Represents the actual number of atoms of each element in compound Not necessary for ionic compounds Necessary for covalent compounds The molecular formula for water is H 2 O ID: 615760
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Slide1
Molecular Formula
Represents the
actual number of atoms of each element in compoundNot necessary for ionic compoundsNecessary for covalent compoundsThe molecular formula for water is H2O, and the empirical formula for water is H2OThe molecular formula for hydrogen peroxide is H2O2,and the empirical formula is HO.
Episode 703Slide2
The empirical formula for glucose is CH2
O
If the molar mass is 180.0 g/mol, find the molecular formula.Find the mass of the empirical formula.Compare the molar mass of molecular formula to molar mass of empirical formula.C 1 x 12 g/mol = 12 g/molH 2 x 1 g/mol = 2 g/molO 1 x 16 g/mol = 16 g/mol30 g/mol180 g/mol30 g/mol= 6
x CH2O
= C
6
H
12
O
6
Episode
703
Episode
703Slide3
The empirical formula for glucose is CH2
O
If the molar mass is 240.0 g/mol, find the molecular formula.Find the mass of the empirical formula.Compare the molar mass of molecular formula to molar mass of empirical formula.C 1 x 12 g/mol = 12 g/molH 2 x 1 g/mol = 2 g/molO 1 x 16 g/mol = 16 g/mol30 g/mol240 g/mol30 g/mol= 8
x CH
2
O
= C
8
H
16
O
8Slide4
Problem Set 1
Empirical Formula
Molar Mass of molecular formulaMolar mass of empirical formulaCompare molar masses Actual molecular formulaCH26 g/molNO2230 g/molC3
H844 g/mol
13 g/mol
230 g/mol
44 g/mol
2
5
1
C
2
H
2
N
5
O
10
C
3
H
8
Episode
703Slide5
Find the molecular formula for a compound with -
4.04
g N11.46 g OMolar mass 108 g/molFind the empirical formula.Find the mass of the empirical formulaCompare the molar mass of molecular formula to molar mass of empirical formula. Empirical formula N2O5N 2 x 14 g/mol = 28 g/molO 5 x 16 g/mol = 80 g/mol108 g/mol108 g/mol108 g/mol
= 1
x N
2
O
5
= N
2
O
5
Episode
703Slide6
Hydrates
Crystals with
water molecules adhering to the ions or moleculesNa2CO3•10 H2OIndicates 10 water molecules adhering to each formula unit of sodium carbonateMass of water = mass of “hydrated” compound minus mass of “dry” compoundAnhydrous means “dry”Episode 703Slide7
Determine the formula of hydrated barium chloride from this data:
Initial mass of hydrated compound = 1.373 g
Mass after heating = 1.175 g1. Determine formula for barium chloride.2. Determine the mass of water removed from hydrate.3. Find the ratio between anhydrous compound and water.BaCl21.373 g – 1.175 g = 0.198 g water1.175g BaCl2
208g BaCl2
1 mol BaCl
2
= 0.0056 molBaCl
2
0.198g H
2
O
1
8g H
2
O
1 mol H
2
O
= 0.012 mol H
2
O
0
.0056
0
.0056
BaCl
2
•2H
2
O
Episode
703Slide8
Determine the formula for the hydrate that is 76.9%
CaSO3 and
23.1% H2O. Remember the trick of changing % to grams!Find the ratio between anhydrous compound and water.76.9g CaSO3 120g CaSO3 1mol CaSO3
= 0.641 mol CaSO3
23.1g H
2
O
18
g H
2
O
1 mol
H
2
O
= 1.28 mol
H
2
O
0.641
0.641
CaSO
3
•2H
2
O
Episode
703