# Study Guide Chapter 3 C = 6.022x10 ```Study Guide Chapter 3 - Stoichiometry
1. A mole is # of atoms in 12 g of 12C = 6.022x1023
2. Molar mass - mass in g of 6.022x1023 molecules
calc by adding up atomic mas of all atoms
Check CoSO4.7H2O
3. % Composition
mass of element in compound
&times; 100%
mass of all elements in compound
4. Determination of formula
A. from % composition
% 6 g/100 g 6mole/100 g ; examine mole ratios of all elements, determine
empirical formula
If given molecular mass MM=N(empirical), determine N
B. From Gas of combustion
g CO2 6 mol CO2 6 mol C in substance
g H2O 6mol H2O; because H2O has 2 atoms H multiply by 2 to get mol H in
substance
g anything else is mass substance - mass substance - mass C - mass H
determine mol of other element
determine empirical formula, molecular formula if possible
5. Chemical Equations
Tells what is going on, tells how much of each compound will react, tells physical
state of each reactant and product
- Be able to balance any equation ie make total # of each atom same on both sides
to the equation
Start most complicated molecule 1 st, save O and H for last
- Stoichiometry
Tell how many moles of each molecule are involved in a reaction
Use as conversion factors to find out how much of other compounds are
needed in a reaction or how much product will be formed
6. Determination of limiting reactant
- Use ratio method
-use table method
7. Theoretical yield
8. % yield = actual/theory x 100%
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