Chapter 10 Chemical Quantities, the Mole, and Conversions
Transcript
Slide 1
Chemical Quantities, the Mole, and Conversions
Slide 2
Measuring Matter -The amount of something is usually determined
one of three ways; by counting, by mass, or by volume. -Groups are
made to make counting easier. (Remember the bean lab?) A bunch, a
dozen, a ream, a can
Slide 3
6.022 x 10 23 of anything! Known as Avogadros number
Representative Particles are the smallest division of a substance
that maintains the characteristics of that substance -For molecular
compounds R.Ps are molecules -For ionic compounds R.Ps are formula
units -For elements R.Ps are atoms
Slide 4
1 mole of any element is equal to the atomic mass on the
periodic table AKA molar mass For example: 18g of H 2 O = 6.022 x
10 23 molecules
Slide 5
Number of particles to moles RP x 1 mole =Moles 6.02 x 10 23
RPs How many moles are 2.80 x 10 24 atoms of silicon? 4.65
moles
Slide 6
Moles to number of particles moles x 6.02 x 10 23 RPs.= RPs 1
mole How many atoms are in 1.14 mol Co? 6.86 x 10 23 atoms Co
Slide 7
The atomic mass is the mass of one atom of an element (in AMU)
Based on the mass of one atom of Carbon-12 AKA relative mass since
each element is compared to C-12 Not very practical when working
with chemicals in real life
Slide 8
Molar Mass is the atomic mass of an element expressed in grams
or, the mass of one mole of atoms of a particular element expressed
in grams Numerically equal to the atomic mass in amus because the
Mole is a constant number!
Slide 9
Formula Mass is the sum of the average atomic masses of all the
elements represented in the formula (in amus) Example: Use the
atomic masses in the periodic table to calculate the formula mass
of SrCl 2 FM = 87.62 + 2(35.453) = 158.526 amu Importance the
formula mass of a compound gives us useful information about the
quantities of reactants needed to make new products (must follow
the Law of Conservation of Mass)
Slide 10
Molecular Mass is the same as formula mass, but this term in
used specifically for molecular compounds Not used for ionic
compounds Example water: H 2 O = 2(1.00797) + 15.9994 = 18.01534
amu
Slide 11
Molar Mass is the mass in GRAMS of one mole (6.02 x 10 23
particles) numerically equal to the formula mass of a compound
(units = g/mol)
Slide 12
Mole-Mass Relationship Use the molar mass of an element or
compound to convert between the mass of a substance and the moles
of a substance Moles to Mass (grams): Multiply by the Molar Mass
Example Find the mass, in grams, of 4.52 x 10 -3 mol C 20 H 42
Slide 13
Step 1: Find the Molar Mass: 20 (12.01115) + 42 (1.00797) =
282.55774 g/mol Step 2: Convert: 4.52 x 10 -3 mol x 282.55774 g C
20 H 42 = 1 mol C 20 H 42 Answer: 1.28 g C 20 H 42
Slide 14
Grams to moles: Divide by the Molar Mass Example Calculate the
number of moles in 75.0 g of Dinitrogen trioxide. Step 1: Write the
correct chemical formula: N 2 O 3 Step 2: Find the Molar Mass: 2(
14.0067) + 3(15.9994) = 76.0116 g/mol Step 3: Convert: 75.0 g N 2 O
3 x 1 mole N 2 O 3 = 76.0116 g N 2 O 3 Answer: 0.987 mole N 2 O
3
Slide 15
# of Moles x Avogadros # # of RPs x molar mass Mass (g)
Slide 16
Avogadros Hypothesis states equal volumes of gases at the same
temperature and pressure contain equal numbers of molecules
Standard Temperature and Pressure (STP) = 0 C (273 K) and 101.325
kPa (1 atm, 760 mmHg, 760 torr)
Slide 17
Molar Volume = the 22.4 L of space occupied by 6.02 x 10 23
representative particles of any gas at STP Example What is the
volume of 3.20 x 10 -3 mol CO 2 at STP? 3.20 x 10 -3 mol CO 2 x
22.4 L = ? 1 mole Answer: 0.0717 L CO 2
Slide 18
Example: How many moles of CO 2 are in 57.0 L of CO 2 ? 57.0 L
CO 2 x 1 mole = 22.4 L Answer: 2.54 mol CO 2
Slide 19
Calculating Molar Mass from Density Background the density of
gases is measured in g/L at a specific temperature, and, the
density of a gas at STP is a constant value (characteristic
property) To find Molar Mass of a gas from density multiply the
density at STP by the molar volume at STP
Slide 20
Example A gaseous compound composed of sulfur and oxygen, which
is linked to the formation of acid rain, has a density of 3.58 g/L
at STP. What is the molar mass of this gas? Molar Mass = 3.58 g of
gas x 22.4 L = 1 L 1 mol Answer: 80.2 g/mol
Slide 21
Definition relative amounts of substances in a compound
expressed as percents by mass Percentage Composition from Mass Data
Equation: Mass of Element x 100 = % Mass Compound
Slide 22
It has been measured in lab that 1.000g of water contains
0.112g Hydrogen. Find the mass percentage of hydrogen Mass %H =
0.112g x 100 = 11.2 %H 1.000g
Slide 23
Percentage Composition from the Chemical Formula (Atomic mass
of elem.) x (# atoms of elem.) x 100 Molar mass of the
compound
Slide 24
What is the percent composition by mass of hydrogen in water?
Mass %H = (1.00797g) x 2 x 100 = 18.01534g Answer: 11.19013 %H
Slide 25
Percentage Composition as a Conversion Factor What is the mass
of hydrogen in 30.0g of water? Mass H = (Mass % 100) x mass sample
= (11.2% 100) x 30.0g = Answer: 3.36gH
Slide 26
Hydrates (compounds containing water): You must include the
water of hydration in the formula mass What are the mass
percentages of the elements in Na 2 CO 3 10 H 2 O? MM = 2(22.98977)
+ 12.01115 + 3(15.9994) + 10(18.01534) = 286.14229g
Slide 27
%Na = 2(22.98977) x 100 = 16.06877 % Na 286.14229 %C = 12.01115
x 100 = 4.197614 % C 286.14229 %O = 3(15.9994) x 100 = 16.77424 % O
286.14229 %H 2 O = 10(18.01534) x 100 = 62.95938 % H 2 O
286.14229
Slide 28
A chemical formula showing the smallest (simplest) whole-number
ratio of atoms in a compound Steps for determining Empirical
Formula from % composition Step 1: Determine the masses of the
elements in a 100g (since its based on a percent by mass) sample
Step 2: Convert these masses in grams to moles by dividing by the
atomic mass
Slide 29
Step 3: Determine the simplest mole ratio between the elements
by dividing each by the smallest number of moles Step 4: Use these
mole ratios to write the subscripts in the formula
Slide 30
Calcium Fluoride occurs as the mineral Fluorite. It contains
51.3% by mass of Ca and 48.7% by mass of F. Determine the Empirical
formula. Step 1: Start with a 100g sample of calcium fluoride 51.3%
of 100g = 51.3 g Ca 48.7% of 100g = 48.7 g F Step 2: Convert to
Moles 51.3 g Ca x 1 mole Ca = 1.28 mole Ca 40.08 g Ca 48.7 g F x 1
mole F = 2.56 mole F 18.998403 g F Step 3: Ratio mole F = 2.56 mole
F = 2 - ratio is 1:2 mole Ca 1.28 mole Ca
Slide 31
CaF 2 is the empirical formula Another Example! Substance X and
Y. Suppose you found the number of moles of X and Y in 100.0g of a
compound to be 1.96 mole X and 2.94 mole Y. What is the empirical
formula? Answer: X2Y3X2Y3
Slide 32
C2H2C8H8C2H2C8H8
Slide 33
The actual formula of a molecular compound which is often a
multiple of the empirical formula (E.g. C 2 H 6 instead of CH 3
)
Slide 34
The Molecular Mass must be determined by experimental analysis
or be given in the problem Divide the Molecular Mass by the Formula
Mass (Molar Mass) and adjust the subscripts in the Empirical
Formula by multiplying by this resulting number
Slide 35
Water the simplest formula is H 2 O, but could the molecular
formula be H 4 O 2 or H 8 O 4 ? Molecular Mass = 18.01 g by
analysis Formula Mass (Molar Mass) = 2(1.00797) + 15.9994 =
18.01534g Ratio of MM/FM = 18.01 18.0534 or 1:1 so this is also the
molecular formula
Slide 36
Hydrogen Peroxide the simplest formula is HO Molecular Mass =
34.0g by analysis (given) Formula Mass (Molar Mass) = 1.00797 g +
15.9994 g = 17.00737 g Ratio of MM/FM = 34.0 17.00737 = 2:1 so the
formula is doubled to H 2 O 2
Slide 37
A compound is found by analysis to consist of 40.1% S and 59.9%
O. The Molecular Mass is 80.1g. Determine the Empirical Formula,
the Molecular Formula, and name the compound Step 1: in a 100.0g
sample 40.1% of 100.0 g = 40.1 g S 59.9% of 100.0 g = 59.0 g O Step
2: convert to moles 40.1 g S x 1 mole S = 1.25 mole S 32.06g S 59.9
g O x 1 mole O = 3.74 mole O 15.9994 g O
Slide 38
Step 3: ratio 3.74 mole O 1.25 mole S = 2.992:1 or 3:1 Step 4:
empirical formula SO 3 Step 5: mass ratios FM* = 32.06 g S +
3(15.9994 g)O = 80.0582 g SO 3 MM = 80.1g (given in problem) 1:1 so
the molecular formula = the empirical