The Mole & Chemical Quantities

The Mole

Mole-the number of particles equal to the number of atoms in exactly 12.0 grams of carbon-12.

1 mol = 6.02 x 1023 particles

The mole is also called Avogadro’s number.

Particles can be:

Atoms = single elementsFormula units = ionically

bonded compoundsMolecules = covalently bonded

compounds

Particle-Mole Problems

Don’t Forget!!! The number of particles in 1 mole of any substance is always the same.

1 mol = 6.02 x 1023 particlesHow many molecules are in 2.2 moles of water?

1 mol H2O = 6.02 x 1023 molecules H2O

1 mol H2O 6.02 x 1023 molecules

6.02 x 10 23 molecules or 1 mol H2O

2.2 mol H20 6.02 x 1023 molecules

1 X 1 mol H2O = 1.3244 x 1024

1.3 x 1024 molecules H2O

How many moles of sodium carbonate contain 7.9 x 1024 formula units?7.9 x 1024 f.u. Na2CO3 1 mol Na2CO3

1 X 6.02 x 1023 f.u. Na2CO3

= 13.1229235 mol 13 mol Na2CO3

Mole-Mass and Mole-Volume Relationships

Mass-Mole ProblemsChange mass to moles or visa

versa using the molar mass.

You need to know the mass of 1 mole of the substance, or the molar mass.

How many moles are in 11.2 g of NaCl?

1. Determine the molar mass.Na: 1 x 22.99 = 22.99gCl: 1 x 35.45 = 35.45g

= 58.44 g/mol 1 mol of NaCl = 58.44 g it can be written as:

1 mol NaCl 58.44g NaCl 58.44 gNaCl or 1 mol NaCl

2. Write what you have been given over 1 and multiply by the equality that will cancel the starting unit (on bottom) and keep the desired unit (on top).

11.2 g NaCl 1 mol NaCl 1 X 58.44 g NaCl = 0.1916495551

mol NaCl 0.192 mol NaCl

What is the mass of 2.50 mol of NaCl?

Find the molar mass.

1 mol of NaCl = 58.44 g

1 mol NaCl 58.44g NaCl

58.44 g NaCl or 1 mol NaCl

2.50 mol NaCl 58.44 g NaCl

1 X 1 mol NaCl = 146.1 g NaCl

146 g NaCl

Mole-Volume Problems

Equal volumes of gases at the same temperature and pressure contain the same number of particles.

Molar volume-the volume of 1 mol of a gas at standard condition (STP).

STP-standard temperature ( 0 ºC) and pressure ( 1 atm).

What is the volume of 0.35 moles of helium gas at STP?

1 mol = 22.4 liters

0.35 mol He 22.4 L He

1 X 1 mol He = 7.84 L He

7.8 L He

Multistep Conversions

Mass-Particleg mol particles

Particle-Massparticles mol g

What is the mass of 8.2 x 1022 atoms of calcium?

1 mol Ca = 6.02 x 1023 atoms Ca1 mol Ca = 40.08 g Ca

What is the mass of 8.2 x 1022 atoms of calcium?

8.2 x 1022 atoms Ca x 1 mol Ca x 40.08 g Ca

1 6.02 x 1023 atoms 1 mol Ca

= 5.459401993 g Ca

5.5 g Ca

Empirical and Molecular Formulas

Remember:Percent composition-the mass of each element in a

compound compared to the entire mass of the compound multiplied by 100.

What is the percent of hydrogen in water?Molar mass of water is 18.02 g.Mass of hydrogen in water is 2.02 g.2.02 g18.02 g X 100 = 11.20976693%

11.2 %

Empirical Formula- formula that gives the simplest whole-number ratio of atoms.

To solve for the empirical formula from percentages:

1. Assume there is a 100 g sample.

2. Convert masses to moles.

3. Find the smallest whole-number ratio between moles. This means divide each number by the smallest moles.

Determine the empirical formula for a compound made of 80 % carbon and 20 % hydrogen.

80. g of carbon20. g of hydrogen80. g C 1 mol C 1 x 12.01 g C = 6.661115737 mol C

6.7 mol C20. g H 1 mol H 1 x 1.01 g H = 19.8019802 mol H

20. mol H

C: 6.7 6.7 = 1

H: 20. 6.7 = 2.98 3

1:3 Ratio

Empirical formula = CH3

Molecular Formula

-formula that gives the actual number of atoms of each element in a molecular compound.

-whole number multiple of the empirical formula

C6H12O6 (molecular formula for glucose)

CH20 (empirical formula)

The molecular formula can be determined by comparing the empirical formula mass to the molecular formula mass.

To solve for molecular formula from masses:

1. Solve for the empirical formula.

2. Solve for empirical formula mass.

3. Compare the empirical formula mass to the molecular formula mass.

A compound has the following composition: 76.54 % carbon, 12.13 % hydrogen, and 11.33 % oxygen. Its molar mass is 282.45 g/mol, what is its molecular formula.

Solve for empirical formula.

76.54 g carbon

12.13 g hydrogen

11.33 g oxygen

76.54 g C 1 mol C

1 x 12.01 g C = 6.373022481 mol C

= 6.373 mol C

12.13 g H 1 mol H

1 x 1.01 g H = 12.00990099 mol H

= 12.0 mol H

11.33 g O 1 mol O

1 x 16.00 g O = 0.708125 mol O

= 0.7081 mol O

6.373 mol C 0.7081 = 8.9 912.0 mol H 0.7081 = 16.9 170.7082 mol O 0.7081 = 1

9:17:1 Ratio

Empirical Formula = C9H17OC: 9 x 12.01 =108.09gH: 17 x 1.01 = 17.17 gO: 1 x 16.00 = 16.00 g

Empirical formula mass = 141.26 g

282.45

141.26 = 1.99 2

2(C9H17O) = C18H34O2