Chemical Formulas and Composition Stoichiometry

Chemical Formulasand Composition

Stoichiometryالصاوى. محمد كريم د

الفيزيائية الكيمياء مساعد استاذالقصيم - جامعة الكيمياء قسم

Stoichiometry

A word derived from the Greek stoicheion, which means “first principle or element,” and metron, which means “measure.”

Stoichiometry describes the quantitative relationships among elements in compounds (composition stoichiometry) and among substances as they undergo chemical changes (reactionstoichiometry).

The chemical formula for a substance shows its chemical composition. This represents the elements present as well as the ratio in which the atoms of the elements occur.

The chemical formula for a single atom is the same as the symbol for the element. Thus, Na can represent a single sodium atom. A subscript following the symbol of an element indicates the number of atoms in a molecule. For instance, H2O is a molecule containing two hydrogen atoms and one oxygen atom.

OH2

Number of hydrogen

atoms

A molecule is a definite group of atoms that are chemically bonded together

Composition is described by the molecular formulaE.g. H2O

Formula unit: the lowest whole number ratio of ions present in an ionic compound

Composition is described by the formula unit. E.g. CaCl2

Formula unit

NaCl(NaCl)n

The chemical formula for a single atom is the same as the symbol for the element.

Subscript in the formula indicates the number of atoms

In a chemical reaction atoms are re-arranged (Dalton’s Theory).

for example, six H2 molecules react with three O2 molecules and rearrange to form six H2O molecules

Equivalently, two H2 molecules react with one O2 molecule rearrange to form two H2O molecules

So, the reaction takes place if we have the correct relative number of reactant PARTICLES (e.g. atoms, molecules,…).

Atoms are so small to count (DIRECTLY), in the lab it is much easier to WEIGH chemicals.

OK, If we get the correct relative weights of reactants, does this mean that we get the correct relative numbers of the reactant particles (atoms)? NOSo. What!!!!!!!!!!!!!!!

Chemists have adopted the mole concept as a convenient way to deal with the enormous numbers of molecules or ions in the samples they work with.

The mole is the SI unit for amount of substance, it is abbreviated mol.

The mole is defined as the amount of substance that contains as many particles (atoms, molecules, ..) as there are atoms in exactly 12 g of pure carbon-12 atoms.

The number of particles in one mole is known as Avogadro’s Number (NA)

Avogadro's number = 6.022x1023 = 602200000000000000000000.

The Mole


One mole of hydrogen (H) contains the same number of particles as 12 g of carbon-12One mole of helium (He) contains the same number of particles as 12 g of carbon-12One mole of neon (Ne) contains the same number of particles as 12 g of carbon-12

The number of particle in one mole of hydrogen (H) =The number of particle in one mole of helium (He) =

The number of particle in one mole of neon (Ne) =..

The number of particle in one mole of any substance =Avogadro's Number=6.022x1023

This means we now have Avogadro's number of carbon atoms (NA) = 6.022x1023 on the scales pan

6.022x1023 carbon atoms

=12.000 g of carbon

(one mole of carbon)



Then, by definition: To get one mole of carbon all what we need to do is to weigh exactly 12 g of carbon.

What if we need to get one mole of some other substance say Hydrogen, Helium, Sodium, …?

How much do we need to weigh of these substance? (what is the molar mass of these substance?)

It will be wrong to weigh 12 grams of each of these substances, 12 grams will not contain the same number of particles since atoms of different substances have different masses (Dalton’s Theory).

Then, to proceed, we need to know how heavy each element is relative to carbon-12. In other words we need to know the relative atomic masses.

Diagram of a simple mass spectrometer (top), showing the separation of neon isotopes (left).Neon gas enters an evacuated chamber, where neon atoms form positive ions when they collide with electrons. Positively charged neon atoms, Ne, are accelerated from this region by the negative grid and pass between the poles of a magnet. The beam of positively charged atoms is split into three beams by the magnetic field according to the mass-to-charge ratios. The three beams then travel to a detector at the end of the tube. ( The detector is shown here as a photographic plate)

DETERMINATION OF RELATIVE ATOMIC MASSES (WEIGHTS)

Thousands of experiments on the composition of compounds have resulted in the establishment of a scale of relative atomic masses (weights) based on the atomic mass unit (amu).

The atomic mass unit (amu), is defined as exactly 1/12 of the mass of an atom of carbon-12.

Carbon-12 isotope was arbitrarily assigned a mass of exactly 12 atomic mass units

On this scale, the atomic weight (mass) of hydrogen (H) is 1.00794 amu Helium (He) is 4.0026 amu sodium (Na) is 22.9897 amu, magnesium (Mg) is 24.3050 amu.

This tells us that Na atoms have nearly 23 times the mass of H atoms, and Mg atoms are about 24 times heavier than H atoms.

Because of this relative scale, we can be sure that if we take 1:4 samples (by weight) of H and He, they will contain the same number of particles.

The Molar Mass (M)

The molar mass of a substance is the mass of one mole of that substance.

Since one mole contains Avogadro's number of particles, then

The Molar mass = the total mass of Avogadro's number of particles.

Since we have established a relative atomic mass scale relative to a common element (C-12), the following relationship is true.

The mass of one mole (Molar Mass) of atoms of a pure element in grams is numerically equal to the atomic weight of that element in

atomic mass units.

The molar mass units are grams/mole, also written as g/mol or g mol-1.

For instance, if you obtain a pure sample of the metallic element titanium (Ti), whose atomic weight is 47.88 amu, and measure out 47.88 g of it, you will have one mole, or 6.022 x1023 titanium atoms.

One mole of atoms of some common elements. Back row (left to right): bromine, aluminium, mercury, copper. Front row (left to right): sulfur, zinc, iron.

Each of these samples has a different mass but contains the same number of atoms; a very large number 6.022x1023 - Avogadro's number

How many moles of atoms does 136.9 g of iron (Fe) metal contain?answer

We are given the total mass of the Fe sample (m) = 136.9 gWe look up the atomic mass of Fe = 55.85 amuThenThe mass of one mole (molar mass M) of Fe= 55.85 g/mol Then

molmolg

gn

M

mn

M

m

massmolarmoleoneofmass

masstotalnmolesofnumber

451.2/85.55

9.136

)()(

How many atoms are contained in 2.451 mol of iron?answer

We are asked about the total number of atoms (N)We are given the number of moles (n) =2.451 mol

We know that one mole of atoms of an element contains Avogadro’s number (NA) of atoms, or 6.022 x1023 mol-1.

Then

24123 10476.110022.6451.2

)(')()(

molmolN

nN

numbersavogadronmolesofnumberNatomsofnumbertotal

A

A

N

N

Calculate the average mass of one iron atom in grams.answer

The atomic mass of Fe = 55.85 amuThenThe mass of one mole (molar mass M) of Fe= 55.85 g mol-1

This is the mass of Avogadro’s number of atomsThen

gmol

molg

N

Mmassmolaratomoneofmass

NNumbersAvogadroatomoneofmassMmassmolar

A

A

23

123

1

10274.910022.6

85.55)(

)(')(

Thus, the average mass of one Fe atom is only 9.274 x10-23 g, that is, 0.00000000000000000000009274 g

FORMULA WEIGHTS, MOLECULAR WEIGHTS AND MOLES

So, we’ve learnt how to get the molar mass of atomic substances using the relative atomic mass scale based on C-12

How about the molar mass of molecules and ionic substances?

The mass of one mole (molar mass) of a substance in grams is numerically equal to the formula weight of that substance in atomic

mass units.

The formula weight = sum of the atomic masses (weights) of the elements in the formula

Calculate the formula weight (mass) and molar mass of each of the following a) chloroform, CHCl3; b). iron(III) sulfate, Fe2(SO4)3; c) water, H2O

Answer

a) chloroform, CHCl3

b) iron(III) sulfate, Fe2(SO4)3

amu

amuamuamu

ClofmassatomicHofmassatomicCofmassatomicweightFormula

4.119

45.353008.110.121

311

amu

amuamuamu

OofmassatomicSofmassatomicFeofmassatomicweightFormula

9.399

)0.1640.32(38.552

)4(32

Then the molar mass of CHCl3= 119.4 g mol-1

Then the molar mass of Fe2(SO4)3= 399.9 g mol-1

c) water, H2O

One molecule of

H2O

One mole of H2O (6.022x1023 molecules of H2O)

Molecule, Mole and Molar Mass

Molar mass of H2O = mass of one mole of H2O =Total mass of 6.022x1023 molecules of H2O =Formula weight in grams of H2O = 18 g mol-1

How many moles of substance are contained in each of the following samples? (a) 18.3 g of NH3

(b) 5.32 g of ammonium bromide NH4Br(c) 6.6 g of PCl5(d) 215 g of Sn

answer

molmolg

g

molgmolg

g

M

mna

M

mn

M

m



NH

NH

NH07.1

0.17

3.18

0.130.14

3.18)

)()(

111

3

3

3

Br 79.9 N 14.0 P 30.9 Cl 35.5 C 12.0 S 32.0 Sn 118.7

What mass, in grams, should be weighed for an experiment that requires 1.54 mol of (NH4)2HPO4?

answer

gmolgmol

Mnm

molgmolg

molg

molg

molgmolgM

Mnm

M

mn

M

mn

M

m



HPONHHPONHHPONH

HPONH

HPONHHPONHHPONH

HPONH

HPONH

HPONH

4.203 132.154.1

132.1 16.04

30.97

0.1

2)0.140.14(

)()(

1

)()()(

11

1

1

11

)(

)()()(

)(

)(

)(

424424424

424

424424424

424

424

424

How many (a) moles of O2, (b) molecules of O2 , and (c) O atoms are contained in 40.0 g of oxygen gas (dioxygen)

2222

2

2

22123

2

22

2

2

2

1050.11052.72)(

)(2)(

)(2)(

)

1052.710022.625.1)(

)()(

)(')()(

)

25.1/)0.162(

0.40

)(

)()(

)()(

)

ON

ONON

OatomsoxygencontainsOmoleculeoxygenEach

c

molmolON

OnON

nN

numbersavogadronmolesofnumberNparticlesofnumbertotal

b

molmolg

g

OM

OmOn

M

mn

M

m



a

A

A

A

N

N

N

Calculate the number of hydrogen atoms in 39.6 g of ammonium sulfate, (NH4)2SO4.

One molecule of (NH4)2SO4 contains 8 hydrogen atoms

One mole of (NH4)2SO4 contains 8 x NA hydrogen atoms

n mole of (NH4)2SO4 contains 8 x NA x n hydrogen atoms

24

1

123

1

1111

)(

)(

)(

1044.1

0.132

6.3910022.68

0.132

0.1640.322)0.140.14(

8

424

424

424

molg

gmolatomshydrogenofnumber

molg

molgmolgmolgmolgM

M

mn

natomshydrogenofnumber

SONH

SONH

SONH

AN

How many hydrogen atoms are contained in 125 grams of butane, C4H10

Answer

One molecule of C4H10 contains 10 hydrogen atoms

One mole of C4H10 contains 10 x NA hydrogen atoms

n mole of C4H10 contains 10 x NA x n hydrogen atoms

25

1

123

111

103.1

0.58

12510022.610

0.580.1100.124

10

104

104

104

molg

gmolatomshydrogenofnumber

molgmolgmolgM

M

mn

natomshydrogenofnumber

HC

HC

HC

AN

What mass of chromium is contained in 35.8 g of (NH4)2Cr2O7?answer

Reading the given chemical formula tells us thatOne mole of (NH4)2Cr2O7 contains 2 moles of Cr We can succinctly write this in terms of molar masses as:

One molar mass of (NH4)2Cr2O7 → 2 x Molar mass of Cr

35.8 g of (NH4)2Cr2O7 → X

gmolg

gmolgX

molgCrofmassmolar

molg

molgmolg

molgmolgOCrNHofmassmolar

OCrNHofmassmolar

gCrofmassmolarX

8.140.252

8.350.522

0.52

0.252

70.1620.52

2)0.140.14()(

)(

8.352

1

1

1

1

11

11

7224

7224

What is the formula…?

When a chemist has discovered a new compound, the first question to answer is, What is the formula?

To answer, you begin by analyzing the compound to determine amounts of the elements for a given amount of compound.

This is conveniently expressed as percentage composition—that is, as the mass percentages of each element in the compound.

But, what do we really mean by a chemical formula?

The simplest, or empirical, formula for a compound is the smallest whole-number ratio of atoms present.

The molecular formula indicates the actual numbers of atoms present in a molecule of the compound. It may be the same as the simplest formula or else some whole-number multiple of it.

The empirical formula for hydrogen peroxide is HOThis tells us that the ratio between the hydrogen and oxygen atoms in the molecule is 1 : 1

E.g.The molecular formula for hydrogen peroxide is H2O2

This tells us that there are two hydrogen atoms and two oxygen atoms in the molecule

How to get these formula?

PERCENT COMPOSITION AND FORMULAS OF COMPOUNDS

If the formula of a compound is known, its chemical composition can be expressed as the mass percent of each element in the compound (percent composition).

Suppose that A is a part of something—that is, part of a whole. It could be an element in a compound or one substance in a mixture.

We define the mass percentage of A as the parts of A per hundred parts of the total, by mass. That is

E.g. Calculate the percent composition by mass of HNO3.Answer

We are not given the mass of HNO3 but we can calculate its molar mass and then derive its percent composition from it.

2.761000.63

0.163%

2.221000.63

0.14%

6.11000.63

0.1%

0.630.1630.140.1

1

1

1

1

1

1

1111

3

3

3

3

molg

molg

M

OofmassO

molg

molg

M

NofmassN

molg

molg

M

HofmassH

molgmolgmolgmolgM

ONH

HNO

HNO

HNO

HNO

Nitric acid is 1.6% H, 22.2% N, and 76.2% O by mass. N.B. All samples of pure HNO3 should have this composition, according to the Law of Definite Proportions.

A 20.882-gram sample of an ionic compound is found to contain 6.072 grams of Na, 8.474 grams of S, and 6.336 grams of O. What is its empirical ( simplest) formula?

answer

DERIVATION OF FORMULAS FROM ELEMENTAL COMPOSITION

To get a chemical formula for a compound we need the number (or the relative number) of atoms in the compound. So if we have the mass of each element we can proceed as follows:

Chemical formula corresponds to the smallest

whole number ratio of atoms

Convert to whole

numbers

Divide by the smallest number

Get the relative number of

atoms for each element

( mass/gram atomic mass

AW)

mass of each

element

A 0.1014 g sample of purified glucose was burned in a C-H combustion train to produce 0.1486 g of CO2 and 0.0609 g of H2O. An elemental analysis showed that glucose contains only carbon, hydrogen, and oxygen. Determine the masses of C, H, and O in the sample.

Answer

1 mole of CO2 → 1 mole of CMolar mass of CO2= 12.0 g mol-1 + 2x16.0 g mol-1 = 44.0 g mol-1

Molar mass of C = 12.0 g mol-1 44 g mol-1 of CO2 → 12.0 g mol-1 of C0.1486 g of CO2 → ? g C

gmolg

molggC 04055.0

44

0.121486.01

1

1 mole of H2O → 2 mole of HMolar mass of H2O= 2x1.0 g mol-1 + 16.0 g mol-1 = 18.0 g mol-1

Molar mass of H = 1.0 g mol-1 18 g mol-1 of H2O → 2.0 g mol-1 of H0.0609 g of H2O → ? g H

gmolg

molggH 00681.0

0.18

0.20609.01

1

ggggO

HofmassCofmasssampleofmassO

0540.0)00681.004055.0(1014.0

)(

In the previous example a 0.1014 -gram sample of glucose was found to contain 0.04055 grams of C, 0.00681 grams of H, and 0.0540 grams of O. What is the empirical ( simplest) formula of glucose?

answer

Chemical formula corresponds to the smallest

whole number ratio of atoms

Convert to whole

numbers

Divide by the smallest number

Get the relative number of

atoms for each element

( mass/gram atomic mass AM)

mass of each

element

DETERMINATION OF MOLECULAR FORMULAS

As we have seen percent composition data yield only simplest (empirical) formulas. To determine the molecular formula for a molecular compound, both its simplest formula and its molecular weight must be known.

Molecular formula = k x simplest formulak = integer 1,2,3,….

massformulasimplist

massmoleculark

If k=1, then the molecular formula and empirical formula are the same E,g. H2O

In the previous example, the simplest formula for glucose was found to be CH2O. Other experiments have shown that its molecular weight is approximately 180 amu. Determine the molecular formula of glucose.

Answer

61262

2

2

6

60.30

180

0.30

0.160.120.12

OHCOCHformulamolecular

amu

amuk

amu

amuamuamumassformulasimplest

massformulasimplest

massmoleculark

OCHkformulamolecular

OCHformulasimplest

formulasimplestkformulamolecular

Deduce the empirical formulas of the following molecular formulasa) C6H12O6 b) H2O2 b) H2O

answer

We start from the definition: Molecular formula = k x simplest formula

k = integer 1,2,3,….

Then we need to find k …k will be the largest common whole number divisor of the numbers of atoms in the molecular formula

a) C6H12O6

The molecular formula tells us that for each C6H12O6 molecule there are:6 C atoms12 H atoms6 O atoms

Common divisors of 6,12 and 6 are 1,2,3 and 6We choose the largest, k=6

The empirical formula is then C6/6 H12/6 O6/6 = CH2O

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Chemical Formulas and Composition Stoichiometry