3.3 chem myp mole formula

M5 Stoichiometry II: M5 Stoichiometry II: Mole ConceptMole Concept

M5 chem Stoichiometry: Mole Concept Slide 2 of 49

Learning ObjectivesLearning Objectives• Concepts:

– mole, Avogadro’s number, molar mass – relative atomic mass, relative molecular mass, relative formula mass

• Skills: – Given a chemical formula be able to state the elements present and their

proportion – Be able to calculate the molar mass of a substance given its formula and

table of relative atomic masses – Given moles of a certain substance (element or compound) determine the

mass of the substance and vice versa – Given moles of a certain gaseous substance (element or compound)

determine the volume of the substance and vice versa – Given moles of a certain substance (element or compound) determine the

number atoms/molecules of the substance and vice versa


Relative Molecular Mass/Formula MassRelative Molecular Mass/Formula Mass

• Relative atomic mass (AR or RAM): Mass of an atom of the element relative to that of hydrogen.

– It essentially tells us much heavier the atom of the element is than a hydrogen atom.

• Relative molecular mass (MR or RMM): This tells us how much heavier the molecule of the compound is than a hydrogen atom.

– It is the sum of RAM of all the atoms that make up the molecule.

• Relative formula mass (or RFM): sum of the relative atomic masses of all the ions in a single formula unit of an ionic substance.

– This tells us how much heavier a formula unit of a compound is than a hydrogen atom.

• Notice then that relative mass, whether atomic or molecular or formula, has no unit.

Obviously because the measurement is relative to the mass of a hydrogen atom.


Mole and the Avogadro numberMole and the Avogadro number

• Mole (abbreviated to mol) is the unit of measurement of amount (number of elementary particles) of chemical substance.

In chemistry amount is measured in numbers.

That is amount refers to the number of elementary particles—the smallest particles.

While a couple of things consists of 2 things; a dozen consists of 12 things; a gross 144; a ream of paper 500 sheets of paper, a mole of anything consists of a very very VERY large number.

A mole of a chemical substance is 6 × 1023 of the elementary particles—atoms, molecules, ions—of that substance.

This number is referred to as the Avogadro Constant (or number).


Magnitude of a moleMagnitude of a mole• That number is such as large number that it is said that if the whole

population of the world wished to count up to this number between them, and they all worked at counting without any breaks at all, it would take six million years for them to finish!

Further, a line 6 × 1023 mm long would stretch from the earth to the Sun and back two million times!!!

If under room temperature and pressure, however, a box the size of 24 dm3 (box of dimension 29 cm × 29 cm × 29 cm) contains that many number of molecules of gas that make up air, then atoms and molecules must be very very small! – That is, at room temperature and pressure, the volume of 1 mole of any

gas (6 × 1023 atoms of He, or 6 × 1023 molecules of H2 or 6 × 1023 molecules of CO2) is always 24 dm3, which is referred to as the molar volume of gas.


Mole of SubstanceMole of Substance

• While 1 dozen apples would represent 12 apples, 1 mole apples would represent 6 × 1023 apples.

Similarly, 1 mole of helium would represent 6 × 1023 atoms of helium.

1 mole of sodium would represent 6 × 1023 atoms of sodium.

1 mole of hydrogen gas (H2) would represent 6 × 1023 molecules of H2.

1 mole of bromine (Br2) would represent 6 × 1023 molecules of bromine.

1 mole of iodine (I2) would represent 6 × 1023 molecules of iodine.

1 mole of carbon monoxide would represent 6 × 1023 molecules of CO.

1 mole of sulfur dioxide would represent 6 × 1023 molecules of SO2.

1 mole of sodium chloride would represent 6 × 1023 pairs of NaCl.

1 mole of magnesium chloride would represent 6 × 1023 units of MgCl2.


Mole and the Avogadro numberMole and the Avogadro number

• Atoms, molecules and ions are very very small, therefore, instead of counting them individually, they are counted by measuring their mass and volume!

– The quantity of a substance which contains the Avogadro constant equivalent of elementary particles (atoms, molecules or ions) is the relative mass of the substance expressed in grams.

– If we were to measure out 1 g of hydrogen atoms it turns out it would contain 6 × 1023 atoms of hydrogen atoms.

• It follows than that,

– since relative molecular mass of hydrogen gas (H2) is 2, 2 g of hydrogen gas will contain 6 × 1023 molecules of H2.

– In other words, since a hydrogen molecules is twice as heavy as a hydrogen atom, twice as many grams of hydrogen molecules would have to me weighed out in order to have the same number of hydrogen molecules as in 1 g of hydrogen atom.


Molar MassMolar Mass– A sodium atom is 23 times as heavier than an atom of hydrogen (ie. Relative tomic

mass of sodium is 23).

– Therefore 23 g of Sodium will contain 6 × 1023 atoms of sodium.

– 28 g of carbon monoxide will contain 6 × 1023 molecules of CO.

– 18 g of water will contain will contain 6 × 1023 molecules of H2O.

– 62 g of Na2O will contain 6 × 1023 units of Na2O

– In other words 2 × 6 × 1023 Na+ ions and 6 × 1023 O2- ions.

• Mass of a mole of a substance is referred to as the molar mass of that substance.

– 1 g is the molar mass of (elemental) hydrogen atoms

– 2 g is the molar mass of hydrogen gas

– 23 g is the molar mass of sodium

– 28 g molar mass of CO etc.


A Mole of WaterA Mole of Water


Changing between the different quantitiesChanging between the different quantities

• Chemical substances can be characterized and quantified in a few different ways.

– Mass, moles, molar mass, number of atoms, (if gaseous) volume, molar volume

• You need to be able to convert between a few of these quantities for elements and compounds.

Though there are whole variety of methods of converting between those quantities, you will be introduced to conversions

– using Factor-label method,

– using ratio,

– using arithmetic and

– using formula.

• We’ll first discuss factor-label method and then look at the rest.


Changing between the different quantities: Changing between the different quantities: ElementsElements

4 g He 1 mole He or 1 mole He 4 g He

– Expressing the relationship as a fraction allows us to use them in calculations requiring the interconversion between the two quantities.

• Let’s take helium: – The relationship between mole, mass, number of atoms and volume for

this gas is as follows: 1 mole of helium weighs 4 g, contains 6 × 1023 atoms, and has a volume of 24 dm3.

• Let’s take the first relationship and express it as an equality: 1 mole of helium gas = 4 g of helium.

– The relationship between mole and mass for Helium can also be expressed as a fraction:


Element: Relationship between mole & massElement: Relationship between mole & mass

4 g He 1 mole He

• The first fraction allows us to go from moles of He to mass of He.

Here’s an example:

• How many grams of helium would have to weight out so that we have 15 moles of the gas?

4 g He15 mole He 60 g He1 mole He

⇒ ÷

• This method of solving quantitative problems in chemistry is variously referred to as Factor- labeled method or dimensional analysis (dimension meaning units).



• The above expression can be derived using ratios or arithmetic.

Using ratio:

4 g He x g He 1 mole He 15 mole He

4 g Hex 15 mole He1 mole He

=

⇒ = ÷

• Using arithmetic: If 1 mole of He weighs 4 g,

Then 15 moles must weigh 4 × 15 g (which is essentially the same as above).



• How many moles of helium does a cylinder with 155 gram of the gas contain?

1 mole He155 g He 38.8 mole He 4 g He

= ÷

1 mole He 4 g He

• The second fraction of course allows us to go from mass of He to moles of helium.



molar mass of A1 mole A

1 mole Amolar mass of A

mole of element A mass of element A

÷

÷

→¬

• The same relationship for molecular elements

2

2

2

2

molar mass of A1 mole A

2 21 mole A

molar mass of A

mole of element A mass of element A

÷

÷

→¬

• 1. mole ⇔ mass

– In general the relationship between the given quantity (mass or mole) of an element and the unknown quantity (the other one of the two) and the conversion factor can be summarized as shown below:


Practice Questions: Mole Practice Questions: Mole ⇒⇒ Mass of Element Mass of Element

• 1. Convert the following into mass: a) 5.0 mole of helium (He)

4 g He5.0 mole He 20 g He1 mole He

= ÷

23 g He0.50 mole Na 11.5 g He1 mole Na

= ÷

c) 0.050 mole of potassium (K)

b) 0.50 mole of sodium (Na)

39 g K0.050 mole K 1.95 g He1 mole K

= ÷



d) 0.00050 mole of lead (Pb)

207 g Pb0.00050 mole Pb 0.104 g Pb1 mole Pb

= ÷

e) 5.0 mole of hydrogen gas (H2)

f) 0.25 mole of chlorine gas (Cl2)

22 2

2

2 g H5.0 mole H 10 g H1 mole H

= ÷

22 2

2

71 g Cl0.25 mole Cl 17.7 g Cl1 mole Cl

= ÷



g) 0.25 mole of bromine (Br2)

h) 0.50 mole of iodine (I2)

i) 0.25 mole of iron (Fe)

22 2

2

160 g Br0.25 mole Br 40 g Br1 mole Br

= ÷

22 2

2

254 g I0.50 mole I 127 g I1 mole I

= ÷

56 g Fe0.25 mole Fe 14 g Fe1 mole Fe

= ÷


Practice Questions: Mass Practice Questions: Mass ⇒⇒ Mole of Element Mole of Element

• 2. Convert the following into mole: a) 25.0 g of helium (He)

b) 25.0 g of sodium (Na)

c) 25.0 g of potassium (K)

1 mole He25.0 g He 6.25 mole He 4 g He

= ÷

1 mole Na25.0 g Na 1.09 mole Na 23 g Na

= ÷

1 mole K25.0 g K 0.64 mole K 39 g K

= ÷



d) 25.0 g of lead (Pb)

1 mole Pb25.0 g Pb 0.12 mole Pb 207 g Pb

= ÷

e) 25.0 g of hydrogen gas (H2)

f) 25.0 g of chlorine gas (Cl2)

22 2

2

1 mole H25.0 g H 12.5 mole H 2 g H

= ÷

22 2

2

1 mole Cl25.0 g Cl 0.35 mole Cl 71 g Cl

= ÷



g) 25.0 g of bromine (Br2)

22 2

2

1 mole Br25.0 g Br 0.156 mole Br 160 g Br

= ÷

h) 25.0 g of iodine (I2)

i) 25.0 g of iron (Fe)

22 2

2

1 mole I25.0 g I 0.0984 mole I 254 g I

= ÷

1 mole Fe25.0 g Fe 0.446 mole Fe 56 g Fe

= ÷


Element: Relationship between Mole & # of Element: Relationship between Mole & # of atomsatoms

23

231 mole He 6 10 atoms He or

1 mole He6 10 atoms He×

×

– Using one of these two fractions, we can easily convert between mole and number of atoms.

• Going back to our relationships between mole, mass, number of atoms and volume for helium: – 1 mole of helium weighs 4 g, contains 6 × 1023 atoms (Avogadro number

equivalent), and has a volume of 24 dm3.

Let’s take the relationship between mole and number of atoms and expressing it as an equality:

1 mole of helium gas = 6 × 1023 atoms of helium.

– Expressed as a fraction:


Two questionsTwo questions

• Which of the two fractions would be appropriate for conversion from moles of a substance to number of particles of the substance? Give the fraction.

Which of the two fractions would be appropriate for conversion from number of particles of a substance to moles of the substance? Give the fraction.


Element: Relationship between Mole & # of Element: Relationship between Mole & # of atomsatoms

2. mole ⇔ number of atoms

• Atomic elementsAvogadro# of atoms of A

1 mole A

1 mole AAvogadro# of atoms of A

mole of element A atoms of element A

÷ ÷

→¬

• Molecular elements

2

2

2

2

Avogadro# of molecules of A1 mole A

2 21 mole A

Avogadro# of molecules of A

mole of element A molecules of element A

÷ ÷

→¬


Practice Questions: Mole Practice Questions: Mole ⇒⇒ # of # of Atoms/Molecules of ElementAtoms/Molecules of Element

• 1. Convert the following into atoms/molecules: a) 2.0 moles of helium (He)

23236 10 atoms He2.0 mole He 12 10 atoms He

1 mole He × = × ÷ ÷

b) 2.0 moles of neon (Ne)

c) 0.30 moles of sodium (Na)

23236 10 atoms Ne2.0 mole Ne 12 10 atoms Ne

1 mole Ne × = × ÷ ÷

23226 10 atoms Na0.03 mole Na 1.8 10 atoms Ne

1 mole Na × = × ÷ ÷


Practice Questions: Mole Practice Questions: Mole ⇒⇒ # of # of Atoms/Molecules of ElementAtoms/Molecules of Element

d) 0.25 mole of bromine (Br2)

23232

2 22

6 10 molecules Br0.25 mole Br 1.5 10 molecules Br1 mole Br

×= × ÷ ÷

e) 0.50 mole of iodine (I2)

f) 0.020 moles of hydrogen gas (H2)

23232

2 22

6 10 molecules I0.50 mole I 3 10 molecules I1 mole I

×= × ÷ ÷

23222

2 22

6 10 molecules H0.020 mole H 1.2 10 molecules H1 mole H

×= × ÷ ÷


Practice Questions: # of Atoms/Molecules of Practice Questions: # of Atoms/Molecules of Element Element ⇒⇒ Mole Mole

• 2. Convert the following into moles: a) 3 × 1024 atoms of He

2423

1 mole He3 10 atoms He 5 moles He6 10 atoms He

× = ÷× b) 3 × 1024 atoms of Ne

c) 9 × 1022 atoms of Na

2423

1 mole Ne3 10 atoms Ne 5 moles Ne6 10 atoms Ne

× = ÷×

2223

1 mole Na9 10 atoms Na 0.15 mole Na6 10 atoms Na

× = ÷×


Practice Questions: # of Atoms/Molecules of Practice Questions: # of Atoms/Molecules of Element Element ⇒⇒ Mole Mole

d) 9 × 1022 molecules of Bromine (Br2)

22 22 223

2

1 mole Br9 10 molecules Br 0.15 mole Br6 10 molecules Br

× = ÷ ÷×

e) 9 × 1024 molecules of iodine (I2)

f) 9 × 1024 molecules of hydrogen gas (H2)

24 22 223

2

1 mole I9 10 molecules I 15 mole I6 10 molecules I

× = ÷ ÷×

24 22 223

2

1 mole H9 10 molecules H 15 mole H6 10 molecules H

× = ÷ ÷×


Element: Relationship between Mole & Volume Element: Relationship between Mole & Volume of gaseous elementof gaseous element

3

31 mole He 24 dm He or

1 mole He24 dm He

– These two fractions allow the interconversions between mole and volume of gas.

• Going back to our given relationships between mole, mass, number of atoms and volume for helium: – 1 mole of helium weighs 4 g, contains 6 × 1023 atoms, and has a volume of

24 dm3.

• Let’s take the relationship between mole and volume expressing it as an equality: 1 mole of helium gas =24 dm3 of helium.

– Expressed as a fraction:


Element: Relationship between Mole & Volume Element: Relationship between Mole & Volume of gaseous elementof gaseous element

3. mole ⇔ volume

• Atomic elements

3

3

24 dm A1 mole A

1 mole A

24 dm A

mole of element A volume of element A

÷

÷

→¬

• Molecular elements

32

2

23

2

24 dm A1 mole A

2 21 mole A

24 dm A

mole of element A volume of element A

÷

÷ ÷

→¬


Practice Questions: Mole Practice Questions: Mole ⇒⇒ Volume of gaseous Volume of gaseous elementelement

• 1. Convert the following into volume: a) 5.0 mole of helium (He)

3324 dm He5.0 mole He 120 dm He

1 mole He

= ÷ ÷ b) 5.0 mole of Neon (Ne)

c) 5.0 mole of hydrogen gas (H2)

3324 dm Ne5.0 mole Ne 120 dm Ne

1 mole Ne

= ÷ ÷

332

2 22

24 dm H5.0 mole H 120 dm H1 mole H

= ÷ ÷


Practice Questions: Mole Practice Questions: Mole ⇒⇒ Volume of gaseous Volume of gaseous elementelement

d) 2.5 mole of Krypton (Kr)

3324 dm Kr2.5 mole Kr 60 dm Kr

1 mole Kr

= ÷ ÷ e) 2.5 mole of Argon (Ar)

f) 2.5 mole of chlorine gas (Cl2)

3324 dm Ar2.5 mole Ar 60 dm Ar

1 mole Ar

= ÷ ÷

332

2 22

24 dm Cl2.5 mole Cl 60 dm Cl1 mole Cl

= ÷ ÷


Practice Questions: Volume of gaseous element Practice Questions: Volume of gaseous element ⇒⇒ Mole Mole

• 2. Convert the following into mole: a) 24.0 dm3 of helium (He)

33

1 mole He24 dm He 1 mole He24 dm He

= ÷

b) 34.0 dm3 of Neon (Ne)

c) 34.0 dm3 of hydrogen gas (H2)

33

1 mole Ne34 dm Ne 1.41 mole Ne24 dm Ne

= ÷

3 22 23

2

1 mole H34 dm H 1.41 mole H24 dm H

= ÷ ÷


Practice Questions: Volume of gaseous element Practice Questions: Volume of gaseous element ⇒⇒ Mole Mole

d) 2.0 dm3 of Krypton (Kr)

33

1 mole Kr2.0 dm Kr 0.083 mole Kr24 dm Kr

= ÷

e) 2.0 dm3 of Argon (Ar)

f) 2.0 dm3 of chlorine gas (Cl2)

33

1 mole Ar2.0 dm Ar 0.083 mole Ar24 dm Ar

= ÷

3 22 23

2

1 mole Cl2.0 dm Cl 0.083 mole Cl24 dm Cl

= ÷ ÷


Quantitative Relationships: CompoundsQuantitative Relationships: Compounds

• In the case of a gaseous compound such as carbon dioxide, the relationship between mole, mass, number of atoms and volume for this gas is as follows: 1 mole of CO2 weighs 44 g (ie. the molar mass is 44 g), contains 6 × 1023 molecules of CO2, and has a volume of 24 dm3.

Be able to perform the following conversion for compounds:

1. mole ⇔ mass

x

x

x

x

molar mass of AB1 mol AB

x x1 mol AB

molar mass of AB

mol of AB mass of AB

÷

÷

→¬


Quantitative Relationships: CompoundsQuantitative Relationships: Compounds

3x

x

x3

x

24 dm AB1 mol AB

x x1 mol AB

24 dm AB

mol AB Volume of AB

÷

÷ ÷

→¬

2. mole ⇔ volume


Practice Questions: Mole Practice Questions: Mole ⇒⇒ Mass of Compound Mass of Compound

• 1. Convert the following into mass: a) 5.0 mole of carbon monoxide (CO)

28 g CO5.0 mole CO 140 g CO1 mole CO

= ÷

b) 0.50 mole of sulfur dioxide (SO2)

c) 0.050 mole of water (H2O)

22 2

2

64 g SO0.50 mole SO 32 g SO1 mole SO

= ÷

22 2

2

18 g H O0.050 mole H O 0.90 g H O1 mole H O

= ÷



d) 0.050 mole of N2O5

2 52 5 2 5

2 5

108 g N O0.050 mole N O 5.4 g N O

1 mole N O

= ÷

e) 5.0 mole of hydrogen chloride gas (HCl)

f) 0.25 mole of magnesium oxide (MgO)

36.5 g HCl5.0 mole HCl 182.5 g HCl1 mole HCl

= ÷

40 g MgO0.25 mole MgO 10 g MgO1 mole MgO

= ÷



g) 0.50 mole of sodium carbonate (Na2SO4)

2 42 4 2 4

2 4

142 g Na SO0.50 mole Na SO 71 g Na SO1 mole Na SO

= ÷

h) 0.25 mole of ammonium phosphate ( (NH4)3PO4 )

i) 0.25 mole of copper sulfate pentahydrate (CuSO4•5H2O)

4 3 44 3 4 4 3 4

4 3 4

149 g (NH ) PO0.25 mole (NH ) PO 37.3 g (NH ) PO

1 mole (NH ) PO

= ÷

4 24 2 4 2

4 2

250 g CuSO 5H O0.25 mole CuSO 5H O 62.5 g CuSO 5H O1 mole CuSO 5H O

•• = • ÷•


Practice Questions: Mass Practice Questions: Mass ⇒⇒ Mole of Compound Mole of Compound

• 2. Convert the following into mole:

a) 25.0 g of methane (CH4)

b) 25.0 g of ammonia (NH3)

c) 25.0 g of sulfur dioxide (SO2)

44 4

4

1 mole CH25.0 g CH 1.56 mole CH16 g CH

= ÷

33 3

3

1 mole NH25.0 g NH 1.47 mole NH

17 g NH

= ÷

22 2

2

1 mole SO25.0 g SO 0.39 mole SO64 g SO

= ÷


Practice Questions: Mass Practice Questions: Mass ⇒⇒ Mole of Compound Mole of Compound

d) 25.0 g of copper sulfate pentahydrate (CuSO4•5H2O)

e) 25.0 g of magnesium chloride (MgCl2)

f) 25.0 g of iron oxide (Fe2O3)

4 24 2 4 2

4 2

1 mole CuSO 5H O25.0 g CuSO 5H O 0.1 g CuSO 5H O250 g CuSO 5H O

•• = • ÷•

22 2

2

1 mole MgCl25.0 g MgCl 0.263 mole MgCl95 g MgCl

= ÷

2 32 3 2 3

2 3

1 mole Fe O25.0 g Fe O 0.156 mole Fe O

160 g Fe O

= ÷


Practice Questions: Mole Practice Questions: Mole ⇒⇒ Volume of gaseous Volume of gaseous CompoundCompound

• 3. Convert the following into volume: a) 5.0 mole of carbon monoxide (CO)

b) 1.2 mole of sulfur dioxide (SO2)

c) 0.25 mole of hydrogen chloride gas (HCl)

3324 dm CO5.0 mole CO 120 dm CO

1 mole CO

= ÷ ÷

332

2 22

24 dm SO1.2 mole SO 28.8 dm SO1 mole SO

= ÷ ÷

3324 dm HCl0.25 mole HCl 6 dm HCl

1 mole HCl

= ÷ ÷


Practice Questions: Volume of gaseous Practice Questions: Volume of gaseous compound compound ⇒⇒ Mole Mole

• 4. Convert the following into mole:

a) 24.0 dm3 of ammonia (NH3)

b) 4.0 dm3 of carbon monoxide (CO)

c) 0.20 dm3 of sulfur dioxide (SO2)

3 33 33

3

1 mole NH24.0 dm NH 1 mole NH

24 dm NH

= ÷ ÷

33

1 mole CO4.0 dm CO 0.167 mole CO24 dm CO

= ÷

3 22 23

2

1 mole SO0.20 dm SO 0.00833 mole SO24 dm SO

= ÷ ÷


Practice Questions: Volume of gaseous Practice Questions: Volume of gaseous compound compound ⇒⇒ Mole Mole

d) 24.0 dm3 of hydrogen sulfide gas (H2S)

e) 2.0 dm3 of methane (CH4)

f) 0.25 dm3 of carbon dioxide (CO2)

3 22 23

2

1 mole H S24.0 dm H S 1 mole H S24 dm H S

= ÷ ÷

3 44 43

4

1 mole CH2.0 dm CH 0.0833 mole CH24 dm CH

= ÷ ÷

3 22 23

2

1 mole CO0.25 dm CO 0.0104 mole CO24 dm CO

= ÷ ÷


Quantitative Relationship: Formula for Atomic Quantitative Relationship: Formula for Atomic ElementsElements

Relationship between mass, moles and molar mass:

Quantitative properties that the element can be characterized by

For exampleElement can be…

Atomic Na (s), Hg (l), He (g) Mass, moles, molar mass, number of atoms, (and if gaseous) volume, molar volume

Mass

MolesMolar mass

Mass = Moles Molar mass ×

MassMolar mass =

Moles

MassMoles =

Molar mass


Quantitative Relationship: Formula for Atomic Quantitative Relationship: Formula for Atomic ElementsElements

• Relationship between number of atoms, moles and Avogadro’s constant.

# of atoms

MolesAvogadro’s constant

# of atoms = Moles Avogadro's constant ×

# of atomsMoles =

Avogadro's constant

# of atomsAvogadro's Constant =

Moles


Quantitative Relationship: Formula for Gaseous Quantitative Relationship: Formula for Gaseous SubstancesSubstances

• If the element is gaseous, relationship between moles, volume and molar volume can be summarized as follows:

Volume

MolesMolar volume

volume = Moles Molar volume ×

Volume of gasMoles =

Molar volume

Volume (of gas)Molar volume =

Moles


Quantitative Relationship: Formula for Molecular Quantitative Relationship: Formula for Molecular Elements & CompoundsElements & Compounds

For exampleElement can also be…

H2 (g), Br2 (l), I2 (s) (S8 (s), P4(s))

Molecular

Quantitative properties that the element can be characterized by

Mass, moles, molar mass, number of molecules, number of atoms, (and if gaseous) volume, molar volume

• In this case, the relationship between mass, moles and molar mass is exactly the same as that for atomic element.

In the case of a gaseous molecular element, relationship between moles, volume and molar volume is also the same as for atomic elements.

As for compounds, the relationship between mole, mass and molar mass as well as that between mole, volume and molar volume is the same as that for elements.


Quantitative Relationship: Formula for Molecular Quantitative Relationship: Formula for Molecular Elements & CompoundsElements & Compounds

• As for the relationship between number of molecules and moles, you just need to change number of atoms to number of molecules.

# of molecules

MolesAvogadro’s constant

# of molecules = Moles Avogadro's constant ×

# of moleculesMoles =

Avogadro's constant

# of atomsAvogadro's Constant =

Moles

3.3 chem myp mole formula

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