Chapter 9

The Mole concept and Chemical Formulas

Law of definite proportions

• In a pure compound elements are always present in the same definite proportion by mass.

Formula Mass

• Sum of atomic masses of all the atoms present in one formula unit of a substance expressed in atomic mass units.

• Formula mass is also called as Molar mass.

Problems

• 1.Calculate the molar mass of LiClO4.

• 2.Calculate the molar mass of (NH4)3PO4 • 3.Calculate the molar mass of C2H6O • Practice Ex: 9.2

• Use sig figures like addition rules.

• Practice Ex: 9.3

Percent composition:

• • • • Percent composition of a compound is the

mass percent

of each element in the compound.

If the formula of a compound is known a two step process is needed to calculate the percent composition.

Step 1 Step 2

Calculate the molar mass of the formula.

Divide the total mass of each element in the formula by the molar mass and multiply by 100.

Percent composition:

total mass of the element molar mass x 100 = percent of the element

Percent Composition From Experimental Data

• • Percent composition can be calculated from experimental data without knowing the composition of the compound.

Step 1

formed.

Calculate the mass of the compound •

Step 2

Divide the mass of each element by the total mass of the compound and multiply by 100.

• 5. A compound containing nitrogen and oxygen is found to contain 1.52 g of nitrogen and 3.47 g of oxygen. Determine its percent composition.

• Practice Ex: 9.4,9.5

The Mole: The Mole

• • • The mass of a single atom is too small to measure on a balance. • •

mass of hydrogen atom = 1.673 x 10-24 g

• 1 mole = 6.022 x 1023 objects • This number is called as Avogadro’s Number

1 mole of any element contains 6.022 x 1023 particles of that substance.

The atomic mass in grams of any element2contains 1 mole of atoms.

This is the same number of particles as there are in exactly 12 grams of

1 mole = 6.022 x 10

23

objects

6.022 x 10

23 is a very

LARGE

number

6.022 x 10

23 is

Avogadro’s Number

number

If 10,000 people started to count Avogardro’s number and counted at the rate of 100 numbers per minute each minute of the day, it would take over 1 trillion years to count the total number.

1 mole of any element contains 6.022 x 10 23 particles of that substance.

The atomic mass in grams of any element 23 contains 1 mole of atoms.

This is the same number of particles 6.022 x 10 23 as there are in exactly 12 grams of 12 6 C

Species

H

Quantity

1 mole

Number of H atoms

6.022 x 10

23

Species

Fe

Quantity

1 mole

Number of Fe atoms

6.022 x 10

23

Species

C

6

H

6

Quantity

1 mole

Number of

6.022 x 10

23

C

6

H

6

molecules

• The molar mass of an element is its atomic mass in grams.

• It contains 6.022 x 10 23 atoms (Avogadro’s number) of the element.

1 mol of atoms = 6.022 x 10 23 atoms 1 mol of molecules = 6.022 x 10 23 molecules 1 mol of ions = 6.022 x 10 23 ions

Element H Mg Na Atomic mass Molar mass 1.008 amu 24.31 amu 22.99 amu 1.008 g 24.31 g 22.99 g Number of atoms 6.022 x 10 23 6.022 x 10 23 6.022 x 10 23

Avogadro’s Number of Particles 6 x 10

23

Particles

1 MOLE Molar Mass

Avogadro’s Number of H

2

O molecules 6 x 10

23 H 2 O

molecules

1 MOLE H

2

O 18.02 g H

2

O

These relationships are present when hydrogen combines with chlorine.

H Cl HCl 6.022 x 10 23 H atoms 6.022 x 10 23 Cl atoms 6.022 x 10 23 HCl molecules 1 mol H atoms 1 mol Cl atoms 1 mol HCl molecules 1.008 g H 35.45 g Cl 36.46 g HCl 1 molar mass H atoms 1 molar mass Cl atoms 1 molar mass HCl molecules

In dealing with diatomic elements ( H 2 , O 2 , N 2 , F 2 , Cl 2 , Br 2 , and I 2 ) , distinguish between one mole of atoms and one mole of molecules.

Problems:

• 6. Convert 1.20 moles of CO to molecules.

• 7. Convert 2.53 moles of Ag to Ag atoms.

• 8. Convert 0.025 mole of magnesium sulfate to formula units.

• Do Practice : Ex: 9.6

• • • • • • • • •

The Mass of a Mole:

The molar mass of an element is its atomic mass in grams.

•It contains 6.022 x 1023 atoms (Avogadro’s number) of the element.

Moles=grams/molar mass.

9. Convert 1.50 moles of CH4molecule to mass in grams.

10. Convert 2.50 moles of NaCl formula units to mass in grams.

11. Convert 1.68 moles of N2 molecule to mass in grams.

12. Convert 1.68 moles of N atoms to mass in grams.

Do Practice Ex: 9.7.

13. What is the mass in grams of a molecule whose mass on the amu scale is 104.00 amu?

Problems

• Do Practice Ex: 9.8

• Do Practice Ex: 9.9

• Do Practice Ex: 9.10

• 14. 56.04 g of N

2

molecules?

contains how many N

2

• 15. How many moles of benzene, C

6

H

6,

present in 390.0 grams of benzene?

are • 16.How many grams of (NH4)3PO4 are contained in 2.52 moles of (NH4)3PO4?

• Do Practice Ex: 9.11,9.12.

Problems

• • 17. Calculate the mass in grams for : a) a single atom of Cu.

• b) a single molecule of CO2 • 18. Caffeine has the formula C8H10N4O2 How many grams of N are present in a 50.0g sample of caffeine.

• Do Practice Ex: 9.14.

Problems

• 19. The compound cholesterol has a formula C27H46O. How many H atoms are present in 2.000 g sample of cholesterol.

• Percent purity: Percent by mass of a specified substance in an impure sample of the substance.

• 20. A 32.00 g sample of HNO3 has a purity of 96.20 % by mass. Calculate the following for this sample of nitric acid.

• a) The mass in grams of HNO3 present,b) the mass in grams of impurities present.

• Do Practice Ex: 9.16.

• 21. How many Fe atoms are present in 1 25.00 g sample of Fe-ore that has a purity of 87.70 % by mass Fe? Assume that there are no Fe containing impurities present.

Empirical Formula versus Molecular Formula

• • • The empirical formula or simplest

formula gives the smallest whole-

number ratio of the atoms present in a

compound.

The molecular formula is the true

formula of a compound.

Two compounds can have identical empirical formulas and different molecular formulas.

Molecular Formula C

2

H

4

Empirical Formula CH

2

Smallest Whole Number Ratio C:H 1:2

Molecular Formula C

6

H

6

Empirical Formula CH Smallest Whole Number Ratio C:H 1:1

Calculating Empirical Formulas

•

Step 1 Assume a definite starting

quantity (usually 100.0 g) of the compound, if if the actual amount is not given, and express the mass of each element in grams.

•

Step 2

Convert the grams of each element into moles of each element using each element’s molar mass.

Calculating Empirical Formulas

•

Step 3

Divide the moles of atoms of each element by the moles of atoms of the element that had the smallest value.

• – If the numbers obtained are whole numbers, use them as subscripts and write the empirical formula.

• – If the numbers obtained are not whole numbers, go on to step 4.

•

Step 4

Multiply the values obtained in step 3 by the smallest numbers that will convert them to whole numbers • Use these whole numbers as the subscripts in the empirical formula.

Some Common Fractions and Their Decimal Equivalents Common Fraction Decimal Equivalent Resulting Whole Number

1 4 0.25

1 1 3 2 3 1 2 Multiply the decimal equivalent by the number in the denominator of the fraction to get a whole number.

0.333… 0.666… 0.5

1 2 1 3 4 0.75

3

Problems:

• 22.The analysis of a salt shows that it contains 56.58% potassium (K); 8.68% carbon (C); and 34.73% oxygen (O). Calculate the empirical formula for this substance.

• 23. The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance. • Do Practice Ex: 9.18, 9.19, 9.20

Calculating the Molecular Formula from the Empirical Formula •The molecular formula can be calculated from the empirical formula if the molar mass is known.

•The molecular formula will be equal to the empirical formula or some multiple,

n,

of it.

•To determine the molecular formula evaluate

n.

•

n

is the number of units of the empirical formula contained in the molecular formula

Problems:

• 24. What is the molecular formula of a compound which has an empirical formula of CH2 and a molar mass of 126.2 g? • Do Practice EX: 9.23, 9.24

Combustion analysis

• A method used to measure the amounts of C and H present in a combustible compound that contains these two elements when burned in O2.

• 25. Ethylene is burned in combustion analysis apparatus and 3.14 g of CO2 and 1.29 g of H2O are produced. What is the formula of ethylene?

• Do Practice Ex: 9.21.