CHEMISTRY SHENANIGANS

Molecular

• P₄O₁₀
• C₁₀H₂₂
• C₆H₁₈O₃
• C₅H₁₂O
• 
  
  N₂O₄
  

• 
  
  P₂O₅
  
  
  
• 
  
  C₅H₁₁
  
  
  
• 
  
  C₂H_6O
  
  
  
• 
  
  C₅H₁₂O
  
  
  
• 
  
  NO₂
  
  
  

ex. (in-class) A sample of an unknown compound is analyzed and found to contain 8.4g of C and 2.1g of H, and 5.6g of O.
C_8.4H_2.1O_5.6  <--------- this is wrong because it is not in whole numbers!

Element	Mass (g)	Atomic Mass	Moles
C	8.4	12.0 g/mol	8.4g ÷ 12.0 g/mol = 0.7 mol
H	2.1	1.0 g/mol	2.1g ÷ 1.0 g/mol = 2.1 mol
O	5.6	16.0 g/mol	5.6g ÷ 16.0 g/mol = 0.35 mol

ex. A compound was analyzed and found to contain 13.5 g Ca, 10.8 g O, and 0.675 g H. What is the empirical formula of the compound?

Element	Mass (g)	Atomic Mass	Moles
Ca	13.5	40.1 g/mol	13.5g ÷ 40.1 g/mol = 0.337 mol
O	10.8	16.0 g/mol	10.8g ÷ 16.0 g/mol = 0.269 mol
H	0.675	1.0 g/mol	0.675g ÷ 1.0 g/mol = 0.675 mol

Here's a helpful hint for putting your ratios into whole numbers:
If the ratio ends in... 
~0.5 multiply by 2
~0.33 or ~0.66 multiply by 3
~0.25 or ~0.75 multiply by 4
~0.2, ~ 0.4, ~0.6, ~0.8 multiply by 5