Chemical Calculations (Stoichiometry)
Upon completion of this lesson, the students should be able to:
- describe a mole
- determine molar and molecular masses
- calculate the percentage composition of a compound
- calculate the empirical and molecular formulas
The branch of chemistry, which deals with the numerical relationships of elements and compounds and the mathematical proportions of reactants and products in chemical transformations, is known as stoichiometry. An understanding of the mole concept together with some skills in writing and balancing chemical equations enable us to solve stoichiometric problems involving mass relations of reactants and products in chemical reactions.
The Mole Concept
In chemical calculations it is necessary to consider quantities of substances in terms of the number of atoms, ions or molecules present. The unit devised by chemists in expressing these numbers of atoms, ions, or molecules is called the mole.
Mole is the unit of measurement used in chemistry in expressing amounts of a chemical substance. It has a value of 6.02214129(27)×1023elementary entities of the substance. Chemists use mole to express the amounts of reactants or of products in chemical reactions instead of units of mass or volume.
A mole is defined as that quantity of a substance that contains the same number of ultimate particles (atom, ions, or units of ions) as present in 12g of Carbon -12.
What is molar mass?
The molar mass is the sum of the masses of the atoms present in one mole of a substance, which can be an element or a compound. It can be expressed as a unit of mass per mole like grams/mol, kg/mol.
1. Find the molar mass of: Al, H2O, Ca(OH)2
a. Al = 27 g/mol
b. H2O = 18 g/mol
H = 2 at x 1g/mol = 2g
O = 1 at x 16g/mol = 16g
c. Ca(OH)2 = 74 g/mol
Ca = 1 at x 40g/mol = 40g
O = 2 at x 16g/mol = 32g
H = 2 at x 1g/mol = 2g
2. How many moles are present in: 5.4 g of Al, 180 g of H2O?
a. 5.4 g x 1 mol/27g = 0.2 mol
b. 180 g x 1 mol/18g = 10 mol
3. What is the mass of: 5 mol of Ca(OH)2 and 2 mol of Al ?
5 mol of Ca(OH)2
Ca = 1 x 40 = 40
O = 2 x 16 = 32
H = 1 x 2 = 2
5 mol x 74g/mol = 370g
2 mol of Al
2 mol x 27g/mol = 54 g
The number of molecules in a mole of any molecular substance is the same as the number of atoms in a gram-atom of any number. This number of atom is called the Avogadro’s number, the accepted value of which is 6.02214129(27)×1023 atoms/gram-atom of any element. In other words, a mole is the amount of substance that contains as Avogadro’s number of particles equal to 6.02214129(27) ×1023 or simply expressed as 6.02 x 1023. The particles can be atoms, molecules, or ions.
There are 6.02 x 1023 atoms in 1 mol of carbon
There are 6.02 x 1023 molecules in 1 mol of H2O
There are 6.02 x 1023 Na
6.02 x 1023 Cl in 1 mol of NaCl
The Mole and Avogadro's Number
International System of Unit (SI base unit)
Amount of substance
1. How many atoms are present in:
a. 5 mol of copper
5 mol x 6.02 x 1023 atoms/mol = 30.10 or 3.01x 1023 atoms
b. 48 g of carbon
48 g x 1 mol/12g x 6.02 x 1023 atoms/mol = 24.08 x 1023 or 2.408 x 1023 atoms
2. How many g of H2O will contain 3.01 x 1024 molecules?
3.01 x 1024 molecules x 1 mol/6.02 x 1023 molecules x 18 g/mol = 90g
- Avogadro's number and the mole
Chem1 Tutorial on chemistry fundamentals Part 2 of 5
Percentage composition is a statement indicating the percentage of each element present in a compound. In chemistry, this composition is always on a weight basis unless specifically stated otherwise. Sometimes the composition of the mixture of gases is given on a volumetric basis. The computation of the percentage composition from the formula of the compound is based upon the meaning of the symbols and formulas. Each symbol stands for one atomic weight’s worth of the element it represents, and each formula stands for one molecular weight’s worth of the compound it represents.
1. What is the percentage composition of water?
% H = 2g/18g x 100% = 11.11%
% O = 18g/18g x 100% = 88.89%
2. What is the percentage composition of H2SO4?
% H = 2g/98.1g x 100% = 2.0%
% S = 32g/98.1g x 100% = 32.7%
% O = 64g/98.1g x 100% = 65.3%
Empirical Formula is a chemical formula denoting the constituent elements of a substance and relative number of atoms of each. It gives the simplest ratio of the number of the atoms.
What is the empirical compound formula of a compound which contains 2.05% H, 32.65% S, and 65.30% O?
Basis: 100 g of the compound
H = 2.05/100 x 100 g = 2.05 g X 1 at/1g = 2.05 at
S = 32.55/100 x 100 g = 32.65 g x 1 at/32g = 1.02 at
O = 65.30/100 x 100 g = 65.30 g x 1 at/16g = 4.08 at
The formula is not H2.05 S1.02 O4.08. Subscripts are small whole numbers, but they are not rounded off but are divided by the smallest number. The correct formula is H2SO4
H2.05 S1.02 O4.08 = H2SO4
Molecular formula is a chemical formula denoting the constituent elements of a substance and the number of atoms of each composing one molecule. It gives the actual ratio of the number of the atoms in a mole of the compound.
A compound is 85.69% C and 14.31% H. If the molar mass is 56 g/mol. What is the molecular formula of the compound?
Basis: 100 g of the compound
Determine the empirical formula:
C = 85.69 g x 1 at/12 g = 7.14 at
H = 14.31 g x 1 at/1 g = 14.31 at
Empirical Formula = C7.14 H14.31/7.14 = CH2
How to Calculate Percent Composition, Empirical Formulas and Molecular Formulas - Chemistry
Solve the following:
- What is the mass in grams of:
- 2 moles of AlCl3
- 8 moles of CaCO3
- 5 moles of (NH4)2SO4
- 4 moles of C6H12O6
3 moles of KNO3
- Calculate the number of moles in each of the following:
- 68.4 g of sucrose C12H22H11
- 80 g of NaOH
- 3.01 x 1023 atoms of Al
- 6.02 x 1024 molecules of CO2
- 12 g of O2
- Find the percentage composition of the following:
- Sugar C12H22O11
- hydrated copper sulfate CuSO4 5HO
- sulfuric acid H2SO4
- hematite Fe2O3