Writing Chemical Equations
Magnesium burns in air.
Burning is a reaction with oxygen, forming an oxide.
Magnesium + oxygen → Magnesium oxide
Mg + O2 → MgO (unbalanced)
2Mg + O2 → 2MgO (balanced
Balancing uses coefficients to change the amounts of different substances reacting. The
purpose of balancing is to conserve atoms. In a chemical reaction, atoms are rearranged, but not
created or destroyed.
Butane, C4H10, undergoes combustion.
When hydrocarbons burn, they react with oxygen, forming
carbon dioxide and water.
C4H10 + O2 → CO2 + H2O (unbalanced)
2 C4H10 + 13O2 → 8CO2 + 10H2O (balanced)
Ammonium dichromate decomposes to yield chromium (III) oxide, nitrogen and water
(NH4)2Cr2O7 → Cr2O3 + N2 + H2O
(unbalanced)
(NH4)2Cr2O7 → Cr2O3 + N2 + 4H2O (balanced)
Stoichiometry
Stoichiometry deals with calculations involving quantities of substances reacting and
forming in a chemical reaction.
The underlying principle of stoichiometry is mole ratio, which relates amounts of reactants
and products in a balanced chemical equation.
Consider the equation for the combustion of magnesium:
2Mg + O2 → 2MgO
The coefficients are 2, 1, 2 (the one in front of O2 is understood).
This means that 2 Mg atoms react with 1 O2 molecule to form 2 MgO molecules.
Or, in terms of moles, 2 mol of Mg react with 1 mol of O2 to form 2 mol of MgO.
But the equation can also be balanced with coefficients 6,3,6 12,6,12, 50,25,50 Indeed,
any 2:1:2 ratio balances the equation.
We should consider the coefficients as ratios of amounts (moles).
If we had 0.80 mol Mg react, then 0.40 mol O2 react, forming 0.80 mol MgO (keeping
2:1:2 ratio).
If we had 1.80 mol O2 react, then 3.60 mol Mg would be needed, forming 3.60 mol MgO
(keeping 2:1:2 ratio).
The mole ratio relates amounts of reactants and products in the equation.
A general approach to solving any stoichiometry problem:
1. Write a balanced equation for the reaction.
2. Take what is known and convert to moles.
3. Use mole ratio in the balanced equation to calculate moles of unknown.
4.
Convert moles of unknown into whatever the problem requires.
A sample of 10.0 g of magnesium reacts with oxygen, forming magnesium oxide.
What mass of oxygen reacts; what mass of magnesium oxide forms?
2Mg + O2 → 2MgO
1 mol
10.0 g Mg x 24.3 g = 0.412 mol Mg
0.412 mol Mg x
1 mol O 2
= 0.206 mol O2
2 mol Mg
g
0.206 mol O2 x 32.0 mol = 6.58 g O2 react
This can be done more efficiently in one line
1 mol
g
1 mol O 2
10.0 g Mg x 24.3g x
x 32.0 mol = 6.58 g O2 react
2 mol Mg
To find the MgO formed:
1 mol 1 mol MgO
g
10.0 g Mg 24.3 g x 1 mol Mg x 40.3 mol
= 16.58 g MgO (16.6 g to 3 sig. fig)
We could have predicted the 16.58 g MgO from our previous answer using the Law of
Conservation of Mass.
10.0 g Mg + 6.58 g O2 must form 16.6 g MgO
How many molecules of O2 are removed from the air when 25.0 g of butane, C4H10,
undergoes combustion?
Equation
2C4H10 + 13O2 → 8CO2 + 10H2O
1 mol
13 mol O 2
23 molecule
24
25.0 g x 58.1g x
x 6.02 x 10
= 1.68 x 10 molecules O2
mol
2 mol C 4 H10
Phosphorus trichloride is converted by a series of chemical reactions to carbon
tetrachloride. Starting with 5.00 mol of phosphorus trichloride, what is the maximum amount of
carbon tetrachloride that can be formed? Assume no additional chlorine atoms are introduced in
the procedure.
Balanced equation: 4PCl3 → 3CCl4
The detailed equations are not needed. Since all chlorine is converted, the 4:3 ratio
balancing chlorine atoms must prevail.
5.00 mol PCl3 x
3 mol CCl4
= 3.75 mol CCl4
4 mol PCl3
Ammonium dichromate decomposes into chromium (III) oxide, nitrogen and water.
Starting with 2.00 g of ammonium dichromate, what mass of chromium (III) oxide forms?
(NH4)2Cr2O7 → Cr2O3 + N2 + 4H2O
1 mol
1mol Cr2 O3
g
2.00 g x 252 g x
x 152
= 1.21 g Cr2O3
mol
1mol NH4 ) 2 Cr2 O7 
A 10.0 g mass of element X reacts completely with 19.0 g of Cl2 forming XCl3. What is
the molar mass of X?
2X + 3Cl2 → 2XCl3
1 mol
2 mol X
19.0 g x 70.9 g x
= 0.179 mol X
3 mol Cl 2
10.0 g X
g
Molar mass = 0.179 mol X = 56.0 mol
Limiting Reactant
When non-stoichiometric amounts of material are mixed and allowed to react, the reaction
goes as far as possible, until one reactant is used up. That reactant is called the limiting reactant.
Suppose we mix 3.0 mol Mg and 1.0 mol O2 and allow them to react to form MgO.
2Mg + O2 → 2MgO
Start 3.0
1.0
React 2.0
1.0
End 1.0
0.0
2.0
O2 is the limiting reactant. There is 1.0 mol Mg left over, and 2.0 mol MgO is formed
stoichiometrically from the 1.0 mol O2 reacting.
In general to find the limiting reactant, compare the actual starting mole ratio (given in the
problem) with the stoichiometric ratio (from the coefficients in the balanced equation).
In this example, comparing
Starting ratio
3.0
1.0 = 3.0
mol Mg
mol O 2
Stoichiometric Ratio
2
1 =2
Since 3.0 > 2, the numerator (Mg) is in excess and the denominator (O2) is the limiting
reactant.
Consider the reaction between aluminum and oxygen to form aluminum oxide.
Analyze the product formed and any reactant left over when we mix 10.00 mol Al with
7.00 mol O2
4Al + 3O2 → 2Al2O3
Compare starting ratios and stoichiometric ratios of
Starting ratio
10.00
7.00 = 1.43
mol Al
mol O 2
Stoichiometric Ratio
4
3 = 1.33
Since 1.43 > 1.33, the numerator (10.00 mol Al) is in excess and the denominator (7.00
mol O2) is the limiting reactant.
Start
React
4Al + 3O2 → 2Al2O3
10.00 7.00
9.33 7.00
End
0.67
0
2
7.00 x 3 = 4.67
The 9.33 mol Al reacting is calculated stoichiometrically from the limiting reactant
4
7.0 x 3 = 9.33
Let’s do the same problem, starting with 10.00 mol Al and 8.00 mol O2
Compare starting ratios and stoichiometric ratios of
Starting ratio
10.00
8.00 = 1.25
mol Al
mol O 2
Stoichiometric Ratio
4
3 = 1.33
Since 1.25 < 1.33, the denominator (8.00 mol O2) is in excess and the numerator (10.00
mol Al) is the limiting reactant.
4Al + 3O2 → 2Al2O3
Start 10.00 8.00
React 10.00 7.50
End
0
0.50
2
10.00 x 4 = 5.00
The 7.50 mol O2 reacting is calculated stoichiometrically from the limiting reactant
3
10.0 x 4 = 7.50
Example using masses
2NH3 + 3CuO → N2 + 3H2O + 3Cu
18.1g 90.4g
___ g
This problem involves a limiting reactant. We cannot assume that both the NH3 and CuO
will get used up.
For a limiting reactant problem, we can modify our general approach to stoichiometry:
1. Write a balanced equation for the reaction.
2. Take what is known and convert to moles.
2a. If necessary, determine the limiting reactant by comparing the starting mol ratio with
the stoichiometric mol ratio, and then use the limiting reactant quantity in all further calculations.
3.
Use mole ratio in the balanced equation to calculate moles of unknown.
4.
Convert moles of unknown into whatever the problem requires.
Example using masses
2NH3 + 3CuO  N2 + 3H2O
18.1g 90.4g ___ g
1 mol
18.1 g x 17.0 g = 1.06 mol NH3
1 mol
90.4 g x 79.5 g = 1.14 mol CuO
Compare starting ratio and stoichiometric ratio of
Starting ratio
1.06
1.14 = 0.930
mol NH3
mol CuO
Stoichiometric ratio
2
3 = 0.67
Since 0.93 > 0.67, the numerator (1.06 mol NH3) is in excess and the denominator (1.14
mol CuO) is the limiting reactant.
Do all further calculations based on the limiting reactant.
1.14 mol CuO x
g
1 mol N 2
x 28.0 mol = 10.6 g N2
3 mol CuO
Percent Yield
In actual practice, doing a reaction may not yield the expected stoichiometric amount of
product. There may be incomplete reaction, incomplete recovery of product, side reactions, or
other problems. The actual yield (the product actually obtained in a real experiment) is often less
than the expected yield (calculated from stoichiometry).
actual yield
Percent yield = expected yield x 100
In the previous problem, we calculated a yield of 10.6 g of N2. Suppose the experimenter
actually obtained 9.00 g of N2. Then the percent yield would be:
9.00
10.6 x 100 = 84.9%
In a certain experiment Cu is obtained from reducing CuO with H2
CuO + H2  Cu + H2O
Starting with 5.00 g of CuO and excess H2, the experimenter gets 3.90 g of Cu. What is the
percent yield?
First calculate the expected (stoichiometric) yield.
1 mol
1 mol Cu
g
5.00 g x 79.5 g x 1 mol CuO x 63.5 mol = 3.99 g Cu
3.90
Percent yield = 3.99 x 100 = 97.7%
Molarity
Solutions are homogeneous mixtures. As mixtures, they do not have a definite
composition, and can have variable amounts of solute and solvent.
The concentration of a solution is a measure relating the quantity of solute to the quantity
of solvent. The most commonly used unit of concentration is molarity.
moles solute
Molarity = liters solution
where “solution” means solute + solvent combined.
Molarity is designated by the symbol M
0.50 M solution means 0.50 mol solute for every liter of solution.
A solution is prepared by dissolving 15.0 g of sodium sulfate in enough water to make 250
mL of total solution. What is the molarity of this solution?
The formula of sodium sulfate is Na2SO4, with molar mass 142.
1 mol
15.0 g x 142g
Molarity =
= 0.423 M
0.250 L
What mass of AgNO3 is present in 25.0 mL of a 0.600 M solution?
mol
g
0.600 L x 0.0250 L x 170 mol = 2.55 g
Blood serum contains 0.14 M NaCl. What volume of blood serum contains 1.0 mg NaCl?
1 mol
1L
0.0010 g x 58.5 g x 0.14 mol = 1.2 x 10-4 L
Describe how you would prepare 250 mL of 0.100 M NaCl.
mol
g
0.250 L x 0.100 L x 58.5 mol = 1.46 g NaCl
Weigh out 1.46 g and dissolve it in some water in a 250 mL volumetric flask, and then add
water to the 250 mL mark. Shake well.
Dilution Problem
How would you prepare 500 mL of 0.30 M HCl starting with concentrated HCl which is
12.0 M?
Moles of solute after dilution = moles of solute before dilution.
mol
Final moles: 0.500 L x 0.30 L = 0.15 mol
Before dilution there must also be 0.15 mol.
1L
0.15 mol x 12 mol = 0.0125 L or 12.5 mL
Alternate equivalent approach:
M1V1 = M2V2
M x V = moles
There is no need to convert to L, as long as V1, V2 have the same units.
0.30(500) = 12.0x
x = 12.5 mL
Book Example
Limestone, which is largely calcium carbonate, reacts with hydrochloric acid according to
the equation;
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O()
What mass of calcium carbonate is needed to react with 225 mL of 3.25 M HCl?
0.225 L x 3.25
mol 1mol CaCO3
g
x
x 100.1
= 36.6 g CaCO3
L
mol
2 mol HCl