Transcript Document
1
Chapters 9 & 10
GOAL
Apply concepts of the mole and Avogadro’s number to conceptualize and calculate empirical and molecular formulas, mass, moles and molecules relationships.
• The mass of one mole of an element or compound.
• 3
Review Clip
Molar Mass
• Examples: – What is the mass of one mole of Carbon? – What is the mass of one mole of Bromine?
– What is the mass of one mole of KCl?
– What is the mass of one mole of NaOH?
– What is the mass of one mole of AgNO 3 ?
4
Determine the molar mass of the following compounds: 1) NaBr 2) PbSO
4
3) Ca(OH)
2
4) Na
3
PO
4
5) (NH
4
)
2
CO
3
1) 102.9 g/mol 2) 303.3 g/mol 3) 74.1 g/mol 4) 164.0 g/mol 5) 96.0 g/mol
5
Law of Definite Proportions & Law of Multiple Proportions
Law of Definite Proportions:
• In a sample of any chemical compound, the masses of the elements are
always in the
same proportions.
Law of Multiple Proportions:
• Whenever the same two elements form more than one compound, the different masses of one element that combine with the same mass of the other elements are in the ratio of small whole numbers
6
Law of Definite Proportions
(sometimes called Proust's Law) A given compound always contains elements in a certain proportion by mass. (Constant composition).
Regardless of the amount, a compound is always composed of the same elements in the same proportion by mass.
The proportions are found by calculating the percent by mass.
MASS compound = sum of MASSES elements % by mass = MASS element x 100% MASS compd
7
Law of Definite Proportions
Atoms combine in whole number ratios, so their proportion by mass will always be the same.
Example:
H 2 O is always made up of 2 atoms of H and one atom of O.
What is the ratio of H to O?
H 1amu (x2)= 2 and O 16 amu The ratio of O to H in water is always 16:2 or 8:1.
MASS compound = sum of MASSES elements % by mass = MASS element x 100% MASS compd
8 The make up of a compound can be expressed as…
Percent Composition
• The relative amounts of the elements in a compound are expressed as percent composition.
9
Percent Composition
Ex. Problem
• When a 13.60g
sample of a compound containing only magnesium and oxygen is decomposed, 5.40g of oxygen is obtained. What is the percent composition of this compound?
10
1
% Composition Practice
2
Calculating %Comp. from the Chemical Formula Example • Propane C 3 H 8 . Calculate the percent composition of propane.
11
Calculating %Comp. from the Chemical Formula 12
13 • Determine
the percentage composition of sodium carbonate, Na 2 CO 3 .
Practice
More practice: Determine the percentage composition of each of the 14 following compounds:
a.
sodium oxalate, Na 2 C 2 O 4 •
ans:
34.31% Na, 17.93% C, 47.76% O
b.
ethanol, C 2 H 5 OH •
ans:
52.13% C, 13.15% H, 34.72% O
c.
aluminum oxide, Al 2 O 3 •
ans:
52.92% Al, 47.08% O
d.
potassium sulfate, K 2 SO 4 •
ans:
44.87% K, 18.40%
15
Law of Multiple Proportions
(John Dalton, 1801) Mass percentages of elements in a compound do NOT depend on amount.
Compounds with the same mass proportions must be the same compound Youtube Clip EX: Carbon combines with oxygen to form CO and CO 2 .
1:1.33
1:2.66
2.66/1.33 =
2
16
Law of Multiple Proportions
(John Dalton)
Empirical Formulas
• Basic ratio of the elements contained in the
compound.
• The lowest whole-number ratio of the atoms of the elements in a compound.
• May or may NOT be the same as the molecular formula.
• Example: Hydrogen Peroxide » Molecular Formula is H 2 » Empirical Formula is HO O 2 17
Determining Empirical Formula of a Compound • Ex: A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula of the compound?
2) Find Mole Ratio 4) Multiply by smallest whole number to get whole # subscripts.
#1: Convert into grams….
3) Divide by smallest # of mol
18
Practice: A solid compound is found to contain K, S, and O with the percent composition listed below.
What is the empirical formula of this compound?
K: 41.09% S: 33.70% O: 25.22% 19
1) Find Mole Ratio 2) Divide by smallest # of mol 3) Multiply by smallest whole number to get whole # subscripts.
Empirical Formula Practice
1) An oxide of chromium is found to have the following % composition: 68.4% Cr and 31.6% O. What is the empirical formula?
(click to see worked out answer)
20 2) The % composition of a compound was found to be 63.5 % Ag, 8.2% N, and 28.3% O. What is the empirical formula?
(click to see worked out answer)
3) A 170g sample of an unidentified compound contains 29.84g Na, 67.49g Cr, and 72.67g O. What is the empirical formula? (hint: find %s 1 st
) (click to see worked out answer)
More problems with soln video: https://www.youtube.com/watch?v=MlaZnRbQN8g
Practice
21
22 formula)
Molecular Formulas
Molecular formula:
-whole-number multiple of its empirical formula (or the same as the empirical • You can calculate the molecular formula if you know… – #1 the empirical formula
AND
– #2 the molar mass of the compound.
Steps
1.Calculate EFM Formula Mass) (Empirical 2.Divide the Molar Mass by the EMF • This tells you the # of empirical formula units in a molecule of the compound. AND it is the multiplier to convert the empirical formula to the molecular formula 3.Multiple the subscripts in the empirical by the answer to step 2.
Example:
Steps
1.Calculate EFM (Empirical Formula Mass)
Hydrogen Peroxide Empirical Formula:
HO
Molar Mass:
34.0 g/mol Step 1 2.Divide the Molar Mass by the EMF • This tells you the # of empirical formula units in a molecule of the compound. AND it is the multiplier to convert the empirical formula to the molecular formula 3.Multiple the subscripts in the empirical by the answer to step 2.
HO EFM= 17.0 g/mol Step 2 34.0/17.0 = 2 Step 3 • H 2 O 2 23
Practice:
24
• Calculate the molecular formula of a compound whose molar mass is 60.0 g/mol & the empirical formula is CH
4
N.
1. Calculate EFM (Empirical Formula Mass) 2. Divide the Molar Mass by the EMF 3. Multiple the subscripts in the empirical by the answer to step 2.
Practice:
(Determining Molecular formula from Empirical Formula) 25 1) What is the molecular formula of ethylene glycol, which is used as antifreeze. The molar mass is 62 g/mol and the empirical formula is CH 3 O.
C 2 H 6 O 2
1) What is the molecular formula of a compound with the empirical formula CClN and a molar mass of 184.5 g/mol?
C 3 Cl 3 N 3
1) What is the molcular formula of a compound that is 56.6% K, 8.7% C, and 34.7% O (MM=138.21) ?
K 2 CO 3
Hydrates
• A
hydrate
is an ionic compound that contains water molecules in its structure • An
anhydrate
is the substance that remains after the water is • removed from a hydrate.
When a hydrate is heated the water molecules are driven off as steam, leaving behind the water-free anhydrate.
MgCO 3 Na 2 CO 3
· ·
5H 2 O 2H 2 O BaCl 2
·
2H 2 O 4 Steps to find formula of a hydrate:
1.Determine mass of water driven off: 2.Determine moles of MgCO 3 and water: 3.Find a whole number molar ratio: 4.Write Formula
Hydrates Problems
#1: A 15.67 g sample of a hydrate of magnesium carbonate was heated, without decomposing the carbonate, to drive off the water. The mass was reduced to 7.58 g. What is the formula of the hydrate?
1) Determine mass of water driven off:
2)
Determine moles of MgCO 3
MgCO 3 H 2 O ⇒ 3) 15.67 minus 7.58 = 8.09 g of water ⇒
and water:
7.58 g / 84.313 g/mol = 0.0899 mol 8.09 g / 18.015 g/mol = 0.449 mol
Find a whole number molar ratio:
MgCO 3 H 2 O ⇒ ⇒ Formula: 0.0899 mol / 0.0899 mol = 1 0.449 mol / 0.0899 mol = 5
MgCO 3 · 5H 2 O
Hydrates Problems
#2: A hydrate of Na 2 CO 3 has a mass of 4.31 g before heating. After heating, the mass of the anhydrous compound is found to be 3.22 g. Determine the formula of the hydrate. 1) Determine mass of water driven off: 4.31 minus 3.22 = 1.09 g of water 2) Determine moles of Na Na 2 CO 3 H 2 O ⇒ ⇒
2 CO 3
and water: 3.22 g / 105.988 g/mol = 0.0304 mol 1.09 g / 18.015 g/mol = 0.0605 mol 3) Find a whole number molar ratio: Na 2 CO 3 ⇒ 0.0304 mol / 0.0304 mol = 1 H 2 O ⇒ 0.0605 mol / 0.0304 mol = 2
Formula:
Na 2 CO 3
·
2H 2 O
Hydrate Starter Problems
1. A hydrate of magnesium sulfate has a mass of 13.52 g. This sample is heated until no water remains. The MgSO4 anhydrate has a mass of 6.60 g. Find the formula and name of the hydrate.
2. A sample of copper (II) sulfate hydrate has a mass of 3.97 g. After heating, the CuSO4 that remains has a mass of 2.54 g. Determine the correct formula and name of the hydrate.
3. A sample of the hydrate of sodium carbonate has a mass of 8.85 g. It loses 1.28 g when heated. Find the formula and the name of the hydrate.
4. A 16.4 g sample of hydrated calcium sulfate is heated until all the water is driven off. The calcium sulfate that remains has a mass of 13.0 g. Find the formula of the hydrate.