Chemistry Term 2: Unit 1

1
Chemistry Term 2 Unit 1

• Chapter 7 - Section 3
• Percent Comp. Empirical Formulas
• Chapters 15/16
• Ionic Covalent Bonding

2
Percent Composition

• Like all percents
• Part x 100 whole
• Find the mass of each component,
• divide by the total mass.

3
Example

• Calculate the percent composition of a compound
  that is 29.0 g of Ag with 4.30 g of S.

29.0 g
Ag
X 100
87.09
33.3 g
4.30 g
S
X 100
12.91
33.3 g

• Percentages should add to 100
• Always round to the hundredths place

4
Getting it from the formula

• If we know the formula, assume you have 1 mole.
• Then you know the pieces and the whole.

5
Examples

• Calculate the percent composition of C2H4?

Molecular Mass (2 x 12.01) (4 x 1.01)
28.06 g
24.02 g
C
x 100
85.60 C
28.06 g
4.04 g
H
x 100
14.40 H
28.06 g
6
Empirical Formula

• From percentage to formula

7
The Empirical Formula

• The lowest whole number ratio of elements in a
  compound.
• The molecular formula the actual ration of
  elements in a compound.
• The two can be the same.
• CH2 empirical formula
• C2H4 molecular formula
• C3H6 molecular formula
• H2O both

8
Calculating Empirical

• Just find the lowest whole number ratio
• C6H12O6
• CH4N
• It is not just the ratio of atoms, it is also the
  ratio of moles of atoms.
• In 1 mole of CO2 there is 1 mole of carbon and 2
  moles of oxygen.
• In one molecule of CO2 there is 1 atom of C and 2
  atoms of O.

9
Calculating Empirical

• Means we can get ratio from percent composition.
• Assume you have a 100 g.
• The percentages become grams.
• Can turn grams to moles.
• Find lowest whole number ratio by dividing by the
  smallest.

10
Example

• Calculate the empirical formula of a compound
  composed of 38.67 C, 16.22 H, and 45.11 N.
• Assume 100 g so
• 38.67 g C x 1mol C 3.22 mole C
  12.01 gC
• 16.22 g H x 1mol H 16.09 mole H 1.01
  gH
• 45.11 g N x 1mol N 3.22 mole N 14.01
  gN

These numbers should be rounded to the
hundredths place
11
Example Contd
3.22 mol C
1 C
/3.22
End result should be a whole number. Only
exception when it ends in .5 you should then
double all of them.
16.09 mol H
/3.22
5 H
1 N
3.22 mol N
/3.22
Divide by the smallest number

• C1H5N1
• CH5N Final Answer

12
Example 2

• A compound is 43.64 P and 56.36 O. What is
  the empirical formula?

43.64 g P / 30.97 1.41
/1.41
1
x 2
2
56.36 g O / 16 3.52
/1.41
2.5
x 2
5
Answer P2O5
13
Empirical to molecular

• Since the empirical formula is the lowest ratio
  the actual molecule would weigh more.
• By a whole number multiple.
• Divide the actual molar mass by the the mass of
  one mole of the empirical formula.

14
Example

• A compound is known to be composed of 71.65
  Cl, 24.27 C and 4.07 H. Its molar mass is known
  to be 98.96 g. What is its molecular formula?

Step 1 Calculate Empirical Formula like before
71.65 g Cl / 35.45 2.02
/2.02
1
24.27 g C / 12.01 2.02
/2.02
1
2
4.07 g H / 1.01 4.03
/2.02
15
Example contd
Step 1 Formula ClCH2
Step 2 Figure out molecular mass (35.45 12.01
21.01) 49.48
Step 3 Divide Mass of molecular formula by mass
of empirical formula round to the whole number
98.96 g
2
49.48 g
16
Example contd

• Step 4 Multiply that number by empirical
  formula to get molecular formula
• ClCH2 x 2 Cl2C2H4

Final Answer