Mass Relationships in Chemical Reactions 3.1 Atomic Mass 3.2 Avogadro’s Number and the Molar Mass of an Element 3.3 Molecular Mass 3.4 The Mass Spectrometer.

Mass Relationships inChemical Reactions

3.1 Atomic Mass3.2 Avogadro’s Number and the

Molar Mass of an Element3.3 Molecular Mass

3.4 The Mass Spectrometer3.5 Percent Composition of

Compounds3.6 Experimental Determination

of Empirical Formulas3.7 Chemical Reactions and

Chemical Equations3.8 Amounts of Reactants and

Products3.9 Limiting Reagents

3.10 Reaction Yield

3.1 Atomic MassThe mass of an atom depends on the number of electrons, protons, and neutrons it contains.By international agreement, atomic mass (sometimes called atomic weight ) is themass of the atom in atomic mass units (amu). One atomic mass unit is defined as amass exactly equal to one-twelfth the mass of one carbon-12 atom. Carbon-12 is the carbon isotope that has six protons and six neutrons.

3.2 Avogadro’s Number (mole) and the Molar Mass of an Element

In the SI system the mole (mol) is the amount of a substance that contains a number of atoms, ( molecules, or other particles) as there are atoms in exactly12 g (or 0.012 kg) of the carbon-12 isotope. The actual number of atoms in 12 g ofcarbon-12 is determined experimentally. This number is called Avogadro’s number

N A = 6.022 x 10 23

1 mole of hydrogen atoms contains 6.022 x 1023 H atoms.Figure 3.1 shows samples containing 1 mole each of several common elements.

3

Molar mass is the mass of 1 mole of in gramseggsshoes

marblesatoms

1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g

1 12C atom = 12.00 amu

1 mole 12C atoms = 12.00 g 12C

1 mole lithium atoms = 6.941 g of Li

For any element atomic mass (amu) = molar mass (grams)

4

M =molar mass in g/mol

NA = Avogadro’s number

Figure 3.2 The relationships between mass (m in grams) of an element and number of moles of an element (n) and between number of moles of an element and number of atoms (N) of an element.

.m is the molar mass (g/mol) of the element and NA is Avogadro’s number

We have seen that 1 mole of carbon-12 atoms has a mass of exactly 12 g and contains 6.022 3 1023 atoms. This mass of carbon-12 is its molar mass ( M), define d as the mass (in grams or kilograms) of 1 mole of units (such as atoms or molecules) of a substance.Note that the molar mass of carbon-12 (in grams) is numerically equal to its mass in amu. Likewise, the atomic mass of sodium (Na) is 22.99 amu and its molar mass is 22.99 g; the atomic mass of phosphorus is 30.97 amu and its molar mass is 30.97 g; and so on. If we know the atomic mass of an element, we also know its molar mass.

Do You Understand Molar Mass?

How many atoms are in 0.32 g of Sodium (Na)?

N (atoms) = 0.32 (g)×6.022×1023/23 (g/mol)

=8.38 x 1021 atoms Na

MmNN

atomsN

atomsN

molgM

gmmoln

A

A

/

)(

)(

)/(

)()(

N (atoms) ?= m =0.32 g

From periodic table

77

x 6.022 x 1023 atoms K1 mol K

=

How many atoms are in 0.32 g of Sodium (Na)?

1 mol K = 23.00 g K

1 mol K = 6.022 x 1023 atoms K

0.32 g K 1 mol K23.00 g K

x

8.38 x 1021 atoms Na

)(

)(

)/(

)()(

atomsN

atomsN

molgM

gmmoln

A

n (mol) = m (g) / M (g/mol)n = 6.46 / 4.003

= 1.61 mol He

From periodic table

n (mole) ?= m =6.46 g

)(

)(

)/(

)()(

atomsN

atomsN

molgM

gmmoln

A

m (g) = n (mol) × M (g/mol)m = 0.356 × 65.39

= 23.28 g Zn

n=0.356 mole m (g)?=

أخرى :Solutionطريقة1 mol Zn = 65.39 g Zn0.356 mole = ? G Znmass of Zn = 0.356x65.39 = 23.28 g 1

)(

)(

)/(

)()(

atomsN

atomsN

molgM

gmmoln

A

N (atoms) = m (g) × NA (atoms) / M (g/mol)N = 16.3 × 6.022×1023 / 32.07

= 3.06×1023 S atoms

N (atom) ?= m =16.3g

Molecular mass (or molecular weight) is the sum ofthe atomic masses (in amu) in a molecule.

SO2

1S 32.07 amu2O +2 x 16.00 amu SO2 64.07 amu

For any molecule

molecular mass (amu) = molar mass (grams)

1 molecule SO2 = 64.07 amu

1 mole SO2 = 64.07 g SO2

Molecular mass = (No. of atoms× atomic mass) +( No. of atoms× atomic mass)

element I element II

a- SO2

=32.07 amu + 2(16.00 amu)

=64.07 amu

b- C8H10N4O2

=8(12.01 amu + )10(1.008 amu) + 4(14.01 amu) + 2(16.00 amu)

=194.20 amu

)(

)(

)/(

)()(

atomsN

atomsN

molgM

gmmoln

A

n (mol) = m (g) / M (g/mol) =6.07) / 12.01 + 4)1.008((

=0.378 mol CH4

No. of moles? m= 6.09 gCH4

)(

)(

)/(

)()(

atomsN

atomsN

molgM

gmmoln

A

N (atoms) = m (g) × NA (atoms) / M (g/mol) =25.6 × 6.022×1023 × 4 / 60.06

=1.03× 1024 H atoms

No. of H atoms? m= 25.6 g (NH2)2CO

M = 60.06 g/mol

Do You Understand Molecular Mass?How many H atoms are in 72.5 g of C3H8O?

N (atoms) = m (g) × NA (atoms) / M (g/mol) =72.5 × 6.022×1023 × 8 / 60.09

=5.81 × 1024 H atoms

m= 72.5 g C3H8ONo. of H atoms?

M g/mol calculate 60.09

)(

)(

)/(

)()(

atomsN

atomsN

molgM

gmmoln

A

No. of H atoms in molecule

22

How many H atoms are in 56.8 g of C3H8O?

1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O

1 mol H = 6.022 x 1023 atoms H

4.56 x 1024 atoms H

1 mol C3H8O molecules = 8 mol H atoms

56.8 g C3H8O1 mol C3H8O

60 g C3H8Ox

8 mol H atoms1 mol C3H8O

x6.022 x 1023 H atoms

1 mol H atomsx =

Molar mass(g/mol)

Atomic mass

molecular mass

Atoms

molecules

Al, As, Ni, H

HCl, H2, O2, NaCl

) الدوري ) الجدول من ذرية كتلة

جزيئية كتلة

It must be noted that:

Percent composition of an element in a compound = is the percent by mass of each element in a compound

n x molar mass of elementmolar mass of compound

x 100%

n is the number of moles of the element in 1 mole of the compound

C2H6O

%C= 2 x (12.01 g)46.07 g

x 100% = 52.14%

%H= 6 x (1.008 g)46.07 g

x 100% = 13.13%

%O= 1 x (16.00 g)46.07 g

x 100% = 34.73%

52.14% + 13.13% + 34.73% = 100.0%

Molar mass= 2(12.01)+ 6(1.008)+ 16.0= 46.07 g

Check the answer!

%H= 3 x (1.008 g)97.99 g x 100% = 3.086%

%P= 30.97 g97.99 g x 100% = 31.61%

%O= 4 x (16.00 g)97.99 g x 100% = 65.31%

Molar mass of H3PO4 = 3(1.008) + 1(30.97) + 4(16.00) =97.99 g

3.3 Molecular MassThe molecular mass (sometimes called molecular weight ) is the sum of the atomic masses (in amu) in the molecule. For example, the molecular mass of H2O is

2(atomic mass of H + )1 atomic mass of Oor 2(1.008 amu) + 16.00 amu = 18.02 amu

What is the formula mass of Al2(SO4)3?

2 Al 2 x 26.983 S 3 x 32.06

12 O +12 x 16.00

342.14 amu

28

Ex. 2In 1 mole of hydrogen peroxide (H 2 O 2 ) there are 2 moles of H atoms and 2 moles of O atoms. The molar masses of H 2 O 2 , H, and O are 34.02 g, 1.008 g, and 16.00 g, respectively. Therefore, the percent composition of H 2 O 2 is calculated as follows:

The sum of the percentages is 5.926% + 94.06% = 99.99%.

% H = 2 x 1.008g

34.02 g H2O2

Hx 100% = 5.926 %

% O = 2 x 16.00g O34.o2 g H2O2

x 100% = 94.06 %

29

Formula mass is the sum of the atomic masses (in amu ) in a formula unit of an ionic compound.

1Na 22.99 amu1Cl +35.45 amuNaCl 58.44 amu

For any ionic compound formula mass (amu) = molar mass (grams)

1 formula unit NaCl = 58.44 amu1 mole NaCl = 58.44 g NaCl

NaCl

30

What is the formula mass of Ca3(PO4)2?

1 formula unit of Ca3(PO4)2

3 Ca 3 x 40.08

2 P 2 x 30.978 O +8 x 16.00

310.18 amu

31

3 ways of representing the reaction of H2 with O2 to form H2O

3.7 Chemical Reactions and Chemical Equations

A process in which one or more substances is changed into one or more new substances is a chemical reaction

A chemical equation uses chemical symbols to show what happens during a chemical reaction

reactants products

الكيميائية المعادلة قراءة

How to “Read” Chemical Equations

الجزئيات- 1 أو الذرات عددالموالت- 2 عدد

)) الجزئيات أو الموالرية- 3للذرات الكتلة

How to “Read” Chemical Equations

2 Mg + O2 2 MgO2 atoms Mg + 1 molecule O2 makes 2 formula units MgO

2 moles Mg + 1 mole O2 makes 2 moles MgO

48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO

IS NOT2 grams Mg + 1 gram O2 makes 2 g MgO

2(24.31) 2(16.0) 2[24.31+16.0]

reactant =80.6 Product = 80.6

√√√

X

Balancing Chemical Equations

1. Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation.

Ethane reacts with oxygen to form carbon dioxide and waterC2H6 + O2 CO2 + H2O

2. Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts.

2C2H6 NOT C4H12

الصحيحه نكتب ( ) الصيغه على ) ناتج ولكل االيسر الطرف على متفاعل لكل) االيمن الطرف

الصيغه بجانب التي االرقام بتغير يكون الكيميائيه المعادله وليست وزنتحتها .التي المعادله طرفي على العدد نفس للعنصر يكون بحيث

Balancing Chemical Equations

3. Start by balancing those elements that appear in only one reactant and one product. C2H6 + O2 CO2 + H2O start with C or H but not O

2 carbonon left

1 carbonon right

multiply CO2 by 2

C2H6 + O2 2CO2 + H2O

6 hydrogenon left

2 hydrogenon right

multiply H2O by 3

C2H6 + O2 2CO2 + 3H2O

ظهورا االقل العناصر أوال توزن

Balancing Chemical Equations

4. Balance those elements that appear in two or more reactants or products.

2 oxygenon left

4 oxygen(2x2)

C2H6 + O2 2CO2 + 3H2O

+ 3 oxygen(3x1)

multiply O2 by 72

= 7 oxygenon right

C2H6 + O22CO2 + 3H2O

72

remove fractionmultiply both sides by 2

2C2H6 + 7O2 4CO2 + 6H2O

2

االكثر العناصر توزن ثمظهورا

37

Balancing Chemical Equations

5.Check to make sure that you have the same number of each type of atom on both sides of the equation .

2C2H6 + 7O2 4CO2 + 6H2O

Reactants Products

4 C12 H14 O

4 C12 H14 O

4 C (2 x 2) 4 C

12 H (2 x 6) 12 H (6 x 2)

14 O (7 x 2) 14 O (4 x 2 + 6)

عنصر لكل الذرات من العدد نفس لديك ان من التأكد هي االخيره الخطوهالمعادله طرفي على

1.Write balanced chemical equation

2.Convert quantities of known substances into moles

3.Use coefficients in balanced equation to calculate the number of moles of the sought quantity

4.Convert moles of sought quantity into desired units

Amounts of Reactants and Products

)()(

Reactant ProductUse gram ratio

of A and B

From balancedequation

Methanol burns in air according to the equation2CH3OH + 3O2 2CO2 + 4H2O

If 209 g of methanol are used up in the combustion, what mass of water is produced?

64 g of CH3OH 72 g of H2O

209 g x g

x = 72×209/64

= 235 g H2O

2×Molar mass = 2[(1 × 12.01)+(4 × 1.008)+ (1 × 16.00)]

4×Molar mass = 4[(2 × 1.008)+ (1 ×16.00)]

41

Methanol burns in air according to the equation

2CH3OH + 3O2 2CO2 + 4H2O

If 123 g of methanol are used up in the combustion, what mass of water is produced?

grams CH3OH moles CH3OH moles H2O grams H2O

molar massCH3OH

coefficientschemical equation

molar massH2O

123 g CH3OH1 mol CH3OH

32.0 g CH3OHx

4 mol H2O2 mol CH3OH

x18.0 g H2O

1 mol H2Ox =

138.38 g H2O

Limiting ReagentExcess Reagents

LIMITING REAGENTS

Limiting Reagents is the reactant used up first In a reaction Excess reagents are the Reactants present in quantitiesGreater than necessary to react With the present quantity of The limiting reagent.

3.9

2NO + 2O2 2NO2

NO is the limiting reagent

O2 is the excess reagent

44

In one process, 124 g of Al are reacted with 601 g of Fe2O3

2Al + Fe2O3 Al2O3 + 2Fe

Calculate the mass of Al2O3 formed.

g Al mol Al mol Fe2O3 needed g Fe2O3 needed

OR

g Fe2O3 mol Fe2O3 mol Al needed g Al needed

124 g Al1 mol Al

27.0 g Alx

1 mol Fe2O3

2 mol Alx

160 .g Fe2O3

1 mol Fe2O3

x = 367 g Fe2O3

Start with 124 g Al need 367 g Fe2O3

Have more Fe2O3 (601 g) so Al is limiting reagent

الوزن- 1. بقسمة وذلك المتفاعلة للمواد الموالت عدد حسابالموالرية الكتلة على بالجرام

موالتها- 2 عدد على متفاعلة مادة كل موالت عدد تقسم . المعادلة في الموجودة

هي- 3 أقل ناتجها تكون اللي المادةالمحدد . الكاشف

الناتجة- 4 المادة كمية لحساب المحدد الكاشف يستخدمالموزونة للمعادلة تبعا المطلوبة

Identify the Limiting Reactant المحدد الكاشف تحديد

Calculate the amount of product obtained from the Limiting Reactant . الكاشف على بناء الناتجة المادة كمية حساب

المحدد

Do You Understand Limiting Reagents?In one process, 124 g of Al are reacted with 601 g of Fe2O3

2Al + Fe2O3 Al2O3 + 2FeCalculate the mass of Al2O3 formed.

n (mol) = m (g) / M (g/mol)nAl = 124 / 26.98

=4.596 mol

nFe2O3 = 601 / 1603.756 mol

مركب كل نسبةAl = 4.596/2 = 2.298

Fe2O3 = 3.756/1 = 3.756المحدد الكاشف هو عدديا limiting reagentاألصغر

فإن النتائج على المحدد Alوبناء الكاشف هو

كمية المحدد Fe2O3لحساب الكاشف نستخدم53.96 g of Al 101.96 g of Al2O3

124 g x gx = 124×101.96/53.96

= 234.3 g Al2O3

48

Use limiting reagent (Al) to calculate amount of product thatcan be formed.

g Al mol Al mol Al2O3 g Al2O3

124 g Al1 mol Al

27.0 g Alx

1 mol Al2O3

2 mol Alx

102 .g Al2O3

1 mol Al2O3

x = 234 g Al2O3

2Al + Fe2O3 Al2O3 + 2Fe

At this point, all the Al is consumed and Fe2O3

remains in excess.

Quantities of product calculated represent the maximum amount obtainable (100 % yield)

Most chemical reactions do not give 100 % yield of product because of:

1-Side reactions (unwanted reactions) 2-Reversible reactions ( reactants products) 3-Losses in handling and transferring

Reaction Yield

Theoretical Yield is the amount of product that would result if all the limiting reagent reacted.

Actual Yield is the amount of product actually obtained from a reaction.

% Yield = Actual YieldTheoretical Yield

x 100

3.10

Reaction Yield

The percent Yield is the proportion of the actually yield to the theoretical yield which can be obtained from the following relation: