unit 4 counting particles
DESCRIPTION
We can weigh a large number of the objects and find the average mass. Counting By Weighing We can weigh a large number of the objects and find the average mass. Once we know the average mass we can equate that to any number of the objects. EXAMPLE: The average mass of a book is 40.0 grams. How many books are present in a sample with a mass of 2000.0 grams? 2000.0g/40.0g = 50.0 booksTRANSCRIPT
Unit 4 Counting Particles We can weigh a large number of the
objects and find the average mass.
Counting By Weighing We can weigh a large number of theobjects and
find the average mass. Once we know the average mass we canequate
that to any number of theobjects. EXAMPLE: The average mass of a
book is 40.0 grams. How many books are present in a sample with a
mass of grams? 2000.0g/40.0g = 50.0 books Counting By Weighing When
we know the average mass of theatoms of an element, as that
elementoccurs in nature, we can calculate thenumber of atoms in any
given sample ofthat element by weighing the sample. The atomic mass
of an element, asfound on the periodic table, allows usto count by
weighing. Avogadros Number Avogadros Number is the number of atoms
in grams of carbon. Its numerical value is 1023. Therefore, a g
sample of carbon contains 6.022 1023 carbon atoms. 1 mol =
Avogadros Number = 6.022 1023 units
The Mole The mole (mol) is a unit of measure for an amount ofa
chemical substance. A mole is Avogadros number of particles, that
is 1023 particles. 1 mol = Avogadros Number = 1023 units We can use
the mole relationship to convert betweenthe number of particles and
the mass of a substance. Section 6.3: The Mole The mole (mol) is
known as the chemists dozen and represents 6.022 x 1023 things
(atoms, particles, molecules, etc). A mole (from the Latin - lump
of stuff), was originally defined as 1.0g of the lightest element
known hydrogen. Not this type of mole! How Big Is a Mole? The
volume occupied by one mole of softballs wouldbe about the size of
the Earth. One mole of Olympic shot put balls has about thesame
mass as the Earth. Molar Mass The atomic mass of any substance
expressed ingrams is the molar massof that substance. The atomic
mass of iron is amu. Therefore, the molar mass of iron is
g/mol(grams in a mole) Since oxygen occurs naturally as a diatomic,
O2, themolar mass of oxygen gas is 2 times g or g/mol. Calculating
Molar Mass
The molar mass of a substance is the sum of themolar masses of each
element. What is the molar mass of magnesium nitrate,Mg(NO3)2? The
sum of the atomic masses is: ( ) = (62.01) = amu The molar mass for
Mg(NO3)2 is g/mol. Molar Mass (g/mol) The molar mass of any
substance is the mass (in grams) of one mole of the substance. The
molar mass of a compound is obtained by summing the masses of ALL
component atoms. Molar Mass: Practice Calculate the following molar
masses: Water H2O
Ammonia NH3 Propane C3H8 Glucose C6H12O6 Mole Calculations We will
be using Dimensional analysis again.
Recall: First we write down the unit asked for Second we write down
the given value Third we apply unit factor(s) to convert the
givenunits to the desired units Mass-Mole Calculations
What is the mass of 1.33 moles of titanium, Ti? We want grams, we
have 1.33 moles of titanium. Use the molar mass of Ti: 1 mol Ti = g
Ti = 63.7 g Ti 1.33 mole Ti 47.88 g Ti 1 mole Ti 6.02 1023
particles = 1 mol = molar mass
Mole Calculations Now we will use the molar mass of a compound
toconvert between grams of a substance and moles orparticles of a
substance. 6.02 1023 particles = 1 mol = molar mass If we want to
convert particles to mass, we must firstconvert particles to moles
and than we can convertmoles to mass. Mole Calculations How many
sodium atoms are in 0.120 mol Na?
Step 1: we want atoms of Na Step 2: we have mol Na Step 3: 1 mole
Na = 6.02 1023 atoms Na = 7.22 1022 atoms Na 0.120 mol Na 1 mol Na
6.02 1023 atoms Na Mole Calculations I How many moles of potassium
are in 1.25 atoms K? Step 1: we want moles K Step 2: we have 1.25
1021 atoms K Step 3: 1 mole K = 6.02 1023 atoms K = 2.08 10-3 mol K
1.25 1021 atoms K 1 mol K 6.02 1023 atoms K Mole Calculations What
is the mass of 2.55 1023 atoms of lead?
We want grams, we have atoms of lead. Use Avogadros number and the
molar mass of Pb 2.55 1023 atoms Pb 1 mol Pb 6.021023 atoms Pb
207.2 g Pb 1 mole Pb = 87.8 g Pb Mole Calculations How many O2
molecules are present in g ofoxygen gas? We want molecules O2, we
have grams O2. Use Avogadros number and the molar mass of O2 0.470
g O2 1 mol O2 32.00 g O2 6.021023 molecules O2 1 mole O2 8.84 1021
molecules O2 How many Oxygen atoms in 8.84 1021 molecules O2?
8.84 1021 molecules O2x 2 oxygen atoms= molecule Mole Video 6.022 X
1023 pennies: would make at least 7 stacks that would reach the
moon.
6.02 X 1023 blood cells: would be more than the total number of
blood cells found in every human on earth. A mol is A LOT of
particles: 602,200,000,000,000,000,000,000 The Mole It is VERY
important to understand that the value listed as the mass number on
the periodic table really tells us TWO things: The mass of one atom
in units of amus. The mass of one mole of atoms in grams. One mol
of Al = g& one atom = amu One mol of Au = g & one atom =
amu One mol of B = g& one atom = amu Remember one mol = X atoms
(or particles or molecules or donuts) Molar Mass Remember: we can
talk about one mole of atoms or one mole of molecules. One mole of
oxygen atoms (O) weighs g. One mole of oxygen molecules (O2) weighs
g. Two moles of O atoms weigh g. Two moles of O2 molecules weigh g.
And so on . . . The Mole To summarize: a sample of any element that
weighs a number of grams equal to the molar mass (from the periodic
table) contains x 1023 atoms (1 mol) of that element. Therefore,
atomic weight of an element = #g in one mol of that element We
write that as g/mol. The Mole: Practice Calculate the number of
moles in a 25.0 g sample of calcium.How many atoms are there?
Calculate the number of moles and the number of atoms in a 57.7 g
sample of sulfur. Calculate the number of atoms in a 23.6 g sample
of zinc. Calculate the number of molecules in a g sample of calcium
carbonate. mol and 3.76 x 1023 atoms mol and 1.08 x 1024 atoms x
10-4 mol and 2.17 x 1020 atoms x 10-3 mol and x 1020 atoms Molar
Mass: Practice 1. Calculate the number of moles of H2CO3 present in
a 7.55 g sample. 2. How many water molecules are present in 10.0 g
of water? (Hint: find moles first) 3. How many molecules of sucrose
(C12H22O11) are present in a 5.0 kg bag of sugar? (Hint: find moles
first) Representative Particles= atoms, molecules, Items, etc
Percent Composition of Compounds
Sometimes it is not enough to know a compounds composition in terms
of numbers of atoms; it may also be useful to know its composition
in terms of the masses of its elements. We can calculate the mass
fraction by dividing the mass of a given element in one mole of a
compound by the mass of one mole of the compound. Percent
Composition of Compounds
Once we know the mass fraction we can multiply by 100 to get the
percent. Remember: percent = part/whole X 100. In one mole of
methane (CH4), there is one mole of C and four moles of H:
1 mol C = g 4 mol H = 4(1.01) = 4.04 g 1 mol CH4 = g g = g %C= mass
of 1mol C x 100 mass of 1 mol CH4 %C = X 100 = %C 16.05 %H= mass of
4 mol Hx 100 %H = X 100 = %H Percent Composition Practice
Determine the mass percent of each element in the following: H2SO4
(sulfuric acid) C3H7OH (isopropyl alcohol) C6H12O6 (glucose)
Formulas of Compounds The object of this section is to do the
opposite of the previous section.Instead of getting the mass from
the formula, we will determine the formula from the mass. To do
this, the mass must be converted to moles using each elements mass
number.How can we convert mass to moles? Formulas of Compounds ex)
An unknown compound with a mass of g is is found to contain: g C g
H g O It must contain: 0.0806g (1 mol C/12.01 g C) = mol C g(1 mol
H/1.008 g H) = mol H 0.1074g(1 mol/16.00 g O) = mol O The numbers
from the previous slide allow us to determine the C:H:O
ratio.
If we divide each number by the smallest number we get 1:2:1 for
the C:H:O ratio. This leads us to a formula of C1H2O1 or CH2O. This
is not necessarily the TRUE formula of the compound, but represents
the RELATIVE numbers of atoms. The multiples represent possible
MOLECULAR FORMULAS.
This represents the lowest whole number ratio of the compound. The
actual formula could be CH2O, C2H4O2, C3H6O3, C4H8O4, C5H10O5,
C6H12O6, . . . Any formula with a C:H:O ratio of 1:2:1 is possible
(in theory, an infinite number). C1H2O1 represents the simplest
possible formula or the EMPIRICAL FORMULA. The multiples represent
possible MOLECULAR FORMULAS. Formulas of Compounds: Practice
Determine the empirical formula from each of the following
molecular formulas: H2O2 (hydrogen peroxide) C4H10 (butane) CCl4
(name?) HC2H3O2 (acetic acid) C6H12O6 (glucose) Calculation of
Empirical Formulas
There are four steps to determine the empirical formula of a
compound: Obtain the mass of each element present (in grams).
Determine the number of moles of each type of atom present (use
atomic mass). Divide each number by the smallest number. Multiply
all numbers by the smallest integer that will make them all
integers Empirical Formulas: Practice
A g sample of a compound containing only carbon and hydrogen is
found to contain g of carbon.Determine the empirical formula for
this compound. A g sample of nitrogen reacts with g of oxygen.
Determine the empirical formula for this compound. When a g sample
of iron metal is heated in air, it reacts with oxygen to achieve a
final mass of 2.573g. Determine the empirical formula for this
compound. If the relative amounts of elements are presented as
percentages, assume we are starting with a 100 g sample (100%).Then
each percentage simply becomes a mass (in grams). For example if
15% of a compound is carbon, we just assume it is 15 g of a 100 g
sample; from there we convert to moles. Empirical Formulas: More
Practice
The simplest amino acid, glycine, has the following mass percents:
32.00% carbon, % hydrogen, 42.63% oxygen, and 18.66%
nitrogen.Determine the empirical formula for glycine. A compound
has been analyzed and found to have the following mass percent
composition: % copper, 10.84% phosphorous, and % oxygen. Determine
the empirical formula for this compound. Calculation of Molecular
Formulas
If we know the empirical formula AND the molar mass, we can
calculate a compounds molecular formula. Note: without the molar
mass, the best you can find is the empirical formula. Once the
molar mass is known, one must ALWAYS find the empirical formula
before one can calculate the molecular formula.It is impossible to
do the reverse. Calculation of Molecular Formulas
It is also important to note that the molecular formula is ALWAYS
an integer multiple of the empirical formula.We can represent the
molecular formula in terms of the empirical formula: (Empirical
Formula)n = molecular formula It should also be noted when n = 1,
the empirical and molecular formulas are identical to each other.
Molecular Formulas: Practice
A compound containing carbon, hydrogen, and oxygen is found to be
40.00% carbon and % hydrogen by mass.The molar mass of this
compound is between 115 and 125 g/mol. Determine the molecular
formula for this compound. Caffeine is composed of 49.47% C, 5.191%
H, % N, and 16.48% O. The molar mass is about 194 g/mol. Determine
the molecular formula for caffeine.