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"When I play with my cat, who knows whether she is not
amusing herself with me more than I with her."
Michel Eyquem De Montaigne
A + 2B  3C + D
SOMETIMES, BUT NOT ALWAYS, WHEN DIFFERENT
SUBSTANCES ARE BROUGHT TOGETHER, A
CHEMICAL CHANGE CAN OCCUR.
CHEMICAL BONDS THAT BIND THE ATOMS
TOGETHER CAN BREAK, THE ATOMS CAN
REARRANGE THEMSELVES, AND NEW CHEMICAL
BONDS CAN FORM. WE CALL THIS A CHEMICAL
REACTION.
WE DESCRIBE THIS IN WORDS OR WE CAN USE
THE CHEMISTS’ SHORTHAND METHOD AND WRITE
A CHEMICAL EQUATION.
LET’S TAKE A LOOK AT A TYPICAL CHEMICAL EQUATION
AND SEE WHAT SORT OF INFORMATION IT GIVES US.
Zn (s) + 2HCl (aq)  ZnCl2 (aq) + H2 (g)
IF WE STATED THIS IN WORDS, WE MIGHT SAY “SOLID
ZINC METAL WHEN ADDED TO AN AQUEOUS SOLUTION
OF HYDROCHLORIC ACID PRODUCES A SOLUTION OF
ZINC CHLORIDE AND HYDROGEN GAS. ONE MOLE OF
ZINC WILL REACT WITH TWO MOLES OF HYDROCHLORIC
ACID TO PRODUCE ONE MOLE OF ZINC CHLORIDE AND
ONE MOLE OF HYDROGEN GAS. THERE IS A GREAT DEAL
OF INFORMATION HERE.
1)THE FORMULAS TELL US WHAT THE REACTANTS AND
PRODUCTS ARE.
2)THE PREFIXES TELL US HOW MANY MOLECULES OR MOLES
OF EACH ARE REACTED OR FORM.
3)THE LETTERS IN PARANTHESIS TELL US THE STATES OF
EACH.
4)KNOWING THE NUMBER OF MOLES AND THE MOLAR MASSES,
WE COULD ALSO INFER THE MASSES INVOLVED.
THE ENERGY LOST OR GAINED COULD ALSO BE INCLUDED IN
THE EQUATION. THE POTENTIAL ENERGIES OF THE CHEMICAL
BONDS INVOLVED IS RARELY THE SAME. SO, YOU OFTEN
HAVE ENERGY LOST OR GAINED.
MASS IS NOT LOST OR GAINED IN A CHEMICAL
REACTION, AND ATOMS OF ONE ELEMENT ARE NOT
CONVERTED INTO ATOMS OF OTHER ELEMENTS.
SO, THE LAW OF CONSERVATION OF MASS ALWAYS
HOLDS.
FOR A CHEMICAL EQUATION TO BE TRUE, IT MUST BE
BALANCED – YOU HAVE THE SAME NUMBER OF
ATOMS OF EACH KIND ON BOTH SIDES OF THE
EQUATION.
YOU ALSO HAVE THE SAME MASS ON EACH SIDE OF THE
EQUATION.
PROBLEMS YOU SHOULD BE ABLE TO SOLVE
WITH EQUATIONS
1)MOLE RELATIONSHIPS IN A REACTION
EXAMPLE: GIVEN THE FOLLOWING
REACTION
2Fe(NO3)3 + 3Na2S  Fe2S3 + 6NaNO3
HOW MANY MOLES OF Na2S WOULD BE
REQUIRED TO MAKE 15 moles OF Fe2S3?
WE WANT 15 moles Fe2S3, AND WE WANT
TO KNOW HOW MUCH Na2S IS NEEDED.
X moles 15 moles
2Fe(NO3)3 + 3Na2S  Fe2S3 + 6NaNO3
3
1
THE EQUATION TELLS US THAT 3 moles OF
Na2S WILL GIVE US 1 mole OF Fe2S3. SO
X / 3 =
15 / 1
X = 3 x 15 = 45 moles
SOLVING FOR X
2) GIVEN THE AMOUNTS OF STARTING MATERIAL
YOU HAVE, HOW MUCH PRODUCT CAN YOU MAKE?
EXAMPLE: UNDER HIGH TEMPERATURE AND PRESSURE,
NITROGEN GAS CAN BE MADE TO REACT WITH HYDROGEN
GAS TO FORM AMMONIA. YOU HAVE 2,000 KG OF HYDROGEN,
HOW MUCH AMMONIA CAN YOU FORM?
FIRST, YOU NEED THE EQUATION, SO YOU’D WRITE
N2 + H2  NH3
IS IT BALANCED?
TO GET IT TO BALANCE
N2 (g) + 3H2 (g)  2NH3 (g)
SO, THIS SAYS THAT ONE MOLE OF NITROGEN GAS REACTS
WITH 3 MOLES OF HYDROGEN GAS TO FORM 2 MOLES OF
AMMONIA GAS.
THE MOLAR MASS OF HYDROGEN IS 2 X 1.01 = 2.02 AND 3 MOLES
ARE INVOLVED OR 3 X 2.02 = 6.06
THE MOLAR MASS OF AMMONIA IS 14 + 3 X 1.01 = 17.0
SO, PLUGGING IN WHAT YOU KNOW AND YOUR UNKNOWN:
2,000 kg
X kg
N2 (g) + 3H2 (g)  2NH3 (g)
6.06 g
34.0 g
OR 2000/6.06 = X/34.0
X = (2000/6.06) X 34.0 = 11,200 kg AMMONIA
3) HOW MUCH STARTING MATERIAL DO YOU NEED
TO PRODUCE A GIVEN AMOUNT OF PRODUCT?
EXAMPLE: YOU WANT TO USE THE THERMITE REACTION
TO PRODUCE 500 g OF IRON FOR A WELD BETWEEN TWO
RAILS ON A RAILROAD. HOW MUCH Fe2O3 DO YOU NEED TO
START WITH?
THE FIRST THING YOU NEED IS A BALANCED EQUATION:
Fe2O3 + 2Al  Al2O3 + 2Fe
THE EQUATION TELLS YOU THAT FOR EVERY MOLE OF
IRON (III) OXIDE YOU START WITH, YOU END UP WITH 2
MOLES OF IRON. SO, LET’S CHANGE OUR MOLE
RELATIONSHIPS TO MASS AND PUT OUR VALUES IN.
FIRST, WE NEED OUR FORMULA MASSES:
2 Fe = 55.8 x 2 = 111.6
3 O = 16.0 x 3 = 48.0
formula mass = 160 g/fm
Xg
500 g
Fe2O3 + 2Al  Al2O3 + 2Fe
160 g
112 g
so, our equation is
X/160 = 500/112 and X = (500/112) x 160 = 714 g
SOME OTHER PROBLEMS YOU SHOULD BE ABLE
TO SOLVE
4) PERCENT COMPOSITION – CALCULATE
PERCENT COMPOSITION FROM FORMULA MASS
EXAMPLE: CALCULATE THE % OF EACH
ELEMENT PRESENT IN Na3PO4
FIRST CALCULATE FORMULA MASS
3 Na = 3 x 23 = 69
1 P = 1 x 31 = 31
4 O = 4 x 16 = 64
formula mass = 164 g/mole
% Na = (69/164) x 100 = 42%
% P
= (31/164) x 100 = 19%
% O = (64/164) x 100 = 39%
5) GIVEN % COMPOSITION, CALCULATE THE
EMPIRICAL FORMULA
EXAMPLE: A molecule with molecular weight of
180.18 g/mol is analyzed and found to contain
40.00% carbon, 6.72% hydrogen and 53.28%
oxygen.
THIS MEANS THAT IF YOU HAD 100 grams,
YOU WOULD HAVE
40.00 g CARBON
6.72 g HYDROGEN
53.8 g OXYGEN
NEXT, TAKE THESE AMOUNTS AND
CALCULATE MOLES
C = 40.0 g/12 g/mole = 3.33 moles
H = 6.72 g/1.0 g/mole = 6.72 moles
O = 53.8 g/16 g/mole = 3.36 moles
NOW WE DIVIDE BY THE SMALLES NUMBER
TO TRY TO GET A WHOLE NUMBER RATIO.
C = 3.33/3.33 = 1
H = 6.72/3.33 = 2
0 = 3.36/3.33 = 1
EMPIRICAL FORMULA = CH2O
5) CALCULATE THE MOLECULAR FORMULA IF
WE KNOW THE EMPIRICAL FORMULA
EXAMPLE: If the empirical formula is CH2O and
the molecular mass is 90 g/mole, what is the
molecular formula.
WE KNOW THE MOLECULAR FORMULA IS
SOME WHOLE NUMBER MULTIPLE OF THE
EMPIRICAL FORMULA, SO CALCULATE THE
EMPIRICAL MASS
1C + 2H + 1O = 12 + 2 + 16 = 30
DIVIDE 90 BY 30 = 90/30 = 3, SO MOLECULAR
FORMULA = C3H6O3