Download 5 pancakes 2 eggs

Chapter 7
Quantitative Composition of Compounds
Making new chemicals is much like following a recipe from a cook book...
1 cup of flour + 2 eggs + ½ tsp baking powder → 5 pancakes
… except you
don’t get to lick
the spoon!
What if you want to make more (or less)?
Suppose you have plenty of flour and
baking powder, but only 8 eggs. How
many pancakes can you make?
You can solve it using conversion factor
that translates eggs into pancakes:
5 pancakes
x
8 eggs
2 eggs
= 20
This is the quantity
you need
5 pancakes
2 eggs
Solve it in your
head:
2 eggs makes 5
pancakes, so four
times more eggs
makes 20 (5x4)
pancakes.
This is the quantity you
want to cancel
Practice using the following mouthwashing, diet-buster recipe:
pancakes
3 blocks cream cheese + 5 eggs + 1
cup sugar = 1 cheese cake.
How many eggs should you use with 9 blocks of cheese?
5 eggs
9 blocks cheese x 3 blocks cheese= 15 eggs
How much sugar do we need for 5 cheese cakes? (5)
Suppose you want to ‘whip’ a batch of hydrogen
iodide, following the balanced chemical equation:
H2 + I2 2 HI
How much H2 and I2 should you use to make 10 g of HI?
A common mistake is that H2 and I2 react in one-to-one mass ratio.
5 g H2 + 5 g I2 10 g HI
The coefficients in a balanced
equation refer to number of atoms
or molecules, not their masses.
Counting atoms is impossible. However, since all atoms of one
type have the same mass, they could be measured by weighing.
The mass of just one atom is too small to be measured on
a balance. Remember: 1 amu = 1.6 x 10-24 g.
Introducing the mole. The mole is like a dozen, but much more.
The mole is Avogadro’s Number of items.
1 mole = 602,214,179,000,000,000,000,000 or 6.022 x 1023.
1 mole of
anything:
donuts, pancakes,
is always 6.022 x 1023
atoms, molecules, ions… of that thing.
1 mole of soft drink cans is enough to cover the surface of
the earth to a depth of over 200 miles.
The mole translates between the number of atoms (or
molecules, ions) and grams of atoms (molecules, ions).
It is defined as the mass of Avogadro’s number of
atoms 126C, which, in turn, weights exactly 12 g.
A mole of atoms weighs the same
number of grams as the atomic mass.
One mole of H atoms weighs 1.008 g.
One mole of C atoms weighs 12.01 g.
Atomic
number
Atomic mass refers to: the sum of protons and
neutrons in a single atom, weighted average mass
of all isotopes of an element and also to the
number of grams in one mole of atoms.
H2
+
I2
1 molecule
2 H atoms
1 molecule
2 I atoms
12 molecules
12 molecules
6.022 x 1023
molecules
1 mole
2.016 g
6.022 x 1023
molecules
1 mole
253.8 g
=
Amadeo Avogadro
Atomic
mass
The mole
1
H
1.008
1 mol of naturally
occurring H atoms has
a mass of 1.008 g
2 HI
2 molecule
2 x (1 atom H, 1 atom I)
24 molecules
1.204 x 1024
molecules
2 mole
255.9 g
or any number of
1 mole of H2 weighs 2 x 1.008 g = 2.016 g
molecules
Conversion factors:
1 mole
6.022 x 1023 species
1 mole
molar mass
Mole - mass - atoms
conversions
Q1: How many atoms in 0.500 mol Au?
6.022 x 1023 atoms Au
0.500 mole Au x
1 mole Au
= 3.011 x 1023 atoms Au
Q1a: How many moles in 7.12 x 1024 atoms of
Cu?
1 mole Cu
7.12 x 1024 atoms Cu x
6.022 x 1023 atoms Cu
= 11.8 mol Cu
Q2: What is the mass of 0.500 mol Au?
197.0 g Au
0.500 mole Au x 1 mole Au
= 98.5g Au
Q3: # molecules in
15.00 g H2O?
2.016
Molar mass H2O = 16.00
18.02 g
15.00 g H2O x
6.022 x 1023 molec. H2O
1 mole H2O
x
1 mole H2O
18.02 g H2O
= 5.013 x 1023 molec. H2O
Percent Composition
Percent composition is % mass that each
element in a molecule contributes to the
total molar mass of the compound.
Assume that you have one mole of the
compound.
Practice: What is the % composition of
glucose, C6H12O6? Check: identical to CH2O!
Types of Formulas
What is the % composition of CH2O?
Total mass = 12.01 g + 2.016 g + 16.00 g
= 30.03 g
12.01 g
%C = 30.026 g x 100
%C = 40.00 %
Rounding to 30.03 g
produces errors:
%H = 6.71 %
percentages add
up to 99.98 % !
+ %O = 53.29 %
100.00 %
CH2O is the empirical formula for glucose, C6H12O6
Empirical Formula: the formula of a Molecular Formula: the formula that states
compound that expresses the
the actual number of each kind of atom found
smallest whole number ratio of the
in one molecule of the compound.
atoms present.
1 molecule of aspirin, C9H8O4 = 9 atoms of
Formulas describe the relative
C, 8 atoms of H and 4 atoms of O.
number of atoms (or moles) of each
element in a formula unit. It’s always 1 mole of C H O = 9 mol of C atoms, 8 mol
9 8 4
a whole number ratio.
of H atoms and 4 mol of O atoms.
If we determine the relative number
of moles of each element in a
compound, we can find the formula
of that compound.
Empirical formulas of organic
compounds (consisting of C,H,
O only) can be found by
combustion analysis.
From the mass of the
products (water and carbon
dioxide) we determine the
number of moles of C, H,
and O, and from them obtain
the empirical formula of the
compound.
Dr. Ent burned 0.5 g of
the sample and
obtained the total of
over 1 g of products.
How is that possible?
Oxygen from
air is a
reactant!
Calculating Empirical and
Molecular Formula
Example 1. Percent composition of
a compound is found to be 31.9%
K, 28.9% Cl, and some O. Find the
empirical formula. If the molar
mass of the compound is 122.55 g
mol-1, find the molecular formula.
1. Determine the mass in grams of
each element present, if necessary.
Remember, % means “out of 100”.
2. Convert grams of each element
into moles.
3. Divide all number of moles with
the smallest to obtain the subscripts
of the empirical formula.
4. Divide the molar mass of the
compound by the molar mass of the
empirical formula to get number of
empirical formula units in the
molecular formula (n). Multiply all
subscripts in empirical formula by n.
1. Assume 100.0 g of the compound.
39.2
31.9 g K
28.9 g Cl
O??g O
g O = 100.0 g – ( 31.9 g + 28.9 g)
= 100.0 g – 60.8 g = 39.2 g
2.
1 mol K
= 0.816 mol K
39.10 g K
1 mol Cl
28.9 g Cl x
= 0.815 mol Cl
35.45 g Cl
1 mol O
39.2 g O x
= 2.45 mol O
16.00 g O
31.9 g K x
3. K: 0.816 / 0.815 = 1.00
Cl: 0.815 / 0.815 = 1.00
O: 2.45 / 0.815 = 3.01
Empirical formula:
K Cl O3
4. Molar mass of empirical formula:
39.1 g + 35.45 g + (3 x 16.00 g) = 122.55
Molar mass compound
122.55
n = Molar mass emp. formula = 122.55 = 1
There is 1 unit of KClO3 in molecular formula.
(KClO3) x 1 = KClO3
Example 2. Find the empirical
and molecular formulas if the
% composition is 40.0% C,
6.71% H, 53.3% O, and the
molar mass of the compound
is 180.16 g/mol.
40.0 g C,
1. Assume that you have 100.00 g sample;
6.71 g H,
the mass of each element is equal to the %
53.3 g O.
composition.
1 mol C
2. 40.0 g C x 12.01 g C = 3.33 mol C
1 mol H
1 mol O
6.70 g H x
53.3 g O x
1.008 g H
16.00 g O
= 6.66 mol H
1. Determine the mass in grams
of each element present, if
necessary. Remember, %
means “out of 100”.
2. Convert grams of each element
into moles of atoms of that
element.
3. Divide all number of moles with
the smallest to obtain the subscripts of the empirical formula.
4. Divide the molar mass of the
compound by the molar mass of
the empirical formula to get
number of empirical formula units
in the molecular formula (n).
Multiply all subscripts in the
empirical formula by n.
3. C: 3.33 / 3.33 = 1
H: 6.66 / 3.33 = 2
O: 3.33 / 3.33 = 1
= 3.33 mol O
Empirical formula CH2O.
Emp. Formula mass = 30.03 g/mol
Molar mass compound
180.2
=
= 6.00
Molar mass emp. formula
30.03
Thus, there are 6 (CH2O) units. Molecular formula: C6H12O6.
4.
Practice (answer in parenthesis):
1. A compound has an empirical formula of NO2. The
colorless liquid used in rocket engines has a molar mass
of 92.0 g mole-1. What is the molecular formula of this
substance? (N2O4)
2. A sample of a brown gas, a major air pollutant, is found to
contain 2.34 g N and 5.34 g O. Determine an empirical
formula for this substance. (NO2)
Chapter 9
Calculations from Chemical Equations
contains 6.022 x 1023 atoms of particles.
is expressed in g/mol.
Molar mass:
of an element is its atomic mass in grams.
of a compound is the sum of the atomic masses of all atoms.
Example: molar mass of NaCl is 22.99 + 35.45 = 65.44 g/mol
For calculations of mole – mass – number_of particles relationship:
Conversions go through moles.
1. Use balanced equation.
∆
2 Al + Fe2O3 Al2O3 + 2 Fe
2 mol 1 mol
1 mol 2 mol
2. The coefficient in front of a
formula represents the number of
moles of the reactant or product.
To quantitatively convert from one quantity
to another we introduce mole ratio.
1 mol Fe2O3
2 mol Al
1 mol Fe2O3
1 mol Al2O3
Mole ratio is found from the coefficients
of the balanced equation:
moles of desired substance
Mole ratio =
moles of starting substance
Which conversion factor will be used depends on starting and desired substance.
Mole – Mole Conversions
Example 1:How many moles of NaCl
result from the complete reaction of
3.4 mol of Cl2? Assume that there is
more than enough Na.
2 Na(s) + Cl2(g) → 2 NaCl(s)
desired substance
1 mole
2 moles NaCl
= 6.8 moles NaCl
3.4 moles Cl2 x
1 mole Cl2
2 moles
These are exact numbers !
starting substance
The following examples refer to the equation:
Ca5(PO4)3F(s) + 5 H2SO4(aq)→ 3 H3PO4(aq) + HF(aq) + 5 CaSO4(s)
1 mole
5 moles
3 moles
1 mole
5 moles
Example 2: Calculate the number of moles of phosphoric acid (H3PO4) formed
by the reaction of 10 moles of sulfuric acid (H2SO4) on phosphate rock:
10 moles H2SO4 x
3 moles H3PO4
5 moles H2SO4
= 6 moles H3PO4
Example 3: Calculate the number of moles of Ca5(PO4)3F needed to produce
6 moles of H3PO4.
6 moles H3PO4 x
1 mole Ca5(PO4)3F
3 moles H3PO4
= 2 moles Ca5(PO4)3F
Mass – Mole conversion
3.024
Example 4: Calculate the
30.97
number of moles of H2SO4
+ 64.00
necessary to yield 784 g of
97.99
H3PO4.
g
97.99
Molar mass of H3PO4 =
mole
Ca5(PO4)3F(s) + 5H2SO4 → 3H3PO4 + HF + 5CaSO4
1 mole
5 moles 3 moles 1 mole 5 moles
1. Convert the starting
1 mole H3PO4
substance into moles.
784 g H3PO4 x
= 8.00 moles H3PO4.
97.99
g
H
PO
3
4
2. Convert moles of
starting substance into
5 moles H2SO4
8.00 moles H3PO4 x
moles of desired
= 13.3 moles H2SO4.
3 moles H3PO4
substance.
3. Convert moles of desired substance into the units specified in the problem.
done.
Ex. 5: Calculate the mass of phosphate rock, Ca5(PO4)3F needed to yield 200. g of HF.
Molar masses: Ca5(PO4)3F = 504.3 g/mol; HF = 20.01 g/mol
Step 1, 200. g HF x 1 mole HF x 1 mole Ca5(PO4)3F = 10.0 moles Ca5(PO4)3F
20.01 g HF
1 mole HF
Step 2
504.3 g ph.r. = 5.04 kg Ca (PO ) F.
Step 3:
5
4 3
10.0 moles Ca5(PO4)3F x
1 mole ph.r.
Step_by_step:
Mass – mass conversion
Ex. 6: Calculate the number of grams of H2SO4 necessary to yield 392 g of H3PO4.
Ca5(PO4)3F(s) + 5H2SO4 → 3H3PO4 + HF + 5CaSO4
1 mole
5 moles
3 moles
1 mole
5 moles
Molar mass H3PO4 = 97.99 g
mole
g
Molar mass H2SO4 = 98.09 mole
1. Convert the starting substance into moles.
1 mole H3PO4
392 g H3PO4 x
= 4.00 moles
97.99 g H3PO4
5 moles H2SO4
2. Convert moles of starting substance
4.00 moles H3PO4 x
3 moles H3PO4
into moles of desired substance.
= 6.67 moles
3. Convert moles of desired substance into the units specified in the problem.
6.67 moles H2SO4 x
392 g H3PO4 x
98.09 g
= 654 g H2SO4.
1 mole H2SO4
Combined steps:
1 mole H3PO4 x 5 moles H2SO4 x 98.09 g
= 654 g H2SO4.
97.99 g H3PO4 3 moles H3PO4 1 mole H2SO4
Example 7: Find the mass of glucose that can be synthesized from 58.5 g of CO2,
assuming that there is more than enough water to react with all the CO2. Molar
masses are 44.01 g (CO2) and 180.2 (glucose).
sunlight
6 CO 2(g) + 6 H 2O(l)  
→ 6 O 2(g) + C6 H12 O 6(aq)
58.5 g CO2 x
1 mole CO2
1 mole glucose 180.2 g glucose
x
x
44.01 g CO2
6 moles CO2
1 mole glucose
= 39.9 g glucose
Conversion – General Case
Mass to moles
of starting compound
Step 1
Moles of starting compound
to moles of desired compound
Step 2
Moles of desired comp.
to units desired.
Step 3
Mass – mass: All 3 steps
Example 8: Calculate the mass of NH3 formed by the reaction of 112 grams of H2.
N2 + 3H2 → 2NH3
grams H2 → moles H2 → moles NH3 → grams NH3
Molar masses: H2: 2.016 g/mol; NH3: 17.03 g/mol
1 mole H2 2 moles NH3 17.03 g NH3
x
x
= 631 g NH3
2.016 g H2 3 moles H2
1 mole NH3
Starting
Step 1
compound
result
Step 2
Step 3
Moles – moles: Step 2 only
Example 9: Calculate the moles of NH3 formed by the reaction of 1.5 moles of H2.
2 moles NH
1.50 moles of H2 x 3 moles H 3 = 1.00 mole NH3.
2
112 g H2 x
Starting
compound
Step 2
result
Moles – mass: Step 2 and Step 3 only
Example 10: Calculate the mass of NH3 formed by the reaction of 1.50 moles of H2.
17.03 g NH3
2 moles NH
1.50 moles of H2 x 3 moles H 3 x 1 mole NH
= 17.0 g NH3.
3
2
Starting
result
Step 2
compound
Step 3
Conversion – General Case (cont’d)
Mass – moles: Step 1 and Step 2 only
Example 11: Calculate the moles of NH3 formed by the reaction of 150. g H2.
1 mole H
2 moles NH
N2 + 3H2 → 2NH3
150. g H2 x 2.016 g H2 x 3 moles H 3 = 49.6 mol NH3.
2
2
Starting
Step 1
Step 2
result
compound
Mass – particles: All 3 steps
Example 12: Calculate the # molecules of NH3 formed by the reaction of 150. g H2.
1 mole H2 2 moles NH3 6.022 x 1023 molecules NH3 = 2.99 x 1025
150. g H2 x
x
x
2.016 g H2 3 moles H2
1 mole NH3
molecules NH3.
Starting
Step 1
Step 3
result
compound
Step 2
Limiting Reactant and Yield Calculations
The amount of the product(s) depends on the reactant
that is used up during the reaction, i.e. limiting reactant.
One bicycle needs 1 frame, 1 seat and 2 wheels,
therefore not more than 3 bicycles can be made.
The number of seats is the limiting part (reactant);
one frame and two wheels are parts in excess; 3
bicycles is the yield.
Limiting Reactant and yield Calculations (cont’d)
Example 13: How many moles of Fe3O4 can be obtained by reacting 16.8 g Fe
with 10.0 g H2O? Which substance is the limiting reactant? Which substance is in
excess? How many grams of the reactant in excess remains unreacted?
Strategy:
1. Write and balance equation.
2. Calculate # moles of a product that can be obtained from each reactant;
3. The reactant that gives the least moles of (the same!) product is the limiting reactant.
4. Find the amount of reactant in excess needed to react with the limiting reactant.
Subtract this amount from the starting quantity to obtain the amount in excess.
5. Find the yield from the limiting reactant.
1. Balanced equation: 3 Fe (s) + 4 H2O (g)
Fe3O4 (s) + 4 H2 (g)
2. # moles of Fe3O4:
1 mol Fe3O4
1 mol Fe
x
= 0.100 mol Fe3O4.
from Fe: 16.8 g Fe x
55.85 g Fe
3 mol Fe
Least moles Fe3O4?
1 mol Fe3O4
1 mol H2O
from H2O: 10.0 g H2O x 18.02 g H O x 4 mol H O = 0.139 mol Fe3O4.
2
2
3. Limiting reactant: Fe
18.02 g H2O
1 mol Fe x 4 mol H2O
x
= 7.23 g H2O.
x
16.8 g Fe
4. Reacted
1 mol H2O
55.85 g Fe
3 mol Fe
H2O:
Yield is 0.100 mol Fe3O4;
Excess H2O: 10.0 g – 7.23 g = 2.77 g H2O.
Answer: Limiting reactant Fe;
5. Yield: 0.1 mol Fe3O4.
Excess H2O is 2.77 g H2O.
Percent Yield
Calculations done so far assumed that the reaction gives maximum (100%) yield.
Many reactions (especially organic) do not give the 100% yield, due to:
side reactions, reversible reactions, product losses due to human factor.
Theoretical yield: Amount calculated from the chemical equation.
Actual yield: Amount obtained experimentally.
Actual yield
Percent yield:
x 100 %
Theor. yield
Strategy:
Find limiting reactant.
Calculate theoretical yield.
Calculate percent yield.
Example 14: If 65.0 g CCl4 was prepared by
CS2 + 3 Cl2
CCl4 + S2Cl2
reacting 100. g CS2 and 100. g of Cl2, calculate
the percent yield. Molar masses: CS2: 76.15; Cl2: 70.90; CCl4: 153.8 g/mol
1 mol CS2
1 mol CCl4
100. g CS2 x
x
= 1.31 mol CCl4.
76.15 g CS2 1 mol CS2
1 mol Cl2
1 mol CCl4
x
= 0.470 mol CCl4.
70.90 g Cl2
3 mol Cl2
Limiting reactant
153.8 g CCl4
0.470 mol CCl4 x
= 72.3 g CCl4. Theoretical yield
1 mol CCl4
100. g Cl2 x
65.0 g CCl4
Actual yield
65.0 g CCl4
x 100 % = 89.9 % Percent yield
72.3 g CCl4
HW, Chapter 7 (p.139):
1 (a-e):Determine the molar masses of
these compounds: KBr, Na2SO4,
Pb(NO3)2; C2H5OH, HC2H3O2
5. Calculate the number of grams
In each of the following:
0.550 mol Au;
15.8 mol H2O;
12.5 mol Cl2;
3.15 mol NH4NO3.
15. 1 molecule of tetraphosphorus
decoxide contains: how many moles? How
many grams? How many P atoms? How
many O atoms? How many total atoms?
26. A 7.52 g sample of ajoene (garlic odor)
was found to contain 3.09 g S,
0.453 g H, 0.513 g O and the rest, C.
Calculate the percent composition.
39. Ethanedioic acid, a compound that
is present in many vegetables, has a
molar mass of 90.04 g/mol and a
composition of 26.7% C, 2.2% H and
71.1% O. What is the molecular formula?
HW, Chapter 9 (p.183):
3. Calculate the number of grams in
these quantities:
2.55 mol Fe(OH)3; 125 kg CaCO3;
10.5 mol NH3; 72 millimol HCl;
500.0 mL of liquid Br2 (d=3.119 g/mL)
7(a-c). Balance the equation for the synthesis
of sucrose: CO2 + H2O C12H22O11 + O2
and set up the mole ratio of:
CO2 to H2O; H2O to C12H22O11;
O2 to CO2.
13. Carbonates react with acids to form
salt, water and carbon dioxide gas. When
50.0 g of CaCO3 are reacted with
sufficient HCl, how many grams of CaCl2
will be produced? (Balance eq. first!)
15. In a blast furnace iron(III) oxide reacts
with carbon to produce molten iron and
CO: Fe2O3 + 3C 2 Fe + 3 CO
How many kilograms of iron would be
formed from 125 kg of Fe2O3?
23. Determine limiting reactant and the one
in excess.
KOH + HNO3 KNO3 + H2O
16.0 g 12.0 g
2 NaOH + H2SO4 Na2SO4 + H2O
10.0 g 10.0 g
29. Elemental silicon can be produced by
the reduction of SiO2 (sand) with carbon:
SiO2 + 2 C Si + 2 CO
When 35.0 kg of SiO2 react with 25.3 kg of C,
and 14.4 kg of Si is recovered, what is the
percent yield of the reaction?