Chapter 7
Reactions
Chemical Changes
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Substances react and form a new substance.
Reactants go in.
Products come out.
Reactants
Products
• Represented by chemical equations
Chemical Equations
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For the process of burning:
Word equation:
Carbon + Oxygen → Carbon dioxide
Chemical equation:
C + O2 → CO2
Law of Conservation of Mass
• Mass is neither created nor destroyed in a
chemical reaction.
• It may change form (solid to liquid or gas).
• If 1 atom of Carbon goes into a reaction, 1
atom of carbon must come out. It can’t be lost
or multiplied.
Chemical Reactions
• C + O2 → CO2
+
→
1 atom of carbon goes in and 1 atom of carbon comes out
2 atoms of oxygen go in and 2 atoms of oxygen come out
Mass is neither created nor destroyed.
Chemical Reactions
N2H4 + O2 → N2 + H2O
+
→
Why is this wrong?
+
Balancing Equations
N2H4 + O2 → N2 + H2O
+
→
+
+
Why is this wrong?
2 atoms of nitrogen go in and 2 come out
4 atoms of hydrogen go in but only 2 come out
2 atoms of oxygen go in but only 1 comes out
Mass appears to be lost and that cannot happen!!
Balancing Equations
• You can balance equations by changing the
coefficients.
• Coefficients are numbers that come in front of
the formulas.
• Never!!!! Change the subscripts in a formula!!
• Changing subscripts makes it a different
formula.
Balancing Equations
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N2H4 + O2 → N2 + H2O
N2H4 + O2 → N2 + 2H2O
The 2 in front of water is a coefficient.
Now 2 atoms of nitrogen go in and 2 come out.
4 atoms of hydrogen go in and (2x2=4) 4 come out.
2 atoms of oxygen go in and (2x1+2) 2 come out.
Steps for Balancing Equations
• Count the atoms of each element on each side
of the equation.
• Then change coefficients 1 at a time until the
equation is balanced.
• REMEMBER: You cannot lose or gain atoms!!
Practice
• Balance the following equations:
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Cu + O2 → CuO
N2 + H2 → NH3
K + Br2 → KBr
Na + Cl → NaCl
Practice Answers
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Cu + O2 → CuO
Cu + O2 → 2CuO
2Cu + O2 → 2CuO
I add a coefficient of 2 to get 2 oxygen atoms
but that also makes me have 2 copper atoms
as products. I then have to add a 2 in front of
copper. Now 2 copper in and out and 2 oxygen
in and out.
Practice Answers
• N2 + H2 → NH3
• N2 + H2 → 2NH3
• First I add a 2 to the product to end with 2
nitrogen atoms, but now I am ending with 6
hydrogen atoms.
• N2 + 3H2 → 2NH3
• Adding a 3 in front of hydrogen gives me 6
atoms in the beginning as well.
Practice Answers
• K + Br2 → KBr
• K + Br2 → 2KBr
• First I add a 2 to the product to end with 2
bromine atoms, but now I am also ending with
2 potassium atoms.
• 2K + Br2 → 2KBr
• Adding a 2 in front of potassium fixes this.
Practice Answers
• Na + Cl → NaCl
• 1 atom of each go in and 1 atom of each come
out
• The equation is already balanced.
Moles
• Because chemical reactions often involve large
numbers of small particles, chemists use a
counting unit called the mole to measure
amounts of a substance.
Moles
• 1 mole = 6.02 X 1023 particles
• 1 mole of flowers = 6.02 X 1023 flowers or
602,000,000,000,000,000,000,000 flowers
• 1 mole of apples = 6.02 X 1023 apples
• 1 mole of oxygen = 6.02 X 1023 atoms of
oxygen
Molar Mass
• For an element, the molar mass is the same as
the atomic mass expressed in grams.
• So for carbon, the atomic mass is 12.0 amu,
and the molar mass is 12.0 grams.
• Element
atomic mass
molar mass
• Oxygen
15.999 amu 15.999 grams
• Nitrogen
14.007 amu 14.007 grams
Mole-Mass Conversions
• Once you know the molar mass of a
substance, you can convert moles of that
substance into mass (or mass into moles).
• Carbon has a molar mass of 12.0 grams.
• How many moles are in 5 grams of carbon?
• 22 g X 1 mol/12.0 g = 1.83 moles or 1.8 moles
in significant digits
• mol is the abbreviation for moles
Mole-Mass Conversions
• How many moles are in 35 grams of oxygen (in
significant digits)?
Mole-Mass Conversions
• How many moles are in 35.0 grams of oxygen
(in significant digits)?
• 35.0 g X 1 mol/15.999 g = 2.19 mol O
Finding the mass of a compound
• To find the mass of a compound:
• CO2
• Find the mass of each element in the
compound.
• C = 12.011
• O = 15.999
Finding the mass of a compound
• C = 12.011 g
• O = 15.999 g
• Then multiply the mass by the number of
atoms of each element.
• There is only 1 atom of carbon so
12.001 x 1 = 12.001
• There are 2 atoms of oxygen (O2) so
15.999 x 2 = 31.998
Finding the mass of a compound
• Then add the masses of each element
together.
• 31.998 g O
• +12.001 g C
• 43.999 g CO2
Practice
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How many moles are in 55.0 grams of CO2?
55.0 g X 1 mol / 43.999 g = 1.25 mol CO2
How many moles are in 25.0 grams of H2O?
How many grams are in 2.3 moles of CO2?
Practice
• How many moles are in 25.0 grams of H2O?
• 1.0079 x 2 + 15.999 = 18.015 (mass of H2O)
• 25.0 g x 1 mol/ 18.015 g = 1.39 mol
• How many grams are in 2.3 moles of CO2?
• 12.011 + 15.999 x 2 = 44.009 (mass of CO2)
• 2.3 mol x 44.009 g/1 mol = 101.22 g
= 1.0 x 102
Calculations
• In chemical reactions, the mass of a reactant
or product can be calculated by using a
balanced chemical equation and molar masses
of the reactants and products.
Calculations
• How much oxygen is required to make 144 grams
of water?
• 2H2 + O2 → 2H2O
• First determine how many moles of water you are
trying to make.
• Molar mass of water = 1.0079 (2) +15.999 =
18.015 grams
• 144 g H2O x 1 mol H2O/ 18.015 g H2O = 7.99 mol
H 2O
Calculations
• 144 g H2O = 7.99 mol H2O
• 2H2 + O2 → 2H2O
• Now you can use the balanced equation to
convert.
• (7.99 mol H2O) x (1 mol O2 )/(2 mol H2O) = 4.00
mol O2
• Finally convert moles back to grams.
• (4.00 mol O2) x (15.999 g O2)/(1 mol O2) =
63.9 g O2
Practice Calculations
• How much magnesium do you need to make
34.5 grams of Magnesium Oxide?
• 2Mg + O2 → 2MgO
• How much sodium chloride will 22.3 grams of
sodium make?
• 2Na + Cl2 → 2NaCl
Practice Calculations
• How much magnesium do you need to make
34.5 grams of Magnesium Oxide?
• 2Mg + O2 → 2MgO
• Molar mass of MgO = 24.305 + 15.999 =
40.304 g/ 1 mol MgO
• 34.5 g MgO x 1 mol MgO/40.304 g MgO x 2
mol Mg/2 mol MgO x 24.305 g Mg/ 1 mol
Mg= 20.8 g Mg
Practice Calculations
• How much sodium chloride will 22.3 grams of
sodium make?
• 2Na + Cl2 → 2NaCl
• 22.3 g Na x 1 mol Na/ 22.990 g Na x 2 mol
NaCl/ 2 mol Na x (22.990 + 35.453=58.443)g
NaCl/ 1 mol NaCl = 56.7 g NaCl