Unit 4: Stoichiometry — Short Questions

Mole Mole is defined in relation to Avogadro's number as follows.
Amount of substance that contain 6.022 × 10²³ no. of particles is known as mole.
One can calculate the number of moles by dividing the number of particles by Avogadro's number

Number of moles = No.of Particles / Avogadro's Number

(a) Molar mass
The mass of one mole of a substance expressed in grams is called molar mass.
It is represented by M.
Its unit is g/mol.
The molar mass is the sum of masses of component atoms.

(b) Molar Volume
The volume of one mole of ideal gas at STP is called molar volume.
Its value is equal to 22.414 dm³ which is rounded to 22.4 dm³. It is denoted by Vm.

(c) Molar concentration
Number of moles of a substance dissolved per volume of a solution in dm³.
Its unit is mol/dm³. It is denoted by C.
C = n / V(dm³)

Molar Volume
The volume of one mole of ideal gas at STP is called molar volume.
Its value is equal to 22.414 dm³ which is rounded to 22.4 dm³. It is denoted by Vm.

Relationship between Avogadro's Number and Molar Volume
One mole of an ideal gas at 273.16 K and 1atm pressure has a volume of 22.414 dm³.
Since one mole of a gas has Avogadro's number of particles so 22.414 dm³ of various ideal gases at STP will have Avogadro's number of particles i.e. 6.02 × 10²³.
22.4dm³ of gas at STP contains 1 mole and 6.02 × 10²³ particles.

39g of K = 1 mole of K = 6.02 × 10²³ atoms
56g of Fe = 1mole Fe = 6.02 × 10²³ atoms
So, 39g of K and 56g of Fe has equal number of atoms in them, because they have equal number of moles.

4g of He = 1mol of He = 22.414 dm³ volume
17g of NH3 = 1 mol of NH3 = 22.414 dm³ volume
96g of SO₂² = 1 mol of SO₂² = 22.414dm³ volume

The above data indicates that all the three gases have different masses but occupy same volume because size and masses do not affect the volume, and the number of moles are same.

1 mole of H2 at 0°C and 1atm = 22.414 dm³ = 6.02 × 10²³ molecules
1 mole of NH3 at 0°C and 1atm = 22.414dm³ = 6.02 × 10²³ molecules.
The above data shows that 1 mole of H2 and 1 mole of NH3 at 0°C and 1atm will have Avogadro's number of particles because 1 mole of any substance have same no. of particles i.e. 6.02 × 10²³.

Definition The branch of chemistry in which the quantitative relationship between reactants and products in a balanced chemical equation is studied.

Assumptions of Stoichiometric calculations are:
(i) All the reactants should be converted into products.
(ii) No side reaction occurs.

Limiting Reactant
The reactant which controls the amounts of products formed in a chemical reaction and is consumed earlier.
In a chemical reaction, it controls the quantity of product formed because limiting reactant consumed first and once it is consumed the reaction cannot proceed further. thus limiting the amount of product.

Theoretical Yield Maximum amount of product produced by given amount of reactant according to balanced chemical equation. It is greater than actual yield.

Actual Yield Amount of product obtained in a chemical reaction. It is smaller than theoretical yield.

Percentage Yield of Reaction
A chemist is usually interested in the efficiency of a reaction. The efficiency of a reaction is expressed by comparing the actual and theoretical yield in the form of percentage (%) yield.

%yield = (Actual Yield / Theoretical Yield) × 100

There are following reasons for that:

• The processes like filtration, separation by distillation, washing etc. if not properly carried out, decrease the actual yield.
• Some of the reactants might take part in a competing side reaction and reduce the amount of desired product. So, in most of the reactions the actual yield is less than the theoretical yield.
• A reaction may be reversible. Therefore, the amount of product will be reduced by the backward reaction.

The mole is the amount of a substance which contains as many elementary entities as there are atoms in 0.012 kg (12 g) of carbon-12. The elementary entities may be atoms, molecules, ions, electrons, and other particles. It is represented by n.
The number of moles of a substance can be calculated by dividing mass in grams by molar mass. The formula for number of moles is:

Number of moles = Given mass / Molar mass

n = m / M

The number of entities present in one mole of a substance is a constant number, named Avogadro's number, i.e. 6.02 × 10²³.
It is represented by NA. This value is attributed to an Italian scientist Amedo Avogadro (1776 -1856).
Avogadro's number is a physical constant representing the molar number of entities.
The exact value of it is 6.02214179 × 10²³ mol. In calculations we use the rounded off value 6.02 × 10²³.

Juglone, is a dye and is produced from the husks of black walnuts. The formula for Juglone is C6H6O3.
Juglone, is a natural herbicide (weed killer). It kills off competitive plants around the black walnut tree but does not affect grass and other noncompetitive plants.

Isopentyl acetate C7H14O2 is the compound responsible for the scent of bananas. Interestingly, bees release about 1 μg (1 × 10⁻⁶ g) of this compound when they sting. The resulting scent attracts other bees to join the attack.

Mass in grams = Moles × molar mass
Moles × Avogadro's number = Number of Particles

Stoichiometric calculations obey law of conservation of mass and law of definite proportions. According to the law of conservation of mass, "matter (mass) can neither be created nor destroyed" It states in terms of stoichiometry that the total mass of reactants is equal to the total mass of products in a balanced equation. According to the law of definite proportions, a pure compound always contains the same element combined in the same ratio by mass.

The balanced chemical equation has the same number of atoms of each element on both sides of equation. It has definite ratios of reactants and products just as compounds have definite ratios of elements. Such ratios are used to calculate the mass or mole of other substances.

The following types of relationship can be studied with the help of a balanced chemical equation involving quantities of reactant(s) and product(s).
i. Mole-Mole Relationship
ii. Mass-Mass Relationship
iii. Volume-Volume Relationship
iv. Mole-Mass Relationship
v. Mole-Volume Relationship
vi. Mass-Volume Relationship

Approach to do Stoichiometric Calculations

• Mass of known solid or volume of a known gas, or molar concentration of know solution.
• Calculate number of moles from the mass of known solid or volume of a known gat or molar concentration of known solution using the relevant formula.
• Find the ratio of the known and the unknown reactant or product from the balanced chemical equation.
• Calculate the number of moles of the unknown reactant or product using the relevant formula.
• Convert the number of moles of the unknown to mass, volume, or concentration of the substance.

If the number of moles of a gas is known, one can calculate its volume by multiplying number of moles of the gas with molar volume.
Volume of a gas = Number of moles x Molar volume
V = n × Vm

Density is defined as the mass per unit volume of a substance
Density = Mass/Volume (d = m/V)
As molar mass of all the gases occupies same volume at STP, therefore, density of a gas depends on its molar mass. A gas having higher molar mass will have higher density and vice versa. If the density of the gas at STP is known, its molar mass can be calculated.
M=d × Vm

Molar concentration of the solution is the defined as the number of moles of substance dissolved per dm³ of the solution.
Units: Mole / dm³

The reactants which are in larger amounts (according to stoichiometry of reaction) and remain unreacted at the end of the reaction are called "excess reagents" (or excess reactants). Consider the reaction between hydrogen and oxygen to form water.

2H₂(g) + O₂(g) → 2H₂O(g)

When we take 2 moles of hydrogen (4g) and allow it to react with 2 moles of oxygen (64g), then we will get only 2 moles (36g) of water.

i. Calculate the number of moles from the given amounts of reactants.
ii. Find out the number of moles of product with the help of a balanced chemical equation.
iii. Identify the reactant which produces the least amount of product as limiting reactant and the other as an excess reactant.

A large amount of inexpensive reactant is supplied because of the following reasons:
a. To ensure that whole of the mass of the expensive reactant is completely converted to the desired product.
b. To produce maximum amount of product.
c. To increase the rate of reaction.

Efficiency of reaction is checked by measuring the percentage yield of the reaction. If the percentage yield of reaction is high than the efficiency of reaction is better.

Stoichiometry is very important in the field of medicine and is used:
i. In the preparation of antibiotics, the stoichiometry ensures that each dose matches the active ingredient and target bacteria.
ii. To determine the cholesterol level in the blood of patients. Cholesterol is a form of fat that is not all bad. However, cholesterol can have harmful effects.
iii. To determine the glucose level in the blood of diabetic patient. Use of insulin relies on the stoichiometry to precise control of blood sugar levels.
iv. To determine the steroid and other stimulants in the urine of athletes. Athletes use steroids and other stimulants to enhance performance and increase strength.
v. To determine the concentration of viral antigens in the preparation of vaccine for effective results.
vi. To determine the amount and number of drugs to give a dosage to a patient. The medicine has no effect when given in small amounts and can cause toxic state or death when given in large amounts.
Paracetamol is used as a pain killer and to decrease fever. An overdose may result a blood thinning, organ damage and severe liver damage.

See Q3. of theory

See Q4. of theory

Data
Mass of NaOH = 2g
Initial Volume of NaOH = 250cm³ / 1000 = 0.250 dm³
Final volume of NaOH = 500cm³ / 1000 = 0.500 dm³

Required
MNaOH = ?
nNaOH = ?
Cinitial = ?
Cfinal = ?

Solution
(a) MNaOH = 23 + 16 + 1 = 40g/mol Answer

(b) nNaOH = mNaOH / MNaOH
= 2g / 40g/ mol = 0.05mol Answer

(c) Initial concentration = Ci = mole / volume
= 0.0500 / 0.250 = 0.200 mol/ dm³ Answer

(d) Final concentration = CF = ?
CF = moles / volume
= 0.0500 / 0.500 = 0.100 mol/ dm³ Answer

Data
4NH3 + 5O2 → 4NO + 6H2O

Mass of NH3 = 34g
Mass of O2 = 96g
Molar mass of NH3 (MNH3) = 17g/mol
Molar mass of NO (MNO) = 14 + 16 = 30g/mol
Molar mass of (MO2) = 32g/mol
Molar mass of H (Ar,H) = 1g
Molar mass of N (Ar,N) = 14g
Molar mass of O (Ar,O) = 16g

Required
(a) Limiting Reactant =?
(b) Mass of NO =?
(c) Mass of excess reagent left behind =?

Solution

(a) nNH3 = mNH3 / MNH3
= 34g / 17g/ mol = 2 moles

nO2 = mO2 / MO2
= 96g / 32g/ mol = 3 moles

To determine limiting reactant, moles of NH3 and O2 are compared with moles of NO as product.

NH3 NO
4 : 4
2 : 2

Ammonia produce 2 moles of NO, now moles of O2 are compared with NO.

O2 NO
5 : 4
1 : 4/5
3 : 4/5 × 3
: 2.4moles

Limiting reactant is NH3 as it produces less moles of product. Answer

(b) Mass of NO
To get the moles of NO, its moles can be calculated from balanced chemical equation by using moles of NH3 because it is limiting reactant.

Stoichiometry Ratio
NH3 : NO
4 : 4
2 : 2

So 2 moles of NO are formed.
mNO = nNO × MNO
= 2 × 30g/mol
mNO = 60g Answer

(c) Mass of excess reactant remaining
To get the excess mass of O2 as non-limiting or excess reactant. First, its number of moles left behind calculated by using moles of ammonia from balanced chemical equation.

NH3 O2
4 : 5
1 : 5/4
2 : 5/4 × 2
: 2.5

So, 2.5 moles of O2 completely react with 2 moles of NH3.
nO2 taken = 3 moles
nO2 used = 2.5
nO2 left behind = 3 – 2.5 = 0.5 moles

mO2 left behind = nO2 × MO2
= 0.5 × 32
mO2 left behind = 16g Answer

Data
Fe2O3 + 3CO → 2Fe+ 3CO2

Mass of Fe2O3 = 1kg = 1000 g
Mass of Fe (Actual yield) = 650 g
Molar mass of Fe (Ar, Fe) = 56 g/mol
Molar mass of Fe2O3 (M, Fe2O3) = 2 × 56 + 3 × 16 = 160 g/mol
Molar mass of O (Ar, O) = 16 g/mol

Required
% yield of Fe = ?

Formula
%yield = (Actual yield / Theoretical yield) × 100

Solution

nFe2O3 = mFe2O3 / MFe2O3
= 1000g / 160g/ mol = 6.25 mol

To get the number of moles of Fe produce in a reaction theoretically, its theoretical yield is calculated from balance chemical equation in the following.

Fe2O3 Fe
1 : 2
6.25 : 6.25 × 2
: 12.5moles

So, 12.5 moles Fe are formed theoretically.
mass of Fe (Theoretical yield) = number of moles of Fe × molar mass of Fe
= 12.5 moles × 56g/mol
mass of Fe (Theoretical yield) = 700g

So, the mass of Fe produced is 700g which is called theoretical yield. Now percentage yield is calculated using formula.

%yield of Fe = (650g / 700g) × 100
= 92.86%

Result
% yield of Fe = 92.86%. Answer

Data Mass of C2H6 = 1.5g
Molar mass of C = 12g/mol
Molar mass of H = 1g/mol
Molar mass of O = 16g/mol
Avogadro's number = 6.022×10²³
Molar volume (Vm) = 22.4dm³
Molar mass of C2H6 = 30.08g/mol

Required
(a) Number of molecules of solid CO2 =?
(b) Number of O2 molecules reacted.
(c) Number of CH bonds of C2H6 are broken =?

Solution

(a) Number of moles of C2H6 = 1.5g / 30.080
= 0.0499moles

2C2H6 + 7O2 → 4CO2 + 6H2O

From balanced chemical equation 2 moles of C2H6 produces moles of CO2 = 4 moles
0.0499 moles of C2H6 produces moles of CO2
= 4/2 × 0.0499 = 0.0998moles

Number of moles of CO2 = 0.0998moles
From balanced chemical equation
2 moles of C2H6 react with moles of O2 = 7 moles
0.499 moles of C2H6 react with moles of O2
= 7/2 × 0.0499 = 1.7465 moles

Number of CO2 molecules
= 0.0998 × 6.02 × 10²³
Number of CO2 molecules = 6.008 × 10²² molecules Answer

(b) Number of O2 molecules
= 1.7465 × 6.022 × 10²³
Number of O2 molecules = 1.051 × 10²⁴ molecules Answer

(c) Number of C2H6 molecules
= 0.04998 × 6.022 × 10²³
= 3.004 × 10²² molecules

One molecule of C2H6 has C – H bonds = 6 bonds
Number of C – H bonds broken
= 3.002 × 10²² × 6
Number of C – H bonds broken = 1.802 × 10²³ bonds Answer