Ch 3 Stoichiometry: Calculations with Chemical Formulas and Equations
Why do we need to balance equations? What is a mole and why do we use them in chemistry?
3.1 Chemical Equations
 Law of conservation of mass: mass is not created or destroyed in a chemical reaction
 Coefficients: give ratio of reactants & products in a chemical reaction
 Subscripts: give ratio of atoms in a compound
 Practice balancing “by inspection” to increase speed.
3.2 Some Simple Patterns of Chemical Reactivity
 Combination (synthesis): 2 or more substances form one product
2Mg(s) + O2(g)  2MgO(s)
 Decomposition: one reactant produces two or more products
2NaN3(s)  2Na(s) + 3N2(g)
 Combustion: rapid reactions that produce a flame; most involve O2(g) from air
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
3.3 Formula Weights / Molecular Weight / Molar Mass
= sum of atomic weights in a formula (ionic) / in a molecular formula / in a mole
ex. Ca(NO3)2
40.1 + 2(14.0) + 6(16.0) = 164.1 amu
Percent Composition: percent by mass of each element in a compound; calculate from formula
Divide mass contributed by each element by molar mass of the compound; multiply by 100.
C12H22O11
C: (12 atoms)(12amu)= 144
/342.0 amu x 100% = 42.1% C
H: (22 atoms)(1.0amu)= 22
/342.0 amu x 100% = 6.4% H
O: (11 atoms)(16.0amu)= 176
/342.0 amu x 100% = 51.5% O
3.4 Avogadro’s Number and the Mole
 Mole (mol): convenient measure of chemical quantities
 1 mole of something = 6.0221421 x 1023 (=Avogadro’s #) of that thing
 1 mol of carbon atoms = 6.022 x 1023 carbon atoms
 1 mole of 12C has a mass of 12g
 Molar mass: grams of 1 mol of a substance; FW in amu = MM in g/mol
 Use dimensional analysis to convert units
How many glucose molecules are in 5.23g of C6H12O6? How many oxygen atoms are in this sample?
5.23g C6H12O6 x 1mol C6H12O6 x 6.02 x 1023 molecules = 1.75 x 1022 molecules C6H12O6
180 g
1 mol
1.75 x 1022 molecules C6H12O6 x 6 atoms O
1 molecule C6H12O6
= 1.05 x 1023 atoms O
3.5 Empirical Formulas: ratio of atoms in compound (subscripts)
 % composition data: Assume 100g of sample  each % = g in a 100g sample
 Convert g into mol for each element; keep 3-4 decimal places to avoid rounding errors
 Lowest whole-number ratio of elements is the empirical formula
o Divide each #mol by the lowest # to get ratio
o Multiply each # by an integer to eliminate fractions, if necessary
 Molecular formula is equal to, or a multiple of, the empirical formula;
o Compare molar mass of empirical formula to molar mass given: divide larger
mass by smaller; multiply subscripts by answer.

Combustion Analysis: sample containing C, H, and O is burned in excess O
o The amount of CO2 gives the amount of C originally present in the sample
o The amount of H2O gives the amount of H;
1 mol H2O = 2 mol H
o The amount of O is given by the difference (combustion must be complete)
Ex. 3.13 Calculating an Empirical Formula
Ascorbic acid (Vitamin C) = 40.92% C, 4.58 % H and 54.50% O by mass. Find the empirical formula.
C: 40.92 g (1 mol C/ 12.01 g C) = 3.407 mol C
/ 3.406 mol = 1.000
x3 = 3
H: 4.58 g (1 mol H/1.008 g H) = 4.54 mol H
/ 3.406 mol = 1.333
x3 = 4
O: 54.50 g (1 mol O/16.00 g O) = 3.406 mol O
/ 3.406 mol = 1.000
x3 = 3
 C3H4O3
Ex. 3.15 Combustion Analysis
Isopropyl alcohol contains C, H and O.
A 0.255g sample is burned, producing 0.561g CO2 and 0.306g H2O. Find the empirical formula.
C: 0.561g CO2 x 1mol CO2/44g CO2 x 1mol C/ 1mol CO2 x 12.0g C / 1mol C = 0.153g C
H: 0.306g H2O x 1mol H2O/18.0g H2O x 2mol H/1mol H2O x 1.01g H/1mol H = 0.0343g H
The rest of the original sample must have been oxygen.
O: 0.255g sample - 0.153gC – 0.0343gH = 0.068g O
C: 0.153g x 1mol/12.01g = 0.0128 mol C
H: 0.0343g x 1 mol/1.01g = 0.0340 mol H
O: 0.068 g x 1mol/16.00g = 0.0043 mol O
/0.0043 mol
= 2.98
= 7.91
=1
 C 3H 8O
3.6 Quantitative Information from Balanced Equations
 Convert grams of one substance to moles using molar mass
 Convert mol of that substance to mol of another substance using mol ratio (coefficients)
 Convert mol to g using molar mass
 Theoretical Yield: calculated amount, ideal, maximum amount that can be produced
 Actual yield: measured amount actually produced; always less than theoretical
o incomplete reaction, side reactions, material stuck to glassware
 Percent yield = (actual / theoretical) x 100
Ex. 3.16 How many grams of water are produced in the oxidation of 1.00 g glucose, C 6H12O6?
C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(l)
1.00 g C6H12O6 x 1 mol
180.0 g
x 6 mol H2O x 18.0 g H2O = 0.600 g H2O
1 mol C6H12O6 1 mol H2O
3.7 Limiting Reactants A reactant that is completely consumed is “limiting”; others are “in excess”
Ex. 3.19
2Na3PO4(aq) + 3Ba(NO3)2(aq)  Ba3(PO4)2(s) + 6NaNO3(aq)
3.50g
6.40g
?g
3.50g Na3PO4 x 1mol Na3PO4
164g
x 1 mol
= 0.01065 mol Ba3(PO4)2
2 mol Na3PO4
Fewer mol of product
6.40g Ba(NO3)2 x 1mol Ba(NO3)2 x 1 mol
= 0.00816 mol Ba3(PO4)2
261g
3 mol Ba(NO3)2
Limiting reactant is Ba(NO3)2
0.00816 mol Ba3(PO4)2 x 602 g
1mol
= 4.91 g Ba3(PO4)2
Theoretical Yield