Mole Concept and Stoichiometry Main Note

/___ Carbon-12: is taken as the standard atom. all other atoms are compared with this standard atom.

Relative Atomic Mass

Definition

Relative Atomic Mass $A_{r}$ of an element is the average mass of one atom of that element relative to $\frac{1}{12}$ the mass of an atom of carbon-12 #ChemistryDefinitions

Relative Atomic Mass = $\frac{a v er a g e ma ss o f o n e a t o m o f t h e e l e m e n t}{\frac{1}{12} ma ss o f an a t o m o f c a r b o n - 12}$

Relative Atomic Mass is represented by $A_{r}$ and has NO unit (found in periodic table)

Calculating Relative Atomic Mass of elements with Isotopes
some elements found in nature as a mixture of isotopes ⇒ with different atomic masses thus relative atomic mass if determined by the average atomic mass of the isotopes.

Eg: relative atomic mass of chlorine ⇒ given in decimal form. → this is because there are only 2 stable isotopes of chlorine - chlorine 35 and chorine 37 sample of chlorine usually has 75% chlorine-35 and 25% chlorine-37

Relative atomic mass of chlorine → (75/100 * 35) + (25/100 * 37) = 35.5

Relative Molecular Mass

Definition

Relative molecular mass, $M_{r}$ of a molecular substance is the average mass of one molecule of that substance relative to $\frac{1}{12}$ the mass of an atom of carbon-12 #ChemistryDefinitions

Relative molecular mass of covalent substance = $\frac{a v er a g e ma ss o f a m o l ec u l e o f t h e co v a l e n t s u b s t an ce}{\frac{1}{12} ma ss o f an a t o m o f c a r b o n - 12}$

Relative molecular mass is represented by $M_{r}$ and has NO unit

To find relative molecular mass $M_{r}$ of a substance, we add together the relative atomic mass $A_{r}$ of all the atoms in its chemical formula. Example:

$N H_{3}$ ⇒ $M_{r}$ = (1 $\times$ 14) + (3 $\times$ 1) = 17

Relative Formula Mass Ionic compounds do not exist as molecules (they are made up of cations and anions) we use the term relative formula mass to refer to the relative mass of an ionic compound.

Definition

Relative formula mass $M_{r}$ of an ionic compound is the average mass of 1 unit of that compound relative to $\frac{1}{12}$ the mass of an atom of carbon-12 #ChemistryDefinitions

⇒ relative formula mass of ionic compound is also represented by $M_{r}$ and has NO unit ⇒ calculated in the same way as relative molecular mass.

*Applications of $A_{r}$ and $M_{r}$

relationship between $A_{r}$ (or $M_{r}$ ) and the rate of diffusion of a gas.

As the $A_{r}$ (Or $M_{r}$ ) of a gas increases, its rate of diffusion decreases (some gasses exist as atoms, eg noble gasses)

2 types of experiments that can be conducted to show how rate of diffusion of a gas is affected by its $M_{r}$

Experiment 1: Explain why the white fumes of ammonium chloride formed closer to the end with concentrated hydrochloric acid

$M_{r}$ of $N H_{3}$ and $M_{r}$ of HCl is 36.5
$N H_{3}$ which has a lower $M_{r}$ diffuses faster than HCl which has a higher $M_{r}$
hence in a fixed duration of time, $N H_{3}$ will travel a longer distance than HCl and thus the reaction between $N H_{3}$ and HCl which produces ammonium chloride takes place closer to the end with hydrochloric acid.

Experiment 2:

EXPT NO.	Gas X	Gas Y	Level of water at point Z after sometime	Explanation
1	carbon dioxide	nitrogen	rise (move up towards the porous pot)	- $M_{r}$ of $N_{2}$ is 28 which is lower than the $M_{r}$ of $C O_{2}$ - $N_{2}$ will diffuse out of the porous pot faster than $C O_{2}$ diffusing into of the porous pot - pressure inside the porous pot is lower than pressure outside ht pot thus causing the water level in the tube at point Z to rise.
2	Nitrogen Monoxide	ethane	remain unchanged	- Gas inside and outside the porous pot are nitrogen monoxide and ethan respectively which has an $M_{r}$ of 30 each. - thus rate of diffusion of both gasses in and out of the porous pot is the same. - since volume of methane in the porous pot remains unchanged, pressure inside the porous pot is the same as pressure outside the porous pot thus causing water level in tube at point z to remain unchanged.

Methods for collecting gasses, with relation to mole

this is related to Method for collecting gasses

Gasses with $A_{r}$ (or $M_{r}$ ) < 30 are less dense than air ⇒ can be collected by upward delivery
Gasses with $A_{r}$ (or $M_{r}$ ) > 30 are more dense than air and can be collected by downward delivery

Percentage Composition of Compounds

refers to percentage by mass of each element in a compound.
allows us to determine chemical formula of the compound

Formula $P erce n t a g e co m p os i t i o n b y ma ss o f an e l e m e n t in a co m p o u n d =$ $\frac{A _{r} o f e l e m e n t / t im es N u mb er o f a t o m s in t h e e l e m e n t in t h e c h e mi c a l f or m u l a o f co m p o u n d}{M _{r} o f co m p o u n d}$ Important Note

Do NOT present working and final answers in the form of fractions but decimals
when dealing with non exact numerical final answers (with decimals) ⇒ correct value to 3 significant figures. Work with 5 S.F beforehand.

Mole and Molar Mass

What is Mole ⇒ Amount of substance containing $6.02 \times 1 0^{23}$ Particles (One Mole represents $6.02 \times 1 0^{23}$ Particles)

Unit of Mole → mol

Recall ⇒ 6 main types of particles.

atoms
molecules
ions
subatomic particles
- electrons
- protons
- neutrons ⇒ note that one mole of any substance will only have the specified number of particles ⇒ regardless of whether there are atoms, molecules, ions, and even sub atomic particles such as electrons, protons, neutrons. Example 1 mol of Helium (He) ⇒ has 1 mol of He Atoms ⇒ and that is $6.02 \times 1 0^{23}$

Formula for Mole $N u mb ero f M o l es = \frac{N u mb er o f P a r t i c l es}{6.02 \times 1 0 ^{23} ( A v o g a d ros C o n s t an t )}$ Note: this value is also known as the Avogadros constant Important Note

Do NOT present working and final answers in the form of fractions but decimals
when dealing with non exact numerical final answers (with decimals) ⇒ correct value to 3 significant figures. Work with 5 S.F beforehand.

Molar Mass

Definition

One Mole of Any substance has a mass equal to its relative atomic mass ( $A_{r}$ ), relative molecular mass ( $M_{r}$ ) or relative formula mass ( $M_{r}$ ) in grams This is the Molar Mass Molar mass is the mass of 1 mole of any substance* #ChemistryDefinitions

unit: g/mol

Calculation of Molar Mass of an element

Molar mass of an element is the mass of one mole of atoms (or molecules) of the element
since molar mass of element is equal to relative atomic mass ( $A_{r}$ ) or relative molecular mass ( $M_{r}$ ) ⇒ calculation of molar mass of an element is the same as the way you would do it to calculate its $A_{r}$ or $M_{r}$

Calculation of Molar Mass of a Compound

Molar Mass of compound is the mass of one mole of molecules (or formula units) of the compound
Molar Mass of compound is equal to its relative molecular mass ( $M_{r}$ ) or relative formula mass ( $M_{r}$ )
thus ⇒ calculate the molar mass of compound in the same way you would calculate its $M_{r}$

Formula 1: $N u mb er o f m o l es o f a s u b s t an ce = \frac{ma ss o f s u b s t an ce in g}{m o l a r ma ss o f s u b s t an ce in g / m o l}$ ⇒ Mass must be in grams (convert if others) ⇒ Do not present working and final answers in form of fraction ⇒ give non exact final answers to 3S.F

Mole and Molar Volume of Gasses

→ According to Avogadro’s Law ⇒ equal volumes of all gasses ⇒ under the same conditions of temperature and pressure ⇒ contain the same number of particles.

Definition

Volume occupied by 1 mole of any gas at room temperature and pressure is equal to $24 d m^{3}$ #ChemistryDefinitions

1dm = 10cm $1 d m^{3}$ ⇒ 1000 $c m^{3}$

Deduce Relationship between number of moles of gas and volume of gas

Equal number of mol of any gas occupies the same volume

Formula 1 $N u mb er o f m o l es o f g a s (m o l) = \frac{v o l u m e o f g a s ( d m ^{3} )}{m o l a r v o l u m e o f g a s ( 24 d m ^{3} )}$ ⇒ volume must be in $d m^{3}$ (conversion required) 1dm3 = 1000cm3 1l = 1000ml = 1000cm3 = 1dm3 ⇒ do not present working and final answer in fraction ⇒ give non exact numerical final answers correct to 3s.f.

Mole and Concentration of Solution

⇒ concentration of solution is the amount of solute dissolved in unit volume of solvent 2 units used ⇒ g/ $d m^{3}$ and mol/ $d m^{3}$

to prepare aqueous solution of substance ⇒ substance (in any physical state) ⇒ is dissolved in water ⇒ substance is called the solute and water is called the solvent.
resulting mixture is called aqueous solution

Steps to prepare 250 $c m^{3}$ of aqueous solution of copper(II)sulfate using a 250ml of volumetric flask in laboratory
electronic balance to measure mass of solute weigh out 39.9g of copper(II)sulfate crystals transfer the $C u S O_{4}$ crystals into the 250ml volumetric flask. add deionised water and swirl to dissolve the solute in the solvent(water) add deionised water until solution just reaches calibration mark on the neck of the volumetric flask. this indicates that the volume of $C u S O_{4}$ solution is 250 $C m^{3}$

Steps to prepare 250

c m^{3}

of aqueous solution of copper(II)sulfate using a 250ml of volumetric flask in laboratory

electronic balance to measure mass of solute
weigh out 39.9g of copper(II)sulfate crystals
transfer the

C u S O_{4}

crystals into the 250ml volumetric flask. add deionised water and swirl to dissolve the solute in the solvent(water)
add deionised water until solution just reaches calibration mark on the neck of the volumetric flask.
this indicates that the volume of

C u S O_{4}

solution is 250

C m^{3}

Concentration of solution
⇒ mass/number of mol of solute per unit volume (in Dm3) of solution

in chapter 8 ⇒ acids and Bases (pH scale) ⇒ we expressed concentration of an aqueous solution of acid in terms of number of particles of acid which dissolved in a fixed volume of water (solvent.)
particles are extremely small and numerous thus when preparing aqueous solutions with different concentrations in laboratory it is not so meaningful to express concentration of aqueous solution in terms of number of particles in the solute that dissolved in a fixed volume of water.

Thus concentration is expressed in 2 ways

concentration in g/Dm3	concentration in mol/Dm3 (molar concentration)
$\frac{ma ss o f so l u t e ( g )}{v o l u m e o f so l u t i o n ( D m ^{3} )}$	$\frac{n o . o f m o l o f so l u t e ( m o l )}{v o l u m e o f so l u t i o n ( D m ^{3} )}$
Basicity of Acids
Define Basicity of acid (in mol)
⇒ No. of mol of $H^{+}$ ions formed per mol of acid ionised in solution
Acids
Calculations involving concentration of solutions

Formula 1 $C o n ce n t r a t i o n o f so l u t i o n (g / d m^{3}) = \frac{ma ss o f so l u t e ( g )}{v o l u m e o f so l u t i o n ( d m ^{3} )}$ Formula 2 $C o n ce n t r a t i o n o f so l u t i o n (m o l / d m^{3}) = \frac{n u mb er o f m o l es o f so l u t e ( m o l )}{v o l u m e o f so l u t i o n ( d m ^{3} )}$ Formula 3 $C o n ce n t r a t i o n o f so l u t i o n (m o l / d m^{3}) = \frac{co n ce n t r a t i o n o f so l u t i o n ( g / d m ^{3} )}{m o l a r ma ss o f so l u t e ( g / m o l )}$

Empirical and Molecular Formula of a Compound

Definition

Simplest ration of elements in a compound is called its empirical formula (Empirical formula is the simplest formula of a compound) #ChemistryDefinitions

⇒ Molecular formula is a multiple of the empirical formula → molecular formula of a compound reflects the exact number of atoms of each element found in one molecule of the compound

Molecular formula/chemical formula ⇒ are sometimes used interchangeably. → However the term ‘molecular’ → only refers to covalent substances. → hence the term molecular formula refers only to the chemical formula of covalent substance

Note: Some elements have the same empirical formula and molecular formula

What does empirical formula of a compound show

elements in a compound
the simplest ratio of the different types of atoms (elements) in the compounds

The Molecular Formula is the actual formula of a compound

What does Molecular Formula of a compound show

elements in a compound
the Actual Number of the different types of atoms (elements) in the compound

Calculating Empirical Formula using mass of elements compound. Y contains 39.1g of carbon, 8.7g of hydrogen and 52.2g of oxygen. calculate the empirical formula of Y.

Steps:

find mass
find molar mass/ $A_{r}$
no. of moles/mol
$\div$ by smallest ratio
Obtain simplest ratio

Element	C	H	O
mass/g	39.1	8.7	52.2
molar mass/g/mol (OR relative atomic mass/ $A_{r}$ )	12	1	16
No. of Moles/mol	$\frac{39.1}{12} = 3.26$	$\frac{8.7}{1} = 8.7$	$\frac{52.2}{16} = 3.26$
$\div$ by smallest no.	$\frac{3.26}{3.26} = 1$	$\frac{8.7}{3.26} = 2.67$	$\frac{3.26}{3.26} = 1$
Simplest Ratio (Make it a whole number)	$1 \times 3$ 3:	$2.67 \times 3$ 8:	$1 \times 3$ 3
Simplest Ratio: 3:8:3
(recall percentage of each compound is directly proportional to its mass in grams. Thus, mass of each element in 100g of the compound is its % in the compound.)
(Recall molecular formula is the formula that shows the exact number of atoms of each element in a molecule)

The molecular formula is a multiple of the empirical formula if the molecular formula and the empirical formula of a compound is different.

Thus to determine molecular formula we need to know the:

Empirical formula of the compound
Relative molecular mass of the compound $n = \frac{re l a t i v e m o l ec u l a r ma ss}{M _{r} f ro m e m p i r i c a l f or m u l a}$

Molecular formula of a compound = $n \times e m p i r i c a l f or m u l a o f a co m p o u n d$ (from this, we can tell that molecular formula is a multiple of the empirical formula)

Calculating Empirical Formula using Percentage Composition by Mass of Elements

example 1: Caffeine is a compound found in coffee and tea. ⇒ percentage composition of caffeine is 49.5% carbon. 5.1% Hydrogen, 16.5% oxygen and 28.9% Nitrogen.

Determine empirical formula of caffeine

Element	C	H	O	N
Percentage/%	49.5	5.1	16.5	28.9
Molar mass/g/mol (OR Relative atomic mass/ $A_{r}$ )	12	1	16	14
No. of Moles/mol	$\frac{49.5}{12} = 4.13$	$\frac{5.1}{1} = 5.1$	$\frac{16.5}{16} = 1.03$	$\frac{28.9}{14} = 2.06$
$\div$ by smallest no.	$\frac{4.13}{1.03} = 4.01$	$\frac{5.1}{1.03} = 4.95$	$\frac{1.03}{1.03} = 1$	$\frac{2.06}{1.03} = 2$
Simplest Ratio	4.01 4:	4.95 5:	1 1:	2 2

Empirical formula of caffeine is $C_{4} H_{5} O N_{2}$

Given that mass of 1 mole of Caffeine is 194g/mol ⇒ determine the molecular formula of caffeine $n = \frac{re l a t i v e m o l ec u l a r ma ss}{M _{r} f ro m e m p i r i c a l f or m u l a}$

Calculations from Chemical Reactions

what important info does a balanced chemical equation tell us:

mole ratio ⇒ the ratio of the number of moles of the reactants and the products
Physical state of the reactants and products

What is stoichiometry ⇒ relationship between number of moles of the reactants and the number of moles of the products involved in that chemical reaction. ⇒ Stoichiometry is based on the fact that particles in matter cannot be destroyed or created in a chemical reaction. in a chemical reaction, the particles in reactants simply changed into another form in the products. thus, sum of mass of reactants is the same as sum of mass of products.

⇒ consider the reaction between nitrogen gas and hydrogen gas

$N_{2} (g)$ +	$3 H_{2} (g)$ +	$2 N H_{3} (g)$
1 molecule $N_{2}$ 1 mol $N_{2}$	3 molecules $H_{2}$ 3 mol $H_{2}$	2 molecules $N H_{3}$ 2 mol $N H_{3}$
→ this balanced equation can be read as 1 mol of nitrogen gas reacts with 3 mol of hydrogen gas to produce 2 mol of ammonia gas.

No. of moles of $N_{2}$ = 1 mol	No. of moles of $H_{2}$ = 3 mol	No. of moles of $N H_{3}$ = 2 mol
Mass of $N_{2}$ = 1 mol $\times$ (2 * 14) = 28g	Mass of $H_{2}$ = 3 mol $\times$ (2* 1) = 6g	Mass of $N H_{3}$ = 2 mol $\times$ [14 + (3 * 1)] = 34g
Total mass of reactants = 28 + 6 = 34g	Total mass of reactants = 28 + 6 = 34g	Total mass of product = 34g

Calculations from Chemical Reactions**

Calculating masses of reactants and products ⇒ by using balanced chemical equation, mass of any reactants used up or mass of any product formed in a reaction. (reactants and products in any physical state as well)

Worked Example

copper metal can be extracted from copper(II)oxide by reacting 8.0g of copper(II)oxide with carbon. Copper and carbon dioxide gas are produced in this reaction. calculate:

mass of carbon, C that will react with 8.0g of copper(II)oxide
mass of copper produced, Cu
Volume of carbon dioxide gas produced, $C O_{2}$ in this reaction

Solution

(i)
Step 1: Write a Balanced Equation for this reaction (include state symbol) 2CuO(s) + C(s) → 2Cu(s) + $C O_{2}$ (g)
Step 2: Identify the reactant/product whose quantity is provided in the question. Convert the quality of the substance to number of moles - in this question ⇒ mass of CuO is provided - Convert mass of CuO to number of moles of CuO - $N o . o f m o l es o f C u O = \frac{ma ss}{m o l a r ma ss} = \frac{8.0}{64 + 16} = 0.1 m o l$
Step 3: Compare Mole ratio from balanced chemical equation $M o l e r a t i o = \frac{n o . o f m o l es o f re a c t an t / p ro d u c t w h ose q u a l i t y i s kn o w n}{n o . o f m o l es o f re a c t an t / p ro d u c t w h ose q u an t i t y i s kn o w n}$ Compare Mole ratio ⇒ $\frac{N o . o f m o l es o f C}{N o . o f m o l es o f C u O} = \frac{1}{2}$ Hence: Number of moles of C = Mole ratio $\times$ No. of moles of CuO from step 2 = 0.5 $\times$ 0.1 = 0.05mol
Step 4: Convert Mole to Molar Mass Mass of C = No. of mole of C $\times$ molar mass of C = 0.05 $\times$ 12 = 0.6g

(ii)
Step 1: Comapre Mole ratio from balanced chemical equation: compare mole ratio, $\frac{n o . o f m o l es o f C u}{n o . o f m o l es o f C u O} = \frac{2}{2}$ Hence ⇒ no. of moles of Cu = mole ratio $\times$ no. of moles of CuO from step 2 in (i) = $\frac{2}{2} \times 0.1 = 0.1 m o l$
Step 2: Convert mole to mass: Mass of Cu = no. of mole of Cu $\times$ molar mass of Cu = 0.1 $\times$ 64 = 6.4g

(iii)
Step 1: Compare mole ratio from balanced equation: Compare mole ratio, $\frac{n o . o f m o l es o f C O _{2}}{n o . o f m o l es o f C u O} = \frac{1}{2}$ Hence, no. of moles of $C O_{2}$ = mole ratio $\times$ no. of moles of CuO from step 2 in (i) = $\frac{1}{2} \times 0.1$ = 0.05 mol
Step 2: Convert mole of gas to volume of gas Volume of $C O_{2}$ = no. of mole of $C O_{2}$ $\times$ molar volume of gas at r.t.p = 0.05 $\times$ 24 = 1.2d $m^{3}$

Thermal Decomposition of some metal carbonates produce metal oxide and carbon dioxide gas. Calculate the mass of calcium carbonate which decomposed upon heating to produce 4.8 $d m^{3}$ of carbon dioxide at r.t.p

Solution
Step 1: Convert volume of carbon dioxide gas to number of moles no. of moles of $C O_{2}$ = $\frac{v o l u m e}{m o l a r v o l u m e} = \frac{4.8}{24} = 0.2 m o l$
Step 2: compare mole ratio Compare mole ratio: $\frac{n o . o f m o l es o f C a C O _{3}}{n o . o f m o l es o f C O _{2}} = \frac{1}{1}$ $∴$ no. of moles of $C a C O_{3}$ = $\frac{1}{1} \times 0.2 = 0.2 m o l$
Step 3: Convert moles to mass mass of $C a C O_{3}$ = no. of moles of $C a C O_{3} \times$ molar mass of $C a C O_{3}$ = 0.2 $\times$ 100 = 20g

Calculations from Chemical Reactions

Calculating Volumes of gaseous reactants and Products ⇒ recall: one mole of any gas occupies 24 $d m^{3}$ at room temperature and pressure. Equal volumes of gasses ⇒ under same conditions of temperature and pressure, contain same number of particles. ⇒ means that volume of gas is proportional to number of moles of a gas.

Consider chemical reaction between Nitrogen gas and hydrogen gas $N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)$ balanced chemical equation can be read ⇒ as 1 mol of nitrogen gas reacts with 3 mol of hydrogen gas to produce 2 mol of ammonia gas

continued on the next table

$N_{2}$	$H_{2}$	$N H_{3}$
No. of moles of $N_{2}$ = 1mol	No. of moles of $H_{2}$ = 3 mol	No. of moles of $N H_{3}$ = 2mol
Volume of $N_{2}$ = 1 mol $\times$ 24 $d m^{3}$ = 24 $d m^{3}$	Mass of $H_{2}$ = 3mol $\times$ 24 $d m^{3}$ = 72 $d m^{3}$	Mass of $N H_{3}$ = 2mol $\times$ 24 $d m^{3}$ = 48 $d m^{3}$

Continued here:
Mole ratio of gaseous reactants and products:
$N_{2}$ : $H_{2}$ : $N H_{3}$
1 : 3 : 2

Volume ratio of gaseous reactants and products: $N_{2}$ : $H_{2}$ : $N H_{3}$ 24 : 72 : 48 1 : 3 : 2 Thus, balanced equation for this reaction can also be read as 1 volume of nitrogen gas reacts with 3 volume of hydrogen gas to produce 2 volume of ammonia gas

> Thus, for chemical reactions that involve gaseous reactants and products, mole ratio of gases is the same as the volume

Using relationship between mole ratio and volume ratio of gases ⇒ volume of any gaseous reactant used up or the volume of any gaseous product formed in a reaction can be determined using mole ratio of gases provided by a balanced chemical equation.

Calculations from Chemical Reactions

Titration in volumetric analysis
Calculating Concentration of aqueous solutions of reactants and products

Volumetric analysis ⇒ titration is a method used to prepare some soluble salts. ⇒ titration can also be used to determine concentration of an aqueous solution by reacting it with a solution of known concentration (standard solution). known as volumetric analysis

Titration involves:

Solution with unknown concentration (titre)
Solution with known concentration (titrant)

Example: Acid base titration ⇒ we can find out volume of titrant (alkali) required to react with fixed volume of titre (acid with unknown concentration)

Steps:

Pipette 25.0 $c m^{3}$ of acid (titre) into a conical flask
Add a few drops of suitable indicator eg: methyl orange
Fill the burette with the alkali (titrant). Measure and record the initial burette reading.
Run the alkali into the conical flask, swirling the flask in the process (to ensure proper mixing of reaction mixture)
Titration is stopped at the end point, i.e. when indicator has changed color permanently.
Measure and record final burette reading. calculate volume of alkali (titrant) added
Titration is repeated until consistent results are obtained.
Average titrant volume is calculated
Volume is used to calculate unknown concentration of the solution Worked Example Question: 25.0 $c m^{3}$ of NaOH is needed to neutralize 20.0 $c m^{3}$ of 0.5mol/ $d m^{3}$ of $H_{2} S O_{4}$ solution. calculate concentration of NaOH solution in (i) mol/ $d m^{3}$ and (ii)g/ $d m^{3}$

Solution 1
Step 1: write balanced chemical equation, include state symbols. $2 N a O H (a q) + H_{2} S O_{4} (a q) \to N a_{2} S O_{4} (a q) + 2 H_{2} O (l)$
Step 2: Calculate no. of moles of sulfuric acid used from values given: No. of moles of $H_{2} S O_{4}$ (aq) = Conc(mol/ $d m^{3}$ ) $\times$ Vol $d m^{3}$ = 0.5 $\times$ $\frac{20}{1000}$ = 0.01mol
Step 3: Compare mol ratio (from equation) to find the no. of moles of NaOH used in the reaction: compare mole ratio, $\frac{n o . o f m o l es o f N a O H}{n o . o f m o l es o f H _{2} S O _{4}}$ = $\frac{2}{1}$ No. of moles of NaOH = $\frac{2}{1} \times 0.01$ = $0.02$ mol (This is the no. of moles of NaOH in 25 $c m^{3}$ of solution)
Step 4: Convert no. of moles in 25 $c m^{3}$ of solution to no. of moles in 1 $d m^{3}$ of solution i.e. concentration in mol/ $d m^{3}$ 25 $c m^{3}$ = 0.025 $d m^{3}$ Concentration of NaOH (mol/ $d m^{3}$ ) = $\frac{n o . o f m o l es o f N a O H ( m o l )}{v o l u m e o f so l u t i o n d m ^{3}}$ concentration of NaOH solution = $\frac{0.02}{0.025}$ = 0.8mol $d m^{3}$ ii) concentration of NaOH (g/ $d m^{3}$ ) = concentration of NaOH (mol/ $d m^{3}$ ) $\times$ molar mass of NaOH = 0.8 $\times$ (23 + 16 + 1) = 32g/ $d m^{3}$

Solution 2
Formula for acid base titration calculation: $\frac{M _{A} \times V _{A}}{M _{B} \times V _{B}}$ = $\frac{n _{A}}{n _{B}}$ $\frac{0.5 \times 0.02}{M _{B} \times 0.025} = \frac{1}{2}$ $\frac{0.5 \times 0.02 2}{0.025}$ = $M_{B}$ $M_{B}$ = concentration of NaOH = 0.8mol/ $d m^{3}$ other than acid base titration ⇒ this formula can be used for redox titrations. Change A and B to the types of reagents used. Note: if you want to use this formula, please include the legend as well. M = Conc(mol/ $d m^{3}$ ) V = Vol( $d m^{3}$ ) A = Acid ( $H_{2} S O_{4}$ ) B = Base (NaOH) $n_{a}$ = no. of moles of acid, $H_{2} S O_{4}$ (from eqn) $n_{b}$ = no. of moles of base, NaOH (eqn) ii) concentration of NaOH (g/ $d m^{3}$ ) = concentration of NaOH (mol/ $d m^{3}$ ) $\times$ molar mass of NaOH = 0.8 $\times$ (23 + 16 + 1) = 32g/ $d m^{3}$
Other than acid-base titration ⇒ this formula can be used for redox titrations, change A and B to the types of reagents used.
Note: if you want to use the formula, include legend as well

Calculations from Chemical Reactions

Calculate the amount of product formed in a reaction with limiting reactant

What is a Limiting Reactant

is used up in a chemical reaction. when it is used up, reaction stops What is a Reactant in Excess
Not used up in reaction so at the end of the reaction there is some left Hence, state importance of limiting reactant
quantity of product obtained is decided by quantity of limiting reactant

Worked Example 1

when 6g of hydrogen is allowed to react with 32g of oxygen, water is produced i) identify limiting reactant in this reaction ii) calculate mass of water formed

i) Step 1: write balanced chemical equation for this reaction (include state symbols) $2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l)$

Step 2: Calculate number of moles of reactants provided for this reaction no. of moles of $H_{2}$ provided for the reaction = $\frac{6}{1 \times 2}$ = 3 mol no. of moles of $O_{2}$ provided for the reaction = $\frac{32}{2 \times 16}$ = 1 mol

Step 3: Determine limiting reactant by calculating the actual amount of each reactant required for complete reaction using mole ratio

Method 1	Method 2
Compare mole ratio, $\frac{n o . o f m o l es o f O _{2}}{n o . o f m o l es o f H _{2}} = \frac{1}{2}$ No. of moles of $O_{2}$ required: = $\frac{1}{2} \times 3$ = 1.5 mol 1.5 mol of $O_{2}$ is required to react completely ⇒ with 3 mol of $H_{2}$ Since ⇒ only 1 mol of $O_{2}$ is provided for this reaction, $O_{2}$ is the limiting reactant and $H_{2}$ is in excess	Compare mole ratio. $\frac{n o . o f m o l es o f H _{2}}{n o . o f m o l es o f O _{2}} = \frac{2}{1}$ No. of moles of $H_{2}$ required = $\frac{2}{1} \times 1$ = 2 mol 2 mol of $H_{2}$ is required to react completely with 1 mol of $O_{2}$ . since ⇒ 3 mol of $H_{2}$ is provided for this reaction, $H_{2}$ is in excess and $O_{2}$ is the limiting reactant

ii)
Use limiting reactant to calculate amount of product
in this reaction ⇒ $O_{2}$ is the limiting reactant.
Compare mole ratio,
$\frac{n o . o f m o l es o f H _{2} O}{n o . o f m o l es o f O _{2}} = \frac{2}{1}$
= no. of moles of $H_{2} O$
= $\frac{2}{1} \times 1$
= 2 mol

mass of $H_{2} O$ = no. of moles of $H_{2} O \times$ molar mass of $H_{2} O$ = $2 \times [(2 \times 1) + 16]$ = 36g

Percentage Yield and Percentage Purity

yield ⇒ refers to the amount of product formed in a reaction

Define theoretical yield:

Definition

Maximum quality of product that can be obtained calculated value from equation #ChemistryDefinitions

In reality, amount of products produced experimentally produced in most reactions is always less than that produced theoretically (by calculation from balanced equation) WHY?

use of reactants which are impure (contains impurity)
Incomplete reaction as some reactants did not react completely
Poor handling of reactants and products ⇒ when conducting experiments ⇒ which resulted loss of reactants or products.

Define Actual Yield:

Actual quantity of product obtained (equal to or less than theoretical yield)

Actual yield < theoretical yield

Percentage Yield ⇒ shows relationship between actual yield and theoretical yield Formula: $p erce n t a g e / y i e l d / o f / p ro d u c t = \frac{a c t u a l y i e l d o f p ro d u c t}{t h eore t i c a l y i e l d o f p ro d u c t} \times 100%$

Worked Example 1

128g of sulfur dioxide, $S O_{2}$ was reacted with oxygen to produce Sulfur Trioxide, $S O_{3}$ 148g of $S O_{3}$ was produced in the reaction. Calculate Percentage yield of $S O_{3}$

Step 1 write balanced chemical equation for this reaction (include state symbol) $2 S O_{2} (g) + O_{2} (g) \to 2 S O_{3} (g)$

Step 2 Identify the actual yield of the product which is provided in the stem of the question Actual Yield of $S O_{3}$ = 140g

Step 3 Calculate theoretical yield of product. No. of moles of $S O_{2}$ = $\frac{ma ss o f S O _{2}}{m o l a r ma ss o f S O _{2}}$ = $\frac{128}{32 + ( 2 \times 16 )}$ = 2mol

compare mol ratio, $\frac{n o . o f m o l es o f S O _{3}}{n o . o f m o l es o f S O _{2}} = \frac{2}{2}$ $∴$ of moles of $S O_{3}$ = mass of $S O_{3}$ = no. of moles of $S O_{3} \times$ molar mass of $S O_{3}$ = 2 $\times$ 80 = 160g

Step 4 Calculate percentage yield of product Percentage yield of $S O_{3}$ = $\frac{a c t u a l y i e l d o f S O _{3}}{t h eore t i c a l y i e l d o f S O _{3}} \times 100%$ = $\frac{140}{60} \times 100%$
= 87.5

Percentage Purity

Often ⇒ percentage yield of a product is less than 100% because reactants are not pure

What is relationship between purity of the reactant and the percentage yield of a product

If purity % reactant increases, % yield of product % and vice versa

Calculating Percentage Purity of a Substance $P erce n t a g e p u r i t y o f s u b s t an ce = \frac{ma ss o f p u re s u b s t an ce in s am pl e}{ma ss o f s am pl e} \times 100%$ ⇒ if percentage purity of substance (sample) is 100% ⇒ we can deduce that the sample is pure

If percentage purity of the substance (sample) is less than 100% → we can deduce that sample is impure

Worked Example

3.2g of impure impure copper was heated in air. 3.8g of copper(II)oxide was formed. i) what was the percentage purity of the copper sample ii) what assumption was made when calculating the percentage purity of impure copper sample.

Step 1 Write a balanced chemical equation for this reaction. (include state symbol) $2 C u (g) + O_{2} (g) \to 2 C u O (s)$

Step 2 identify mass of sample in which is provided in the stem of the question Mass of impure copper = 3.2g

Step 3 Calculate mass of pure substance in sample. $N o . o f m o l es o f C u O = \frac{ma ss o f C u O}{m o l a r ma ss o f C u O}$ = $\frac{3.8}{64 + 16}$ = 0.0475 mol

Compare mol ratio, $\frac{n o . o f m o l es o f C u}{n o . o f m o l es o f C u O} = \frac{2}{2}$ $∴$ No. of moles of pure Cu = $\frac{2}{2} \times 0.0475$ = 0.0475 mol

Mass of pure Cu = no. of moles of Cu $\times$ molar mass of Cu = $0.0475 \times 64$ = 3.04g

Step 4 Calculate percentage purity of sample Percentage purity of copper sample = $\frac{ma ss o f p u re co pp er in C u S am pl e}{ma ss o f ∣ C u S am pl e} \times 100%$ = $\frac{3.04}{3.2} \times 100$$ = 95.0%

Assumption: copper is the only substance that will react to give CuO. eg: if $C u C O_{3}$ was an impurity, heating could have given off CuO as well

if mass is same on both sides, then mol ratio is 1 : 1

Important

mol ratio can exist between compounds only.

Quartz 4

Explorer

Mole Concept and Stoichiometry Main Note

Methods for collecting gasses, with relation to mole

Percentage Composition of Compounds

Mole and Molar Mass

Molar Mass

Mole and Molar Volume of Gasses

Mole and Concentration of Solution

Empirical and Molecular Formula of a Compound

Calculations from Chemical Reactions

Calculations from Chemical Reactions**

Calculations from Chemical Reactions

Calculations from Chemical Reactions

Calculations from Chemical Reactions

Worked Example 1

Percentage Yield and Percentage Purity

Worked Example 1

Percentage Purity

Worked Example

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