Calculate the empirical formula of vitamin C from the given data.

1. STOICHIOMETRY:
CALCULATIONS WITH
CHEMICAL FORMULAS
AND EQUATION
QBA Miguel A. Castro Ramírez

2. CHEMICAL EQUATIONS
• Chemical reactions are presented in a concise way by chemical
equation
• Example: Methane + Oxygen  Carbon Dioxide + Water
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Reactant
Product

3. CHEMICAL EQUATIONS

4. CHEMICAL EQUATIONS
Balancing Equation
• Balance the equation by determine the coefficients that provide
equal numbers of each type of atom on each side of the equation.
• Balance equation must contain the smallest possible whole
number coefficient.
• Need to understand the coefficient in front of the formula and
subscript inside the formula.
• Two steps in balancing equation:
1. Write the unbalanced equation
2. Adjust the coefficients to get the equal number of each
kind of atom on both sides of the arrow.

5. CHEMICAL EQUATIONS
Balancing Equation
• Example :
CH4 (g) + O2 (g) CO2 (g) + H2O (g) (unbalanced)
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) (balanced)
• Need to understand the coefficient in front of the formula and
subscript inside the formula. (3H2O)
• Never change subscripts when balancing an equation.
• Don’t forget to put the state of reactants and products in the
balanced equation. (gas-g : liquid-l : solid-s : aqueous(water)-aq)
• Δ : indicating heat

6. CHEMICAL EQUATIONS
Example :
Sodium hydroxide, NaOH and phosphoric acid H3PO4, react
as aqueous solutions to give sodium phosphate, Na3PO4 and
water. The sodium phosphate remains in solution. Write the
balanced equation for this reaction.

7. CHEMICAL EQUATIONS
EXAMPLE:
This molecular scene depicts an important reaction in nitrogen
chemistry (nitrogen is blue, oxygen is red). Write a balanced
equation.

8. CHEMICAL EQUATIONS
EXERCISE:
When aqueous solutions of calcium chloride, CaCl2 and potassium
phosphate K3PO4, are mixed, a reaction occurs in which solid
calcium phosphate, Ca3(PO4)2, separates from the solution. The
other product is KCI(aq). Write the balanced equation.

9. SOME SIMPLE PATTERNS OF CHEMICAL
REACTIVITY
Combination Reaction
• Combination reaction: 2 or more
substances react to form 1 product.
• Example: Magnesium metal burns
in air with a dazzling brilliance to
produce magnesium oxide.
• Mg(s) + O2 (g)  2MgO(s)
• Combination reaction between metal & non-metal : Produced
ionic solid.
• In the above reaction: Mg loses e- to form Mg2+ while oxygen gains
electron to produce O2-.

10. SOME SIMPLE PATTERNS OF CHEMICAL
REACTIVITY
Decomposition Reaction
• Decomposition Reaction: One substance
undergo a reaction to produce two or
more other substances.
• Heating plays an important roles for this
reaction

11. SOME SIMPLE PATTERNS OF CHEMICAL
REACTIVITY
Combustion in Air
• These are generally rapid reactions that
produce a flame.
• Most often involve hydrocarbons
reacting with oxygen in the air.
• The amount of O2 required in the
reaction and the number of CO2 and H2O
produced are depend on the composition
of the hydrocarbon.
• Example:
CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (g)
C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (g)

12. FORMULA WEIGHT
Formula and Molecular Weight
• Formula weight of a substance: Sum of the atomic weights of
each atom in its chemical formula.
Example: FW of H2SO4 = 2(AW of H) + (AW of S) + 4(AW of O)
2(1 amu) + 32.1 amu + 4(16 amu)
98.1 amu.
• Formula weights are generally reported for ionic
compounds.
• Molecular weight: Sum of the atomic weights of the atoms
in a molecule.
Example: MW of C12H22O11 = 12(AW of C) + 22(AW of H) +
11(AW of O)
= 342.0 amu

13. FORMULA WEIGHT
Percentage Composition from Formulas
• One can find the percentage of the mass of a compound that
comes from each of the elements in the compound by using this
equation
• Purpose of calculating: To verify the purity of a compound-
Compare the calculated percentage composition of the substance
with that found experimentally. Example: Forensic Chemist
•
(number of atoms)(atomic weight)
% element = x 100
(FW of the compound)

14. FORMULA WEIGHT
EXAMPLE
A sample of a liquid with a mass of 8.657 g was decomposed into
its elements and gave 5.217 g of carbon, 0.9620 g of hydrogen
and 2.478g of oxygen. What is the percentage composition of this
compound?

15. AVOGADRO’S NUMBER AND THE MOLE
•The unit for dealing with the number of atom , ions or molecules
in a common-sized sample is the mole, abbreviated mol.
• 1 mole of atoms is the Avogadro number of the atoms.
• No Avogadro: 6.02 x 1023 : Defined as wide variety of ways. But
it is simply understand as the conversion factor between amu and
gram.
• A mole of atoms, a mole of molecules or a mole of anything else
all contain Avogadro’s number of these objects:
1 mol 12C atoms = 6.02 x 1023 12C atoms
1 mol H2O molecules = 6.02 x 1023 H2O molecules

16. AVOGADRO’S NUMBER AND THE MOLE
Molar Mass
• Molar mass is the mass of 1 mol of a substance in gram
- Thus, the mass of a substance (atom, element, molecule
or ions) in amu is numerically equal to the mass in gram of 1 mol of
that particular substance.
1 atom of 12C has a mass of 12 amu - 1 mol 12C has a mass of 12g
1 NO3- ion has a mass of 62 amu – 1 mol NO3- has a mass of 62 g
1 NaCl unit has a mass of 58.5 amu – 1 mol of NaCl has a mass of
58.5 g

17. AVOGADRO’S NUMBER AND THE MOLE
Interconverting Masses, Moles and Numbers of Particles
• Frequently encountered in calculations using the moles concept.
Moles provide a bridge from the molecular scale to the
real-world scale.

18. AVOGADRO’S NUMBER AND THE MOLE
• One mole of atoms, ions, or molecules contains Avogadro’s
number of those particles.
• One mole of molecules or formula units contains Avogadro’s
number times the number of atoms or ions of each element in
the compound.

19. AVOGADRO’S NUMBER AND THE MOLE
EXAMPLE
1)Calculate the quantity in moles of carbon dioxide which
contains 1.505 X 1022 molecules.
2)What is number of atoms in 0.6 mole of Li.
EXAMPLE
1)Calculate the number of sodium ions and oxide ions in 0.4 mole
of sodium oxide.

20. AVOGADRO’S NUMBER AND THE MOLE
EXERCISE
For a 250g sample of SO2, determine
a)The number of moles of SO2 molecules present
b)The number of moles of the sulfur atoms and oxygen atoms
present
c)The number of SO2 molecules present
d)The number of sulfur and oxygen atoms present
e)The average mass of one SO2 molecule.

21. EMPIRICAL FORMULA FROM ANALYSES
Determining Empirical Formula
• Empirical formula : The relative number of atoms of each
element it contains.
EXAMPLE
1)Determine the empirical formula of the compound formed from
2.44 g of Ti and 3.60 g of Cl.
2)Determine the empirical formula of an alcohol that has a %
composition by mass of 60.0% C, 13.4% H and 26.6% O.
What is the name of the alcohol.

22. EMPIRICAL FORMULA FROM ANALYSES
Molecular Formula from Empirical Formula
• Formula obtained from percentage composition: Empirical
Formula
• Can obtain the molecular formula from the empirical formula if
we know the molecular weight.
• The subscripts in the molecular formula of a substance are always
a whole-number multiple of the corresponding subscripts in its
empirical formula.
• The multiple is found by comparing the formula weight of the
empirical formula with the molecular weight.

23. EMPIRICAL FORMULA FROM ANALYSES
EXAMPLE
Ascorbic acid (vitamin C), found in citrus fruits, tomatoes and
green vegetables is composed of 40.9% C, 4.6% H and 54.5% O.
Determine:
a) The empirical formula of ascorbic acid
b) The molecular formula if the experimentally determined
molar mass of an ascorbic acid molecule is 176.13 g/mole.

24. EMPIRICAL FORMULA FROM ANALYSES
EXERCISE
Elemental analysis of a sample of an ionic compound showed
2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. What is the empirical
formula and name of the compound?

25. EMPIRICAL FORMULA FROM ANALYSES
Combustion Analysis
• Empirical = Based on observation and experiment
• The empirical formula of a compound is based on experiments
that give the number of moles of each element in a sample.
• Compound containing carbon and hydrogen burn, the C will
converted into CO2 while the H will converted into H2O.
• Amounts of H2O and CO2 are determined by measuring the
increased volume of the absorber.

26. EMPIRICAL FORMULA FROM ANALYSES
• From mass, we can calculate the number of moles of each
element.
•3rd element’s mass : Can be determine by subtracting the masses
of C and H from the compound's original mass.

27. EMPIRICAL FORMULA FROM ANALYSES
EXAMPLE
Acetic acid contains only C,H and O. A 4.24 mg sample of acetic
acid is completely burned. It gives 6.21 mg of carbon dioxide
and 2.54 mg of water. What is the mass percentage of each
element in acetic acid with 3 sig. fig?

28. EMPIRICAL FORMULA FROM ANALYSES
EXERCISE
When a 1.000 g sample of vitamin C (M = 176.12 g/mol) is placed
in a combustion chamber and burned, the following data are
obtained:
mass of CO2 absorber after combustion = 85.35 g
mass of CO2 absorber before combustion = 83.85 g
mass of H2O absorber after combustion = 37.96 g
mass of H2O absorber before combustion = 37.55 g
What is the molecular formula of vitamin C?

29. QUANTITATIVE INFORMATION FROM
BALANCED EQUATION
• The coefficients in a chemical equation represent the relative
numbers of molecules in a reaction.
• Allow us to convert this information to the masses of the
substances.
• The coefficients in a balanced chemical equation indicate both
relative numbers of molecules (or formula units) in the reaction
and the relative numbers of moles

30. QUANTITATIVE INFORMATION FROM
BALANCED EQUATION
• The quantities: 2 mol H2, 1 mol O2 and 2 mol H2O :
Stoichiometrically equivalent quantities.
• The relationship can be represented as
2 mol H2 = 1 mol O2 = 2 mol H2O
• = symbol means is stoichiometrically equivalent to.
• Can be used to convert between quantities of reactants and
products in a chemical reaction.

31. QUANTITATIVE INFORMATION FROM
BALANCED EQUATION
Example
2 C4H10(g) + 13 O2(g)  8 CO2(g) + 10 H2O
Calculate the mass of CO2 produced when 1.00 g of C4H10 is
burned.

32. LIMITING REACTANTS
• Limiting reactant: The reactant that is entirely consumed
when a reaction goes to completion.
• Excess reactant: Reactant that is not completely consumed.

33. LIMITING REACTANTS
• You can make cookies until
you run out of one of the
ingredients.
• Once this family runs out
of sugar, they will stop
making cookies (at least
any cookies you would
want to eat).
• In this example the sugar would be the limiting reactant,
because it will limit the amount of cookies you can make.

34. LIMITING REACTANTS
Example
Zinc metal reacts with hydrochloric acid by the following
reaction:
Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)
If 0.30 mol Zn is added to HCl containing 0.52 mol HCl, how
many moles of H2 are produced?

35. LIMITING REACTANTS
EXERCISE
2 Na3PO4(aq) + 3Ba(NO3)2 (aq)  Ba3(PO4)2 (s) + 6NaNO3 (aq)
Suppose that a solution containing 3.50 grams of Na3PO4 is mixed
with a solution containing 6.40 grams of Ba(NO3)2. How many
grams of Ba3(PO4)2 can be formed?

36. LIMITING REACTANTS
Theoretical Yields
• The theoretical yield is the maximum amount of product that
can be made.
– In other words it’s the amount of product possible as
calculated through the stoichiometry problem.
• This is different from the actual yield.
• The actual yield is never excess than the theoretical yield.

37. LIMITING REACTANTS
One finds the percent yield by comparing the amount
actually obtained (actual yield) to the amount it was
possible to make (theoretical yield).
Actual Yield
Percent Yield = Theoretical Yield x 100%

38. LIMITING REACTANTS
EXERCISE
Adipic acid, H2C6H8O4, is used to produce nylon. The acid is made
commercially by a controlled reaction between cyclohexane
(C6H12) and O2:
2 C6H12(l) + 5 O2(g) → 2 H2C6H8O4(l) + 2 H2O(g)
(a) Assume that you carry out this reaction starting with 25.0 g of
cyclohexane and that cyclohexane is the limiting reactant. What
is the theoretical yield of adipic acid?
(b) If you obtain 33.5 g of adipic acid from your reaction, what is
the percent yield of adipic acid?