Chapter 8: Atomic Masses: counting by weighing

Objective: To understand atomic mass and its experimental

determination

Atomic Mass: Atomic Mass Unit

C(s) + O2 (g) CO2 (g)

Atomic mass unit = unit of mass

1 amu = 1.66 x 10-24

Remember C has several isotopes: 12C, 13C, and 14C.

Average atomic mass unit=12.01

So we can calculate the mass of any of the following:

1) Calculate the mass (in amu) of a sample of Carbon that contains 62 atoms.

2) Calculate the mass (in amu) of a sample of iron that contains 15 atoms.

3) Calculate the mass (in amu) of a sample of aluminum that contains 75 atoms.

Calculate the number of atoms in

• Calculate the number of copper atoms present in a sample that has a mass of 1779.4 amu.

• Calculate the number of argon atoms present in a sample that has a mass of 3755.3 amu.

• Calculate the number of sodium atoms present in a sample that has a mass of 1172.49 amu.

Chapter 8.3: The Mole

• Objectives: To understand the mole concept and Avogadro’s number.

• To learn to convert among moles, mass, and number of atoms in a given sample.

The Mole: Revisited

• A mole (mol) is the unit a chemist uses to describe the numbers of atoms. It is defined as the number = to the number of carbon atoms in 12.01 g of carbon.

• Avogadro’s number is this number=6.022 x 1023

Just like a dozen = 12 eggs…..A mole of water = 6.022x 1023

Average atomic mass contains 1mol of atoms.

The mole: Revisited

Sample AMass=1.008 g

Sample BMass=0.500 g

1 mol of H atoms ? Number of H atoms.

Calculating Moles and Numbers of Atoms

• Calculate the number of moles of atoms and the number of atoms in a 25.0g sample of calcium.

• Calculate the number of moles of atoms and the number of atoms in a 57.7 g sample of sulfur.

0.624 mol Ca; 3.76 x 1023 Ca atoms

1.80 mol S; 1.08 x 1024 S atoms

Calculating the Number of Atoms

• Calculate the number of atoms in a 23.6 mg sample of Zinc.

• Calculate the number of atoms in a 128.3 mg sample of silver.

2.17 x 1020 Zinc atoms

7.16 x 1020 Ag atoms

8.4: Molar Mass

• To understand the definition of molar mass

• To learn to convert between moles and mass of a given sample of a chemical compound.

Mass of 1 mol of C= 1 x 12.01 g=12.01 gMass of 4 mol of H= 4 x 1.008 g= 4.032 gMass of 1 mol of CH4= 16.04 g.

Molar Mass of any substanceIs the mass ( in grams) of 1 mol of the substance.

Calculate the molar mass

Water H2O

Ammonia NH3

Propane C3H8

Glucose C6H12O6

18.02 g/mol17.03 g/mol44.09 g/mol180.2 g/mol

Calculating Mass from Moles

Calculate the mass of 1.48 mol of potassium oxide (K2O)

Calculate the mass of 4.85 mol of acetic acid, HC2H3O2.

Calculating Moles from Mass

• Formaldehyde has the formula H2CO. How many moles of formaldehyde does a 7.55 g sample represent?

• How many moles of tetraphosphorous decoxide does a 250.0 g sample represent?

0.251 mol

0.8805 mol

Calculating Number of Molecules

• How many water molecules are in a 10.0g sample of water?

• Sucrose or table sugar has the formula C12H22O11. How many molecules of sugar are in a 5.00 lb bag of sugar?

3.34 x 1023

3.99 x 1024

8.5: Percent Composition of Compounds

• Objective: To learn to find the mass percent of an element in a given compound.

Percent Composition:

• Mass fraction

for a given element = mass of the element present in 1 mol of compound

Mass of 1 mol of compound

Mass percent = mass fraction x 100

C2H5OH Mass of C= 2 mol x 12.01 g/mol = 24.02Mass of H= 6 mol x 1.008 g/mol = 6.048Mass of O= 1 mol x 16.00 g/mol= 16.00 g

Compounds containing H2O

• The crystals of a hydrate are made of solid substances combined chemically w/ water.

Sodium carbonate decahydrate

Na(CO3). 10 H2O

1) First find the weight of NaCO3

2) Find the weight of 10 molecules of water

3) (Mass of water/ mass of hydrate) x 100

Percent Composition:Mass Percent

1) Determine the mass percent of each element in sulfuric acid (H2SO4)

2) Rubbing alcohol is an aqueous solution of isopropyl alcohol. Isopropyl alcohol has the formula C3H7OH. Determine the mass percent of each element in isopropyl alcohol.

8.6: Formulas for Compounds

Objective: To understand the meaning of the empirical formula of compounds.

Empirical Formulas

Empirical formula: smallest whole number ratio of the atoms. (simplest formula)

Molecular formula: actual formula

(CH2O)6= C6H12O6

Give the empirical formula for the following:

H2O2 hydrogen peroxide

C4H10 butane

Calculation of Empirical Formulas

• To learn to calculate empirical formulas

Calculation of Empirical Formulas

Total massof nickel

oxide

Mass of nickel

originallypresent

=Mass of oxygen

that reactedwith the

nickel

0.3354 g - 0.2636 = 0.0718 g

Calculate the amount of moles0.2636 g Ni x 1mol Ni atoms = 0.00491 58.69 g Ni

0.0718 g O x 1 mol O atoms = 0.00449 mol O atoms 16.00 g O

Calculation of Empirical Formulas

Steps for Determining the Empirical Formula of a Compound

STEP 1: Obtain the mass of each element present (in grams).

STEP 2: Determine the number of moles of each type of atoms present.

STEP 3: Divide the # of moles of each element by the smallest # of moles to convert the smalles # to 1. If all of the #s so obtained are integers, they are the

subscripts in the empirical formula. If one or more are not integers, go to

STEP 4: Multiply the #s you derived in step 3 by the smallest integer that will convert all of them to whole numbers. This set of whole numbers represents the subscripts in the empirical formula.

Calculation of Empirical Formulas

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 w/ 1.970 g of oxygen. Determine the empirical formula for this compound.

CH3

N2O

Calculation of Empirical Formulas for Binary Compounds

1) When a 2.000 g sample of iron metal is heated in air, it reacts with oxygen to achieve a final mass of 2.573 g. Determine the empirical formula for this Iron oxide.

2) A 4.550 g sample of cobalt reacts with 5.475 g chlorine to form a binary compound. Determine the empirical formula for this compound.

FeO [iron (II) oxide]

CoCl2 [cobalt(II)chloride]

8.8 Calculation of Molecular Formula

Objective: To learn to calculate the molecular formula of a compound, given its empirical formula and molar mass.

Molecular Formula

• Molecular formula = (empirical formula)n

Molar mass= n x empirical formula

So to solve for n= molar mass

Empirical formula mass

8.8: Molecular Formula

Example: A white powder is analyzed and found to have an empirical formula of P2O5. The compound has a molar mass of 283.88 g. What is the compound molecular formula?

Empirical formula weight: Molar Mass2 P: 2 x 30.97 g= 61.94 g 283.88 g =25 O: 5 x 16.00 g= 80.00 g 141.94 g 141.94 g

MOLECULAR FORMULA is P4O10

EXAMPLES

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 g/mol and 125 g/mol. Determine the molecular formula for this compound.

C4H8O4

Example #2

• Caffeine is a compound containing carbon, hydrogen, nitrogen, oxygen. The mass percent composition of caffeine is 49.47% carbon, 5.191% hydrogen, 28.86% nitrogen and 16.48% oxygen. The molar mass of caffeine is approximately 194 g/mol. Determine the molecular formula for caffeine.

C8H10N4O2