The Mole

The Mole

Stoichiometry: Cookbook Chemistry

The Mole

A mole is a number Avogadro’s number = 6.02x1023

Named after Amadeo Avogadro Loschmidt determined the number

of particles in one cubic centimeter of a gas at ordinary temperature and pressure

Counting atoms by counting moles

By counting moles, atoms or molecules are counted

Counting atoms by using moles eliminates waste in chemical reactions

Coefficients in chemical equations represent mole quantities

Counting atoms by counting moles

2Na + Cl2 2NaCl “Two moles sodium and one mole

chlorine gas react to give two moles sodium chloride”

4 moles sodium require 2 moles Cl2 5.2 moles sodium require 2.6 moles Cl2 3.1 moles Cl2 require 6.2 moles sodium 2:1 is the sodium/chlorine mole ratio

Counting atoms by counting moles

Counting atoms allows prediction of product quantities

2Fe + 6HCl 2FeCl3 + 3H2

How many moles iron (III) chloride can be made using 4.3 moles HCl?

Set up a proportion Coeff. 2mol FeCl3 = x mol FeCl3 prob. side 6mol HCl 4.3mol HCl side x=2(4.3)/6=1.43 mol FeCl3

Limiting reagents

If mole quantities are not exact, one of the reactants will run out first – this reactant is the limiting reagent

2H2 + O2 2H2O

If 3 moles H2 are reacted with 1 mole O2, what is the limiting reagent?

Limiting reagents

Divide each mole quantity by the coefficient to find equivalents.

H2 3/2=1.5eq

O2 1/1=1eq limiting reagent The reactant with the fewest

equivalents (O2) is the limiting reagent. The other (H2) is “in excess”.

Limiting reagents

2Fe + 6HCl 2FeCl3 + 3H2

0.0037 mol Fe is reacted with 0.017 mol HCl. What is the limiting reagent?

Fe 0.0037/2 = 0.00185eq Fe HCl 0.017/6 = 0.00283eq HCl

Using limiting reagents

The quantity of product obtained is limited by the amount of the limiting reagent

2H2 + O2 2H2O If 4.5 moles hydrogen gas and 1.9

moles oxygen are reacted, how many moles water will be formed?

Using limiting reagents

Solution: First determine the limiting reagent.

H2: 4.5/2=2.25eq

O2: 1.9/1=1.9eq limiting reagent Then set up a proportion between

the limiting reagent and the desired product.

O2 1 = 1.9

H2O 2 xx=3.8 moles

Molar Mass

Molar mass is the mass of one mole of particles

Atomic mass – found in the bottom of each square of the periodic table – units are grams/mole

Atomic mass is the weighted average of the mass numbers of all the isotopes of an element.

Molecular mass – the mass of one mole of molecules

It is equal to the sum of the atomic masses of all the atoms in the molecule.

Molar mass

H2O – (H) 2x1 = 2 (O) 1x16 = 16 total = 18g/mol NH3 – (N) 1x14 = 14 (H) 3x1 = 3 total = 17g/mol glucose (C6H12O6) (C) – 6x12 = 72 (H) – 12x1 = 12 (O) – 6x16 = 96 total = 180g/mol

Molar mass

Formula mass is the sum of all the atomic masses in a formula unit (for salts)

NaCl – (Na) 23 (Cl) 35.5 total = 58.5g/mol Mg(NO3)2 (Mg) 1x24.3 = 24.3 (N) 2x14 = 28 (O) 6x16 = 96 total = 148.3g/mol

Using molar mass

Mass to moles conversions mass/(molar mass) = moles

g g/mol = g x mol/g = moles

Example: How many moles are represented by 2.5 grams of water?

Solution: 2.5g/(18g/mol) = 0.14mol

Using molar mass

Moles to mass conversionsmolesx(molar mass) = mass

mol x g/mol = g Example: What is the mass of

0.094 moles sodium chloride? Solution: 0.094mol x 58.5g/mol =

5.5g

Mass-mass stoichiometry

2HNO3 + H2O2 + 2Fe(NO3)2 2Fe(NO3)3 + 2H2O

How many grams hydrogen peroxide (H2O2) are needed to make 2.43 grams iron (III) nitrate (Fe(NO3)3) according to the reaction below?

x g H2O2 2.43 g iron (III) nitrate

moles H2O2 moles iron (III) nitrate

mole ratio(2:1)

by molar mass

x by molar mass

Per cent yield

Mass obtained from calculations is “theoretical yield” – never obtained in practice

Per cent yield = actual yield x 100%

theoretical yield

Per cent yield

Jorma makes drugs for a hobby (aspirin, that is) and expects to obtain 2.13g aspirin from his synthesis reaction. In reality he only gets 1.89g. What is his % yield?

(1.89/2.13)x100% = 88.7%

Per cent composition by mass

% composition by mass is a tool for compound identification

To calculate: divide the molar mass contribution of each element by the total molar mass and multiply by 100%

Per cent composition by mass

Example: H2SO4 (sulfuric acid) total molar mass H: 1x2=2 S: 32x1=32 O: 16x4=64 sum=98g/mol %H=2(100%)/98=2.04% %S=32(100%)/98=32.65% %O=remainder=65.31%=64(100%)/98

Determining formulas from % composition

Formulas are a mole ratio of elements Empirical formula: simplest mole ratio of

elements, like NaCl or Ca(NO3)2 Applies to any type of compound Molecular formula: mole ratio of elements

in an actual molecule (all nonmetals), like H2O or NH3

Often the molecular formula and the empirical formula are the same, but not always

Determining formulas from % composition

Hydrazine, a rocket fuel molecular formula – N2H4

empirical formula – NH2

Hydrogen peroxide molecular formula – H2O2

empirical formula – HO Glucose, a sugar

molecular formula – C6H12O6

empirical formula – CH2O

Determining formulas from % composition

% composition is a mass ratio – so by converting mass to moles, the empirical formula can be determined.

Example: Laboratory analysis finds a compound to consist of 28.05% Na, 29.27% C, 3.67% H, and 39.02% O. What is the empirical formula?

Determining formulas from % composition

Treat the % like grams Convert grams to moles Na: 28.05g/(23g/mol) = 1.22 mol C: 29.27g/(12g/mol) = 2.44 mol H: 3.67g/(1g/mol) = 3.67mol O: 39.02g/(16g/mol) = 2.44 mol

Determining formulas from % composition

Convert to simplest whole number ratio – divide all mol quantities by the smallest one. These results become the subscripts in the formula.

Na: 1.22/1.22 = 1 C: 2.44/1.22 = 2 H: 3.67/1.22 = 3 O: 2.44/1.22 = 2 So the empirical formula is NaC2H3O2

(sodium acetate).

Ideal gas law

Boyle’s Law: PV = C (P1V1 = P2V2) Factors that affect pressure/volume:

Temperature (T) Amount (moles) of gas (Avogadro’s

Principle) (n) Ideal Gas Law: PV nT Constant of proportionality = R (gas

constant)

Ideal Gas Law

Ideal gas Law: PV = nRT V must be liters, T is Kelvins, n is

moles Values for gas constant (depends on

pressure units) P in atm: R = 0.08206Latm/molK P in kPa: R = 8.314LkPa/molK P in torr: R = 62.4Ltorr/molK

Ideal Gas Law

Example: Find the moles of oxygen in a balloon of 2.3L volume and 1.3atm pressure if the temperature is 45ºC.

Solution: PV = nRT 1.3(2.3) = n(0.08206)(45+273) n = 1.3(2.3)/(0.08206)(45+273) n = 0.115 mol

Ideal Gas Law

Example 2: Find the molar volume of a gas at STP.

Solution: STP = standard temperature and pressure (273K and 1 atm)

PV = nRT 1V = 1(0.08206)(273) = 22.4L/mol