MOLAR MASS
The mass of one mole of any substance (atoms, molecules, ions, or formula units) expressed in grams is called molar mass. The standard unit of molar mass is g/mol, whereas the SI unit is kg/mol. For example,
- H2O has molar mass =18g/mol
- H2 has molar mass = 2g/mol
- OH– has a molar mass =17g/mol
ATOMIC MASS UNIT (amu)
The atomic mass unit (amu), which is also called unified atomic mass unit (u) or Dalton (Da), is a standard unit in which the masses of atoms, molecules, ions, or subatomic particles are expressed. The amu, or 1 amu, is exactly equal to one-twelfth ( 1/12th ) of the mass of a 6C12-atom in its ground state. Or simply, if a 6C12-atom is cut into twelve equal parts, and one part is equal to 1 amu. Thus, 1 amu is 1/12th of a 16C12-atom. It is approximately equal to 1.66×10-24 g or 1.66×10-27kg.
RELATIVE ATOMIC MASS OR SIMPLE ATOMIC WEIGHT
The relative atomic mass, which is also called atomic weight, is a unitless quantity obtained by comparing the average mass of atoms of an element to the one part or 1/12th of 6C12-atom. For example,
- 7N14-atom has a relative atomic mass of 14 amu. It means it is 14 times heavier compared to I part of 6C12-atom or it is 14 times heavier compared to 1/12th of 6C12-atom. Similarly, 1H1 has 1 amu, and 12Mg24 has 24 amu.
MOLE
The molar mass of atoms, ions, or molecules is expressed in grams equal to one mole of atoms, ions, or molecules, respectively. For example,
- The molar mass of H2O is 18g =1 mole of H2O
- The molar mass of H2 is 2g = 1 mole of H2
- The molar mass of OH– is 17g = 1 mole of OH–
It is the SI unit of substance that contains a specific number of atoms, ions, or molecules present in it. The number of entities present in one mole of substance is equal to Avogadro’s number (Na), i.e., 6.022×1023.
The mass of any entity given in amu can be converted into grams by multiplying by 1.66×10-24g i.e., 24 amu of Mg =24×1.66×10-24g.
The mass of any entity (atoms, ions, or molecules) expressed in amu can be converted into molar mass (mole) by the following sample interconversion.
For, e.g. 1 atom of 6C12 = 12 amu = 12×1.66×10-24g or 12×1.66×10-24g = 1 atom ⇒
1g = 1atom/12×1.66×10-24g ⇒ 12g = 12gx1atom/12x1.66×10-24g =6.02×1023 atoms (1 mole)
Similarly, 2 g H2 molecules = 1mole H2 molecules = 6.02×1023 molecules
17 g OH– ions = 1 mole OH– ions = 6.02×1023 ions
The number of moles of any substance can be calculated as
- No. of moles = Mass/Molar mass
- No. of moles = Volume (L)/22.414 (L or dm3)
- No. of moles = No. of particles/6.022×1023
In one mole of a molecule different number of moles of atoms/ions are present. For example, in one mole of H2SO4, 2 moles of H+ ions and one mole of SO4-2 ions are present (H2SO4 → 2H+ + SO4-2). Similarly, in one mole of H2SO4 molecule, 2 moles of H atoms, one mole of S atoms, and 4 moles of O atoms are present.
In any chemical reaction, the mole ratios of reactants and products can be determined by knowing the quotients involved in a balanced chemical equation. For example,
C3H8 + 5O2 → 3CO2 + 4H2O
In this equation, 1 mole of propane reacts with 5 moles of O2, and as a result, 3 moles of CO2 and 4 moles of H2O are formed as products.
AVOGADRO’S NUMBER (NA)
Avogadro’s number is the number of particles (ions, atoms, or molecules) present in one mole of a substance, and it is equal to 6.02×1023. It is represented by NA.
The number of atoms, ions, or molecules present in any substance is calculated by the following expression.
- Number of atoms in a substance = moles of atoms x NA
- Number of molecules in a substance = moles of molecules x NA
- Number of ions in a substance = moles of ions x NA
In one mole of a molecule different number of atoms/ions are present. For example, in one mole of H2SO4, 6.02×1023 molecules of H2SO4 are present. Whereas 2×6.02×1023 ions of H+ and 6.02×1023 ions of SO4-2 present (H2SO4 → 2H+ + SO4-2). Similarly, in one mole of H2SO4, 6.02×1023 molecules of H2SO4, 2 x 6.02×1023 atoms of ‘H’, 6.02×1023 atoms of ‘S’, and 4x 6.02×1023 atoms of ‘O are present.
MOLAR VOLUME (VM)
The volume occupied by one mole of a substance at STP is called the molar volume. Its value is 22.414dm3. Its unit is m3/mol or dm3/mol (gases) or cm3/mol (solid or liquid). For example,
1 mole of O2 gas = 32g = 6.02×1023 molecules =22.414dm3
1 mole of H2 gas = 2g =6.02×1023 molecules =22.414dm3
The general formulas used for the calculations of molar volume are:
Vm=V/n, Vm=M/ρ, Vm = RT/P (for ideal gas)
STOICHIOMETRY
The word stoichiometry originates from Greek words: stoichion means ‘elements’, and metron means ‘measurement’.
Instoichiometry, we study the quantitative relationship between the number of reactants and products. From this study,
- Quantitative information about a chemical reaction is obtained
- The yield of a product is obtained
For example,
2H2 + O2 → 2H2O
In the following chemical reaction, the following information is obtained:
- 2 moles of H2 and 1 mole of O2 form 2 moles of H2O
- 2×6.02×1023 molecules of H2 react with 6.02×1023 molecules of O2 and 2×6.02×1023 molecules of H2O form.
- 2×22.414 dm3 of H2 react with 22.414dm3 of O2 and 2×22.414dm3 of H2O form.
Thus, from the stoichiometric calculation, mole-mass, mass-mass, mole-Avogadro’s, mass-Avogadro, mole-volume, mass-volume relationships can be calculated. To calculate any quantity, first of all, the mole of a reactant/product is to be calculated.
IMPORTANT NUMERICALS
Q: In the equation, 2H₂ + O₂ → 2H₂O, how many moles of H₂O are formed from 4 moles of H₂?
Sol: First of all, the mole ratios of reactants and products are compared as given in the balanced chemical equation,
2 mol H₂ : 2 mol H₂O ⇒ 1 mol H₂ : 1 mol H₂O ⇒ So, from the 4 mol H₂, 4 moles of H₂O are formed.
Q: In the equation, C + O₂ → CO₂, how many grams of CO₂ are formed from 12 g of C?
Sol: First of all, the mole is determined, and then the mole ratios of reactants and products are compared as given in the balanced chemical equation.
12 g C = 1 mol ⇒1 mol C: 1 mol CO₂ ⇒ The mass of CO₂ = moles x molar mass ⇒ mass=1×44 ⇒44g.
Q: in the reaction 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O, what’s the O₂:CO₂ mole ratio?
Sol: 13:8.
Q: How many Liters of CO₂ (at STP) are obtained from 50 g of CaCO₃ in the following chemical reaction? CaCO₃ → CaO + CO₂.
Sol: First of all, the mole is determined, and then the mole ratios of reactants and products are compared as given in the balanced chemical equation
50 g CaCO₃ = 0.5 mol of CaCO₃ ⇒ from the balanced chemical equation, 1mole of CaCO₃ =1 mole of CO₂ ⇒ So, from 0.5 mol of CaCO₃, 0.5 mol of CO2 is obtained ⇒ as from the molar volume, it is obtained that 1 mole of any gas contain 22.414dm3 (or 22.414L, as 1dm3=1L) ⇒ Thus, from 0.5 mole of CO2, 11.2 L CO2 is obtained.
Q:1 mol Mg burns: 2 Mg + O₂ → 2 MgO. How many mol O₂ are consumed and MgO produced?
Sol: First of all, the mole ratios of reactants and products are compared as given in the balanced chemical equation. From the equation, it is obtained that, 2 moles of Mg react with 1 mole of O2 and 2 mole of MgO are formed i.e. Mg: O₂:MgO = 2:1:2. So 1 mol Mg uses 0.5 mol O₂ and makes 1 mol MgO.
Q: How many molecules are in 0.5 mol CO₂?
Sol: As the number of molecules in 1 mole of any substance is 6.022×10²³ ⇒ thus, in 0.5 moles, the number of molecules is 0.5×6.022×10²³ = 3.01×10²³
Q: How many moles in 2.4×10²⁴ H atoms?
Sol: As one mole of any substance contains 6.02×1023 particles. ⇒ 6.02×1023 particles = 1 mole ⇒ 1 particle = 1 mole/6.02×1023 ⇒ 2.4×10²⁴ particles = (1 mole/6.02×1023) x2.4×10²⁴ ≈ 3.99 mol
Q: How do you find moles from mass?
Sol: The general formula for the relation between mass and moles is:
Number of moles = mass (g) ÷ molar mass (g/mol). E.g., 18 g H₂O → 18 ÷ 18 = 1 mol.
Q: How many liters does 2 mol of ideal gas occupy at RTP (~25 °C)?
Sol: From the standard relation between moles and molar volume, it is obtained that, 1 mole = 22.414dm3 ⇒ 2 moles = 44.828 L
Q: How are molar volume and density related?
Sol: Molar volume Vm=M/ρ, where M is molar mass and ρ is density
Q: How many moles are in 20 g of NaCl?
Sol: From the standard relation between moles and molar mass, and mass, it is obtained that the number of moles = mass/molar mass.
So, no. of moles = 20/58.5 ⇒ 0.342
Q: What is the mass of 2 moles of CO₂?
Sol: From the standard relation between moles and molar mass , and mass, it is obtained that no. of moles = mass/molar mass ⇒ 2 moles = mass/44 ⇒ mass = 44×2 ⇒ 88g.
Q: How many molecules are there in 2 moles of H₂O?
Sol: From the standard relation between moles and number of particles, it is obtained that
no. of moles = no. of particles/6.022×1023 ⇒ 2 = no. of particles /6.022×1023 ⇒ no. of particles = 6.022×1023 x 2 = 1.2044×1024
Q: How many moles are there in 3.011×1023 atoms of Na?
Sol: From the standard relation between moles and number of particles, it is obtained that
no. of moles = no. of particles/6.022×1023 ⇒ Moles=3.011×1023/6.022×1023=0.5 mol
Q: Find the volume occupied by 3 moles of CH₄ at STP.
Sol: From the standard relation between moles and volume, it is obtained that,
No. of moles = Volume (L)/22.414 (dm3 or L) ⇒ 3 = Volume /22.414 ⇒ Volume = 3×22.4=67.2 L
Q: How many moles are present in 11.2 L of O₂ at STP?
Sol: From the standard relation between moles and volume, it is obtained that.
No. of moles = Volume (L)/22.414(dm3 or L) ⇒ Moles =11.2/22.414=0.5 mol
Q: How many molecules are in 44.8 L of CO₂ gas at STP?
Sol: This problem is solved in two steps: firstly, the relation between moles and volume is used as shown below:
No. of moles = Volume (L)/22.414(dm3 or L) ⇒ number of moles = 44.8/22.4 = 2 mol. Now, in the second step, the relation between the number of molecules and moles is used, as shown by the following equation.
No. of moles = No. of particles/6.022×1023 ⇒ No. of particles = 2×6.022×1023 = 1.2044×1024
Q: How many atoms are in 2 moles of H₂O?
Sol: 1 H₂O molecule = 3 atoms ⇒ Total molecules = 2×6.022×1023=1.2044×1024 ⇒ Total atoms =
1.2044×1024×3=3.6132×1024
Q: Find moles in a mixture of 4 g H₂ and 32 g O₂?
Sol: Firstly, the number of moles of H2 and O2 is calculated by using the following formula. No of moles = mass/molar mass
For H2, no. of moles = 4/2 = 2
For O2, no. of moles = 32/32 = 1 ⇒ total moles = 2+1 = 3
Q: Calculate the volume of O₂ required to completely react with 4.48 L of H₂ at STP in the following reaction: 2H2 + O2 → 2H2O₂.
Sol: From the following reaction, first of all, the number of moles of H2 is calculated, then these moles are compared with O2 to evaluate the moles of O2. As we know that, no of moles = volume (L)/22.414 ⇒ no of moles = 4.48/22.414 = 0.199
The balanced chemical equation shows that 2 moles of H2: 1 mole of O2
2 moles H2 = 1 mole O2 ⇒ 1mole H2 = ½ O2 ⇒ 0.199 mole H2 =1/2 x0.199 =0.099
For the calculation of the volume of O2, we use the relationship between moles and volume
No of moles = Volume (L)/22.414 ⇒ 0.099 = Volume /22.414 ⇒ Volume = 22.414x 0.099 = 2.24 L
Q: What volume will 88 g of CO₂ occupy at STP?
Sol: In this question, first of all number of moles of CO₂ is calculated by using the formula of the relationship between moles and mass
No. of moles = mass/ molar mass ⇒ no. of moles = 88/44 =2
Now, the volume is calculated by using the relationship between moles and volume
No of moles = Volume/ 22.414 ⇒ 2= Volume/22.414 ⇒ Volume =22.414×2 =44.828 L
Q: What is the mass of 11.2 L of SO₂ at STP?
Sol: In this problem, first, the number of moles is calculated by using the volume-mole relationship, then, mass is calculated by using the mass-mole relationship.
No. of moles = volume/22.414 ⇒ No. of moles = 11.2/22.414 =0.5
No. of moles = mass/molar mass ⇒ 0.5 = mass/64 ⇒ mass = 64×0.5 =32g
Q: A compound contains 2 mol C, 6 mol H, and 1 mol O. What is the empirical formula?
Sol: C₂H₆O
Q: If 1 L of gas contains 0.045 moles, what volume would 1 mole occupy under the same conditions?
Sol: This problem is solved by a simple method as shown below:
0.045 moles = 1 L ⇒ 1 mole = 1/0.045 = 22.22 L
Q: Calculate the number of moles of oxygen atoms present in 4.8 g of ozone (O3).
Sol: First of all, the number of moles of O3 molecules is calculated, because the mass of an O3 molecule is given, no. of moles = mass/molar mass ⇒ no. of moles = 4.8/48 ⇒ 0.1. Since each molecule of O3 contains 3 oxygen atoms, moles of oxygen atoms = 0.1 mol x 3 = 0.3 mol.
Q: How many grams of CO2 are produced when 1 mole of carbon is burned in excess oxygen? C + O2 → CO2
Sol: First of all, the mole relations between reactant (C) and product (CO2) is established, from the above equation, it is obtained that, 1mole of C = I mole of CO2
Thus, the moles of CO2 formed are also 1. Now, the mass in grams of CO2 is calculated by using the mass-mole relationship
No. of moles = mass/ molar mass ⇒ 1 = mass/ 44 ⇒ mass = 44×1 = 44 g.
Q: Calculate the mass of Fe2O3 produced when 5 moles of Fe react with excess oxygen. 4Fe + 3O2 → 2Fe2O3.
Sol: From the balanced equation, firstly, the number of moles of Fe and Fe2O3 are compared.
4 moles of Fe = 2 moles of Fe2O3 ⇒ now, by calculating as
4 moles of Fe = 2 moles of Fe2O3 ⇒ 1 mole of Fe = 2/4 moles of Fe2O3 ⇒ 5mole of Fe = (2/4) x 5 moles of Fe2O3 = 2.5 moles of Fe2O3, Now, the mass of Fe2O3 can be calculated by using moles-mass relationship,
No. of moles = mass/ molar mass ⇒ 2.5 = mass/160 ⇒ mass= 160 x 2.5 = 400g
Q: A mixture contains 2 moles of CH4 and 3 moles of O2. Calculate the number of moles of CO2 produced when the mixture is ignited. CH4 + 2O2 → CO2 + 2H2O.
Sol: In this equation, the molesofCO2 will be compared to CH4, because O2 acts as an excess reagent. From the balanced equation, it is indicated that, 1mole of CH4 = 1 mole of CO2
Thus, 2 moles of CH4 = 2 moles of CO2. Moles of CO2 produced = 2 moles
Q: What is the mass of 1 molecule of oxygen (O2) in grams?
Sol: The mass of one molecule is calculated by firstly, by the moles-Avagadro’s number relationship, and then by the mole-mass relationship.
No. of moles = no of particles/6.02×1023 ⇒ No. of moles = 1/6.02×1023 = 0.166×10-23
To calculate the mass, use this relationship: no. of moles = mass/molar mass ⇒ 0.166×10-23 = mass/32 ⇒ mass = 0.166×10-23 x 32 = 5.31×10-23 g.