The document discusses different types of stoichiometry calculations encountered in AS Organic Chemistry, including: 1) Percentage yield calculations from chemical reactions. 2) Determining empirical formulas from elemental analysis data involving masses, percentages, or combustion analysis. 3) Calculating molecular formulas using empirical formulas and experimentally determined molar masses.

1. Stoichiometry
AS Organic Chemistry
Mr.Scobie

2. Stoichiometry: What is it?
• Includes all the quantitative relationships in chemical
reactions.
What type will I encounter in AS Organic Chemistry?
• Percentage yield calculations
• Empirical formula from: - masses
- percentage composition
- elemental analysis data
(combustion analysis)
• Molecular formula

3. Working out Percentage Yield
• Organic preparations = rarely give 100% yield
• Often get yields of 60% or less!
Why?
• Occurrence of other reactions (& intermediates)
• Reversible reactions leading to equilibrium
• Losses involved in separation

4. Percentage Yield…
Remember three formulae:
• % yield = actual mass (yield) of purified product x 100
theoretical yield (mass)
• % purity = actual mass x 100
mass of impure sample
• NO. Moles used = Mass used .
relative molecular mass (Mr)

5. Worked example….
• When 15.0g of butan-1-ol (A) and 10.0g of ethanoic acid (B)
were refluxed with conc. H2SO4, 17.8g of 1-butylethanoate (C)
were formed.
Butan-1-ol + Ethanoic acid ------- 1-butylethanoate + Water
Calculate the percentage yield.

6. Answer….
When 15.0g of butan-1-ol (A) and 10.0g of ethanoic acid (B) were refluxed
with conc. H2 SO4, 17.8g of 1-butylethanoate (C) were formed.
1. Work out: Molecular Mass and NO. of Moles used in reaction
Mr of A = 74 15.0 / 74 = 0.203
Mr of B = 60 10.0 / 60 = 0.167
2. Look at equation:
tells us that 1 mole of (A) reacts with 1 mole of (B) to form 1 mole of (C).
3. Determine which one is the controlling factor: (A) is in excess
[larger amt.] & hence it is (B) which will determine amt. of product
(C).
Thus: 0.167 moles of (B) gives 0.167 moles of (C).
4. Therefore theoretical mass of (C) = 0.167 x Mr
= 0.167 x 116
= 19.4g
5. % Yield = 17.8g x 100
19.4g
= 91.8 %

7. Now Try These…
1. When lithium chlorate (VII) is strongly heated, oxygen gas and lithium
chloride are formed:
LiClO4 (s) LiCl (s) + 2O2 (g)
If 1.06g lithium chlorate (VII) was heated and produced 0.40g lithium
chloride. What is the yield of lithium chloride ?
2. 10g of 1-bromobutane was hydrolysed and 3.9g of butan-1-ol obtained.
Calculate the percentage yield of butan-1-ol.
C4H9Br + NaOH C4H6OH + NaBr
Remember to show all working !

8. Empirical Formula (from masses)
An 18.3g sample of a hydrated compound contained 4.0g Ca, 7.1g Cl and 7.2g of
water only. Calculate its empirical formula.
1. List the masses of each component and its molar mass.
Ca Cl H2O
Mass (g) 4.0 7.1 7.2
Molar Mass 40.0 35.5 18.0
2. Now calculate the amount of each substance present
Ca Cl H2O
Amount/mol 4.0/40.0 7.1/35.5 7.2/18.0
= 0.10 0.20 0.40
3. Calculate the relative amts. of each by dividing
each amt. by the smallest amt.
Amt/smallest 0.10/0.10 0.20/0.10 0.40/0.10
= 1.0 2.0 4.0
Ratio = 1:2:4
Thus formula =
CaCl2.4H2O

9. Now try these….
1. A sample of hydrated compound was analysed and found to
contain 2.10g of cobalt, 1.14g of sulphur, 2.28g of oxygen &
4.50g of water. Calculate its empirical formula.
2. 10.0g of hydrated barium chloride is heated until all the water
is driven off. The mass of anhydrous compound is 8.53g.
Determine the value of x in BaCl2.xH2O
3. When 585mg of the salt UO(C2O4).6H2O was left in a vacuum
desiccator for 48 hours, the mass changed to 535mg. What
formula would you predict for the resulting substance ?
Remember to show all working!

10. Empirical Formula (from % composition)
These are similar to calculations involving masses.
An organic compound was analysed and found to have 48.8% C, 13.5% H and
37.7% N. Calculate the empirical formula of the compound.
1. Assume 100g sample and convert percentages to grams.
2. As before set up and fill in your table
C H N
Mass (g) 48.8 13.5 37.7
Molar mass (g/mol) 12.0 1.00 14.0
Amount/mol 48.8/12 = 4.07 13.5 2.69
Amt/smallest 4.07/2.69 = 1.51 5.02 1.00
Simplest ratio 3 10 2
3. Use simplest ratio to write empirical formula = C3H10N2

11. Now try these ….
1. A compound of carbon, hydrogen and oxygen contains 40.0% C,
6.6% H and 53.4% O. Calculate its empirical formula.
2. Determine the formula of a mineral with the following mass
composition: Na = 12.1%, Al = 14.2%, Si = 22.1%, O = 42.1%, H2O
= 9.48%
3. A 10.0g sample of a compound contains 3.91g C, 0.87g H and the
remainder is oxygen. Calculate the empirical formula of the
compound.
Remember to show all working!

12. Empirical Formula from combustion
• Combustion analysis = burning a sample.
• A known mass of an organic compound is completely
burnt in excess oxygen.
• Excess O2 needed to avoid incomplete combustion.
• CO2 and H2O is produced
• Products collected and masses measured to determine
amounts present

13. Combustion example……
A 1.00g sample of hydrocarbon was burnt in excess oxygen. The CO2 and H2O
produced were absorbed in previously weighed tubes containing magnesium
chlorate (VII) & soda-lime. The tubes were found to have gained 3.83g of
CO2 & 0.692g of H2O. Determine the empirical formula of the compound.
1. Calculate the amount of C in the sample from the mass of CO2 absorbed.
Amt of CO2 = Mass / Molar mass = 3.38g / 44.0 = 0.0768 mol.
In 1 mol of CO2 there is 1 mol of C.
Thus the amount of C = amount of CO2 = 0.0768 mol.
2. Calculate amount of H from the water absorbed
Amt of H = Mass / Molar mass = 0.692g / 18.0 = 0.0384 mol.
In 1 mol of H2O there are 2 mol of H atoms.
Thus the amount of H = 2 x amount of H2O = 0.0768
3. Empirical formula = Ratio of C:H = C1H1 (i.e. CH.)

14. Another combustion one….
A 1.00g sample of compound A was burnt in excess oxygen producing 2.52g
of CO2 and 0.443g of H2O. Calculate the compounds empirical formula.
1. Amount of CO2 = Mass / Molar mass = 2.52g / 44.0 = 0.0573 mol
Thus, the amount of C = 0.0573 mol
2. Amount of H2O = 0.443g / 18.0 = 0.0246 mol
Thus, the amount of H = 2 x 0.0246 mol = 0.0492 mol
3. Now calculate the mass of carbon and mass of hydrogen
Mass of C = amt x molar mass = 0.0573 x 12.0 = 0.688g
Mass of H = amt x molar mass = 0.0492 x 1.0 = 0.0492g
4. The mass of Oxygen = Mass of sample – ( C + H) = 1.00g – 0.7372g
= 0.263g
5. Now use the table (as before) to work out ratios and formula

15. Example continued ….
5. Use the table to work out ratios:
C H O
Mass (g) 0.263
Molar mass 16.0
Amount per mol 0.0573 0.0492 0.0164
Amt / smallest 0.0573/0.0164
= 3.49 3.00 1.00
Simplest ratio 7 6 2
Empirical formula = C7H6O2

16. Now try these …
1. Complete combustion of a hydrocarbon, Z, gives 0.66g of
carbon dioxide and 0.154g of water. What is the empirical
formula of Z ?
2. The combustion of 0.146g of compound B gave 0.374g of carbon
dioxide and 0.154g of water. Assuming B contains carbon,
hydrogen and oxygen only, determine its empirical formula.
Remember to show all working!

17. Molecular Formulae
• Remember that the empirical formula = Simplest ratio
of atoms
• Molecular formula = Actual numbers of atoms present
• The molecular formula is thus a simple multiple of the
empirical formula.
• For example: the empirical formula CH2 can be used
to represent C2H4, C3H6, C4H8, etc.
• Molecular formulae are therefore more useful.

18. Molecular Formulae
To work out Molecular Formula, you need to know the
empirical and then use the equation:
NO. of empirical units in
molecular formula =
Mr found by experiment
Mr for empirical formula
Mr from experimental
data is often given to you
in the question.

19. Molecular Formula Example
The empirical formula of compound X is found from quantitative analysis to
be C2H5. Mass spectrometry gave a relative molecular mass of 58.
What is the molecular formula of X ?
1. Determine the Mr using the formula, C2H5.
Mr = (2 x 12) + (5 x 1) = 29
2. Use the equation and the information in the question.
Mr of compound (by experiment)
Mr from empirical (C2H5) =
58
29 = 2
3. Molecular formula = (C2H5) 2 = C4H10

20. Another example….
A polymer of empirical formula CH2 has a molar mass of 28 000g mol.
What is its molecular formula ?
Empirical formula mass = 12 + (1 x 2) = 14 g mol.
Molar mass = 28 000 g mol.
28 000 g mol
14 g mol = 2000
The molar mass is 2000 times the empirical
formula mass; therefore the molecular mass =
(CH2) 2000

21. Now try these….
1. An organic compound A has a relative molecular mass of 178
and contains 74.2% C, 7.9% H and 17.9% O. Determine:
a) the empirical formula of A
b) the molecular formula of A.
2. Substance Q has the elemental composition C = 60.0%, H = 13.3%,
O = 26.7% (by mass) and a relative molecular mass of 60.1.
a) Calculate the empirical formula of Q
b) What is its molecular formula ?
Remember take your time and show all working!