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Chapter 9. Chemical Calculations I Chemical Formulas

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Chapter 9. Chemical Calculations I Chemical 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. - PowerPoint PPT Presentation
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Chapter 9. Chemical Calculations I Chemical 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.
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Page 1: 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.

Page 2: Chapter 9. Chemical Calculations I Chemical Formulas

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

Page 3: Chapter 9. Chemical Calculations I Chemical Formulas

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

Page 4: Chapter 9. Chemical Calculations I Chemical Formulas

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.

Page 5: Chapter 9. Chemical Calculations I Chemical Formulas

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

Page 6: Chapter 9. Chemical Calculations I Chemical Formulas

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

Page 7: Chapter 9. Chemical Calculations I Chemical Formulas

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.

Page 8: Chapter 9. Chemical Calculations I Chemical Formulas

Chemical Quantities

How many atoms does it take to get meas-urable quantities of an element?

A WHOLE LOT!!!

Page 9: Chapter 9. Chemical Calculations I Chemical Formulas

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

Page 10: Chapter 9. Chemical Calculations I Chemical Formulas

Moles

We count atoms in groups called moles.

1.000 mole = 6.022 x 1023 atoms

(atoms, or anything else)

Page 11: Chapter 9. Chemical Calculations I Chemical Formulas

Moles

Page 12: Chapter 9. Chemical Calculations I Chemical Formulas

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

Page 13: Chapter 9. Chemical Calculations I Chemical Formulas

Molar Masses

Page 14: Chapter 9. Chemical Calculations I Chemical Formulas

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.

Page 15: Chapter 9. Chemical Calculations I Chemical Formulas

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

Page 16: Chapter 9. Chemical Calculations I Chemical Formulas

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.

Page 17: Chapter 9. Chemical Calculations I Chemical Formulas

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

Page 18: Chapter 9. Chemical Calculations I Chemical Formulas

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!

Page 19: Chapter 9. Chemical Calculations I Chemical Formulas

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

Page 20: Chapter 9. Chemical Calculations I Chemical Formulas

Molar Masses

Examples:

(a) What is the mass of 2.500 mol of carbon?

(b) How many atoms of carbon are present?

Page 21: Chapter 9. Chemical Calculations I Chemical Formulas

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?

Page 22: Chapter 9. Chemical Calculations I Chemical Formulas

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.

Page 23: Chapter 9. Chemical Calculations I Chemical Formulas

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

Page 24: Chapter 9. Chemical Calculations I Chemical Formulas

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

Page 25: Chapter 9. Chemical Calculations I Chemical Formulas

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!

Page 26: Chapter 9. Chemical Calculations I Chemical Formulas

Molar Masses of Compounds

Examples:

Find the molar masses of

(a) N2

(b) NaCl

(c) CaCO3

(d) Mg(NO3)2

Page 27: Chapter 9. Chemical Calculations I Chemical Formulas

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.

Page 28: Chapter 9. Chemical Calculations I Chemical Formulas

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.

Page 29: Chapter 9. Chemical Calculations I Chemical Formulas

Percent Composition

Examples:

Calculate the percent composition of the following compounds.

(a) H2O

(b) Mg(NO3)2

Page 30: Chapter 9. Chemical Calculations I Chemical Formulas

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.

Page 31: Chapter 9. Chemical Calculations I Chemical Formulas

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

Page 32: Chapter 9. Chemical Calculations I Chemical Formulas

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

Page 33: Chapter 9. Chemical Calculations I Chemical Formulas

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

Page 34: Chapter 9. Chemical Calculations I Chemical Formulas

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)

Page 35: Chapter 9. Chemical Calculations I Chemical Formulas

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


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