Chapter 3: stoichimetry of formulas and equations, Cheat Sheet of Chemistry

Download Chapter 3: stoichimetry of formulas and equations and more Cheat Sheet Chemistry in PDF only on Docsity!

CHEM 1A: GENERAL CHEMISTRY

Chapter 3:

Stoichiometry

of Formulas

and Equations

Instructor: Dr.

Orlando E.

Raola

Santa Rosa

Junior College

Amount - Mass Relationships in Chemical Systems

3.5 Fundamentals of Solution Stoichiometry

3.1 The Mole

3.2 Determining the Formula of an Unknown Compound

3.3 Writing and Balancing Chemical Equations

3.4 Calculating Quantities of Reactant and Product

One mole (6.022x10^23 entities) of some familiar substances.

Figure 3.

The molar mass (M) of a substance is the mass per

mole of its entites (atoms, molecules or formula units).

For monatomic elements , the molar mass is the

same as the atomic mass in grams per mole. The

atomic mass is simply read from the Periodic Table.

The molar mass of Ne = 20.18 g/mol.

Molar Mass

Table 3.1 Information Contained in the Chemical Formula of Glucose C 6 H 12 O 6 ( M = 180.16 g/mol)

Carbon (C) Hydrogen (H) Oxygen (O)

Atoms/molecule of compound

6 atoms 12 atoms 6 atoms

Moles of atoms/mole of compound

6 mol of atoms 12 mol of atoms 6 mol of atoms

Atoms/mole of compound

6(6.022x10^23 ) atoms 12(6.022x10^23 ) atoms 6(6.022x10^23 ) atoms

Mass/molecule of compound

6(12.01 u) = 72.06 u

12(1.008 u) = 12.10 u

6(16.00 u) = 96.00 u

Mass/mole of compound

72.06 g 12.10 g 96.00 g

Interconverting Moles, Mass, and

Number of Chemical Entities

Sample Problem 3.1 Calculating the Mass of a Given Amount of an Element

PROBLEM:

SOLUTION:

amount (mol) of Ag

mass (g) of Ag

Silver (Ag) is used in jewelry and tableware but no longer in U.S. coins. How many grams of Ag are in 0.0342 mol of Ag?

PLAN: To convert mol of Ag to mass of Ag in g we need the molar mass of Ag.

multiply by M of Ag (107.9 g/mol)

0.0342 mol Ag x 1 mol Ag

107.9 g Ag = 3.69 g Ag

Sample Problem 3.2 Calculating the Number of Entities in a Given Amount of an Element

PROBLEM: Gallium (Ga) is a key element in solar panels, calculators and other light-sensitive electronic devices. How many Ga atoms are in 2.85 x 10-3^ mol of gallium?

To convert mol of Ga to number of Ga atoms we need to use Avogadro’s number.

PLAN:

mol of Ga

atoms of Ga

multiply by 6.022x10^23 atoms/mol

Sample Problem 3.3 Calculating the Number of Entities in a Given Mass of an Element

PROBLEM: Iron (Fe) is the main component of steel and is therefore the most important metal in society; it is also essential in the body. How many Fe atoms are in 95.8 g of Fe?

The number of atoms cannot be calculated directly from the mass. We must first determine the number of moles of Fe atoms in the sample and then use Avogadro’s number.

PLAN:

mass (g) of Fe

amount (mol) of Fe

atoms of Fe

divide by M of Fe (55.85 g/mol)

multiply by 6.022x10^23 atoms/mol

95.8 g Fe x

SOLUTION:

55.85 g Fe

1 mol Fe = 1.72 mol Fe

1.72 mol Fe x 6.022x (^23) atoms Fe 1 mol Fe

= 1.04 x 10^24 atoms Fe

Sample Problem 3.

Sample Problem 3.4 Calculating the Number of Chemical Entities in a Given Mass of a Compound I PROBLEM: Nitrogen dioxide is a component of urban smog that forms from the gases in car exhausts. How many molecules are in 8.92 g of nitrogen dioxide?

number of NO 2 molecules

PLAN: Write the formula for the compound and calculate its molar mass. Use the given mass to calculate first the number of moles and then the number of molecules.

divide by M

multiply by 6.022 x 10^23 formula units/mol

mass (g) of NO 2

amount (mol) of NO 2

SOLUTION: NO 2 is the formula for nitrogen dioxide.

M = (1 x M of N) + (2 x M of O) = 14.01 g/mol + 2(16.00 g/mol) = 46.01 g/mol

8.92 g NO 2 x

= 1.17 x 10^23 molecules NO 2

1 mol NO 2 46.01 g NO 2

6.022x10^23 molecules NO 2 1 mol NO 2

= 0.194 mol NO 2

0.194 mol NO 2 x

Sample Problem 3.

number of (NH 4 ) 2 CO 3 formula units

mass (g) of (NH 4 ) 2 CO 3

amount (mol) of (NH 4 ) 2 CO 3

SOLUTION: (NH 4 ) 2 CO 3 is the formula for ammonium carbonate.

M = (2 x M of N) + (8 x M of H) + (1 x M of C) + (3 x M of O) = (2 x 14.01 g/mol) + (8 x 1.008 g/mol)

• (12.01 g/mol) + (3 x 16.00 g/mol)

= 96.09 g/mol

divide by M

multiply by 6.022 x 10^23 formula units/mol

number of O atoms

3 O atoms per formula unit of (NH 4 ) 2 CO 3

Sample Problem 3.

41.6 g (NH 4 ) 2 CO 3 x

= 2.61x10^23 formula units (NH 4 ) 2 CO 3

1 mol (NH 4 ) 2 CO 3 96.09 g (NH 4 ) 2 CO 3

6.022x10^23 formula units (NH 4 ) 2 CO 3 1 mol (NH 4 ) 2 CO 3

= 0.433 mol (NH 4 ) 2 CO 3

0.433 mol (NH 4 ) 2 CO 3 x

2.61x10^23 formula units (NH 4 ) 2 CO 3 x 3 O atoms 1 formula unit of (NH 4 ) 2 CO 3

= 7.83 x 10^23 O atoms

Sample Problem 3.