The Change in Mass Is Due to a Change in the Number of Neutrons

Chapter 3

Chemical Stoichiometry

Isotopes

· Atoms with different mass but the same atomic number are called Isotopes.

· The change in mass is due to a change in the number of neutrons.

· We Identify them by the following notation:

Has 12 e- and 12p+
Has 12 no / Has 12 e- and 12 p+
Has 13 no

Avogadro’s Number

· One mole of atoms is 6.022x1023 atoms.

· One mole of anything is 6.022x1023 of them.

– For the reaction:

2H2

+

O2

/ /

2H2O

we can say that two moles of hydrogen gas reacting with one mole of Oxygen gas gives us two moles of water.

– This would be 6.022 x 1023 x 2 atoms of water.

Atomic Mass

· The textbook uses Atomic Mass Units (amu’s) which are the same as the atomic mass numbers found on the periodic chart.

· For our purposes, amu’s will have the same label as the atomic mass which is grams/moles.

Molecular Mass

· This term is synonymous with formula weight.

· To determine the molecular mass of a compound, add the atomic mass for each atom in the compound.

Moles of Molecules

· In chemistry, we are unable to solve most problems unless we can convert masses to a comparable unit (The Mole)

· To convert grams of a substance to moles we divide grams by moles:

· In other words, the percent of hydrogen in water is the same percent everywhere.

ex) 5 grams of Iron is how many moles?

Percent Composition from Formula

· All samples of a pure compound contain the same elements in the same proportion by mass.

– Percent by mass = F.W. of X/F.W. of sample x 100

Ex) what percent of water is hydrogen?

Derivation of Formula

· To derive an empirical formula from percent, we set up a mole ratio of the constituents that make up the formula

Ex)What is the formula of a compound that is 27.29% Carbon and 72.71% Oxygen?

So: CO2

Solutions

· A solution is a homogeneous mixture composed of:

– Solute: that which is dissolved

– Solvent: that in which the solute dissolves

– A small amount of solute dissolved in a solvent is said to be dilute.

– The more solute added will increase the concentration.

· The concentration of a solution is measured in Molarity (M).

– Molarity is defined as:

ex) What is the concentration of 4.3x10-5 grams of PbCrO4 dissolve in one liter of water?

– The formula weight of PbCrO4 = 323.2 g/mole

Mole Relationships Based on Equations

· A chemical Reaction gives us a lot of information.

· One of these is the Mole relationships of each of the reactants and products with each other.

· The Reaction:

2H2O / / 2H2 / + / O2

tells us that the ratio of Water to Hydrogen gas to Oxygen is 2:2:1

Calculations Based on Equations

· Stoichiometry is the method of solving reaction based problems.

· All calculations proceed from the law of conservation of matter.

· Also, the fact that balanced equation give the ratios of moles of reactants and products.

· And we can determine the masses of reactants and products from the numbers of moles involved.

· The general strategy for solving all of these types of equations is as follows:

Write
What’s
Given / Convert
To
Moles / Mole
To
Mole
Ratio / Convert
To
What’s
Asked For

· Grams to Grams

Ex) Calculate the grams of oxygen produced during the thermal decomposition of 100.0 grams of potassium chlorate to form potassium chloride and oxygen in the following reaction:

2KClO3(s)

/ /

2KCl(s)

3O2(g)

Grams
KClO3 / Moles
KClO3 /

Moles

O2(g) / Grams
O2(g)
= 39.16 Grams O2(g)

· Grams to Molarity

ex) What volume of a 0.75 M solution of hydrochloric acid can be prepared from the HCl produced by the reaction of 25.0 g of NaCl with an excess of sulfuric acid?

NaCl(s)

H2SO4(l)

/ /

HCl(g)

NaHSO4(s)

Grams
NaCl(s) / Moles
NaCl(s) /

Moles

HCl(g) / Volume
Of
Solution

· Molarity to Molarity

ex) Potassium Iodide, KI, reacts with copper nitrate, Cu(NO3)2, to form a mixture of CuI and KNO3. What volume of a 0.2089 M solution of KI will contain enough KI to react exactly with the Cu(NO3)2 in 43.88 mL of a 0.3842 M Cu(NO3)2 solution according to the following equation:

2Cu(NO3)2(aq) / + / 4KI(aq) / / 2CuI(s) / + / I2(s) / + / KNO3(aq)
Concentration
Cu(NO3)2 / Moles
Cu(NO3)2 /

Moles

KI / Volume
KI

· Theoretical Yield, Actual Yield, and Percent Yield

– ex) A general chemistry student, preparing copper metal by the reaction of 1.274 g of copper sulfate with zinc metal, obtained a yield of 0.392 g of copper. What was the percent yield?

CuSO4(aq) / + / Zn(s) / / Cu(s) / + / ZnSO4(aq)
Mass
CuSO4 / Moles
CuSO4 /

Moles

Cu

/ Theoretical
Mass
Cu

· Limiting Reagents

– under certain conditions, not all of the reactants may be consumed

– ex)

2AgNO3

/ /

2Ag

Cu(NO3)2

• the equation indicates that 2 moles of silver (I) nitrate reacts with 1 mole of copper

• If there are two moles of silver(I) nitrate and 2 moles of copper present, one mole of copper will not react because only one mole of copper can react.

• The reactant that is completely consumed is called the Limiting Reagent

ex) A mixture of 5.0 g of H2(g) and 10.0 g of O2(g) is ignited. Water forms according to the following reaction:

2H2(g) / + / O2(g) / / 2H2O(g)

Which reactant is limiting? How much water will the reaction produce?

– First, determine the number of moles of water that 5 grams of H2 will produce and the number of moles of water that 10 grams of O2 will produce

Water Produced by Hydrogen:

Water Produced by Oxygen:

– 2.5 moles of Water can be produced by 5 grams of H2, but only 0.625 moles of water can be produced by 10 grams of O2.

– O2 must be the limiting reagent.

– So, the grams of water produced is: