Chapter 11 : Matter Notes

Mole (mol) is equal to 6.02x1023

The mole was named in honor of Amedeo Avogadro. He determined the volume of one mole of gas.

Molar Mass: mass in grams of 1 mole of any pure substance. Equal to the atomic mass for an element. Unit is g/mol.

Converting moles to Mass:

Number of moles number of grams = mass(g)

1 mole

Converting moles to particles.Number of moles_ 6.02x1023_ particles = particle

1 mole

Example: How many molecules are in 3.50 moles of sucrose?

3.50 mol sucrose 6.20 x 1023 molecules of sucrose

1 mole sucrose

= 2.11 x1023molecules of sucrose

Converting Particles to Moles:Number of particles 1 mole = number of moles

6.02 x1023 particles

Example: How many moles are in 4.5 x 1023 atoms of Zinc?

4.5 x 1023 atoms Zn 1 mole Zn

6.02 x1023 atoms Zn

= 7.48 moles Zn

Converting Moles to Mass:Element: How many grams are in 3 moles of

Manganese?

3.00 mol Mn 54.9g Mn = 165g Mn

1 mol Mn

Converting Moles to Mass:Compound: How many grams are in 3 moles of K2CrO4? 1. Find molar mass of compound.

2 mol K 39.10 g K = 78.20 g K1 mol K

1mol Cr 52.00g Cr = 52.00 g Cr1 mol Cr

4 mol O 16.00 g O = 64.00 g O

1mol O 194.20 g K2CrO4

Converting Moles to Mass:Compound:1. Multiply moles by molar mass:

3 moles K2CrO4 194.20 g K2CrO4

mol K2CrO4

= 582.6 g K2CrO4

Converting Mass to Moles:Number of grams 1 mole = moles

mass (g)

Element: How many moles are in 525g of Ca?

525g Ca 1 mol Ca = 13.1 mol Ca

40.06g Ca

Converting Mass to Moles: Compound: How many moles are in 325g

of Calcium Hydroxide [Ca(OH)2]?

1. Find the molar mass of compound:

1mol Ca 40.08g Ca = 40.08 g

1mol Ca

2mol O 16.00g O = 32.00 g

1mol O

2mol H 1.008g H = 2.026g

1mol H 74.096 g Ca(OH)2

Converting Mass to Moles:Compound:1. Convert grams to moles:

325g Ca(OH)2 1mol Ca(OH)2

74.096 g Ca(OH)2

= 4.39 mol Ca(OH)2

Converting Mass to Particles:Number of grams 1 mol 6.02 x 1023 particles

mass(g) 1 mol

= particles

Element: How many atoms are in 25 g of Gold?

25g Au 1 mol Au 6.02 x 1023 atoms Au

196.97g Au 1 mol Au

= 7.65 x 1023 atoms Au

Converting Mass to Particles: Compound: How many formula units are in

35.6 g of aluminum chloride? [AlCl3]

1. Find the molar mass of compound:

1 mol Al 26.98 g Al = 26.96 g

1mol Al

3 mol Cl 35.45g Cl = 106.35g

1 mol Cl 133.33g AlCl3

Converting Mass to Particles:Compounds:1. Convert grams to formula units:35.6 g AlCl3 1 mol AlCl3 6.02 x 1023 formula units

133.33g AlCl3 1 mol AlCl3= 1.61 x1023 formula units AlCl3

Chapter 11 Matter Notes

Section 11.4

Percent Composition:

Percent composition is the percent by mass of any element in a compound.

Mass of element X 100% = percent

composition

Mass of compound

Percent Composition:Example: Calculate the percent composition of

water. H2O2 mol H 1.01 g H =2.02 g H

1 mol H 18.02 g H2O

1 mol O 16.00 g O = 16.00 g O

1 mol O 18.02 g H2O

Percent Composition: con’t.

2.02 g H

H X 100% = 11.2% H

18.02g H2O

16.00 g O

O X 100% = 88.8% O88.8% O

18.02 g H2O

Empirical Formula

Empirical formula is a formula with the smallest whole-number ratio of the element.

Example: Determine the empirical formula for methyl acetate which has 48.64% Carbon, 8.16% Hydrogen, 43.20 % Oxygen.

Empirical Formula: con’t.

1. Find the number of moles of each element

48.64g C 1 mol C = 4.05 mol C

12.01 g C

8.16 g H 1 mol H = 8.10 mol H

1.008 g H

43.20 g O 1 mol O = 2.7 mol O

16.00 g O

Empirical Formula: con’t.1. Divide each mole by the smallest

number:

4.05 mol C = 1.5 mol C

2.7

8.10 mol H = 3 mol H

2.7

2.7 mol O = 1 mol O

2.7

Empirical Formula: con’t.1. Multiply each mole by the smallest

number that will produce whole numbers:

1.5 mol C x 2 = 3 C

3 mol H x 2 = 6 H

1 mol O x 2 = 2 O

1. Use the new whole numbers as subscripts for each element in the compound.

C3H6O2

Molecular Formula:

Experimental molar mass = n

empirical formula mass

Determine the molecular formula for a compound composed of 40.68% carbon, 5.08 % Hydrogen, and 54.24 % Oxygen which has a molar mass of 118.1 g/mol.

Molecular Formula:

1. Find the empirical formula:

40. 68 g C 1 mol C = 3.387 mol C

12.01 g C

5.08 g H 1 mol H = 5.04 mol H

1.008 g H

54.24 g O 1 mol O = 3.39 mol O

16.00 g O

Molecular Formula:

( finding the empirical formula con’t.)

3.387 mol C = 1 mol C x 2 = 2

3.387

5.04 mol H = 1.5 mol H x 2 = 3

3.387

3.39 mol O = 1 mol O x 2 = 2

3.387 empirical formula: C2H3O2

Molecular Formula:

1. Find the molar mass of the empirical formula:

2 mol C 12.01 g C = 24.2 g

1 mol C

3 mol H 1.008 g H = 3.024 g

1 mol H

2 mol O 16.00 g O = 32.00g

1 mol O 59.04 g C2H3O2

Molecular Formula:

1. Divide the molar mass given in the problem by the molar mass of the empirical formula to find n:

n = 118.1 g

59.04g

n = 2

Molecular Formula:

1. Multiply all subscripts of the empirical formula by n:

(C2H3O2)2

molecular formula: C4H6O4