Chapter 6: Chemical Composition •Understand the concept of average mass and explore how counting can be done by weighing. •Understand atomic mass and its experimental determination. •Understand the mole concept and Avogadro’s number and convert among moles, mass, and number of atoms in a sample. 6-1

Objectives • Understand the definition of molar mass. • Learn to convert between moles and mass of a given sample of a chemical compound. • Learn to find the mass percent of an element in a compound. • Understand the meaning of empirical formulas. • Learn to calculate empirical formulas. • Learn to calculate the molecular formula of a compound, given its empirical formula and molar mass. 6-2

Section 6.1: Counting By Weighing • We can weigh a large number of the objects and find the average mass. • Once we know the average mass we can equate that to any number of the objects. • EXAMPLE:

– The average mass of a book is 40.0 grams. – How many books are present in a sample with a mass of 2000.0 grams? – 2000.0g/40.0g = 50.0 books

6-3

Counting By Weighing • When we know the average mass of the atoms of an element, as that element occurs in nature, we can calculate the number of atoms in any given sample of that element by weighing the sample. • The atomic mass of an element, as found on the periodic table, allows us to count by weighing.

6-4

Section 6.2: Atomic Masses – • Atoms are very small, a single carbon atom has a mass of 1.99 x 10-23 grams (0000000000000000000000199g). So, scientists had to come up with a smaller unit to measure their mass: the atomic mass unit (amu). 1 amu = 1.66 x 10-24 grams

• According to the periodic table, the average mass of a carbon atom is 12.01 amu. 6-5

Section 6.3: The Mole

• Not this type of mole!

• The mole (mol) is known as the “chemists dozen” and represents • 6.022 x 1023 things (atoms, particles, molecules, etc). • Just as we could have a dozen donuts, we could also have a mole of donuts; that’d be A LOT of donuts!

6-6

The Mole: Interesting Mole Facts • 6.022 X 1023 watermelon seeds: would be found inside a melon slightly larger than the moon.

6-7

• 6.02 X 1023 grains of sand: would be more than all of the sand on Miami Beach.

• 6.022 X 1023 pennies: would make at least 7 stacks that would reach the moon. • 6.02 X 1023 blood cells: would be more than the total number of blood cells found in every human on earth. • A mol is A LOT of particles: • 602,200,000,000,000,000,000,000

The Mole • It is VERY important to understand that the value listed as the mass number on the periodic table really tells us TWO things: – The mass of one atom in units of amu’s. – The mass of one mole of atoms in grams. – One mol of Al = 26.98 g & one atom = 26.98 amu – One mol of Au = 196.97 g & one atom = 196.97 amu – One mol of B = 10.81 g & one atom = 10.81 amu 6-10

The Mole • To summarize: a sample of any element that weighs a number of grams equal to the average atomic mass (from the periodic table) contains 6.022 x 1023 atoms (1 mol) of that element. • This is an important concept to understand before attempting calculations.

6-11

Using the mol in calculations • How many mols are in 7.0 g of N? (7.0 g)(1 mol/14.0 g)= 0.5 mol

How many atoms are in 7.0 g of N? 23 atoms 6.022 x 10 (0.5mol)( /1 mol)=

3.011 x 1023

atoms. OR 23 atoms 1 mol 6.022 x 10 (7.0 g)( /14.0 g)( /1 mol)= 3.011 x 1023 atoms

The Mole: Practice 1) Calculate the number of moles of atoms and the number of atoms in a 25.0 g sample of calcium. 2) Calculate the number of moles of atoms and the number of atoms in a 57.7 g sample of sulfur. 3) Calculate the number of atoms in a 23.6 mg sample of zinc. 4) Calculate the number of atoms in a 128.3 mg sample of silver. 6-13

The Mole: More Interesting Mole Facts • A one-liter bottle of water contains 55.5 moles H20 molecules. • A five-pound bag of sugar contains 6.6 moles of C12H22O11 (sucrose). • We have 3 types of moles that live underground in North America: Eastern Mole, Hairy-Tailed Mole and Star-Nosed Mole • The "Mexican" Mole is a chocolate sauce or turkey stew. It comes from the Aztec word "molli."

6-14

Section 6.4: Molar Mass • A chemical compound is a collection of atoms. • One methane (CH4) molecule contains one C atom and four H atoms. • It follows then that one mole of methane molecules contains one mole of C atoms and four moles of H atoms.

6-15

Figure 6.3: Various numbers of methane molecules.

6-16

Molar Mass • The molar mass of any substance is the mass (in grams) of one mole of the substance. • The molar mass of a compound is obtained by summing the masses of ALL component atoms. • The term “formula weight” may be used for ionic compounds and is analogous to molecular weight; both are molar masses.

6-17

Molar Mass • If we know how many atoms and how many moles are present, we can calculate the mass of one mole of a compound. • This is called the “molar mass” or sometimes you may also see it referred to as “molecular weight”. • Since 1 mol C = 12.01 g and 4 mol H = 4(1.008) or 4.032 g, 1 mol CH4 = 12.01 + 4.032 or 16.04 g (sig figs). • Remember to use mass numbers from the periodic table. 6-18

Molar Mass

• Remember: we can talk about one mole of atoms or one mole of molecules. • One mole of oxygen atoms (O) weighs 16.00 g. • One mole of oxygen molecules (O2) weighs 32.00 g. • Two moles of O atoms weigh 32.00 g. • Two moles of O2 molecules weigh 64.00 g. • And so on . . . 6-19

Molar Mass: Practice 1) Calculate the following molar masses: a) b) c) d)

Water – H2O Ammonia – NH3 Propane – C3H8 Glucose – C6H12O6

2) Calculate the mass of 1.48 mol C3H8. 3) Calculate the mass of 4.85 mol HC2H3O2.

6-20

Molar Mass: Practice 4. Calculate the number of moles of H2CO present in a 7.55 g sample. 5. Calculate the number of moles of tetraphosphorous decoxide present in a 250.0 g sample. 6. How many water molecules are present in 10.0 g of water? (Hint: find moles first) 7. How many molecules of sucrose (C12H22O11) are present in a five pound bag of sugar? 6-21

Representative Particles= atoms, molecules, Items, etc

Section 6.5: Percent Composition of Compounds • Sometimes it is not enough to know a compound’s composition in terms of numbers of atoms; it may also be useful to know its composition in terms of the masses of its elements. • We can calculate the mass fraction by dividing the mass of a given element in one mole of a compound by the mass of one mole of the compound. 6-23

Percent Composition of Compounds • Once we know the mass fraction we can multiply by 100 to get the percent. • Remember: percent = part/whole X 100.

6-24

ex) In one mole of methane (CH4), there is one mole of C and four moles of H: 1 mol C = 12.01 g 4 mol H = 4(1.008) = 4.032 g 1 mol CH4 = 12.01 g + 4.032 g = 16.042 g %C= mass of 1mol C/ mass of 1 mol CH4 x 100 %C = 12.01/16.042 X 100 = 74.87% C %H= mass of 4 mol H/mass of 1 mol CH4 x 100 %H = 4.032/16.042 X 100 = 25.13% H

Percent Composition Practice • Determine the mass percent of each element in the following: – H2SO4 (sulfuric acid) – C3H7OH (isopropyl alcohol) – C6H12O6 (glucose)

• Note: the sum of the percentages should always equal 100% exactly. Be careful when rounding. 6-26

Section 6.6: Formulas of Compounds

• The object of this section is to do the opposite of the previous section. Instead of getting the mass from the formula, we will determine the formula from the mass. • To do this, the mass must be converted to moles using each element’s mass number. How can we convert mass to moles?

6-27

Formulas of Compounds • ex) An unknown compound with a mass of 0.2015 g is is found to contain: – 0.0806 g C – 0.01353 g H – 0.1074 g O

• It must contain: 0.0806g (1 mol C/12.01 g C) = 0.00671 mol C 0.01353g(1 mol H/1.008 g H) = 0.01342 mol H 0.1074g(1 mol/16.00 g O) = 0.00671 mol O 6-28

• The numbers from the previous slide allow us to determine the C:H:O ratio. • 0.00671 (C) : 0.01342 (H) : 0.00671 (O) • If we divide each number by the smallest number we get 1:2:1 for the C:H:O ratio. • This leads us to a formula of C1H2O1 or CH2O. • This is not necessarily the TRUE formula of the compound, but represents the RELATIVE numbers of atoms.

6-29

• This represents the lowest whole number ratio of the compound. • The actual formula could be CH2O, C2H4O2, C3H6O3, C4H8O4, C5H10O5, C6H12O6, . . . • Any formula with a C:H:O ratio of 1:2:1 is possible (in theory, an infinite number). • C1H2O1 represents the simplest possible formula or the EMPIRICAL FORMULA. • The multiples represent possible MOLECULAR FORMULAS.

6-30

Formulas of Compounds: Practice • Determine the empirical formula from each of the following molecular formulas: – H2O2 (hydrogen peroxide) – C4H10 (butane) – CCl4 (name?) – HC2H3O2 (acetic acid) – C6H12O6 (glucose)

6-31

•

1. 2. 3. 4.

Section 6.7: Calculation of Empirical Formulas

There are four steps to determine the empirical formula of a compound: Obtain the mass of each element present (in grams). Determine the number of moles of each type of atom present (use atomic mass). Divide each number by the smallest number. Multiply all numbers by the smallest integer that will make them all integers

6-32

Empirical Formulas: Practice 1) A 1.500 g sample of a compound containing only carbon and hydrogen is found to contain 1.198 g of carbon. Determine the empirical formula for this compound. 2) A 3.450 g sample of nitrogen reacts with 1.970 g of oxygen. Determine the empirical formula for this compound. 3) When a 2.000 g sample of iron metal is heated in air, it reacts with oxygen to achieve a final mass of 2.573g. Determine the empirical formula for this compound. 6-33

• If the relative amounts of elements are presented as percentages, assume we are starting with a 100 g sample (100%). Then each percentage simply becomes a mass (in grams). • For example if 15% of a compound is carbon, we just assume it is 15 g of a 100 g sample; from there we convert to moles.

6-34

Empirical Formulas: More Practice 1) The simplest amino acid, glycine, has the following mass percents: 32.00% carbon, 6.714% hydrogen, 42.63% oxygen, and 18.66% nitrogen. Determine the empirical formula for glycine. 2) A compound has been analyzed and found to have the following mass percent composition: 66.75% copper, 10.84% phosphorous, and 22.41% oxygen. Determine the empirical formula for this compound. 6-35

Section 6.8: Calculation of Molecular Formulas • If we know the empirical formula AND the molar mass, we can calculate a compound’s molecular formula. • Note: without the molar mass, the best you can find is the empirical formula. • Once the molar mass is known, one must ALWAYS find the empirical formula before one can calculate the molecular formula. It is impossible to do the reverse. 6-36

Calculation of Molecular Formulas • It is also important to note that the molecular formula is ALWAYS an integer multiple of the empirical formula. We can represent the molecular formula in terms of the empirical formula: (Empirical Formula)n = molecular formula

• It should also be noted when n = 1, the empirical and molecular formulas are identical to each other. 6-37

Molecular Formulas: Practice 1) A compound containing carbon, hydrogen, and oxygen is found to be 40.00% carbon and 6.700% hydrogen by mass. The molar mass of this compound is between 115 and 125 g/mol. Determine the molecular formula for this compound. 2) Caffeine is composed of 49.47% C, 5.191% H, 28.86% N, and 16.48% O. The molar mass is about 194 g/mol. Determine the molecular formula for caffeine. 6-38