TOPIC 1 STOICHIOMETRIC RELATIONSHIPS 1.2 THE MOLE CONCEPT

ESSENTIAL IDEA The mole makes it possible to correlate the number of particles with the mass that can be measured. NATURE OF SCIENCE (2.3) Concepts – the concept of the mole developed from the related concept of “equivalent mass” in the early 19 th century.

UNDERSTANDING/KEY IDEA 1.2.A The mole is a fixed number of particles and refers to the amount, n, of substance.

A chemical species may be an atom, a molecule or an ion. This is the same as using the terminology that “particles” refer to atoms, molecules and formula units. (Note that formula units are made up of ions.)

The exact definition of a mole is the amount of a substance that contains the same number of chemical species as there are atoms in exactly 12g of the isotope carbon-12. Avogadro’s constant (L) has the value 6.02x10 23 particles per mole.

The mole is the chemist’s counting unit. It contains 6.02 x particles of a substance. Example: 1 dozen = 12 of something 1 mole = 6.02 x of something What are species (particles)? Atoms – smallest species of an element Atoms – smallest species of an element Molecules – smallest species of a covalent cmpd Molecules – smallest species of a covalent cmpd Ions – smallest species of an ionic cmpd Ions – smallest species of an ionic cmpd When we want to count atoms or molecules, we use Avogadro’s # as a conversion factor. There are 6.02 x number of atoms in one mole of an element. There are 6.02 x number of molecules in a covalent compound.

UNDERSTANDING/KEY IDEA 1.2.B Masses of atoms are compared on a scale relative to 12 C and are expressed as relative atomic mass (A r ) and relative formula/molecular mass (M r ).

1 The relative atomic mass (A r ) is the average mass of an atom, taking into account the relative abundances of all the naturally occurring isotopes of the element. 2 The relative molecular mass (M r ) of a covalent or molecular compound is the sum of all the atomic masses of the elements in the compound. 3 The relative formula mass (M f ) of an ionic compound is the sum of all the atomic masses of the ions in the formula.  Notice that A r, M r and M f have no units because they are “relative” terms.

UNDERSTANDING/KEY IDEA 1.2.C Molar mass (M) has the units g/mol.

The mass of one mole of a species is called the molar mass (M) and it has the units of g/mol or g mol -1.

APPLICATION/SKILLS Be able to calculate the molar masses of atoms, ions, molecules and formula units.

To calculate molar mass, add up the atomic masses of the separate atoms from the periodic table. CO 2 1-C = 1 x 12 = 12 2-O = 2 x 16 = g/mol CH 4 1-C = 1 x 12 = 12 4-H = 4 x 1 = 4 16 g/mol

1. NaHCO 3 2. Sr(CN) 2 3. Al 2 (SO 3 ) 3 4. C 12 H 22 O (NH 4 ) 2 CO 3 FIND THE FOLLOWING MOLAR MASSES

APPLICATION/SKILLS Be able to solve problems involving the relationships between the number of particles, the amount of substance in moles and the mass in grams.

Converting from moles to particles or particles to moles To convert from moles to particles or particles to moles, use Avogadro’s number as the conversion factor.

VOLUME (L) MOLE MASS (g) PARTI CLES ATOMS MOLECULES FORMULA UNITS DIVIDE BY 6.02X10 23 PART MOL MULTIPLY BY 6.02X10 23 PART MOL

Mole/Particle Conversion Problems  1. How many atoms are in 3.0 mol of Sn?  2. How many formula units are in 0.40 mol KCl?  3. How many molecules are in 7.50 mol SO 2 ?  4. How many formula units are in 4.80x10 -3 mole NaI?  5. How many moles are in 1.50x10 23 molecules of NH 3 ?  6. How many moles are in 1.0x10 9 molecules of O 2 ?  7. How many moles are in 6.02x10 23 molecules of Br 2 ?  8. How many moles are in 4.81x10 24 atoms of Li?

Converting from moles to mass or mass to moles To convert from moles to mass or mass to moles, use the molar mass as the conversion factor.

VOLUME (L) MOLE MASS (g) PARTI CLES ATOMS MOLECULES FORMULA UNITS DIVIDE BY MOLAR MASS MULTIPLY BY MOLAR MASS

MASS/MOLE CONVERSIONS  1. What is the mass of 2.40 mol N 2 ?  2. How many moles is 5.00g H 2 ?  3. What is the mass of 3.32 mol K?  4. How many moles is 11.0g CH 4 ?  5. What is the mass of mol H 2 O 2 ?  6. How many moles is 847g (NH 4 ) 2 CO 3 ?  7. What is the mass of 5.08 mol Ca(NO 3 ) 2 ?  8. How many moles is 333g SnF 2 ?  9. What is the mass of 10.0 mol Cr?  10. How many moles is g Li 2 HPO 4 ?

UNDERSTANDING/KEY IDEA 1.2.D The empirical formula and molecular formula of a compound give the simplest ratio and the actual number of atoms present in a molecule respectively.

Empirical formula – the simplest whole number ratio of the atoms it contains. Molecular formula – multiple of the empirical formula showing the actual number of atoms of each element present.

APPLICATION/SKILLS APPLICATION/SKILLS Be able to determine the empirical formula when given the percentage composition by mass.

Rules for empirical formula 1. Find the % composition. 2. Assume 100 g and convert %’s to g’s. 3. Convert the grams to moles by dividing by the molar mass. 4. Divide all mole quantities by the smallest number of moles found in step If the numbers found in step 4 are not whole numbers, multiply by a factor to make whole numbers.

Sample Problem A hydrocarbon contains 85.7% by mass of carbon. What is the empirical formula? Step 1: Find %’s Step 1: Find %’s 85.7 % C85.7 % C = 14.3% H (Hydrocarbons contain H and C) = 14.3% H (Hydrocarbons contain H and C) Step 2: Convert to grams Step 2: Convert to grams 85.7 g C85.7 g C 14.3 g H14.3 g H Step 3: Convert to moles Step 3: Convert to moles 85.7 g C x mol/12.01 g C = 7.14 mol C85.7 g C x mol/12.01 g C = 7.14 mol C 14.3 g H x mol/1.01 g H = mol H14.3 g H x mol/1.01 g H = mol H

Step 4: Divide by the smallest number of moles. Step 4: Divide by the smallest number of moles mol / 7.14 mol = mol / 7.14 mol = mol / 7.14 mol = 1.98 = mol / 7.14 mol = 1.98 = 2 Step 5: The numbers in Step 4 are whole numbers so this is the end of the problem. Step 5: The numbers in Step 4 are whole numbers so this is the end of the problem. The empirical formula is CH 2The empirical formula is CH 2

APPLICATION/SKILLS APPLICATION/SKILLS Be able to determine the molecular formula of a compound from its empirical formula and molar mass.  Hint: You have to be given the experimental molar mass in order to solve a molecular formula problem.

Rules for molecular formula 1. Solve for empirical formula. (It may be given to you or you may have to solve for it.) 2. Find the empirical molar mass. 3. Divide the given molecular or molar mass by the empirical molar mass – this answer should be a whole number. 4. Multiply the empirical formula by the number found in step 3.

Sample problem The compound with the empirical formula CH 2 is analyzed by a mass spectrometer and found to have a relative molecular mass of Deduce its molecular formula. Step 1: The empirical formula is CH 2. Step 1: The empirical formula is CH 2. Step 2: Find the empirical molar mass. Step 2: Find the empirical molar mass. CH 2 has a molar mass of g/mol.CH 2 has a molar mass of g/mol. Step 3: Divide the given molar mass by the empirical molar mass. Step 3: Divide the given molar mass by the empirical molar mass / = / = 3 Step 4: Multiply the empirical formula by the number found in Step 3. Step 4: Multiply the empirical formula by the number found in Step 3. (CH 2 ) 3 = C 3 H 6(CH 2 ) 3 = C 3 H 6

APPLICATION/SKILLS APPLICATION/SKILLS Be able to obtain and use experimental data for deriving empirical formulas from reactions involving mass changes.

EMPIRICAL FORMULAS FROM COMBUSTION 1. Sometimes you are not given the percentages, but have to calculate them from a combustion problem. 2. The problems usually give you an amount of original substance and amounts of CO 2 and H 2 O that are formed from the combustion reactions. 3. Make these assumptions: All of the carbon in the original sample is converted completely to CO 2. All of the carbon in the original sample is converted completely to CO 2. All of the hydrogen in the original sample is converted completely to H 2 O. All of the hydrogen in the original sample is converted completely to H 2 O.

EXAMPLE You have a compound composed of carbon, hydrogen and nitrogen. When.1156g of this compound is burned,.1638g of CO 2 and.1676g of H 2 O is produced. Determine the empirical formula of the compound.

Citations Brown, Catrin, and Mike Ford. Higher Level Chemistry. 2nd ed. N.p.: Pearson Baccalaureate, Print. Most of the information found in this power point comes directly from this textbook. The power point has been made to directly complement the Higher Level Chemistry textbook by Catrin and Brown and is used for direct instructional purposes only.