Download Unit 3 - sotochem

Unit 3: Stoichiometry
Be able to write chemical equations in words and symbols:
➢ Example​
: Fe​
S​
(s) + 6HCl (g) → 2FeCl​
(s) + 3H​
S (g)
2​
3​
3​
2​
○ Solid iron (III) sulfide reacts with gaseous hydrogen chloride to form solid iron (III) chloride and
hydrogen sulphide gas
Be able to balance chemical equations and know why balancing is important:
➢ Equations should be balanced in order to satisfy the ​
law of conservation of mass​
(matter can neither be created
nor destroyed, just changed in form) - same # of each kind of atom on both sides of the equation
➢ When balancing: never start w/ H or O​
, treat polyatomic ions as chunks not separate atoms, and if there is an
2​
odd # of atoms on one side and an even # of those atoms on the other, place a 2 as a coefficient next to the
chemical formula with the odd #s to make it even;
Example​
: __Cr + __S​
→ __Cr​
S​
8​
2​
3
Understand and be able to apply the concept of the mole in chemical calc. (Avogadro's number):
➢ 1​
Mol​
= 6.022x1023​
​representative particles (element = atoms, molecular compound = molecules, ionic
compound = formula units) = ​
Avogadro’s number ​
(6.022x1023​
​)
➢ Mass of 1 mol of substance = molar mass (g); for elements, it is the amount in grams numerically equal to the
atomic mass in amu’s
○ Atomic Mass (AMU)​
(mass of 1 atom of substance) = ​
Molar Mass (g)​
(mass of 1 mol of substance)
➢ Molar Mass = M = grams/mol
○ Example​
: How many hydrogen atoms are in 72.5g of C​
H​
O?
3​
8​
■ Convert g of C​
H
​
O
to
mol
of
C
​
H
​
O
and
then
there
are 8 mols of H in 1 mol of C​
H​
O and
3​ 8​
3​ 8​
3​
8​
there are 6.022x1023​
​atoms of H in 1 mol of H;
Ans​
.​
5.81x1024
​​
atoms H
➢ Percent Composition​
: percent that each element in a compound is composed of
○ Calculate the M of the compound and then, divide the total mass of each element by the total mass of
the compound, multiply by 100
○ Example​
: Calculate the percent composition of water in the hydrate, CaSO​
.​
2H​
0.
4​
2​
■ Hydrate = ​
ionic compound + water trapped in crystal​
; Anhydrate = ​
ionic compound on the
product side of the equation​
; Hydrate-Anhydrate = ​
mass of H​
O
Ans​
​
.​
20.93%
2​
Understand and be able to apply the concept of stoichiometric coefficients relating to reacting ratios
➢ Meaning of coefficients in a chemical equation: ​
used to describe a reaction in moles, and particles (molecules,
formula units and atoms) but NOT grams.
➢ Mole ratio​
: conversion factor between 2 different amts. in a balanced chemical equation; ​
2KClO​
→2KCl+3O​
3​
2
○ Mole ratios: 2 mol KClO​
=
2
mol
KCl
​
OR
​
2
mol
KCl
=
3
mol
O
​
O
​
R
​
2
mol
KClO
​
=
3​
2​
3​ 3 mol O​
2
➢ Stoichiometry​
: given an amount of either reactant or product, you can determine the other quantities in a
reaction using dimensional analysis
➢ Example​
: A 25.5g sample of potassium chlorate is decomposed. How many moles of O​
are produced?
2​
○ Write a balanced chemical equation, convert g of KClO​
to
moles
of
KClO
​
and
use mole ratio from
3​
3​
coefficients of compounds to get moles of O​
Ans
.
​
​
0.312 mol O​
2
2
Be able to calculate empirical formula from percentage by mass data:
➢ Empirical Formula: ​
the lowest whole number ratio of atoms in a chemical formula
○ From % composition, empirical formula can be determined using mole ratios
○ Percent to mass, mass to mole, divide by small, multiply till whole
■ Use the percentages as mass measurements out of 100 g. Convert these masses to mole
amounts. Divide each mole amount by the smallest mole value present. If the answers are not
already within .1 of a whole number, multiply them by whole numbers until they are within .1
of a whole number.
○ Example​
: A compound has the following percent composition: 31.9% K, 29.0% Cl, and 39.1% O.
What is the empirical formula?
Ans. KClO₃
Be able to convert empirical formulae to molecular formulae:
➢ Molecular Formula: ​
actual ratio of the molecule that exists
○ Find the molar mass of the molecular formula and the empirical formula and divide the molecular
formula mass by the empirical formula mass to get a whole number to multiply the subscripts of the
empirical formula
○ Example​
: A compound has an empirical formula of NO​
. A liquid has a molar mass of 92.0 g/mole.
2​
What is the molecular formula of this substance?
Ans. N₂
O₄
Be able to use combustion data to calculate empirical formulae of compounds:
➢ Empirical Formula from Combustion Analysis:
○ Combustion equation: C​
H​
+ O​
→ CO​
+ H​
O
x​
y​
2​
2​
2​
○ Use the mass of CO​
to
convert
to
moles
of
CO
​
to moles of C to grams of C
2​
2​
○ Use the mass of H​
O
​
to
convert
to
moles
of
H
​
O
2​ ​
2​ to moles of H to grams of H
○ Subtract the combined mass of C and H from the mass of the sample to get the mass of oxygen and
convert it to moles
○ Simplify the mole ratios of C, H and O to get the empirical formula
○ Example​
: Calculate the empirical formula for the combustion of 11.5g ethanol. 22.0g CO​
and 13.5g
2​
H​
O
are
produced
in
the
reaction.
2​
Understand, and be able to apply, the concept of a limiting reagent:
➢ Limiting Reagent​
- reactant that determines the amount of product formed (one that you run out of first)
○ Limiting reagent problems deal with 2 given amounts of reactants (given in mole or g)
○ Example: ​
124.0g of Al are reacted with 601.0g of Fe​
O​
. Calculate the mass of Al​
O​
formed.
2​
3​
2​
3​
Understand, and be able to apply, the concept of percentage yield:
➢ Percent Yield​
: the ratio in percent of amount of product you produced in the lab compared to the amount of
product that you should have produced if the reaction went to completion and no problems arose in the lab.
○ Must ALWAYS use Limiting Reagent to determine the product produced
○ Theoretical yield​
: the amount you would produce if everything went perfectly
■ Use the balanced chemical equation to calculate
○ Experimental Yield​
: the actual amount you make in lab under imperfect conditions
■ Percent Yield = Actual (what you got) / Theoretical (what you should have got) * 100
■ Example​
: Al burns in Br​
to produce AlBr​
. In a lab, 6.0g of Al reacts with excess Br​
. 50.3g of
2​
3​
2​
AlBr​
are
produced.
What
is
the
percent
yield?
​
Ans.
85%
3​
➢ Hydrate = ​
ionic compound + water trapped in crystal​
; Anhydrate = ​
ionic compound on the product side of the
equation​
; Hydrate-Anhydrate = ​
mass of H​
O
2​
➢ A gas occupies 22.4L of Volume at STP
➢ Example: ​
CuCl​
when heated to 100o​
​
C is dehydrated. If 0.235g of CuCl​
.​
xH​
O gives 0.185g of CuCl​
on heating,
2​
2​
2​
2​
what is the value of x?
➢ Example: ​
Solutions of sulfuric acid and lead (II) acetate react to form solid lead (II) sulfate and acetic acid. If
7.50g of acid and 7.50g of lead (II) acetate are mixed, calculate the number of grams of each reactant and product
present after the reaction is completed.
➢ Example: ​
When H​
S gas is bubbled into a soln. of NaOH, the reaction forms Na​
S and H​
O. How many grams
2​
2​
2​
of Na​
S
are
formed
if
the
soln.
used
contains
2.00g
of
NaOH
and
the
reaction
has
a
92.0%
yield.
How many
2​
molecules of H​
S
reacted
2​