Chapter 9. Chemical Calculations I Chemical Formulas
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Transcript of Chapter 9. Chemical Calculations I Chemical Formulas
Chapter 9.
Chemical Calculations IChemical Formulas
The Law of Definite Proportions states that in a pure compound, the elements are always present in the same definite proportion by mass.
This is consistent with Atomic Theory.
It allows us to make useful calculations.
Chemical Formulas
Two samples of NH3 from different sources:
# Mass of Mass of N Mass of H Sample
A 1.840 g 1.513 g 0.327 g
B 2.000 g 1.644 g 0.356 g
A % N = 1.513 g x 100% = 82.23 %1.840 g
B % N = 1.644 g x 100% = 82.20 %2.000 g
Chemical Formulas
Two samples of N2H4 from different sources:
# Mass of Mass of N Mass of H Sample
A 3.245 g 2.836 g 0.409 g
B 2.950 g 2.578 g 0.372 g
A % N = 2.836 g x 100% = 87.40 %3.245 g
B % N = 2.758 g x 100% = 87.39 %2.950 g
Chemical FormulasA property of NH3 (ammonia) is that it
always has 82.2% nitrogen, regard-less of the source or amount of the sample.
Hydrazine (N2H4), a different compound than ammonia, always has 87.4% nitrogen, regardless of the source or amount of the sample.
Percent composition is a property of a compound.
Chemical Formulas
If you try to make a compound by combining its elements, you must combine them in the correct mass ratios, or you'll have some of one element left over:
Make CaS (calcium sulfide)
Mass Mass Mass Excess Mass Ratioof Ca of S of CaS S Ca/S
55.6 g 44.4 g 100.0 g 0.0 g 1.25/1.00
55.6 g 50.0 g 100.0 g 5.6 g 1.25/1.00
Chemical Formulas
What's going on? At the atomic level:
4 atoms 4 atoms 4 units of Ca of S of CaS
4 atoms 6 atoms 4 units 2 units of S of Ca of S of CaS left over
Chemical Formulas
And no matter what happens, the mass ratio of Ca to S in the compound is 1.25 to 1.00. Therefore, the individual atoms of Ca must be 1.25 times the mass of the S atoms.
Experiments with relative masses of elements in compounds were used to determine atomic masses for the Periodic Table.
Chemical Quantities
How many atoms does it take to get meas-urable quantities of an element?
A WHOLE LOT!!!
Chemical Quantities
Since we can't see or work with individual atoms, we work with them en mass, so to speak. Just like we do with lots of common things:
flour, sugar, produce, other staples
We also count things in groups:
pairs, dozens, scores, gross, reams
Moles
We count atoms in groups called moles.
1.000 mole = 6.022 x 1023 atoms
(atoms, or anything else)
Moles
Moles
1 mole = 6.022 x 1023 atoms (or anything else)
This is huge!
602,200,000,000,000,000,000,000
volume of Earth's oceans in liters
age of earth in seconds
population of earth, individuals
cost of a car, US dollars
Molar Masses
Molar Masses
The molar mass of an element is the mass of 6.022 x 1023 atoms of that element.
1.008 g is mass of 6.022 x 1023 hydrogen atoms
12.01 g is mass of 6.022 x 1023 carbon atoms
238.0 g is mass of 6.022 x 1023 uranium atoms
The atomic mass of an element is the mass of 1 atom of that element.
1.008 amu is mass of one hydrogen atom, etc.
Molar Masses
Relationship between grams and amu's (atomic mass units):
1.000 gram = 6.022 x 1023 amu
1.000 amu = 1.661 x 10-22 gram
Molar Masses
6.022 x 1023 is also called Avogadro's number for Amedeo Avogadro, who first deduced a relationship between the number of molec-ules of a gas and its volume.
Avogadro's number can be used as a conver-sion factor, like "dozen" or "score".
It allows us to "count" atoms by determining the mass of a group of them.
Molar Masses
The molar mass is a conversion factor between moles of a substance and its mass.
Molar mass has units grams g mole mol
12.01 g C 1 mol C mol C 12.01 g C
Molar Masses
Example:
In 27.43 grams of iron,
(a) How many moles of iron are there?
(b) How many atoms of iron are there?
Competency I-2
Note: When working with molar mass, always use enough significant figures in the molar mass to match or exceed the significant figures in other terms. Don't limit the accuracy of your work with molar mass!
Molar Masses
Conversion between molar mass, moles, and atoms (or molecules).
Use the correct conversion, and don't use one you don't need!
Grams Moles Atoms
MolarMass
Avogadro'sNumber
Molar Masses
Examples:
(a) What is the mass of 2.500 mol of carbon?
(b) How many atoms of carbon are present?
Molar Masses
Examples:
In 1.00 x 1022 atoms of gold,
(a) What mass of gold is present?
(b) How many moles of atoms are present?
Molar Masses of Compounds
At the atomic level, the formula of a compound gives the number of each type of atom that makes up a formula unit or molecule of the compound.
At the macroscopic level, the formula of a com-pound gives the number of moles of each type of atom that makes up a mole of for-mula units or molecules of the compound.
Molar Masses of Compounds
One molecule of H2O contains
2 hydrogen atoms
1 oxygen atom
One mole of H2O contains
2 moles of hydrogen atoms
1 mole of oxygen atoms
1 mole of water molecules
Molar Masses of Compounds
What is the mass of one mole of water?
2 mol H x 1.01 g H = 2.02 g H mol H
1 mol O x 16.00 g O = 16.00 g O mol O 18.02 g
H2O
Molar Masses of Compounds
What is the mass of one mole of water?
18.02 g H2O mol H2O
There are 6.022 x 1023 molecules in there, and you can swallow it in one gulp!
Molar Masses of Compounds
Examples:
Find the molar masses of
(a) N2
(b) NaCl
(c) CaCO3
(d) Mg(NO3)2
Percent Composition
The percent composition of a compound is the percent of its mass contributed by each element in its formula.
Percent composition can be calculated and checked against an experimental value to confirm the identity of a compound.
Percent Composition
Steps:
(a) calculate molar mass of compound, writing out mass contributions of
the elements
(b) divide mass contribution of each element by molar mass,
express result as a percentage.
Percent Composition
Examples:
Calculate the percent composition of the following compounds.
(a) H2O
(b) Mg(NO3)2
Empirical FormulasEmpirical formulas show the smallest
whole-number ratio of the elements found in a compound.
One can obtain a percent composition by experiment, and use it to calculate the empirical formula of a compound.
The calculation is essentially the reverse of determining a percent composition.
Empirical Formulas
Steps:
(a) assume 100.00 g of compound, so mass percent of each element
can be expressed in grams
(b) calculate number of moles of each element present in that mass
(c) determine mole ratios of elements
(d) write empirical formula
Empirical FormulasExample:
The empirical formula for Freon-12, a refrigerant, is given below. De- termine its empirical formula.
9.933 % C
58.63 % Cl
31.44 % F
Competency I-3
Molecular Formulas
Molecular formulas show the number of atoms present in a molecule of a compound. A mol-ecular formula is a whole number multiple of an empirical formula.
Empirical Formula Molecular Formula
NH2 N2H4
CH2 C2H4, C4H8, C6H12
CH2O C5H10O5, C6H12O6
Molecular Formulas
To determine a molecular formula, one needs:
(a) The empirical formula of the compound.
(b) The molar mass of the compound.
Steps:
(a) Use empirical formula to determine formula mass of compound.
(b) Divide molar mass by formula mass.
(c) Multiply empirical formula by result from (b)
Molecular Formulas
Example:
Uracil, a component of ribonucleic acid (RNA), has the empirical formula C2H2NO. Its
molar mass is 112.09 g/mol. What is its molecular formula?
Competency I-4