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Chemistry 1210: General Chemistry. The Mole and Stoichiometry . Dr. Gina M. Florio. 13 September 2012 Jespersen, Brady, Hyslop , Chapter 4. Conversion Factors. Conversion Factor. – relates one quantity to another. – used to convert between two units in chemistry.

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Chemistry 1210:

General Chemistry

The Mole and Stoichiometry

Dr. Gina M. Florio

13 September 2012

Jespersen, Brady, Hyslop, Chapter 4

Conversion Factors

Conversion Factor

– relates one quantity to another

– used to convert between two units in chemistry

What is my height in centimeters (cm) if I am 5 feet 4 inches tall?

1. How many inches are in a foot?

12 inches = 1 foot

2. How many inches are in a centimeter?

1 inch = 2.54 cm

Factor Label Method

The factor-label method, or dimensional analysis lets us treat a numerical problem as one involving a conversion from one kind of units to another using conversion factors.


A recipe calls for 32 grams of cheese. My kitchen scale only displays weight in ounces. How many ounces of cheese do I need to use?

Conversion Factor

Factor Label Method

Atomic Mass and Molecular Mass

Atomic Mass

– the mass of an individual atom measured in atomic mass units, u

– listed in the periodic table beneath the element symbol

– the mass of an individual molecule measured in u

Molecular Mass

– the sum of the atomic masses of the atoms in the molecular formula


The molecular mass of water, H2O, is twice the mass of hydrogen (2 x 1.008 u) plus the mass of oxygen (15.999 u)

= 18.015 u


Formula Mass

Formula Mass

– the mass of an individual formula unit measured in atomic mass units, u

– used for ionic compounds

– calculated the same as a molecular mass


The formula mass of calcium oxide, CaO, is the mass of calcium (40.08 u) plus the mass of oxygen (15.999 u)

= 56.08 u


The Mole: Connecting the macroscopic & molecular

The mole is a conversion factor

– relates mass to the number of atoms or molecules for a chemical substance (element, molecular compound, ionic compound)

– one mole of a substance has a mass (g) equal to its formula (or molecular or atomic) mass (u)



Molecular Mass = 2(H) + (O)

= 2(1.0 g) + 16.0 g

= 18.0 g

1 mole H2O = 18.0 g H2O


How many water molecules are in a mole of water?

1 mole H2O = 18.0 g H2O


1 molecule H2O = 18.0 u (molecular mass)

Need the conversion factor between mass (g) and u:

1 u = 1.66 x 10-27 kg

Need the conversion factor between g and kg:

1 kg = 1000 g



Grains of sand



Avogadro, how many ______ are in a mole?

6.023 x 1023 ______ are in a mole.


The “blank” can be anything:


This conversion factor (6.023 x 1023 objects/mole) is known as Avogadro's number.


Avogadro’s Number

Two important points:

1. Notice the magnitude of Avogadro's number (6.023 x 1023):


Why is this number soHUGE?

Because atoms and molecules are sotiny.

A huge number of atoms or molecules are needed to make a lab-sized sample.

2. Avogadro’s number links moles and atoms, or moles and molecules, and provides an easy way to link mass and atoms or molecules.


Example Problem

Assuming that pennies are 100% Cu, how many Cu atoms are found in a penny weighing 3.00 g?

1. Convert mass (g) to moles using molar mass

2. Convert moles to number of atoms (or molecules) using Avogadro’s number


Mole-to-Mole Ratios: Chemical Formulas


  • Water (molar mass 18.015 g/mol) as an example:
      • 1 mole H2O  6.023 x 1023 molecules H2O
      • 1 mole H2O  18.015 g H2O
      • 18.015 g H2O 6.023 x 1023 molecules H2O

Mole-to-Mole Ratio:

Within chemical compounds, moles of atoms always combine in the same ratio as the individual atoms themselves.

  • 1 mole H2O  2 moles H
  • 1 mole H2O  1 mole O



– the study of the mass relationships in chemical compounds and reactions


– relates the masses of reactants needed to make a compound

– MOLE to MOLE ratios


Example Problem

How many grams of iron are in a 15.0 g sample of iron(III) oxide?

  • ANALYSIS: 15.0 g Fe2O3 ? g Fe
  • KNOWN: 1 mol Fe2O3 2 mol Fe
  • 1 mol Fe2O3 159.7 g Fe2O3
  • 1 mol Fe  55.85 g Fe


Percent Composition

Percentage composition (or percentage composition by mass)

The percentage by mass is the number of grams of the element in 100 g of the compound and can be calculated using:

From our previous example:


Hmmm…what in the world is this stuff?

Detective, send that sample back to the lab for analysis!

Percent Composition, Why Should I Care?

Sample Problem

A sample was analyzed and found to contain 0.1417 g nitrogen and 0.4045 g oxygen. What is the percentage composition of this compound?

ANALYSIS: Find sample mass and calculate %

  • KNOWN: Mass of the whole sample


Molecular Formula & Empirical Formula

Hydrogen peroxide consists of molecules with the formula H2O2.

This is called the molecular formula.

However, the simplest formula for hydrogen peroxide is HO and is called the empirical formula.

The empirical formula thus contains the simplestmole-to-mole ratio of the atoms in the compound.

We can calculate the empirical formula for a compound from mass data.


Example Problem

A 2.012 g sample of a compound contains 0.522 g of nitrogen and 1.490 g of oxygen. Calculate its empirical formula.

  • ANALYSIS: We need the simplest whole number mole ratio between nitrogen and oxygen

Convert to moles

Find the simplest whole number mole-to-mole ratio


Empirical Formula (a few final notes)

The formula for an ionic compound is the same as the empirical formula.

For molecules, the molecular formula and empirical formula are usually different.

The molecular formula will be a common multiplier times all the coefficients in the empirical formula.


The empirical formula of hydrazine is NH2, and its molecular mass is 32.0 g/mol. What is its molecular formula?


Indirect Analysis of Empirical Formula

Combustion Analysis: an indirect method used to determine the empirical formula of a unknown compound containing only C, H, and O.

When a compound made only from C, H, and O burns completely in pure oxygen (O2) only carbon dioxide (CO2) and water (H2O) are produced:

We can collect and quantify all of the carbon dioxide (CO2) and water (H2O) produced to find the empirical formula of the original compound.


Example Problem

The combustion of a 5.217 g sample of a compound of C, H, and O gave 7.406 g CO2 and 4.512 g of H2O. Calculate the empirical formula of the compound.

ANALYSIS: This is a multi-step problem.

  • First determine the mass of each element.
  • Then use the masses of the elements to calculate the empirical formula of the compound.


Example Problem

The combustion of a 5.217 g sample of a compound of C, H, and O gave 7.406 g CO2 and 4.512 g of H2O. Calculate the empirical formula of the compound.



Stoichiometry & Chemical Equations

The coefficients of a balanced chemical equation provide the mole-to-mole ratios for the substances involved in the reaction.

Whenever a problem asks you to convert between different substances, the calculation usually involves a mole-to-mole relationship.

  • Example:
  • If 0.575 mole of CO2 is produced by the combustion of propane, C3H8, how many moles of oxygen are consumed? The balanced equation is:
    • C3H8 + 5 O2 3 CO2 + 4 H2O


Balancing Chemical Equations

Chemical equations provide quantitative descriptions of chemical reactions.

Conservation of mass is the basis for balancing equations.

  • To balance an equation:
    • 1. Write the unbalanced equation.
    • 2. Adjust the coefficients to get equal numbers of each kind of atom on both sides of the arrow


Balancing Chemical Equations: some tips

Guidelines for Balancing Equations:

  • 1. Balance elements other than H and O first
  • 2. Balance as a group any polyatomic ions that appears unchanged on both sides of the arrow
  • 3. Balance separately those elements that appear somewhere by themselves

4. As a general rule, use the smallest whole-number coefficients when writing balanced chemical equations


  • _ NH3 + _ O2 _ NO + _ H2O
  • _ NH3 + _ O2 _ NO + _ H2O


Limiting Reactants (aka Limiting Reagent)

All reactions eventually use up a reactant and stop.

The reactant that is consumed first is called the limiting reagent (reactant) because it limits the amount of product that can form.

Any reagent that is not completely consumed during the reactions is said to be in excess.

The computed amount of product is always based on the limiting reagent.


Example Problem


How many grams of NO can form when 30.0 g NH3 and 40.0 g O2 react according to:

  • 4 NH3 + 5 O2 4 NO + 6 H2O


Percentage Yield

The amount of product isolated from a chemical reactions is almost always less than the calculated, or maximum, amount.

The actual yieldis the amount of the desired product isolated.

The theoretical yield is the amount that would be recovered if no loss occurred (the calculated, maximum amount).

The percentage yield is the actual yield as a percentage of the theoretical yield: