Calculate the number of carbon atoms in 0.350 mol of C6H12O6.
Let’s look at this more mathematically now!
Avo’s number provides a conversion factor between the number of moles and molecules of a given compound.
1 mole sugar = 6.022 x 1023 molecules of sugar.
So we can use the “GIVEN and DESIRED” formula! (Given: mol, Desired molecules!)
unitDesiredunitGivenunitGiven
unitDesired
26
The mass in grams of 1 mol of substance (units g/mol).
Molar Mass
27
The molar mass (in g) of any substance is always numerically equal to its formula weight (in amu).
One HCl molecule weighs 36.46 amu One mol of HCl weighs 36.46 g.
Can you see where this comes from?
Remember we said that amu and g.mol-1 are equivalent and can be used interchangeably?
So….. HCl weighs 36.46 amu = 36.46 g.mol-1.36.46 g.mol-1 x 1 mol = 36.46 g.
1 mol of N2
1 mol of H2O
1 mol of NaCl
18 g
28 g
58.45 g
28
MAKE SURE YOU UNDERSTAND THESE RELATIONSHIPS!!!
29
Inter-converting masses, moles and number of particles:
Molar Mass
30
Converting from mass to moles and moles to mass is straight forward:
1.Calculate the molar mass of the substance2.Use the mole concept
number of moles = mass / molar mass n = m / Mr
Once the number of moles of the substance is calculated, use Avogadro’s number to attain number of molecules:
number of molecules = n x AVO
Now we use the number of molecules to solve for number of atoms, or ions within the substance.
TAKE NOTE OF THESE PARTICLES AND HOW THEY RELATE TO EACH OTHER!
Molar Mass
31
Sugar C6H12O6
This is 1 MOLECULE of sugar.
This molecule contains 24 ATOMS; 6 carbon, 12 hydrogen and 6 oxygen.
Sulphuric Acid H2SO4
Example 5
Calculate the number of H atoms in 20.00 g of C6H12O6.
Mass Mols Molecules
Strategy
n = molesm = mass (grams)MW = molecular weight or molar mass (g/mol)
n = molesNA = Avogadro’s Number
32
MUST know two things to figure this out…(1)mass (g) and (2)molecular weight (molar mass) (g.mol-1)
…of the compound in question.
33
AMOUNT number of moles (must always compare moles!)
n = mass / molar mass, or n = concentration/volume.
PERCENT COMPOSITION
r
r
No. of atoms of Element A% Element 100
F of Compound
Definitions
Calculate by dividing the atomic weight (Ar) for each element by the formula weight (Fr) of the compound, and
express as a percentage:
34
What is the % by mass (% composition) of O in K2Cr2O7?
r
r
No. of atoms of Element A% Element 100
F of Compound
First determine atomic and formula weights (or MWs).
Then apply equation (works the same in amu or g/mol).
Example 6
35
36
Percent Composition
If you are battling to understand what is meant by percentage (by mass) contributed by each element in the
compound then just think about a box of colourful smarties!
What percentage of the box of smarties is yellow?
100#
#%
smartiesTotal
smartiesyellowYellow
So it’s the same as ELEMENTAL composition, except that we need to consider mass when we use elements!!!
Mass %of elements
Grams ofeach element
in sample
Empirical formula
Strategy
Assumehave 100 g
sample Moles of each element
n = mMr
37
Empirical Formulas from Analyses
Mole ratio
Mr of that specific element!
Example 7 (p. 85 8th Ed; p. 93 9th Ed; p. 98 10th Ed)
Information given in problem
Ethylene glycol: 38.7% C 9.7% H
51.6% O by mass
Molar mass (from mass spectrometry): 62.1 g mol-1
Calculate:(a) empirical formula(b) molecular formula
Ethylene glycol, the substance used in automobile antifreeze,is composed of 38.7% C, 9.7% H, and 51.6% O by mass. Itsmolar mass is 62.1g/mol.
38
C H O
38.7% 9.7% 51.6%
38.7g 9.7g 51.6g
3.22 mol 9.62 mol 3.23 mol
3.22/3.22 9.62/3.22 3.23/3.22
1 2.99 1.003
assume 100 g
number of moles
38.7g 9.7g 51.6g12.01 g mol-1 1.008 g mol-1 16.00 g mol-1
mole ratio (divide by smallest no. mols)
39
Experimental error is to be expected, so
C H O1 2.99 1.003
C H O1 3 1
is likely to be
The empirical formula is therefore CH3O
which has the formula weight 1 x 12.01 + 3 x 1.008 + 1 x 16.00= 31.03 g/mol
But molecular weight = 62.1 g/mol, which is twice this value
The molecular formula must be C2H6O2.40
CxHyOz Catalysis: completeoxidation of C to CO2
Hydrogen in CxHyOz isoxidized to H2O and absorbed in the H2Oabsorber
C in CxHyOz isabsorbed here
O in CxHyOz is determined by mass difference
Combustion Analysis to Determine Experimental Data
41
Example 88th ed., p. 99, 3.54(a); 9th ed., p. 107, 3.48(a); 10th ed., p. 114, 3.52(a)
Combustion of 2.78 mg of ethyl butyrate produced 6.32 mg CO2 and 2.58 mg of H2O. What is the empirical formula of this compound?
Consider the given information separately. Let’s look at CO2 first. Recallthat we can only determine C and H content from information given!
42
Ask yourself: How much of the 6.32 mg of CO2 is due to C?
Combustion of 2.78 mg of ethyl butyrate produced 6.32 mg CO2 and 2.58 mg of H2O. What is the empirical formula of this compound?
Now let’s deal with H2O, noting we can only determine H content.
43
Example 8
How much of the 2.58 mg of H2O is due to H?
Combustion of 2.78 mg of ethyl butyrate produced 6.32 mg CO2 and 2.58 mg of H2O. What is the empirical formula of this compound?
Finally, same strategy as before…we have all three masses and can determine the
empirical formula.
Now, determine how much O is present from the difference in mass.
44
Example 8
START HERE
THIS TIME!
45
C H O
1.73 x 10-3 g 0.289 x 10-3 g 0.76 x 10-3 g12.01 g mol-1 1.008 g mol-1 16.00 g mol-1
1.44 x 10-4 mol 2.87 x 10-4 mol 4.75 x 10-5 mol4.75 x10-5 mol 4.75 x10-5 mol 4.75 x10-5 mol
Determine moles of each element
Divide each by smallest # moles to determine ratios
46
Quantitative Information from Balanced Equations
In a balanced equation the coefficients can be interpreted as the relative numbers of molecules involved in the
reaction AND as the relative number of moles.
When given any chemical equation you always ensure that you BALANCE it 1st!
Then you use the given information (usually a mass in grams, or a number of moles in mol), together with your calculated molar masses, to solve for
the unknown.
47
Quantitative Information from Balanced Equations
Example 9
What mass of CO2 is produced when 1.00 g of propane, C3H8, is burned?
First, write the balanced equation
Then, look at what is given…grams of reactant. In orderto determine how much CO2 is produced, you need to
consider reaction STOICHIOMETRY.
48
STRATEGY
49
What mass of CO2 is produced when 1.00 g of propane, C3H8, is burned?
C3H8 + 5O2 → 3CO2 + 4H2O
1.00 g ??? g
Calculate moles of propane:
Next, consider the MOLAR relationship between propane and carbon dioxide.It is NOT 1:1, but rather 1:3 (THIS is stoichiometry!).
Stoichiometric coefficients are MOLE relationships!!50
Finally, calculate mass (g) of CO2 produced…need molar mass!
Follow-up problem: What mass of O2 was consumed in the process? Ans: 3.63 g
51
Notice how we link or compare the number of moles of each substance!! Not their masses,
or anything else!!
If the reactants are not present in stoichiometric amounts, then at the end of a reaction some reactants are still present.
EXCESS!
52
Limiting Reactants
Limiting ReactantOne reactant that is consumed
completely in the reaction.
The limiting reactant/reagent hampers us from continuing
with the reaction!
Non-Chemistry Limiting Reagent
Cheese Sandwich
2B + 1C = B2C
Ideally…
More likely…
4B + 1C = B2C + 2BYou’ve run out of cheese,
time to go to the store.
53
The cheese limits us from making another sandwich! Therefore we call it the limiting reagent!
Suppose we have 10 mol H2 and 7 mol O2
H2(g) + O2(g) → H2O(l)
54
Strategy
Example10Al reduces Fe2O3 to Fe. How much Fe is produced from the reaction of 30.0 g of Al and 100 g Fe2O3?
Mass of product
Moles of product basedon limiting reagent
Massreactants
Identify limitingreagent
molesreactants
55
Solving Limiting Reagant Problems
Mass ofreactants
Identify limitingreagent
Mass of product
Moles ofreactants
Moles product based on
limiting reagent
1
2
3
4
5
Example 10 Al reduces Fe2O3 to Fe. How much Fe is produced from the reaction of 30.0 g of Al and 100 g Fe2O3?
56
Massreactants
Identify limitingreagent
Mass of product
molsreactants
Mols of product based on
limiting reagent
1
2
3
4
5
From reaction stoichiometry
2 Al : 1 Fe2O3
__________is in excess, and __________ is the limiting reagent!
57
Massreactants
Identify limitingreagent
Mass of product
molsreactants
Mols of product based on
limiting reagent
1
2
3
4
5
Now compare stoichiometry of Fe to Al.
m = n x Mr
=
Finally, calculate mass of Fe produced.
58
= of Fe
100yield lTheoretica
yield ActualYield %
Theoretical yield
The percent yield relates the actual yield (amount of material recovered in the laboratory) to the theoretical
yield:
59
Percentage Yield
So, we conduct an experiment in the lab:Let us make a metal oxide, calcium oxide!
From theory we now know that we need to do a combination reaction using a metal and O2 as our reactants.
The equation: Ca + O2 CaOBalance: 2Ca + O2 2CaO
60
Percentage Yield
We are given 2.053 g of calcium, and the oxygen used is atmospheric oxygen, and therefore is in excess (unlimited
quantity).
From this information we can calculate how much product the stoichiometry of our equation predicts we will produce
theoretical yield.
61
Percentage YieldExample 11Preparation of calcium oxide is achieved by reacting 2.053 g of calcium with an excess of oxygen. Calculate the theoretical yield of calcium oxide, and solve for the percentage yield, if the real mass of the product was measured to be 2.26 g.
2Ca + O2 2CaO
m = 2.053 g m = excess m = desired quantity!
Mr = 40.078g/mol Mr = 32.00 g/mol Mr = 56.08 g/mol
n= mass/molar mass n= n(Ca) /2 n(CaO) = n(Ca)
n = 2.053 g = 0.05123 mol
40.078 g.mol-1n = 0.05123 mol
= 0.02561 mol 2
n(CaO) = n(Ca)
n = 0.05123 mol
62
2Ca + O2 2CaO
Example 11Calculate the theoretical yield of calcium oxide, and solve for the percentage yield, if the real mass of the product was measured to be 2.26 g.
m = 2.053 g m = excess m = desired quantity!
Mr = 40.078g/mol Mr = 32.00 g/mol Mr = 56.08 g/mol
n= 0.05123 mol n= 0.02561 mol n= 0.05123 mol
mass(CaO) = n x Mr
= 0.05123 mol x 56.08 g/mol= 2.873 g
100
yield lTheoreticayield Actual
Yield %
63
Expressing Concentration
There are many ways in which you can express the concentration of a solution.
Qualitatively OR Quantitatively
•Mass percentage•Mole fraction•Molarity•Molality
DILUTE or CONCENTRATED
MANY WAYS TO DESCRIBE THIS!
•Dilute = Low concentration of solute.•Concentrated = High concentration of solute.
64
Quantitative Concentration
Mass Percentage
A 24.00 % NaCl solution by mass contains 24.00 g of NaCl in a 100.0 g solution.
So if our solution has a total mass of 634.0 g, then what mass of NaCl does it contain?
100% solutionmassTotal
solutionincomponentofmasscomponentofMass
24.00 % of the total mass is due to the NaCl.
Ans: 152.2 g of NaCl
The remaining mass is due to the SOLVENT (water).
65
Parts per Million (ppm)
Example: A 3 ppm solution
contains 3 particles of solute, for every 1 million (106) particles of solution.
3 g of solute for every million grams of solution.
610solutionmassTotal
solutionincomponentofmasscomponentofppm
very dilute solutions!
The concentration of a solution in grams of solute per 106 (million) grams of solution.
This is equivalent to milligrams (mg) of solute, per litre of solution (for aqueous solutions).
66
Parts per Billion (ppb)
Example: 329 ppb corresponds to 329 µg of solute per litre of solution.
even more dilute solutions!
910solutionmassTotal
solutionincomponentofmasscomponentofppb
The concentration of a solution in grams of solute per 109 (billion) grams of solution.
This is equivalent to micrograms (µg) of solute, per litre of solution (for aqueous solutions).
67
Example 12Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water.
100% solutionmassTotal
solutionincomponentofmasscomponentofMass
68
Example 13A bleaching solution contains 3.62 mass percent sodium hypochlorite. Calculate the mass of sodium hypochlorite in a bottle containing 2500 g of bleaching solution.
ASIDE what is the chemical formula of sodium hypochlorite?
NaOCl
69
Example 14In an experiment 2.569 mg of CuSO4 was dissolved in water to form a 2.000 kg solution. Express the concentration of the solution in parts per million.
610solutionmassTotal
solutionincomponentofmasscomponentofppm
RULE: Both masses MUST be in the same unit of mass!
Convert 2 kg into mg:1st convert kg g and then g mg
70
Mole Fraction
Quantitative Concentrations in terms of moles
componentsallofmolesTotal
componentofmolesComponentoffractionMole
We often express mole fraction using an X.Example: the mole fraction of HCl in a solution of HCl XHCl. 1.00 mol of HCl, dissolved in 8.00 mol of water, gives a mole fraction of HCl of 0.111.
71
Mole Fraction
Quantitative Concentrations in terms of moles
The sum of the mole fractions of all the
components in a solution must equal 1!
Let’s test this theory using our previous example!1.00 mol of HCl dissolved in 8.00 mol of water gives a
mole fraction of HCl of 0.111. So lets calculate the mole fraction of the water!
72
Molarity
Quantitative Concentrations in terms of moles
Example 15: Dissolving 2.36 g of CuSO4 in enough water to fill a 250 ml volumetric flask.
solutionofLitres
soluteofMolesMolarity Units mol.L-1(or more
commonly: M).
1. Weigh out 2.36 g of these crystals.
2. Place in volumetric flask.
3. Fill to the line with water.
73
Molarity
Quantitative Concentrations in terms of moles
Example 15: Dissolving 2.36 g of CuSO4 in enough water to fill a 250 ml volumetric flask.
1. Calculate the number of copper sulphate moles (n) 2. Remember to convert your volume of solution (ml L).
74
Molality
Quantitative Concentrations in terms of moles
There are two major differences between Molality and Molarity!
They both appear in the denominator of the equation:
NB: Notice molaLity not molaRity!
solventofkg
soluteofMolesMolality
MOLARITYConsider the VOLUME of the entire solution
(always in L)!
MOLALITYConsider the MASS of the solvent (always in kg)!
75
Be sure that you are able to
calculate concentrations using each
of these methods.
Example 16:A solution containing equal masses of glycerol (C3H8O3) and water has a density of 1.10 g/ml. Calculate the a)Molality of glycerolb)Mole fraction of glycerolc)Molarity of glycerol
76
What information has been given?
•We have 1 solution. It contains 2 components.•Each component has the same mass.
We don’t know what mass though!•The solution’s density is 1.10 g/mol.
Because we know the masses of glycerol and water are equal….we can make an assumption!
Volume
MassDensity
Assume that each component has a mass of 10.00 g, and therefore the mass of the entire solution must be 20.00 g.
77
Example 16:A solution containing equal masses of glycerol (C3H8O3) and water has a density of 1.10 g/ml. Calculate the a)Molality of glycerolb)Mole fraction of glycerolc)Molarity of glycerol
We can now calculate the number of moles of glycerol.
From our assumption:Mass of solvent = 10.00 g
= 20.00 x 10-3 kg
a) Molality involves the number of moles of the component in question, and the mass in kg of the solvent.
78
b) Calculating the mole fraction of glycerol involves calculating the number of moles of glycerol, and the total number of moles within the solution (nglycerol + nwater).
Example 16:A solution containing equal masses of glycerol (C3H8O3) and water has a density of 1.10 g/ml. Calculate the a)Molality of glycerolb)Mole fraction of glycerolc)Molarity of glycerol
79
c) To calculate the molarity we need the number of moles of glycerol, and the volume of the solution in litres!
Example 16:A solution containing equal masses of glycerol (C3H8O3) and water has a density of 1.10 g/ml. Calculate the a)Molality of glycerolb)Mole fraction of glycerolc)Molarity of glycerol
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