PUJI LASTARIA1C312075

KHUSNUL HATIMAHA1C312021

NOVITASARIA1C312030

ALFIA RIANIA1C312080

WENNY REZKY AMELIAA1C312032

MIFTAH FARIDA1C312013

After learning this chapter, you are expcted to

be able to:

a. Prove based on experiment that substance mass before and after

reaction is constant (law of conservation of mass/Lavoisier’s law).

d. Use experimental data to prove volume ratio law (Gay Lussac’s

Law).

b. Prove in accordance with experiment and interpret data

concerning two elements mass that compounds (Proust’s law).

e. Estimate gases volume of reactants or products based on Gay

Lussac’s law.

c. Prove the applicable of ratio multiplication law (Dalton’s law) on

several compound.

f. Find gases volume relation with number of molecules

measured at the same temperature and pressure (Avogadro’s

Law).

g. Explain mole definition as a unit amount of substance.

i. Determine emprical formula, molecular formula and water

crystal and percent composition of compounds.

h. Converse total mole with number of particles, mass and

substances volume.

j. Determine limiting reagent in a reaction.

Fundamental Laws of Chemistry

Chemical equation

Avogadro’s Hypothesis

Mole Concept

Reaction Stoichiometry

Compound of Stoichiometry

Fundamental Laws of Chemistry

Law of

Conservati

on of Mass

Proust’s

Law

Dalton’s

Law

Gay

Lussac’s

Law

The Law of Conservation of Mass

“In every chemical transformation, an equal quantity of matter exists before and after the reaction.”

Fundamental Laws of Chemistry

At 2,4 grams magnesium burning in air, magnesium oxide produce 4 grams. How many grams of oxygen are used up in the reaction?(Ar Mg = 24 Ar O = 16 )

Example:

Menu

The Law of Definite Proportions

“ In a given chemical compound the proportion by mass of the elements that compose it are fixed, independen of the origin of the compound or its mode of

preparation.”

Example :

Analysis of the salt from various

regions.

Check to see it meets law of proust?

Solution :

As shown in the calculation above, the mass

ratio of Na to Cl apparently fixed. Namely 1 :

1,54. so fulfill the law of compound proust

Salt of Mass of

sodium

(Na)

Mass of

chlorine

(Cl)

Mass

Na : Cl

Indramayu

Madura

Impor

0,786 g

0,59 g

0,983 g

1,214 g

0,91 g

1,517 g

0,786 g : 1,214 g = 1 : 1,54

0,59 g : 0,91 g = 1 : 1,54

0,983 g : 1,517 g = 1 : 1,54

Menu

The Law of Multiple Proportions

“When two elements form a series of

compounds, the masses of one that combine with a fixed mass of the other are in the ratio of smallintegers yo each other.”

Example :

Sulfur (S) and oxygen (O) formed two types of compounds. Levels of sulfur in compounds I and II in a row is 50% and 40%. Whether the law applies to the compound dalton.

Solution :

Compound I consists of 50% sulfur, the mass oxygen is 50%.Compound II consists of 40% sulfur, the mass oxygen is 60%.Mass S : O of compound I = 50 : 50 = 1 : 1Mass S : O of compound II = 40 : 60 = 2 : 3 or 1 : 1,5If mass S in the compound I = compound II, as equally as 1 gram, then mass O of compound I : II = 1 : 1,5 = 2 : 3.Comparison is a simple integer and two compounds hat fulfill the law of dalton.

Menu

The Law of Combining Volumes

“When two gases are allowed to react, such that the goes are at the

same temperature and pressure, the volumes of each gas consumed will be

in the ratio of small integers. Morever, the ratio of the volumes of

each product gasto the volume of either reacting gas will be ratio of

simple integers.”

Example :

Two researchers independently studied thedecomposition reaction of dinitrogen pentaoksidabe nitrogen dioxsyde and oxygen. The researcherfound that the decompotion of 50mL (100°C, 1atm) nitrogen pentaoksida produce 100mL(100°C, 1 atm) nitrogen dioxyde and 25mL(100°C, 1 atm) oxygen. Whether the result ofthese trials fulfill the law of combining volumes?

Solution : 50mL N2O5 100mL NO2 + 25mL O2

2 4 1Then

2N2O5 4NO2 + O2

This statement is meet

Menu

Chemical Equations

Lavoisier, 1788

Because of the principle of the law conservation of mass,

an equation must be balanced.

It must have the same number of atomsof the same kind

on both sides.

Balancing Equations

Depict the kind of reactants and products and their relative amounts in a reaction.

___ Al(s) + ___ Br2(liq) ---> ___ Al2Br6(s)

2 Al(s) + 3 Br2(g) ---> Al2Br6(s)

The numbers in the front are called

stoichiometric coefficients

The letters (s), (g), and (l) are the physical states of compounds.

Equation is said equivalent if the type and number of atoms reacted substances (reactants) together with the type and number of atoms of the reaction (products)

Reactants are written on the left followed by an arrow and then the product

2H2 (g) + O2 (g) → 2H2O (l)

AVOGADRO’S HYPOTHESIS

"At the same temperature

and pressure, all gases

with the same volume

will contain the same

number of molecules as well."

Exercise Problem

A hydrocarbon compounds (C XHY) by burning gaseous

reaction:

C x H y (g) + O 2 (g) → CO 2 (g) + H 2 O (g) (not equal)

Of the experiment is known that 2 liters of gas to burn C x

H y (T, P) required 5 liters of oxygen gas (T, P) and

produced 4 liters of carbon dioxide gas (T, P). Determine

the molecular formula of the hydrocarbon?

Answer

Due to volume ratio is a reaction coefficient, then theequation becomes:2 C x H y (g) + 5 O 2 (g) → 4 CO 2 (g) + .... H 2 O (g) (notequal)For equality oxygen atom, then the coefficient of H 2 Ois 2 (10-8), thus the equations become:2 C x H y (g) + 5 O 2 (g) → 4 CO 2 (g) + 2 H 2 O (g)For C and H atoms equality as follows.Thus, the hydrocarbon molecular formula is C 2 H 2.

Mole Concept

Mole definition

Relation with the

number of moles of particles

Molar mass and

Molar volume

Calculating Mass and volume of Product

Kemolaran Solution

Mole Definition

In Chemistry, number of atomic particles or elements that involved in chemical reaction explained by mole.

Mole is a unit to express the number of particles.

Menu

Relationship Between Moles and Number of Particles

The number of particles in the 12 gram 12Catom specified based experiment result is

6,02 x 1023

Relationship Between the number of moles (n) and the number of particles (x) can be formulated as follows :

x = n × 6,02 × 1023

E AX M

PL

E

How many particles are found in 3 mol of Iron metal?

Answer :1 mol of Iron metal (Fe) = 6,02 × 1023

3 mol of Iron metal = 3 × 6,02 × 1023

= 1,806 × 1024

Menu

M

O

AL R

Mass &Volume

Molar mass in one mol mass of substance expressed in grams.

Molar Mass

m = mass (g)m = n × Mm n = number of mole (mol)

Mm = molar mass(g/mol)

E AX M

PL

E

Known Ar Fe = 56. Calculate the molar mass of Fe ?

Answer :

Mass of 1 mol of Fe = 1 × 56 = 56 g

Thus, Molar Mass of Fe = 56 g/mol.

Molar volume expresses volume for each1 mol of gas. The gas molar volume is 22,4L at STP and 24,5 L at room temperature.

Molar Volume

If expressed at STP molar volume is symbolized by Vm . Relationship between

volume, number of moles, and the gas molar volume is as follows

V = Volume of the gas V = n × Vm n = Number of moles

Vm = The gas molar volume

When volume of the gas is measuredat certain temperature andpressure, then equation used is isideal gas equation.Mathematically, the ideal gasequation is as follows :

P = pressure of the gas (atm)PV = nRT V = volume of the gas (L)

n = number of molesV = nRT R = gas constant (0,0821 atm L mol-1 K-1

P T = absolute temperature (K)

E AX M

PL

E

At certain temperature and pressure,5 mole of SO2 gas have 100 L volume. At the same temperature

and pressure . What is the volume of 3 mole of No2 gas?

Comparison of mole of SO2 gas and NO2 gas = 5 : 3

Volume of 5 mole of SO2 gas is 100 L then thevolume of 3 mol of NO2 is

Thus, volume of NO2 gas is 60 L

A SN W

ER

Menu

Calculating Mass and Volume of Product

There following steps can determine mass or volume of product.

a. Calculate the number of moles of elementary unit (atoms, molecules or ions) from elements, compounds, or ions from known mass or volume of subtance.

b. Calculate the number of moles of the unknown subtances using subtance coefficient in balanced equation.

c. Determine mass of volume of the unknown subtances based on the number of moles calculated.

Calculate mass of carbon dioxide gas (CO2) produced when 108 g ethane (C2H6) burned in O2

gas.

Answer :Chemical equation :

2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(g)

known 108 g of C2H6, Mr C2H6 = 30 and Mr CO2

= 44

E AX M

PL

E

Three following steps can determine mass ofcarbon dioxide :Step I : Calculate moles of 108 g of ethane

1 mol of C2H6 = 30 g then 108 g C2H6 = 108 g = 3,6 mol30 g/mol

Step II : Calculate the number of moles of carbon dioxide produced.

Based on equation : 2 mol C2H6 produced 4 mol CO2. 3,6 mol C2H6 produce 4 x 3,6 mol = 7,2 mol CO2

2

Step III : Calculate the number of carbon dioxide produced in gram,1 mol CO2 = 44 g7,2 mol CO2 = 7,2 x 44 g = 316,8 g

Thus, the mass of carbon dioxide gas produced in thecombustion of 108 g ethane is 316,8 g. Menu

One way to express concentration of solution used in chemistry is kemolaran (M).

kemolaran expressed in number of moles of solute per liter of solution, or the amount of

solute in mmol per mL of solution.

M = Kemolaran solutionM = n = number of moles of solute

V = volume of solution

nV

Calculate the number of moles and the mass of urea (Mr = 60) were present in 200 mL of 0.4 M urea.

Answer :determine the mass of solute,n = V x M• number of moles of urea (n) = V x M

= 0,2 L x 0,4 Mol L-1

= 0,08 Mol

mass urea (m) = n x Mm

= 0,08 mol x 60 g mol-1` = 4,8 gram

E AX M

PL

E

Menu

Reaction Stoichiometry

Stoichiometry

coefficients of

the reaction as

the basis of the

reaction

Limiting

Reagent

Chemical

Formula of

hydrates

As the reaction stoichiometry coefficients Elementary Reactions

Reaction coefficient is the ratio of particles of substances involved in the reaction. Therefore 1 mole of any substance contains the same number of particles, the reaction coefficient is also a comparison of the number of moles of substances involved in the reaction.

Number of moles = (coefficient of substances were asked) / (coefficient of known substances) x number of substances known

Reaction stoichiometry

Aluminum dissolves in sulfuric acid produce aluminum sulfate and hydrogen gas.

2AL (S) + 3H2SO4 (aq) AL2 (SO4) 3 (aq) + 3H2 (g)

How many moles of hydrogen gas can be produced when 0.5 moles of aluminum used

EXAMPLE

ANSWER

In the know: Coefficient substances were = 3Coefficient of substances known to = 2The number of moles of Al = 0.5 mol

In question: the number of moles of H2 = ...?

Number of moles = (coefficient of substances were) / (coefficient of unknown substances) x number of substances knownThe number of moles of H2 = (coefficient H2) / (coefficient Al) x number of moles of Al

= 3/2 x 0.5 mol= 0.75 mol

ANSWER

Menu

The reactans used up first a rection andrestricts the running reaction so that nofurther one called limiting reagent.

Methane burns (reacts with oxygen) by the equation:

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)

In an experiment, a total of 8 grams of methane burned with 40 grams of oxygen.(Ar C = 12; H = 1; O = 16)Determine the limiting reagent.

EXAMPLE

ANSWER

In the know: Mass CH4 = 8 gramsMass O2 = 40 grams

In question: limiting reagent = ...?

The number of moles of methane (CH4) = (8 g) / (16 g / mol) = 0.5 molThe number of moles of oxygen (O2) = (40 g) / (32 g / mol) = 1.25 mol

When compared with the coefficient of the reaction, methane is multiplied by the number 0.5 / 1 or 0.5, while the number of oxygen with 1.25 / 2 or 0.625.Because pengalinya smaller, the limiting reagent is methane.

ANSWER

Menu

Chemical formula Hydrant

Hydrants are solids that bind several molecules of water as part of its crystal structure.

British salt, MgSO4, 7H2O: Magnesium sulphate pentahydrate

If a hydrate in the heat, some or all of the water crystals can be separated (yawn).

CuSO4. 5H2O (S) CuSO4. (S) + 5H2O (g)If a hydrate is dissolved in water, it will loose its crystalline water.

CuSO4. 5H2O (S) CuSO4. (Aq) + 5H2O (l)

A total of 10 grams of hydrated iron (II)sulphate crystals are heated so that allwater is evaporated. The remaining solidmass was 5.47 grams. How is the formulathat hydrates?

(Ar H = 1; O = 16 S = 32; Fe = 56)

EXAMPLE

ANSWER

Differences existing mass is the mass of the crystal water.For example the amount of water crystals is x, so it is FeSO4.xH2O formula hydrates.

FeSO4.xH2O mass = 10 gramsFeSO4 mass = 5.47 grams

Mass of water = 10 to 5.47 = 4.53 grams.

ANSWER

FeSO4.xH2O (s) FeSO4 (aq) + xH2O (g)

The number of moles of FeSO4 = (5.47 g) / (152 g / mol) = 0.036 molThe number of moles of H2O = (4.53 g) / (18 g / mol) = 0.252 molFeSO4 mol: mol of water = 0.036: 0.252 = 1: 7Meaning, one molecule binds FeSO4 7 water molecules.The formula hydrates is FeSO4.7H2O.

Menu

Compound of Stoichiometry

Composition

Percentage

Empirical

and

Molecular

Formulas

Composition percent is determined bydividing mass of each element in one molof compound with molar mass ofcompound and multiply by 100 percent.

Composition Percentage

Compound stoichiometry

Example

Determine percentage of each element in ether anaesthesia, C4H10O. (Ar C = 12, H = 1, O = 16).

Menu

Empirical and Molecular Formulas

A molecular formula shows the exactnumber of atoms of each element in thesmallest unit of a substance.

Empirical formula is the simplest ratioof atoms that compose a molecule, hence,empirical formula is also called comparisonformula.

A compound consists of 84 % carbon and 16 % hydrogen. If Ar C = 12, Ar H = 1, and Mr =

100, determine the empirical and molecular formulas of the compound.

Answer:

Menu