Ch. 10- The Mole Why- Within the next 2 months we will be working with chemical quantities....
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Transcript of Ch. 10- The Mole Why- Within the next 2 months we will be working with chemical quantities....
Ch. 10- The Mole
Why- Within the next 2 months we will be working with chemical quantities. Determining what a reaction will produce or how much we need to go into a reaction with. These math skills can be applied across all science, ALL
professions and many day to day life activities.
Measuring MatterCount, Mass, Volume……
Units indicating a specific number of itemsExamples????
Remember your dimensional analysis….well lets get those gears agrindingWhat is the mass of 90 apples if 1 dozen of the
apples has a mass of 2.0kg
Jan 25- “Mole Day”Hey you….don’t be miserable today…you don’t
want to be labeled as being dismole
Why is it hard to count atoms, ions, or molecules 1x1?
We as chemists use the mole (mol) in the same way we use dozen1 mole = 6.02 x 1023 representative particles
Avogadro’s number- named after him to honor his work on molecular theories
How big is a mole?
If you count to 6.02 x 1023, one number per second, how long will it take?
6.02 x 1023 x 1min/60sec x 1hr/60min x 1 day/24 hr x 1 yr/365 days =
1.9 x 1016 years = 19 quadrillion years!
The earth is 4.6 billion years old, so that’s 4 million times the earth’s age!
Representative particlesRefers the the species present in a substance
Na = atomsCO2 = molecules
NaCl = formula unit
So……
1 mole of Na = 6.02 x 1023 atoms1 mole of CO2 = 6.02 x 1023 molecules
1 mole of NaCl = 6.02 x 1023 formula units
Mole Calculations
Cont.
Practice1. What is Avogadro’s #?
2. What is the representative particle for MgS?
3. How many formula units are in 2.34 moles of NaNO3?
4. How many moles are in 3.48 x 1024 atoms of Li?
Remember our friend amu?Decimal from periodic table
Relative values based on the mass of Carbon-12 isotopeSince ratio is relative, changing the unit will not
change the ratioGrams is convenient to use in lab so….lets us it
Molar Mass (IMPORTANT)
Molar mass of CompoundsSimply…add the masses of all the elements in
the compound togetherEx. SO3
EX. C9H11NO2
On your own…try
1.) K3PO4
2.) Al2(SO4)3
Mole Land…. Where the fun is multiplied by 6.02 x 1023 times
Mole Mass problems
Mass Mole Problems
Molar VolumeAvogadro’s hypothesis- equal volumes of
gases at the same temp. and pressure contain equal number of particles
Volume measured at STP (Standard Temp. and Pressure)0°C and 101.3kPa (1 atm)
Molar volume = At STP, 1 mole or 6.02 x 1023 representative particles, of any gas occupies a volume of 22.4 L.
Mole Volume Problems
Volume Mole Problems
% CompositionRelative amounts of the elements in a
compoundHow important in the real world
% mass of element = _mass of element_ x 100%
mass of compound
% Composition from mass data When a 13.60g sample of a compound containing only Mg
and O is decomposed, 5.40g of O is obtained. What is the % composition of this compound?
% O = _mass of O_ x 100% = 5.40 g x 100% = 39.7 %
mass of MgO 13.60g
% Mg= _mass of Mg__ x 100% = 8.20 g x 100% = 60.3%
mass of MgO 13.60g
% Composition from Chemical Formula
% mass = mass of element in 1 mol compound molar mass of compound
Propane (C3H8) is commonly used in gas grills. What is the percent composition of Propane?
% C = 36.0g x 100% = 81.8%
44.0g
% H = 8.0g x 100% = 18%
44.0g
% PracticeWhat is the percent composition for each
element in the following compounds
C11H12N2O2 -Tryptophan
C6H8O6 – Vitamin C
Magnesium Phosphate
Empirical FormulaThe lowest whole number ratio of the atoms of
the elements in a compoundBasic ratio (ie 1:2:1)May or may not be the same as the molecular
formulaTells the actual # of each kind of atom in the
compound
Determining empirical formula of a compound A compound is found to contain 25.9% N and 74.1% O.
What is the empirical formula of the compound Since % means part per 100, we can assume that 100.0g of
the compound contains 25.9g and 74.1g.
Step 1 – convert to moles
25.9g N x 1 mol = 1.85 mol N
14.0g
74.1g O x 1 mol = 4.63 mol O
16.0g
Step 2- Divide each molar quantity by the smaller molar quantity
1.85 mol N = 1.00 mol N 4.63 mol O = 2.50 mol
1.85 1.85
Step 3- Multiple each part of the ratio by the smallest whole # that will convert both subscripts to whole #’s
1 mol N x 2 = 2 mol N
2.5 mol O x 2 = 5 mol O
Cont
Calculate the empirical formula for a compound containing 67.6% Hg, 10.8% S, and 21.6% O.
Step 1
67.6g Hg x 1 mol Hg = 0.337 mol Hg
200.6g Hg
10.8g S x 1 mol S = 0.336 mol S
32.1g S
21.6g O x 1 mol O = 1.35 mol O
16.0 g
Step 2
0.337mol Hg = 1.00mol Hg 0.336mol S = 1.00mol S 1.35mol O =4.02mol O
0.336 0.336 0.336