STPM Form 6 Chemistry Solids

1. Mole and Avogadro Constant
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2. The Mole
• A counting unit
• Similar to a dozen, except instead
of 12, it’s 602 billion trillion
602,000,000,000,000,000,000,000
• 6.02 X 1023 (in scientific notation)
• This number is named in honor of
Amedeo _________ (1776 – 1856), 1856)
who studied quantities of gases
and discovered that no matter what
the gas was, there were the same
number of molecules present

3. Avogadro’s Number
• Avogadro’s number (symbol N) is the number of
atoms in 12.01 grams of carbon.
• Its numerical value is 6.02 × 1023.
• Therefore, a 12.01 g sample of carbon contains
6.02 × 1023 carbon atoms.
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4. The Mole
• The mole (mol) is a unit of measure for an amount of a
chemical substance.
• A mole is Avogadro’s number of particles, which is 6.02 ×
1023 particles.
1 mol = Avogadro’s number = 6.02 × 1023 units
• We can use the mole relationship to convert between
the number of particles and the mass of a substance.
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5. Analogies for Avogadro’s Number
• The volume occupied by one mole of softballs would be
about the size of the Earth.
• One mole of Olympic shotput balls has about the same
mass as the Earth.
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6. One Mole of Several Substances
C12H22O11 H 2O
lead mercury
K2Cr2O7
sulfur
copper NaCl
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7. Example 1
• How many sodium atoms are in 0.120 mol Na?
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8. Example 2
• How many moles of potassium are in 1.25 × 1021
atoms K?
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9. 9

10. Molar Mass
• The atomic mass of any substance expressed in grams is
the molar mass (MM) of that substance.
• The atomic mass of iron is 55.85 amu.
• Therefore, the molar mass of iron is 55.85 g/mol.
• Since oxygen occurs naturally as a diatomic, O2, the
molar mass of oxygen gas is 2 times 16.00 g or 32.00
g/mol.
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11. 11

12. Calculating Molar Mass
• The molar mass of a substance is the sum of the molar
masses of each element.
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13. Example 3
Calc. the molar mass:
1. Fe2O3
2. C6H8O7
3. C16H18N2O4
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14. Mole Calculations
• Now we will use the molar mass of a compound to
convert between grams of a substance and moles or
particles of a substance.
6.02 × 1023 particles = 1 mol = molar mass
• If we want to convert particles to mass, we must first
convert particles to moles and then we can convert
moles to mass.
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15. Example 4
• What is the mass of 1.33 moles of titanium, Ti?
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16. Example 5
• What is the mass of 2.55 × 1023 atoms of lead?
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17. Example 6
• How many O2 molecules are present in 0.470 g of
oxygen gas?
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18. Example 7
• What is the mass of a single molecule of sulfur
dioxide? The molar mass of SO2 is 64.07 g/mol.
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19. 19

20. Molar Volume
• At standard temperature and pressure, 1 mole of
any gas occupies 22.4 L.
• The volume occupied by 1 mole
of gas (22.4 L) is called the molar
volume.
• Standard temperature and
pressure are 0°C and 1 atm.
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21. Molar Volume of Gases
• We now have a new unit factor equation:
1 mole gas = 6.02 × 1023 molecules gas = 22.4 L gas
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22. Gas Density
• The density of gases is much less than that of
liquids.
• We can calculate the density of any gas at STP
easily.
• The formula for gas density at STP is:
molar mass in grams/mol
= density, g/L
molar volume in liters/mol
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23. Example 8
• What is the density of ammonia gas, NH3, at STP?
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24. Example 9
• We can also use molar volume to calculate the molar
mass of an unknown gas.
• 1.96 g of an unknown gas occupies 1.00 L at STP. What
is the molar mass?
• We want g/mol; we have g/L.
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25. Mole Unit Factors
• We now have three interpretations for the mole:
– 1 mol = 6.02 × 1023 particles
– 1 mol = molar mass
– 1 mol = 22.4 L at STP for a gas
• This gives us 3 unit factors to use to convert
between moles, particles, mass, and volume.
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26. Example 10
• A sample of methane, CH4, occupies 4.50 L at STP.
How many moles of methane are present?
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27. Example 11
• What is the mass of 3.36 L of ozone gas, O3, at
STP?
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28. Example 12
• How many molecules of hydrogen gas, H2, occupy
0.500 L at STP?
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29. Concentration
Which beaker of
tea is stronger?
Which beaker of
tea is more
concentrated?

30. Concentration
Concentration = amount of solute in a
specific volume
More solute = higher concentration

31. Concentration
Concentration can be measured in three ways:
1. Percent by Volume
2. Percent by Mass
3. Molarity

32. Percent Composition
• The percent composition of a compound lists the mass
percent of each element.
• For example, the percent composition of water, H2O is:
– 11% hydrogen and 89% oxygen
• All water contains 11%
hydrogen and 89% oxygen
by mass.
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33. Calculating Percent Composition
• There are a few steps to calculating the percent
composition of a compound. Let’s practice using H2O.
– Assume you have 1 mole of the compound.
– One mole of H2O contains 2 mol of hydrogen and 1
mol of oxygen.
– 2(1.01 g H) + 1(16.00 g O) = molar mass H2O
– 2.02 g H + 16.00 g O = 18.02 g H2O
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34. Calculating Percent Composition
• Next, find the percent composition of water by
comparing the masses of hydrogen and oxygen in
water to the molar mass of water:
2.02 g H
%H= × 100% = 11.2% H
18.02 g H2O
16.00 g O
% O = 18.02 g H O × 100% = 88.79% O
2
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35. Example 13
• TNT (trinitrotoluene) is a white crystalline
substance that explodes at 240 °C. Calculate the
percent composition of TNT, C7H5(NO2)3.
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36. Percent Composition of TNT
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37. Empirical Formulas
• The empirical formula of a compound is the simplest
whole number ratio of ions in a formula unit or atoms of
each element in a molecule.
• The molecular formula of benzene is C6H6.
– The empirical formula of benzene is CH.
• The molecular formula of octane is C8H18.
– The empirical formula of octane is C4H9.
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38. Calculating Empirical Formulas
• We can calculate the empirical formula of a compound
from its composition data.
• We can determine the mole ratio of each element from
the mass to determine the formula of radium oxide, Ra?
O?.
• A 1.640 g sample of radium metal was heated to
produce 1.755 g of radium oxide. What is the empirical
formula?
• We have 1.640 g Ra and 1.755-1.640 = 0.115 g O.
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39. Calculating Empirical Formulas
• The molar mass of radium is 226.03 g/mol, and the
molar mass of oxygen is 16.00 g/mol.
1 mol Ra
1.640 g Ra × = 0.00726 mol Ra
226.03 g Ra
1 mol O
0.115 g O × = 0.00719 mol O
16.00 g O
• We get Ra0.00726O0.00719. Simplify the mole ratio by dividing
by the smallest number.
• We get Ra1.01O1.00 = RaO is the empirical formula.
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40. Example 14
• Determine the empirical formula of sodium
sulfate given the following data
A 3.00 g sample of NaxSyOz is comprised of 0.9722
g of sodium, 0.678 g of sulfur and 1.35 g of
oxygen
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41. Empirical Formulas from Percent Composition
• We can also use percent composition data to calculate
empirical formulas.
• Assume that you have 100 grams of sample.
• Acetylene is 92.2% carbon and 7.83% hydrogen. What is
the empirical formula?
• If we assume 100 grams of sample, we have 92.2 g
carbon and 7.83 g hydrogen.
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42. Empirical Formula for Acetylene
• Calculate the moles of each element:
1 mol C
92.2 g C × = 7.68 mol C
12.01 g C
1 mol H
7.83 g H × = 7.75 mol H
1.01 g H
• The ratio of elements in acetylene is C7.68H7.75. Divide by
the smallest number to get the formula:
7.68 7.75
C 7.68 H 7.68 = C1.00H1.01 = CH
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43. Example 15
• Find the empirical formula of Iron(III) chloride
from the following data
• In Iron(III) chloride
– Fe 34.43 % by mass
– Cl 65.57 % by mass
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44. Molecular Formulas
• The empirical formula for acetylene is CH. This
represents the ratio of C to H atoms on acetylene.
• The actual molecular formula is some multiple of the
empirical formula, (CH)n.
• Acetylene has a molar mass of 26 g/mol. Find n to find
the molecular formula:
(CH)n 26 g/mol n = 2 and the molecular
=
CH 13 g/mol formula is C2H2.
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45. Example 16
• Find the empirical and molecular formula for
hydrazine, NxHy
• The molar mass of hydrazine is 32.046 g/mol
• In hydrazine
– N is 87.42% by mass
– H is 12.58 % by mass
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46. Chapter Summary
• We can use the following flow chart for mole
calculations:
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