Chemistry 103 Lecture 13
Chemistry 103 Lecture 13. Outline I. The MOLE continued…. II. Determining Chemical Formulas Percent Composition (review) Empirical/Molecular Formulas.
Dec 20, 2015
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Chemistry 103
Lecture 13
Outline
I. The MOLE continued….
II. Determining Chemical Formulas Percent Composition (review) Empirical/Molecular Formulas
Atomic masses in Grams
107.9g of Ag. How many Ag atoms?
12.01g of C. How many atoms of C?
32.07g of S. How many S atoms?
Avogadro’s Number
Solution = 6.022 x 1023
Avogadro’s number is equal
to 1 mole Makes working with large numbers
easier
Molar Mass from Periodic TableMolar mass • Is the atomic
mass expressed in grams
Molar Mass
The molar mass • Is the mass of one mole of
an element or compound• Is the atomic mass
expressed in grams
Copyright © 2005 by Pearson Education, Inc.Publishing as Benjamin Cummings
Some One-Mole Quantities
32.07 g 55.85 g 58.44 g 294.20 g 342.30 g
Conversion Factors
1 mole = 6.022 x 1023
(when the question asked for “individual” numbers of something)
1 mole = molar mass
(relationship between mass and number)
The Mole
Molar mass 1 mole 6.022 x 1023
Mass in grams Individual particles
Conversion Factors
1 mole = 6.022 x 1023
1 mole = molar mass
Mole ratio = (“n” mole part)/ (1 mole total)
Subscripts State Atoms and Moles
1mole aspirin 9 mol C 8 mol H 4 mol O
Learning Check
Calculate the number of moles of aspirin in 52.1 g of aspirin (C9H8O4).
52.1g C9H8O4 (1 mol C9H8O4)
180.16g C9H8O4.
Learning Check
Calculate the number of moles of hydrogen in 52.1 g of aspirin (C9H8O4).
52.1g C9H8O4 x (1 mol C9H8O4) (8moles H)
180.16g C9H8O4 1 mol C9H8O4
Learning Check
Calculate the number of grams of hydrogen in 52.1 g of aspirin (C9H8O4).
Learning Check
Calculate the individual number of hydrogen atoms in 52.1 g of aspirin (C9H8O4).
Learning Check
Allyl sulfide C6H10S is a compound that has the odor of garlic. How many moles of C6H10S are in 225 g?
Learning Check
Allyl sulfide C6H10S is a compound that has the odor of garlic. How many moles of C atoms are in 225g of C6H10S?
Learning Check
Allyl sulfide C6H10S is a compound that has the odor of garlic. How many grams of C are in 225g of C6H10S?
Learning Check
How many H2O molecules are in 24.0 g H2O?
a) 4.52 x 1023
b) 1.44 x 1025
c) 8.02 x 1023
Learning Check
How many H atoms are in 24.0 g H2O?
a) 4.01 x 1023
b) 1.60 x 1024
c) 8.02 x 1023
Learning Check
Determine the mass of 6 molecules of O2 in gram units.
Percent Composition
Percent composition
• Is the percent by mass of each element in a formula.
Mass element in compound x 100%
Mass of compound
Percent Composition
Example:
Calculate the percent composition of CO2.
CO2 = 1 C(12.01g) + 2 O(16.00 g) = 44.01 g/mol
12.01 g C x 100 = 27.29 % C 44.01 g CO2
32.00 g O x 100 = 72.71 % O 44.01 g CO2 100.00 %
What is the percent composition of lactic acid, C3H6O3, a compound that appears in the blood after vigorous activity?
Learning Check
STEP 13C(12.01) + 6H(1.008) + 3O(16.00) = 90.08 g/mol
36.03 g C + 6.048 g H + 48.00 g O
STEP 2%C = 36.03 g C x 100 = 40.00% C
90.08 g
%H = 6.048 g H x 100 = 6.714% H 90.08 g
%O = 48.00 g O x 100 = 53.29% O 90.08 g
Solution
The molecular formula Is the true or actual number of the atoms in a
moleculeThe empirical formula Is the simplest whole number ratio of the atoms
(this is the formula for ionic compounds)
H2O2 HOmolecular formula empirical formula
Types of Formulas
Some Molecular and Empirical Formulas
The molecular formula is the same or a multiple of the empirical.
1. What is the empirical formula for C4H8?
a) C2H4 b) CH2 c) CH
2. What is the empirical formula for C8H14?
a) C4H7 b) C6H12 c) C8H14
Learning Check
1. What is the empirical formula for C4H8?
a) C2H4 b) CH2 c) CH
2. What is the empirical formula for C8H14?
a) C4H7 b) C6H12 c) C8H14
Learning Check
A compound has an empirical formula SN. If there are four atoms of N in one molecule, what is the molecular formula?
1) SN
2) SN4
3) S4N4
Learning Check
A compound has an empirical formula SN. If there are four atoms of N in one molecule, what is the molecular formula?
1) SN
2) SN4
3) S4N4
Learning Check
Empirical Formula
Using Experimental data to find the empirical formula of a compound.
Law of Definite Proportions
Experimental studies (decomposition reactions) led to the conclusion that the percentage of each element present in a given compound does not vary - Law of Definite Proportions (Dalton)
Experiment - Law of Definite Proportions Measure Mass of a compound (an oxide of tin)
Experimentally “decompose” into oxygen and tin. Measure mass of each.
Law of Definite Proportions
4 different experiments with a compound that contains both Sn and Oxygen only.
Mass Sn(g) Mass tin oxide(g) Mass O(g) Mass Sn/Mass O
5.00g (78.7%) 6.35g 1.35g (21.3%) 3.70
10.0g (78.7%) 12.7g 2.7g (21%) 3.7
23.7g (78.7%) 30.1g 6.4g (21%) 3.7
73.4g (78.8%) 93.2g 19.8g (21.2%) 3.71
Compound of Sn and O Row 1: mass Sn = 5.00 g mass O = 1.35g
Row 3: mass Sn = 23.7g mass O = 6.4g
Empirical Formula Problems
1). Convert percentages given to grams 2). Convert grams to moles 3). Divide by the smallest number in order to
ascertain the whole number ratio of atoms of different elements in the compound
4). Clear any obvious fractions in step 3.
A compound contains 7.31 g Ni and
20.0 g Br. Calculate its empirical (simplest) formula.
Learning Check
A particular compound is 60.0% C, 4.5% H and 35.5% O. Calculate its empirical (simplest) formula.
Learning Check
Percentages given
Assume 100 g sample (percentages easily translate into gram quantities)
Convert grams into moles Divide by the smallest number of moles
present. If you do not have whole numbers at this
point, clear the fraction by multiplying every number by the whole number that accomplishes this end.
Percentages given
Assume 100 g sample (percentages easily translate into gram quantities)
Convert grams into moles Divide by the smallest number of moles
present. If you do not have whole numbers at this
point, clear the fraction by multiplying every number by the whole number that accomplishes this end.
Percentages given
Assume 100 g sample (percentages easily translate into gram quantities)
Convert grams into moles Divide by the smallest number of moles
present. If you do not have whole numbers at this
point, clear the fraction by multiplying every number by the whole number that accomplishes this end.
Percentages given
Assume 100 g sample (percentages easily translate into gram quantities)
Convert grams into moles Divide by the smallest number of moles
present. If you do not have whole numbers at this
point, clear the fraction by multiplying every number by the whole number that accomplishes this end.
Converting Decimals to Whole NumbersWhen the number of moles for an element is a decimal,
all the moles are multiplied by a small integer to obtain
whole number.
Empirical Formula for Aspirin
Learning Check
A pure phosphorus/oxygen compound is
43.7% “P” and the remainder “O”.
What is the empirical formula of this compound?
A molecular formula Is a multiple (or equal) of its empirical
formula Has a molar mass that is the empirical
mass multiplied by a whole numbermolar mass = a whole number empirical mass
Is obtained by multiplying the empirical formula by a whole number
Relating Molecular and Empirical Formulas
Some Compounds with Empirical Formula CH2O
Molecular Formula from Empirical If the molar mass of the compound with
empirical formula P2O5 is 284g/mol, what is the molecular formula?
A compound has a molar mass of 176.1g and an empirical formula of C3H4O3. What is the molecular formula?
A) C3H4O3
B) C6H8O6
C) C9H12O9
Learning Check
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