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- Concepts:
- subscript, empirical formula, empirical weight, molecular formula,
structural formula
- Skills:
- Identify a given formula as empirical or molecular
- Given masses or percentage mass composition of elements or masses of
combustion products, determine the empirical formula
- Given empirical formula and molecular weight or formula weight of the
compound, determine the molecular formula
- Given a structural formula determine the molecular formula
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- Molecular Formula: formula that indicates the actual number (actual
ratio) and types of atoms that make up the molecule (in the case of a
covalent compound) or ions that make up the formula unit (in the case of
an ionic compound)
- Empirical formula: gives only the relative number (smallest whole number
ratio) of atoms
- Structural formula: formula that shows groups of atoms in a molecule
that are bonded together.
- For example: CH3COOH which indicates that the first C is
bonded to 3 H’s, the second C to 2 O’s etc. (the molecular formula
being C2H4O2)
- CH3CH2CH2OH shows that the first C is
bonded to 3 H’s the second C to 2 H’s, and the third C to 2 H’s and 1 O
(the molecular formula being C3H8O).
- Structural formula becomes important when you start looking at molecular
geometry and also when studying organic chemistry, and therefore we will
not look at it in great details.
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- Molecular formula Empirical
formula
- N2O4 NO2
- P4O6 P2O3
- C2H4 CH2
- Mn2O3 Mn2O3
- C6H12O6 CH2O
- H2O H2O
- CO2 CO2
- HNO3 HNO3
- (NH4)2SO4 (NH4)2SO4
- Fe2O3 Fe2O3
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- If all one knew about a compound was its empirical formula then the
molecular formula cannot be determined. The form the molecular formula
would take is the best one can do.
- Empirical formula Form of
molecular formula in the absence of other info
- NO2 NxO2x (could be one of NO2 ,
N2O4 , N3O6 etc.)
- P2O3 P2xO3x (could be
one of P2O3 , P4O6 , P6O9
etc.)
- CH2 CxH2x
- Mn2O3 Mn2xO3x
- CH2O CxH2xOx
- H2O H2xOx
- CO2 CxO2x
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- Since the subscripts in a formula (any formula) such as N2O4,
C2H6, Fe2O3 indicate the
ratio in which the elements are combined to form the compound,
determination of a formula is just a matter of determining that ratio.
- A formula such as N2O4 tells us that a molecule
of dinitrogen tetroxide contains 2 atoms of nitrogen and 4 of oxygen.
- Or, a pair of dinitrogen tetroxide molecules contains 2 pairs of
nitrogen atoms and 4 pairs of oxygen atoms.
- Or, 1 dozen molecules of dinitrogen tetroxide contains 2 dozen nitrogen
atoms and 4 dozen oxygen atoms.
- Or, 1 mole of dinitrogen tetroxide contains 2 moles of nitrogen atoms
and 4 of oxygen atoms.
- Similarly, one mole of C2H6 contains 2 moles of
carbon and 6 of hydrogen atoms.
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- There are other ways of expressing the relative amount of components of
the fundamental particles that a chemical substance is made up of.
- Let’s look at N2O4 again.
- A molecule of dinitrogen tetroxide which has a RMM of 92.02 is made up
of 28.02 nitrogen and 64.00 of oxygen.
- Or, 92.02 g of N2O4 is made up of 28.02 g
nitrogen (element NOT nitrogen gas N2) and 64.00 g of oxygen
(element NOT oxygen gas, O2)
- Or, 30.4% nitrogen and 69.6% oxygen by mass.
- Notice that, the atomic composition, mole composition, mass composition,
and % mass composition are all interconvertible.
- So, the determination of empirical formula from available elemental
analysis data (% mass composition or actual mass composition or
combustion mass data) involves the determination of the molar ratio of
the elements in the compound.
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- Find the mass percentage of its elements for the following compounds:
- (a) NH3 (b) C2H4O2
- (c) K2SO4 (d) Ca(OH)2
- (e) Zn(NO3)2·6H2O (f) Streptomycin, C22H20O12N7
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- a) From (weight/weight)% composition of the compound:
- Elemental analysis of a compound of mercury and chlorine showed the
composition to be 73.9% mercury and 26.1% chlorine by mass. What is the
empirical formula of the compound?
- Step 1. Convert the percentage composition to mass ratio.
- If the composition is 73.9% mercury and 26.1% chlorine by mass, then a
100-gram sample of the compound would contain 73.9 g mercury and 26.1 g
chlorine.
- Therefore, mass of Hg: mass of Cl = 73.9:26.1
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- Step 2. Convert the mass ratio to mole ratio by using molar masses.
- Convert the masses into moles.
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- Step 3. Convert the fractional absolute mole ratio to relative whole
number ratio, since a formula must have whole number ratios.
- That is, pick one element and determine how many moles of each of the
other elements combines with 1 mole of that element.
- Since, we are interested in finding the whole number ratio, it would
make sense to pick as the first element the one with the smaller (or
smallest) number of moles in the sample analyzed, which in this case is
mercury.
- To get one mole of mercury, all that needs to be done is divide 0.3684
by 0.3684 itself.
- Relative # of moles of Hg =
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- Notice how when the final results are finally rounded off to the
appropriate number of significant figures, they yield whole numbers.
- Empirical formula of the mercury compound is HgCl2 (with the
molecular formula being of the form HgxCl2x.)
- Should one or more of the relative ratios come out not to be a whole
number, then the ratios should be multiplied by the smallest whole
number to covert it (or them) to one.
- If one of them for example is 1.33 then multiplying throughout by 3
will result in this one being equal to 4.
- Similarly, if one of ratios were 1.754, the ratio must be multiplied by
4, while 1.50 would have to be multiplied by 2.
- However, a number such as 1.12 can be/should be rounded off to 1 and
1.88 can be/should be rounded off to 2.
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- Step 1. Convert the percentage composition to mass ratio.
- Step 2. Convert the mass ratio to mole ratio by using molar masses.
- Step 3. Convert the fractional ratio to relative whole number ratios.
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- b) From mass composition of a sample of compound:
- Elemental analysis of 3.51 g of compound X is found to contain only 3.00
g of carbon and 0.51 g of H. What is the empirical formula?
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- In the absence of any other information, if only percentage composition
or mass composition is given, then empirical formula is all that can be
determined for an unknown compound.
- Molecular mass (or molar mass, boils down to the same thing really) must
be given in order to determine the molecular formula and therefore the
identity of the compound.
- For the two compounds in the examples, it has already been noted that
the formula would be of the generic form HgxCl2x
and CxH2x.
- What that means is that the molar mass of the compounds will be a
multiple (x) of the empirical formula.
- That is, x (mass of empirical
formula) = mass of the molecular formula
- [mass = relative mass or molar mass, depending on what is given]
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- Given that the molar mass of the mercury compound in the first example
is 271.5 g, and that of the hydrocarbon in the second is 56.07 g,
determine “x” and then the molecular formula of the two compounds.
- Mercury compound:
- x (molar mass of empirical formula) = molar mass of the compound
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- x (molar mass of empirical formula) = molar mass of the compound
- Þ x (14.03 g) = 56.07 g
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- c) From masses of reactants and products:
- A 1.367 g sample of an organic compound was combusted in a stream of
air (in a setup similar to the one below) to yield 3.002 g of CO2
and 1.640 g of H2O. If the original compound contained only
C, H, and O, what is its empirical formula?
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- Step 1. Determine mass of an element in the compound using the given
mass data for the products.
- Moles and mass of carbon in CO2 (and therefore in the
compound):
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- Mass of O in the compound = mass of compound - (mass of C + mass of H)
- = 1.367 g -
(0.8192 + 0.1838) g = 0.364 g
- Moles of O:
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- Moles of C = 0.068211; moles of H = 0.18201; moles of O = 0.02275
- Since O is the smallest, the number of moles of C and H for every mole
of O is:
- C
H O
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- 1. From mass of products find corresponding mass of element(s) that
originated from the sample analyzed using molar ratio in the formula of
the product.
- Mass of carbon and hydrogen in the example.
- 2. From mass of element(s) in sample determined from quantities of
product, determine mass of remaining element in the sample analyzed.
- Mass of oxygen in the example.
- 3. Convert grams of the constituent element in the analyzed sample into
moles using molar mass.
- 4. Convert fractional molar ratio into relative whole number ratio.
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- Hydrated salts lose their water of crystallization when heated.
- CuSO4·5H2O(s)
à CuSO4 (s)
+ 5H2O(g)
- The number of moles of water combined with each mole of the anhydrous
salt can be determined by gravimetric analysis.
- It involves heating a known mass of the original hydrated sample,
driving off all the water by repeated heating and measuring the mass
until there is no change in the mass of the anhydrous residue left
behind.
- The residue of course will be the anhydrous salt, and the difference in
mass between the original hydrated salt and that of the residue will
give the mass of the water in the sample.
- Example: A 0.520 g of NiSO4·xH2O after a couple of cycles of heating, cooling
and measuring the mass gave a residue of 0.306 g. Determine the formula
of the salt (ie determine x).
- NiSO4·xH2O(s)
à NiSO4 (s)
+ xH2O(g)
- 0.520 g 0.306 g
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- Notice that 1 mol of the hydrated salt produces 1 mol of the anhydrous
salt as well.
- In other words, molar mass equivalent grams of NiSO4·xH2O will produce
154.77 g of NiSO4.
- Let’s therefore, first find the moles of NiSO4 involved here.
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- Mass of water in 1 mole of NiSO4·xH2O = molar mass of NiSO4·xH2O - molar mass of NiSO4
(since molar ratio of NiSO4·xH2O: NiSO4= 1:1)
- = 263 - 154.77 = 108 g
- Number of moles of water
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- 1. MS03/1. Which compound has the empirical formula with the greatest
mass?
- A. C2H6 B.
C4H10 C. C5H10 D.
C6H6
- 2. NS03/3. A hydrocarbon contains 90% by mass of carbon. What is its
empirical formula?
- A. CH2 B. C3H4 C.
C7H10 D. C9H10
- 3. MS04/2. The percentage by mass of the elements in a compound is
- C = 72%, H = 12 %, O = 16%.
- What is the mole ratio of C : H in the empirical formula of this
compound?
- 1 : 1 B. 1 : 2 C. 1 : 6 D. 6 : 1
- 4. NS04/1. Which of the following compounds has/have the empirical
formula CH2O?
- I. CH3COOH
- II. C6H12O6
- III. C12H22O11
- A. II only B. III only C. I and II only D. II and III only
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- 5. MS05/1. Which is a correct definition of the term empirical formula?
- A. formula showing the numbers of atoms present in a compound
- B. formula showing the numbers of elements present in a compound
- C. formula showing the actual numbers of atoms of each element in a
compound
- D. formula showing the simplest ratio of numbers of atoms of each
element in a compound
- 6. A hydrocarbon contains 81.8% by mass of carbon. What is the empirical formula of this
hydrocarbon?
- A. C3H8 B. C2H3 C. CH3 D. CH2
- 7. Which of the following pairs represent(s) an empirical formula
followed by a molecular formula?
- I. O2 and O3 II. NO2 and N2O4 III. C2H4 and C4H8
- A. II only B. I and II only C. II and III only D. I, II and III only
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- 8. Elemental analysis of a
compound showed that it consisted of 54.53% carbon, 9.15% hydrogen and
36.32% oxygen. How many hydrogens appear in the empirical formula of the
compound?
- Use the following information to answer questions 10 and 11.
- Dihydroxytartaric acid contains 26.38% carbon, 3.32% hydrogen, and
70.29% oxygen.
- 9. The number of hydrogen atoms
in the simplest formula for dihydroxytartaric acid is
- 10. If the molecular formula of
dihydroxytartaric acid is twice the empirical formula, the molar mass,
in g mol-1, would be
- A. 91 B. 122 C. 148 D. 182
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- 1. How many atoms of each kind are represented in the following
formulas?
- (a) Na2CO3 (b) (NH4)3PO4
- (c) Na3Ag(S2O3)2 (d) NH4NO3
- 2. How many moles of atoms of each kind are represented in a mole of
- Na2CO3 (b) (NH4)3PO4
- (c) Na3Ag(S2O3)2
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- 3. Determine the total number of moles of atoms in 2.00 g each of
- (a) Na2CO3
- (b) (NH4)3PO4
- (c) Na3Ag(S2O3)2
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- 1. MS03/A3. The relative molecular mass of aluminium chloride is 267 and
its composition by mass is 20.3% Al and 79.7% chlorine.
Determine the empirical and molecular formulas of aluminium chloride.
[4]
- 2. NS04/1. An oxide of copper was reduced in a stream of hydrogen as
shown below.
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- After heating, the stream of hydrogen gas was maintained until the
apparatus had cooled.
- The following results were obtained.
- Mass of empty dish = 13.80 g
- Mass of dish and contents before heating = 21.75 g
- Mass of dish and contents after heating and leaving to cool = 20.15 g
- (a) Explain why the stream of hydrogen gas was maintained until the
apparatus cooled. [1]
- (b) Calculate the empirical formula of the oxide of copper using the
data above, assuming complete reduction of the oxide. [3]
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- (c) Write an equation for the reaction that occurred. [1]
- (d) State two changes that would be observed inside the tube as it was
heated. [2]
- 3. MS05/A2. The percentage composition by mass of a hydrocarbon is C =
85.6 % and H = 14.4 %.
- (a) Calculate the empirical formula of the hydrocarbon. [2]
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- (b) A 1.00 g sample of the hydrocarbon at a temperature of 273 K and a
pressure of 1.01 × 105 Pa (1.00 atm) has a volume of 0.399
dm3.
- (i) Calculate the molar mass of the hydrocarbon. [2]
- (ii) Deduce the molecular formula of the hydrocarbon. [1]
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- 4. A sample of organic compound containing C, H, and O, which weighs
12.13 mg, gives 30.6 mg of CO2 and 5.36 mg of H2O
when combusted. The amount of oxygen in the original sample is obtained
by difference. Determine the empirical formula of this compound.
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- 5. Weighed samples of the following hydrates are heated to drive off the
water, and then the cooled residues are weighed. From the data given,
find the formulas of the hydrates:
- (a) 0.895 g of MnI2·xH2O gave a residue of 0.726 g
- (b) 0.654 g of MgSO4·xH2O gave a residue of 0.320 g
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- (c) 1.216 g of CdSO4·xH2O gave a residue of 0.988 g
- (d) 0.783 g of KAl(SO4)2·xH2O gave a residue of 0.426 g
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Notes
