The Mole, part II Glencoe, Chapter 11 Sections 11.3 & 11.4.

The Mole, part II

Glencoe, Chapter 11

Sections 11.3 & 11.4

Mole Relationships

Section 11.3

You will have to determine the mole relationships required to solve stoichiometry problems.

Mole relationships can come from chemical formulas and from balanced chemical equations.

Freon: CCl2F2

The ratio of atoms in a freon molecule are 1:2:2.

That is, for every carbon atom, there are two chlorine atoms and two fluorine atoms.

In one mole of freon, there is one mole of carbon atoms, two moles of chlorine atoms, and two moles of fluorine atoms.

How many fluorine atoms are in 2.5 mol freon?

Glucose: C6H12O6

Identify the ratio of atoms in glucose.

If you have 1.8 mol of glucose, how many hydrogen atoms do you have? How many oxygen atoms?

What is the mass of 1.8 mol of glucose?

Cu + 2AgNO3 Cu(NO3)2 + 2Ag

The ratio of coefficients in the above reaction are 1:2:1:2.

For every mole of copper used, two moles of silver nitrate are required.

One mole of copper (II) nitrate and two moles of silver are produced.

List all the mole relationships from this equation.

1 mol Cu/ 2 mol AgNO3

1 mol Cu/ 1 mol Cu(NO3)2

1 mol Cu/ 2 mol Ag

2 mol AgNO3/ 1 mol Cu

1 mol Cu(NO3)2/ 1 mol Cu2 mol Ag/1 mol Cu

2 mol AgNO3/1 mol Cu(NO3)2

2 mol AgNO3/ 2 mol Ag

1 mol Cu(NO3)2/ 2 mol AgNO3

2 mol Ag/ 2 mol AgNO3

1 mol Cu(NO3)2/ 2 mol Ag

2 mol Ag/ 1 mol Cu(NO3)2

2H2 + O2 2H2O

What is the ratio of coefficients in the above equation?

List all the mole relationships from this equation.

Complete p. 326, #31-35. You can check your answers in the back of the text.

Empirical & Molecular Formulas

Section 11.4

Percent Composition

In analytical chemistry, it is important to identify the elements in a compound, and determine their percent by mass.The percent by mass of each element in a compound is called percent composition.

mass of element X 100 mass of compound

Water: H2O

To determine the percent composition of water, you need its molar mass: 18.02g/mol.Hydrogen has a mass of 1.01, and there are two H atoms in a molecule of water. Therefore, hydrogen’s percent by mass is

2.02 g X 100 = 11.2% 18.02g

Oxygen has a molar mass of 16.00g/mol. There is one oxygen atom in a molecule of water. Oxygen’s percent by mass is:

16.00g X 100 = 88.8%

18.02g

So the percent composition of water is 11.2% H and 88.8% O.

Empirical Formula (EF)

The empirical formula of a compound is the formula with the smallest whole-number ratio. It may or may not be the same as the molecular formula.Data may be given in percent composition, or in mass. Convert this number to moles, and then divide by the lowest value. This will give you the simplest whole number ratio.

Given: 40.05% S and 59.95% O, determine the EF.

Convert the percentages to grams by assuming there is 100g of the compound. Then convert it to moles.

40.05g S/32.07 g S = 1.25 mol S

59.95g O/16.00 g O = 3.75 mol O

1.25 mol S/1.25 = 1 mol S

3.75 mol O/1.25 = 3 mol O

The ratio is 1:3, so the EF is SO3

Molecular Formula (MF)

For these problems, you will have to calculate the EF and determine its molar mass.You will be given an experimentally determined molar mass of a compound.Divde the determined molar mass of a compound by the molar mass of the EF to get “n”. N is the multiplier by which you will determine the molecular formula.MF = n(EF)

G: EF is CH. Molar mass of compound is 78.12 g/mol. Determine the MF.

The molar mass of CH is 13.02 g/mol.

78.12 g/mol/13.02 g/mol = 6

“n” is 6

6(CH) = C6H6

This is the formula for benzene.

fin