Atomic Mass (Weight)
Chapter Three
Stoichiometry
Atomic Mass (weight)
It is important to know the mass of the atoms especially for the lab work. However; atoms are very very small particles and we can not count it or weight it easily that because it contains huge number of atoms. For example the smallest thing we can see by our nicked eyes contains about 1016 atom, it is huge number is not?!!!!
It is clear that we can not weight a single atom, but as we said it is important to know the weight of the atom so how we can do that? The answer to this question is that we can determine the mass of one atom relative to another atom experimentally. To do that we need to assign (نحدد) a value to the mass of one atom of a given element so we can use it as standard.
By international agreement, an atom of the carbon isotope 12C that has six protons and six neutrons has a mass of exactly 12 atomic mass unit (amu). One atomic mass unit is defined as a mass exactly equal to 1/12 the mass of one 12C atom.
i.e. mass of one 12C atom = 12 amu
1 amu = mass of one 12C atom /12
The units in which the mass of an atom are expressed are atomic mass units amu. At one time, the lightest atom was assigned a mass of 1 amu and the mass of any other atom was expressed in terms of this standard. Today atomic mass units are defined in terms of the 12C isotope, which is assigned a mass of exactly 12.000 amu.
Average Atomic Weight (average atomic weight)
The atomic weight of an element is the average weight of the atomic masses of the different isotopes of that element. Naturally occurring carbon, for example, is a mixture of two isotopes, 12C (98.89%) and 13C (1.11 %). Individual carbon atoms therefore have a mass of either 12.000 or 13.03354 amu. But the average mass of the different isotopes of carbon is 12.011 amu.
Please note that (98.89%) is the natural abundance of 12C and (1.11 %) is the natural abundance of 13C, and in order to be able to make an accurate calculation we need to convert the percentage to fraction as you have seen in the above example.
N.P. There are other isotopes for carbon like C14, but these isotopes are not stable and its natural abundance is very little.
Molecular Weight
The molecular weight of a compound is the sum of the atomic weights of the atoms in the molecules that form these compounds.
Example: The molecular weight of the sugar molecule found in cane sugar is the sum of the atomic weights of the 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms in a C12H22O11 molecule.
12 C atoms = 12(12.011) amu = 144.132 amu
22 H atoms = 22(1.0079) amu = 22.174 amu
11 O atoms = 11(15.9994) amu = 175.993 amu
342.299 amu
C12H22O11 has a molecular weight of 342.299 amu. A mole of C12H22O11 would have a mass of 342.299 grams. This quantity is known as the molar mass, a term that is often used in place of the terms atomic weight or molecular weight.
The Mole
The term mole literally means a small mass. It is used as the bridge between chemistry on the atomic and macroscopic scale. If the mass of a single 12C atom is 12.000 amu, then one mole of these atoms would have a mass of 12.000 grams. By definition, a mole of any substance contains the same number of elementary particles as there are atoms in exactly 12 grams of the 12C isotope of carbon.
Example: A single 12C atom has a mass of 12 amu, and a mole of these atoms would have a mass of 12 grams.
A mole of any atoms has a mass in grams equal to the atomic weight of the element. The term mole can be applied to any particle: atoms, a mole of atoms, a mole of ions, a mole of electrons, or a mole of molecules. Each time we use the term, we refer to a number of particles equal to the number of atoms in exactly 12 grams of the 12C isotope of carbon.
Molecular Mass
What is molecular mass?
Molecular mass or formula mass or molecular weight of any compound is the summation of the atomic masses of each element in the compound.
Example1
What is the molecular mass of NaOH ?
Answer
As you know from the structure that NaOH consist of three elements which are 23Na and 16O and 1H.
From the definition of the molecular mass of any compound we now know that the molecular mass of any compound is the summation of the atomic mass of each element in the compound. That mean
Molecular mass of NaOH = (1 x 23) + (1x 16) + (1x 1) = 40
Example 2
What is the molecular mass of ascorbic acid C6H8O6?
Answer:
Molecular mass of C6H8O8 = (6 x 12) + (8 x 1) + (6 x 16) = 176
Example 3
What is the molecular mass of Ca3(PO4)2?
Molecular mass of Ca3(PO4)2 = (3 x 40) + (2 x 31) + (8 x 16) = 310
Percentage Composition
We always need to know what the percentage of each element is in a given compound. This is very important to understand the exact composition of for example drugs, food, and many other things.
The Percentage Composition of a compound can be calculated from the relative atomic masses of each element present in that compound.
Let us see how we calculate the Percentage Composition of any compound.
1) Calculate the percentage composition of Na2CO3
Answer:
There are three steeps to answer this question
1) First of all you must calculate the molecular mass of the compound as you learned in the previous section.
Molecular mass of Na2CO3 = (2 x 23) + (1 x 12) + (3 x 16) = 106
2) Now to calculate the Percentage Composition of each element in Na2CO3 you
need to calculate the mass of each element in Na2CO3.
Mass of Na in Na2CO3 = (2 x 23) = 46
Mass of C in Na2CO3 = (1 x 12) = 12
Mass of O in Na2CO3 = (3 x 16) = 48
3) Now the percentage composition of Na2CO3 can be calculated by applying the
Following
%Na = 46/106 x 100 = 43%
%C = 12/106 x 100 = 11%
%O = 48/106 x 100 = 45%
Please note that the summation of the percentage of Na + C + O SOULD BE equal 100. But some times it is not exactly 100 because of the rounding off when we write the numbers.
Exercises
1) What is the percentage composition of C6H8O6
2) What is the percentage composition of Ca3(PO4)2
The Mole (mol)
A mole (also known as Avogadro's number) is the number that is used in making calculations involving atoms and molecules. it is difficult to calculate amounts in terms of numbers of atoms. So, we use the mole. A mole, (mol for short) is equal to 6.022 x 1023 atoms or molecules
One mol of any thing- molecule, atoms, ions, electrons always contains the same number of partials which is equal to 6.022 x 1023 Avogadro's Number.
One mol of any substance has a mass in gram equal to its molecular weight (if it is compound) or atomic mass if it is element.
Examples
a) 1 mol of water H2O contains 6.022 x 1023 molecules and has a mass of 18g
(1x2) + (1x16) = 18
b) 1 mol of carbon C contains 6.022 x 1023 atoms and has a mass of 12g
c) 1 mol of glucose C6H12O6 contains 6.022 x 1023 molecules and has a mass of 180g
Mol = Mass per gram / Molecular mass per gram
Example:
What is the mass of 1mol of Ammonia NH3? How many molecules are present in one mole of Ammonia NH3?
Answer
Mol = Mass per gram / Molecular mass per gram
1 = mass per g / (1x14) + (3 x 1)
mass per gram = 17g
By definition; any mol of any substance contain a constant number of atoms equal to 6.022 x 1023 Avogadro's number.
Molecular Formula and Empirical Formula
There are two kinds of chemical formulas which are Empirical formulas & Molecular formula.
Empirical Formula
Empirical formulas give the lowest whole number ratio of the atoms in a compound
Molecular Formula
The molecular formula gives the exact composition of atoms in a compound.
Look at the table and try to see the difference.
How we can create the empirical and molecular formulas from an experimental data?
Empirical formulas can be calculated using experimental data.
To understand how we can do that let us look at this example.
Example 1
A compound contains 69.58% Ba, 6.09% C and 24.32% O, calculate the empirical formula of this compound?
Solution
First step:
Assume that you have 100g of the compound. That mean you have 69.58 g (69.58 %) Ba, and 6.09 g C (6.09 %C), and 24.32g O (24.32% O).
Second step:
Now convert all of the grams to mol.(i.e. for each element)
Mol of Ba = weight of Ba/atomic mass of Ba → mol of Ba = 69.58/137.33 = 0.5067
mol of Ba
Mol of C = weight of C/atomic mass of C → mol of C = 6.09/12 = 0.5071 mol of C
Mol of O = weight of O/atomic mass of O → mol of O = 24.32/16 = 1.520 mol of O
Final step:
Now to find the empirical formula you need to divide each of the above numbers by the smaller number i.e. 0.5071 so
For Ba = 0.5067/0.5067 = 1
For C = 0.5071/0.5067 ≈ 1
For O = 1.520/0.5067 = 2.9998 ≈3
That mean we have one atom of Ba and one atom of C and 3 atoms of O. so the empirical formula is BaCO3
Please ask me if you did not understand the above example.
Now how we can create the molecular formula?
Well let us see another example.
Example 2
A compound of molar mass 56 contains 85.6 % C, and 14.4% H. What is its empirical formula? And what is its molecular formula?
Solution
1) Assume that you have 100g of the compound. That means you have 85.6g C (85.6
%) and 14.4 g H (14.4 %H).
2) Convert weight to mol
mol C = 85.6/12 = 7.1 mol of C
mol H = 14.4/1= 14.4 mol H
3) Divide each of the above numbers by the smallest number which is 7.1
For C = 7.1/7.1 = 1
For H = 14.4/7.1 = 2
Now the empirical formula is CH2
4) Now to know the molecular formula all what you need to do is to calculate the
Molecular mass of the compound CH2 in this example, and then to divide it by
the given molar mass which is 56 in this example.
molecular mass of CH2 = (1 X 12) + (2 X 1) = 14
56/14 = 4
so the empirical formula is 4 x CH2 which is C4H8
The empirical formula is CH2
The molecular formula is C4H8
Home work
1) A compound contains 11.2 % H, 88.8 % O, what is the empirical formula? and if
the molecular mass is 18 what is the molecular formula?
2) A compound of molar mass 270 contains 17.0% Na, 47.4 % S, and 35.6% O. what is the empirical and molecular formula of this compound?
Chemical Reaction & Chemical Equations
Chemical Reaction
A chemical reaction is the process by which one or more substances (reactants) are transformed into one or more new substances (products). Energy is released or is absorbed, but no loss in total molecular weight occurs.
To be understandable internationally; chemists are representing any chemical reaction by using chemical symbols. The general form of a chemical reaction takes the following form
A + B → C + D
In which A and be are the reactants. The singe + means react. That means A reacts with B. The sign → means to produce C and D (products). Remember that all substance before the → are reactants (starting materials) and the substances after the → are the products. In all chemical reactions some number may occurs in front of the reactant and the products (if the number is 1 we ignore it as it its understandable without being written), these numbers indicates the number of molecules (mol) that reacted or produced. These numbers are very important to satisfy the law of mass conservation. Other important figures that may appear in a chemical equation are the reaction conditions i.e. the conditions under which the chemical reaction was carried out for example, temperature, catalysts, etc. The reaction conditions are normally written on the → singe. In many cases the physical states (gas, solid liquid) of the reactant and products are indicated in the chemical equation. Normally abbreviations are used (gas = g, solid = s, liquids = l) The chemical equation can be written as the following example.
CH4(g) + 2O2 (g) → CO2(g) + 2H2O(g)
We read equation 2 as the following. One molecule (or one mol) of methane gas has reacted with 2 moles of oxygen gas to produce 1 mol of carbon dioxide gas and two moles of water gas (steam).
Please go back to your text book to see other examples.
Balancing Chemical Equations
Balancing chemical equations is absolutely essential if you want to determine quantities of reactants or products. An unbalanced Chemical Equation gives only the identify of the beginning reactants and the final products using the appropriate formulas as well as the conditions of temperature, physical state, and pressure conditions under which the reaction is to operate under. However an unbalanced equation can say nothing about the quantities involved until the equation has been balanced. A balanced equation assures that the Conservation Law of matter is obeyed. The total mass of reactants must equal the total mass of products.
The following principles should be employed when balancing a Chemical Equation by inspection:
1) Never touch subscripts when balancing equations since that will change the
composition and therefore the substance itself. For example (3 CO2 never touch the
2 you only can play with the number 3 to balance the equation)
2) Check to be sure that you have included all sources of a particular element that you
are balancing on a particular side since there may be two or more compounds that
contain the same element on a given side of an equation.
3) I would suggest that you adjust the coefficient of mono atomic elements near the
end of the balancing act since any change in their coefficient will not affect the
balance of other elements
4) When there are a group of atoms that are acting as a unit such as a polyatomic ion
and they appear intact on both sides of the equation, it is best to balance them as a
self contaned group. For example, if there are Phosphate groups, PO4-3, that appear
on both sides, balance the phosphates as a group instead of separating the
Phosphorus and Oxygens. It can be done either way, but there is less likely of a
mistake if they are balanced as Phosphate groups.
Example 1:
(NH4)2CO3 → NH3 + CO2 + H2O
The first thing to do is choose a starting point. If we choose the carbon as the starting point, we go nowhere, since there is already one carbon on each side. Choosing oxygen is not good, because the oxygen is divided between two species on the right side. The interesting part is in the ammonium (NH4) ion. If we start with the nitrogen on the left (in the ammonium ion), we see that there are two nitrogens and therefore we must put at "2" in front of the ammonia on the right side.
So the equation at this stage becomes like
(NH4)2CO3 → 2 NH3 + CO2 + H2O
Since we are already working with the ammonia, now consider the hydrogens that make up the rest of the molecule: the six hydrogens in the ammonia plus the two hydrogens in the water make eight hydrogens. Bouncing back across to the left side, we count eight hydrogens; since no other atomic species have been changed, we can count up everything and see that the reaction is now balanced:
2 N, 8 H, 1 C, and 3 O on each side
Notice how the reaction itself led us through the steps of the balancing process. A more complex reaction would have involved more steps, but the process would have been the same.
Example 2:
Fe2(SO4)3 + K(SCN) → K3Fe(SCN)6 + K2SO4
Look at K3Fe(SCN)6 it is pretty complicated. Now look at it and see what ions are in it that are not shared on the same side of the reaction. These are SCN and Fe.
We start to balance these
Fe2(SO4)3 + 6 K(SCN) → K3Fe(SCN)6 + K2SO4
Now we go left to right. The irons are messed up, two on the Left, and one on the right. We adjust
Fe2(SO4)3 + 6 K(SCN) → 2 K3Fe(SCN)6 + K2SO4
Now this messes up the SCN, we adjust.
Fe2(SO4)3 + 12 K(SCN) → 2 K3Fe(SCN)6 + K2SO4
Now we go back to the left and work right. The irons are OK, the sulfates are not. Three on the left, one on the right. We adjust
Fe2(SO4)3 + 12 K(SCN) → 2 K3Fe(SCN)6 + 3 K2SO4
Now we go back left to right. Fe is ok, SO4 is OK, K is ok, and so forth, this is the final balanced equation.
Exercises
Balance the following reaction
1) Cu + O2 → Cu2O
2) CaCl2 + AgNO3 → AgCl + Ca(NO3)2
3) Mg + P4 → Mg3P2
4) H2SO4 + NaOH → H2O + Na2SO4
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Amounts of reactants and products
Please read this section form your text book. It is better to understand this section by solving problem
Problem 1:
Given the following chemical reaction between aluminium sulfide and water, if we are given 65.80 g of Al2S3:
a) How many moles of water are required for the reaction?
b) What mass of H2S & Al(OH)3 would be formed?
solution
Limiting Reactant
In a chemical reaction where some amounts of reactants are mixed and allowed to react, the one that is used up first is the limiting reactant. A portion of the other reactants remains. In other words the limiting reactant is the substances that present in a lees amount with respect to other reactants.
There is a systematic procedure for finding the limiting reagent based on the reactant ratio (RR) defined as the ratio of the number of moles of a reactant to its coefficient in a balanced chemical equation. The reagent with the smallest reactant ratio is the limiting reactant.
For a Reaction of the Form:
aA + bB → cC + dD
If the Reactant Ratios of all reactants are equal, then there is no limiting reactant, and all reactants will be consumed.
If all reactant ratios are not equal, then the reactant with the smallest reactant ratio is the limiting reactant.
To proceed, first calculate the reactant ratios for all of the reactants:
From among these, choose (RR)min, the smallest reactant ratio. This identifies the limiting reactant.
Once (RR)min is identified, it can be used to calculate the actual moles of other reactants and products actually consumed, e.g.
Actual moles of B consumed = (RR)min b
Actual moles of C produced = (RR)min c
Theoretical, Actual, and Percent Yields
Theoretical yield: The amount of product indicated by the stoichiometrically equivalent molar ratio in the balanced equation.
Side Reactions: These form different products that take away from the theoretical yield of the main product.
Actual yield: The amount of product that is actually obtained.
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