Chemistry SS 1 Week 3
Topic: Mole Concept in Terms of Masses Numbers, Volumes of Reactants and Products
Molar Mass
The molar mass of any molecular compound is the mass in grams numerically equivalent to the sum of the atomic masses of the atoms in the molecular formula.
The mole is just a number; it can be used for atoms, molecules, ions, electrons, or anything else we wish to refer to. Because we know the formula of water is H2O, for example, then we can say one mole of water molecules contains one mole of oxygen atoms and two moles of hydrogen atoms. One mole of hydrogen atoms has a mass of 1.008 g and 1 mol of oxygen atoms has a mass of 16.00 g, so 1 mol of water has a mass of (2 x 1.008 g) + 16.00 g = 18.02 g. The molar mass of water is 18.02 g/mol.
Example 1: The formula mass of methane, CH4, calculates its molarmass.
Solution: C + H x 4
12.01 + (1.008 x 4) = 16.04 u and its molar mass is 16.04 g/mol.
Example 2: What is the molar mass of glucose, C6H12O6?
Solution:
| element | atomic mass | mass formula | Contribution | ||
| C | 12.01 g/mol | x | 6 | = | 72.06 g/mol |
| H | 1.008 g/mol | x | 12 | = | 12.10 g/mol |
| O | 16.00 g/mol | x | 6 | = | 96.00 g/mol |
| total = 180.16 g/mol |
The molar mass of ionic compounds can be calculated similarly, by adding together the atomic masses of all atoms in the formula to get the formula mass, and expressing the answer in units of g/mol.
Example 3: Calcium chloride, CaCl2, has a molar mass of?
Solution: Ca + Cl x 2
40.08 + (2 x 35.45) = 110.98 g/mol.
Example 4: How many moles are there in 0.326 g of barium hydroxide?
Solution:
Molarmass of Ba(OH)2 = 171
Mass of Ba(OH)2 = 0.326g
Number of moles = 0.326/171 = 0.00191 mol
The Molar Volume
The volume occupied by one mole of a substance is called the molar volume for that substance. It may be easily calculated from the density of the substance:
Density = Mass/Volume
If one works in moles, this becomes
Density = Molar mass /Molar volume
The molar volume is of particular interest when dealing with gases. Since the density of a gas is greatly dependent on the temperature and pressure, the density should be measured at the standard temperature and pressure (STP), which is a temperature of 273K (0ºC) and pressure of 1.0 atm or 760 mm of Hg or 760 torr or 76 cm of Hg.
It turns out that the molar volume (measured at STP) is nearly the same for all gases, and has the value of 22.4dm3
To put it in another way, one mole of any gas at STP will occupy a volume of 22.4 dm3
Percentage Composition of Compounds
Assume 1 mole of the compound
Percent = mass of the element / molar mass of compound x 100
Using the molar masses of the elements and the compound, we can express the composition in terms of mass percentage of the elements. For example, carbon dioxide has a formula weight of 44.01 u, made up of 12.01 u for the average mass of 1 carbon atom and 32.00 u for 2 oxygen atoms. One mole of CO2 has a mass of 44.01 g made up of 12.01 g of carbon and 32.00 g of oxygen. The composition of carbon dioxide is calculated as follows:
12.01g carbon / 44.00 carbon dioxide x 100 = 27.30% carbon
32.00g oxygen / 44.00g carbon dioxide x 100 = 72.70% oxygen
The formula for a compound, and its composition expressed as percentage by mass, are fixed and unchanging properties of the compound. Any pure sample of carbon dioxide is 72.70 % oxygen by mass.
This property allows us to relate the amount of an element in a compound to the mass of the compound.
Example 1: Calculate the number of moles of iron in 2.98 g of iron (III) oxide.
Solution:
First we need the formula for iron (III) oxide. Since iron (III) is Fe3+ and oxide is O2- the formula must be Fe2O3. Calculate the molar mass as (2 x 55.85) + (3 x 16.00) = 111.7 + 48.00 = 159.7 g/mol.
The percentage by mass of iron in iron (III) oxide can then be calculated
= 111.7 g iron / 159.7 g iron (III) oxide x 100 = 69.94% iron
This percentage by mass can be used to calculate the mass of iron in the sample
= 2.98 g iron (III) oxide x 69.94/100 = 2.08g iron
Finally, the mass of iron can be converted to an amount in moles using the molar mass
= 2.08 g iron / 55.85 g/mol = 3.73 x 10-2 mol
The mole of iron in 2.98 g of Fe2O3 is 3.73 x 10-2 mol.
Example 2: Calculate the average atomic weight of silicon having 92.2% of Si-28 isotope of relative mass 27.98 amu, 4.7% of Si-29 isotope of relative mass 28.98 amu and 3.1% of Si-30 isotope of relative mass 29.97 amu.
Solution:
Data given: 92.20% of silicon of mass 27.98 amu
4.7% of silicon of mass 28.98 amu
3.1 % of silicon of mass 29.97 amu
Average atomic weight
= 25.80 + 1.36 + 0.93
= 28.09 amu
Average atomic weight of silicon is 28.09 amu
Example 3: Calculate the % of copper in copper sulphate, CuSO4
Solution:
Relative atomic masses: Cu = 64, S = 32 and O = 16
Relative formula mass = 64 + 32 + (4×16) = 160
Only one copper atom of relative atomic mass 64
% Cu = 64 x 100 / 160 = 40% copper by mass in the compound
To calculate the % of the other elements in the compound
% sulfur = (32/160) x 100 = 20% S
% oxygen = (64/160) x 100 = 40% O
Also note that if you haven’t made any errors, they should add up to 100% !
Example 4: Calculate the % of oxygen in aluminium sulphate, Al2(SO4)3
Solution:
Relative atomic masses: Al = 27, S = 32 and O = 16
Relative formula mass = 2×27 + 3(32 + 4×16) = 342
Giving a total mass of oxygen in the formula of 12 x 16 = 192
% O = 192 x 100 / 342 = 56.1% oxygen by mass in aluminium sulphate
Example 5: Calculate the % of water in hydrated magnesium sulphate MgSO4.7H2O
Solution:
Relative atomic masses: Mg = 24, S = 32, O = 16 and H = 1
Relative formula mass = 24 + 32 + (4 x 16) + [fusion_builder_container hundred_percent=”yes” overflow=”visible”][fusion_builder_row][fusion_builder_column type=”1_1″ background_position=”left top” background_color=”” border_size=”” border_color=”” border_style=”solid” spacing=”yes” background_image=”” background_repeat=”no-repeat” padding=”” margin_top=”0px” margin_bottom=”0px” class=”” id=”” animation_type=”” animation_speed=”0.3″ animation_direction=”left” hide_on_mobile=”no” center_content=”no” min_height=”none”][7 x (1 + 1 + 16)] = 246
7 x 18 = 126 is the mass of water
So % water = 126 x 100 / 246 = 51.2 % H2O
Empirical and Molecular Formulae
Molecular Formula is a formula indicating the actual number of atoms of each element making up a molecule. The molecular formula must accurately state the exact number of atoms of all of the elements in one molecule of the substance.
Empirical formula is the formula giving the simplest ratio between the atoms of the elements present in a compound. You must find the ratio of atom to atom in a molecule, and then reduce it.
Facts on Empirical and Molecular Formulae
- Empirical Formula of a compound shows the ratio of elements present in a compound.
- Molecular Formula of a compound shows how many atoms of each element are present in a molecule of the compound.
- The empirical formula mass of a compound refers to the sum of the atomic masses of the elements present in the empirical formula.
- The Molecular Mass (formula mass, formula weight or molecular weight) of a compound is a multiple of the empirical formula mass.
MM = n x empirical formula mass - Empirical Formula can be calculated from the percentage (or percent) composition of a compound.
Creating an Empirical Formula From Mass or Percent Composition
From Percent Composition:
- Step 1: Create a chart with six columns and a number of rows equal to the number of elements in the compound.
- Step 2: Write the elements in the first column.
- Step 3: Write the percent composition of each element in the second column.
- Step 4: Using the percent composition as the mass, divide each by the molecular mass of the respective element.
- Step 5: Divide each of those numbers by the smallest of the numbers in that column to reduce the ratio. If one or more numbers in the ratio is still distant from a whole number, try multiplying the entire ratio by a whole number. If the number is close, then round it to the nearest whole number. Each one of the numbers is the subscript for the corresponding element.
Example 1: Determine the empirical formula for a compound containing 74.0% carbon, 8.65% hydrogen, and 17.3% nitrogen by mass.
| C | 74.0% | 74.0 g / 12.0 g | 6.16 / 1.24 | 4.96 | 5 |
| H | 8.65% | 8.65 g / 1.01 g | 8.56 / 1.24 | 6.90 | 7 |
| N | 17.3% | 17.3 g / 14.0 g | 1.24 / 1.24 | 1 | 1 |
Therefore the empirical formula is C5H7N
From Mass:
Follow the same procedure as fir Percent Composition but first divide the mass of each element by the mass of the total sample. The quotient is the percent composition.
Example 2: 9.2 g sample of a compound is 2.8 g of nitrogen and 6.4 g of oxygen. Find the empirical formula of the compound.
| N | 2.8 g / 9.2 g | 30% | 30 g / 14 g | 2.143 g / 2.143 g | 1 | 1 |
| O | 6.4 g / 9.2 g | 70% | 70 g / 16 g | 4.375 g / 2.143 g | 2.04 | 2 |
- Therefore, the empirical formula is NO2.
Examples 3: If carbon and hydrogen are present in a compound in a ratio of 1:2 and the molecular mass of the compound is 28.0 g/mol.
Solution: The empirical formula for the compound is CH2
The empirical formula mass of this compound is: 12.0 + (2 x 1.0) = 14.0 g/mol
Since the molecular mass of the compound is 28.0 g/mol then we can find the molecular formula for the compound.
MM = n x empirical formula mass
28.0 = n x 14.0
n = 2
So the molecular formula for the compound is 2 x empirical formula, i.e., 2 x (CH2) which is C2H4 – Ethene
Example 4: A compound containing 92.3 weight percent of carbon and 7.7 weight percent of H. What is the empirical formula?
Solution:
Assume that you have 100 g of the compound, then you have 92.3 g of carbon and 7.7 g of hydrogen. Thus the mole ratio of C to H should be
C H
92.3/12 7.7/1.08
7.69/7.13 7.13/7.13
1.07 1.0
Thus, the empirical formula is CH.
A compound with an empirical formula of CH has a molecular weight of 78 g/mol. What is the molecular formula?
Solution:
The formula weight of CH is 13.0.
Since 78/13 = 6,
The molecular formula is C6H6, the formula for benzene.
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